Revision

Since pictures are laid on a graph paper, a bar in graph paper has a measure of 1 unit (Here, 1 bar = 1 cm) and is square in shape.

Using the information that, 1 bar = 1 cm and square in shape.

For shape 1:

We need to find the number of complete smallest square regions enclosed by the figure.

Complete smallest square regions is equal to 1 bar of the graph.

Note that, in the figure

There are 6 bars enclosed by the figure.

These 6 bars are completely enclosed within the boundary of the figure.

⇒ There are 6 number of complete smallest square regions enclosed by the figure.

We need to find number of half of smallest square regions enclosed by the figure.

As we already know that, 6 complete square regions are enclosed by the figure 1. There are no other regions enclosed by the figure other than these 6 squares.

⇒ There are 0 number of half of smallest square regions enclosed by the figure.

We need to find number of more than half of smallest square regions enclosed by the figure.

There are no square (be it half or more than half) regions enclosed by the figure than the 6 complete square regions.

⇒ There are 0 number of more than half of smallest square regions enclosed by the figure.

We need to find number of less than half of smallest square regions enclosed by the figure.

There are no square (be it half, more than half or less than half) regions enclosed by the figure than the 6 complete square regions.

⇒ There are 0 number of less than half of smallest square regions enclosed by the figure.

Now, we need to find the total area of the figure, given that area of one smallest square region is 1 cm².

From the previous data, we can conclude that

Complete smallest square regions = 6

As,

Area of 1 smallest square region = 1 cm²

Then, area of 6 smallest square regions = 6 × 1 cm²

⇒ Area of 6 smallest square regions = 6 cm²

Hence, we have taken out the data of shape 1.

For shape 2:

We need to find number of complete smallest square regions enclosed by the figure.

Complete smallest square regions is equal to 1 bar of the graph.

Note that, in the figure

There are many regions, some are half of a bar while some are more than or less than half of a bar.

But there is only one complete bar.

This 1 bar is completely enclosed by the figure.

⇒ There is only 1 complete smallest square region enclosed by the figure.

We need to find number of half of smallest square regions enclosed by the figure.

Out of all regions enclosed by the figure, there are only 3 regions which are half of square.

These 3 bars are completely enclosed by the figure.

⇒ There are 3 half of smallest square regions enclosed by the figure.

We need to find number of more than half of smallest square regions enclosed by the figure.

Apart from 1 complete square region and 3 half square regions, we can note that there is only 1 region which is more than half of square but less than a complete square.

This 1 bar is completely enclosed by the figure too.

⇒ There is 1 more than half of smallest square region enclosed by the figure.

We need to find number of less than half of smallest square regions enclosed by the figure.

Apart from 1 complete square region, 3 half square regions and 1 more than half square region, we can note that there are 3 regions which are less than half of square.

These 3 bars are completely enclosed by the figure.

⇒ There are 3 less than half of smallest square regions enclosed by the figure.

Now, we need to find the total area of the figure, given that area of one smallest square region is 1 cm².

To find area of the figure, enclose the figure into a rectangle or square (whichever is possibly fit).

Here, we have a rectangle, PBQA.

We need to find the area of this rectangle.

Area of this rectangle = number of complete bars × 1 cm² [∵, 1 cm² is the area of 1 bar]

Number of complete bars in this rectangle are 12.

So,

Area of rectangle = 12 × 1 cm²

⇒ Area of PBQA = 12 cm² …(i)

Now, there are 2 right angled triangles apart from the shape 2 enclosed in this rectangle.

We have right-angled triangles namely, ∆PBQ and ∆PAR.

We can find area of these two triangles.

Area of triangle is given by,

For ∆PBQ,

∠PBQ = 90°

Base = PB

∵ 1 side of each 6 bars cover length PB

⇒ PB = 6 cm (length of one side of a bar = 1 cm)

⇒ Base = 6 cm

Height = BQ

∵ 1 side of each 2 bars cover length BQ

⇒ BQ = 2 cm (length of one side of a bar = 1 cm)

⇒ Height = 2 cm

Thus,

⇒ Area of ∆PBQ = 6 cm² …(ii)

For ∆PAR,

∠PAR = 90°

Base = RA

∵1 side of each 2 bars cover the length RA

⇒ RA = 2 cm (length of one side of a bar = 1 cm)

⇒ Base = 2 cm

Height = PA

∵ 1 side of each 2 bars cover the length PA

⇒ PA = 2 cm (length of one side of a bar = 1 cm)

⇒ Height = 2 cm

Thus,

⇒ Area of ∆PAR = 2 cm² …(iii)

Add these two triangles’ area, then subtract from the area of the rectangle. We have

Area of shape 2 = Area of PBQA – (Area of ∆PBQ + Area of ∆PAR)

⇒ Area of shape 2 = 12 – (6 + 2) [from (i), (ii) & (iii)]

⇒ Area of shape 2 = 12 – 8

⇒ Area of shape 2 = 4 cm²

Hence, we have taken out the data of shape 2.

For shape 3:

We need to find number of complete smallest square regions enclosed by the figure.

Complete smallest square regions is equal to 1 bar of the graph.

Note that, in the figure

There are 8 complete squares enclosed by the figure.

⇒ There are 8 complete smallest square regions enclosed by the figure.

We need to find number of half of smallest square regions enclosed by the figure.

Apart from 8 complete square regions, we can note that there are 2 regions which are (approx.) half of square.

The small extra region can be ignored.

⇒ There are 2 half of smallest square regions enclosed by the figure.

We need to find number of more than half of smallest square regions enclosed by the figure.

Apart from 8 complete square regions and 2 half square regions, we can note that there are 4 regions which are more than half of square region.

⇒ There are 4 more than half of smallest square regions enclosed by the figure.

We need to find number of less than half of smallest square regions enclosed by the figure.

Apart from 8 complete square regions, 2 half square regions and 4 more than half of square regions, we can note that there are 2 regions which are less than half of square region.

⇒ There are 2 less than half of smallest square regions enclosed by the figure.

Now, we need to find the total area of the figure, given that area of one smallest square region is 1 cm².

To find area of the figure, enclose the figure into a rectangle or square (whichever is possibly fit).

Here, we have a rectangle, GHIJ.

We need to find the area of this rectangle.

Area of this rectangle = number of complete bars × 1 cm² [∵, 1 cm² is the area of 1 bar]

Number of complete bars in this rectangle are 18.

So,

Area of rectangle = 18 × 1 cm²

⇒ Area of GHIJ = 18 cm² …(i)

Now, there are 4 right angled triangles apart from the shape 3 enclosed in this rectangle.

We have right-angled triangles namely, ∆AGB, ∆FHE, ∆EID and ∆BJC.

We can find area of these four triangles.

Area of triangle is given by,

For ∆AGB,

∠AGB = 90°

Base = BG

[∵, 1 side of one bar and 4/5^th side of another bar covers BG ⇒ (length of one side of a bar = 1 cm, so length of fourth part of a side of a bar = 4/5 since the bar has 5 divisions)]

Height = AG

[∵, 1 side of each 2 bars cover length AG ⇒ AG = 2 cm (length of one side of a bar = 1 cm)]

⇒ Height = 2 cm

Thus,

…(ii)

Similarly,

For ∆FHE,

∠FHE = 90°

…(iii)

For ∆BJC,

∠BJC = 90°

Base = JC

[∵, 1 side of each 2 bars cover the length JC ⇒ RA = 2 cm (length of one side of a bar = 1 cm)]

⇒ Base = 2 cm

Height = BJ

[∵, 1 side of each one bar and 1/5^th side of another bar covers BJ ⇒ (length of one side of a bar = 1 cm)]

Thus,

…(iv)

Similarly,

For ∆EID,

∠EID = 90°

…(v)

Add these four triangles’ area, then subtract from the area of the rectangle. We have

Area of shape 3 = Area of GHIJ – (Area of ∆AGB + Area of ∆FHE + Area of ∆BJC + Area of ∆EID)

[from (i), (ii), (iii) & (iv)]

⇒ Area of shape 3 = 18 – 6

⇒ Area of shape 3 = 12 cm²

Hence, we have taken out the data of shape 3.

For shape 4:

Let the figure be ABCDEFG.

We need to find number of complete smallest square regions enclosed by the figure.

Complete smallest square regions is equal to 1 bar of the graph.

Note that, in the figure

There are many regions, some are half of a bar while some are more than or less than half of a bar.

But there is only one complete bar.

⇒ There is 1 complete smallest square region enclosed by the figure.

We need to find number of half of smallest square regions enclosed by the figure.

Apart from 1 complete square region, we can see only 1 region that looks like half of a square region.

⇒ There is 1 half of smallest square region enclosed by the figure.

We need to find number of more than half of smallest square regions enclosed by the figure.

Apart from 1 complete square region and 1 half of square region, we can see 7 regions that are more than half of smallest square regions enclosed by the figure.

⇒ There are 7 more than half of smallest square regions.

We need to find number of less than half of smallest square regions enclosed by the figure.

Apart from 1 complete square region, 1 half of square region and 7 more than half of square region, we can see 11 less than half of square regions.

The rest of the region less than half of a square is the region we are looking for. Neglect the regions which has negligible area.

⇒ There are 11 less than half of smallest square regions enclosed by the figure.

Now, we need to find the total area of the figure, given that area of one smallest square region is 1 cm².

To find area of the figure, divide the figure into possible general figures as follows:

See figure 1: It is a rectangle, ABCG.

Area of this rectangle is determined by the number of bars it encloses.

So, figure 1 encloses 3 complete bars and 1 half bar.

∵, area of 1 complete bar = 1 cm²

Then,

Area of 3 complete bars = 3 × 1 = 3 cm²

Area of ABCG = Area of 3 complete bars + Area of 1 half bar

…(i)

See figure 2: It is a triangle, FEG.

Area of triangle is given by

Here, base = EG

And height is given by the arrow, QF.

EG = 2 cm [∵, one side of each 2 bars cover distance of EG, and length of a side of a bar = 1 cm ⇒ Length of side of 2 bars = 2 cm]

⇒ Base = 2 cm

2 complete bars and 3/5^th part of a bar cover the distance QF.

Thus,

…(ii)

See figure 3: It is a triangle, EGC.

Area of triangle is given by

Here, base = CG

And height is given by the arrow.

One side of 1 bar covers the distance of CG, and length of side of a bar = 1 cm.

Then, CG = 1 cm

⇒ Base = 1 cm

And one side of each 2 bars cover the distance of height.

⇒ Height = 2 cm

⇒ Area of ∆EGC = 1 cm² …(iii)

Further dividing figure 4 into 4 and 5.

For figure 4: We have the triangle, CHD.

Area of triangle is given by

Here, base = HC

One side of complete bar and 1/5^th part of side of another bar covers the distance HC.

If length of a side of complete bar = 1 cm

Then,

Height = DS

One side of each 2 complete bars and 4/5^th part of side of a bar covers the distance DS.

Thus,

…(iv)

For figure 5: We have triangle, EHC.

Area of triangle is given by

Here, base = HC

As we know,

Height = TE

One side of a complete bar and 1/5^th part of side of a bar covers distance of TE.

Thus,

…(v)

∴, Area of the whole figure = Area of figure 1 + Area of figure 2 + Area of figure 3 + Area of figure 4 + Area of figure 5

[from (i), (ii), (iii), (iv) & (v)]

⇒ Area of whole figure = 9.5 cm²

Thus,

Area of shape 4 = 9.5 cm²

Hence, we have taken out the data of shape 4.

For shape 5:

Let the figure be ABC.

We need to find number of complete smallest square regions enclosed by the figure.

Complete smallest square regions is equal to 1 bar of the graph.

Note that, in the figure

There are many regions, some are half of a bar while some are more than or less than half of a bar.

But there are 6 complete bars.

⇒ There are 6 complete smallest square regions enclosed by the figure.

We need to find number of half of smallest square regions enclosed by the figure.

Apart from 6 complete bars, we can see that there is just one nearly half of the square.

⇒ There is 1 half of the smallest square regions enclosed by the figure.

We need to find number of more than half of smallest square regions enclosed by the figure.

Apart from 6 complete square region and 1 half of square region, we can see 3 more than half of square region.

⇒ There are 3 more than half of smallest square region enclosed by the figure.

We need to find number of less than half of smallest square region enclosed by the figure.

Apart from 6 complete square region, 1 half of square region and 3 more than half of square region, we can see 3 less than half of square region.

⇒ There are 3 less than half of smallest square region enclosed by the figure.

Now, we need to find the total area of the figure, given that area of one smallest square region is 1 cm².

To find area of the figure:

Since, this figure is ∆ABC.

Since, sides of each four bars cover distance BC, and length of 1 side of a bar = 1 cm.

⇒ Length of BC = 4 × 1

⇒ Length of BC = 4 cm

Sides of each five bars cover distance AB, and length of 1 side of a bar = 1 cm.

⇒ Length of AB = 5 × 1

⇒ Length of AB = 5 cm

Thus,

⇒ Area of ∆ABC = 2 × 5

⇒ Area of ∆ABC = 10 cm²

Hence, we have taken out the data of shape 5.

For shape 6:

We have from the question:

We need to find the number of complete smallest square region enclosed by the figure.

Here, either we can draw the figure by hand on a graph paper, or we need to imagine the bar lines which are missing behind the rigid figure.

Clearly, the figure encloses only 1 complete square region (that is in the center of the figure).

⇒ There is 1 complete smallest square region enclosed by the figure.

We need to find the number of half of smallest square region enclosed by the figure.

Here, apart from the 1 complete square region, we need to see the bar which is almost half occupied by the figure.

We can say, the bar (2, 3) is almost half filled.

⇒ There is 1 half of smallest square region enclosed by the figure.

We need to find the number of more than half of smallest square region enclosed by the figure.

Here, apart from the 1 complete square region and 1 half of square region, we can see that 6 bars looks like more than half of smallest square regions.

⇒ There are 6 more than half of smallest square region enclosed by the figure.

We need to find the number of less than half of smallest square region enclosed by the figure.

Here, apart from the 1 complete square region, 1 half of square region and 6 more than half of square regions, we can see that only 1 bar seems to look like less than half of square region.

⇒ There is 1 less than half of smallest square region enclosed by the figure.

Now, we need to find the total area of the figure, given that area of one smallest square region is 1 cm².

To find area of the figure:

The figure consists of a central circle and four semi-circular petals surrounding it.

The radius of the central circle can be out as:

Draw a straight line from the mid of the circle and measure the line, comes out to be 1.4 cm (approx.).

Since, the line covers one complete bar and 2/5^th part of another bar.

So, length of one complete bar = 1 cm

And,

This length is actually the diameter of the central circle.

If , then

Area of circle is given by

Area = πr²

⇒ Area = 1.54 cm² …(i)

The four surrounded petals are semi-circular. So, joining any two petals would make a circle. We have 2 circles.

Radius of these circle = 1 cm

As the petals approximately covers one complete bar.

So,

Area of one circle of radius 1 cm is given by,

⇒ Area = 3.14 cm²

So, area of 2 similar circles = 2 × 3.14 cm²

⇒ Area of 2 circles = 6.28 cm² …(ii)

Adding (i) and (ii) to get area of the whole figure, we have

Area of whole figure = 1.54 + 6.28

⇒ Area of whole figure = 7.82 cm²

Hence, we have taken out the data of shape 6.

For shape 7:

Let us name the figure as ABCDEFGHIJKLMNOP.

We need to find number of complete smallest square regions enclosed by the figure.

We have drawn the figure roughly.

Note, the central bar is complete and is enclosed by the figure.

⇒ There is 1 complete smallest square region enclosed by the figure.

We need to find number of half of smallest regions enclosed by the figure.

Apart from the 1 complete square region, we can see that there are 4 regions that can be considered as half of square region.

⇒ There are 4 half of smallest square regions enclosed by the figure.

We need to find number of more than half of smallest regions enclosed by the figure.

Apart from 1 complete square region and 4 half of square region, we can see that there are 4 more than half of square region.

⇒ There are 4 more than half of square regions enclosed by the figure.

We need to find number of less than half of smallest regions enclosed by the figure.

Apart from 1 complete square region, 4 half of square regions and 4 more than half of square regions, we can see no other region less than half of square region.

⇒ There are 0 less than half of square region enclosed by the figure.

Now, we need to find the total area of the figure, given that area of one smallest square region is 1 cm².

To find area of the figure:

Dissipate the figure into smaller general shapes.

It can be broken down to several squares and triangles for ease of calculation.

Take BNJF as a square, since it is approximately a square.

BNJF seems to be one completely bar approximately, and we know area of a complete bar = 1 cm²

So, Area of BNJF = 1 cm² …(i)

Take ABNO, CBFE, FGIJ and JKMN as squares with same dimensions.

So, Area of ABNO = Area of CBFE = Area of FGIJ = Area of JKMN = 1 cm² …(ii)

[∵, they are approximately equal to a complete bar]

Take ∆OPA, ∆CDE, ∆GHI and ∆KLM as triangles with same dimensions (as they are approximately).

Let us take out are of ∆OPA.

Base = OA = 1 cm [∵, BN = OA = 1 cm]

[taken approximately]

Area is given by,

So,

…(iii)

Add all the area to get a total area of the figure.

Adding (i), (ii) & (iii), we get

⇒ Total Area = 5.8 cm²

Hence, we have taken out the data of shape 7.

For shape 8:

We have from the question:

Since, the shape given is not of uniform nature, we shall find the answer to approximations.

We need to find the number of complete smallest square region enclosed by the figure.

Here, either we can draw the figure by hand on a graph paper, or we need to imagine the bar lines which are missing behind the rigid figure.

Clearly, the figure encloses 2 complete square regions.

⇒ There are 2 complete smallest square regions enclosed by the figure.

We need to find the number of half of smallest square region enclosed by the figure.

Apart from the 2 complete square regions, we can see that there are 4 regions that covers approximately half of square regions.

⇒ There are 4 half smallest square regions enclosed by the figure.

We need to find the number of more than half of smallest square regions enclosed by the figure.

Apart from 2 complete square regions and 3 half of square regions, we can see that there are 5 regions which can be said that they are covering more than half of smallest square regions.

⇒ There are 5 more than half of smallest square regions enclosed by the figure.

We need to find the number of less than half of smallest square regions enclosed by the figure.

Apart from 2 complete square regions, 3 half of square regions and 5 more than half of square regions, we can see that there are 8 regions which are approximately less than half of square regions.

⇒ There are 7 less than half of smallest square regions enclosed by the figure.

Now, we need to find the total area of the figure, given that area of one smallest square region is 1 cm².

To find area of this disfigured shape:

We need to merge halves together to form one complete bar.

We need to merge more than halves and less than halves together to form one complete bar.

Finding area by this way only leads us to an approximate area.

Data:

Complete bars = 2

⇒ Area of 2 complete bars = 2 cm² [∵, Area of 1 complete bar = 1 cm²] …(i)

Half bars = 4

And by merging 2 half bars, we get 1 complete bar.

So, by merging 4 half bars, we get 2 complete bars.

⇒ Area of 2 complete bars = 2 cm² …(ii)

More than half bars = 5

Less than half bars = 7

And by merging 1 more than half bar and 1 less than half bar, we get 1 complete bar.

So, by merging 5 more than half bars and 5 less than half bars, we get 5 complete bars.

⇒ Area of 5 complete bars = 5 cm² …(iii)

Now, we are left with 2 less than half bars, which can be merged to make a half bar.

If area of 1 complete bar = 1 cm²

Then,

⇒ Area of 1 half bar = 0.5 cm² …(iv)

Adding all these areas, we get

Area of the figure = 2 + 2 + 5 + 0.5 [∵, from (i), (ii), (iii) & (iv)]

⇒ Area of the figure = 9.5 cm²

Hence, we have taken out the data of shape 8.

Filling the derived data of shape 1, 2, 3, 4, 5, 6, 7 and 8 in the table, we get

We have

Let ABCD be the rectangular courtyard.

AB = CD = 6 m

BC = AD = 4.2 m

Let the mattress be EFGH.

EF = GH = 3.5 m

FG = HE = 2.5 m

We need to find the area of the courtyard without the mattress, that is, the area of the space between ABCD and EFGH.

Let area of the space between ABCD and EFGH denoted by A.

Then, it is given as

A = Area of ABCD (full courtyard) – Area of EFGH (full mattress) …(A)

[Since, if we remove the area of mattress from the area of courtyard, we are left with the area of the courtyard without the mattress]

Let us find the area of ABCD.

Since, ABCD is a rectangle, then

Area of rectangle = length × breadth

⇒ Area of ABCD = AB × BC

given that, AB = 6 cm and BC = 4.2 cm

⇒ Area of ABCD = 6 × 4.2

⇒ Area of ABCD = 25.2 cm² …(i)

And let us find the area of EFGH.

Since, EFGH is a rectangle, then

Area of rectangle = length × breadth

⇒ Area of EFGH = EF × FG

given that, EF = 3.5 cm and FG = 2.5 cm

⇒ Area of EFGH = 3.5 × 2.

⇒ Area of EFGH = 8.75 cm² …(ii)

Putting (i) & (ii) in equation (A), we get

A = 25.2 – 8.75

⇒ A = 25.20 – 8.75

⇒ A = 16.45 cm²

Hence, area of the courtyard without the mattress is 16.45 cm².

We have

The Ajanta Housing Complex is ABCD, which is square shaped.

The path around it is 3 m wide uniformly.

We can conclude that,

If ABCD is a square, then

EFGH is also a square.

In ABCD:

Let AB = p

Then, AB = BC = CD = DA = p [∵ all sides are equal in a square]

In EFGH:

EF = AB + 3 + 3 [∵, 3 m width is around the square park]

⇒ EF = AB + 6

⇒ EF = p + 6

Then, EF = FG = GH = HE = p + 6 [∵all sides are equal in a square]

Given that,

Perimeter of park including the path = 484 m

Or

Perimeter of ABCD + Perimeter of EFGH = 484 m

⇒ (4 × AB) + (4 × EF) = 484 [∵, Perimeter of a square = 4 × length]

⇒ (4 × p) + (4 × (p + 6)) = 484

⇒ 4p + 4p + 24 = 484

⇒ 8p + 24 = 484

⇒ 8p = 484 – 24

⇒ 8p = 460

⇒ p = 57.5 m

Thus, AB = BC = CD = DA = 57.5 m

And, EF = FG = GH = HE = 57.5 + 6 = 63.5 m

We need to find the area of the path.

Area of the path = Area of EFGH – Area of ABCD …(A)

Area of EFGH:

Area of square is given as,

Area = (length)²

⇒ Area of EFGH = (EF)²

⇒ Area of EFGH = (63.5)² …(i)

Area of ABCD:

⇒ Area of ABCD = (AB)²

⇒ Area of ABCD = (57.5)² …(ii)

Substitute equations (i) and (ii) in (A), we get

Area of the path = (63.5)² – (57.5)²

⇒ Area of the path = 4032.25 – 3306.25

= 726 m²

Hence, the area of the path is 726 m².

According to the question,

Let Mihir’s garden be ABCD.

Then, AB = 50 m

BC = 30 m

A road EFGH is in the middle of this garden, of width 4 m.

⇒ EF = AB = 50 m

And FG = HE = 4 m

The road is of dimension (50 × 4) m².

We need to find the area of the road.

Area of EFGH = EF × FG [∵, Area of rectangle = length × breadth]

⇒ Area of EFGH = 50 × 4

⇒ Area of EFGH = 200 m²

Hence, area of the road is 200 m².

(A). We have,

Given that, the road EFGH is parallel to the breadth of ABCD.

Then, HE = 4 m [as given]

EF = AD = 30 m

Then, area of the road EFGH is given as,

Area of EFGH = EF × HE [∵, Area of rectangle = length × breadth]

⇒ Area of EFGH = 30 × 4

⇒ Area of EFGH = 120 m²

Hence, area of the road is 120 m².

(B). We have

According to the question,

There are two roads of width 4 m, parallel to each length and breadth.

Since, these roads divide the rectangle ABCD into four equal parts.

We can find QE and MQ.

If DA = 30 m and PM = 4 m. Then,

AM + PD = DA – PM

⇒ AM + PD = 30 – 4

⇒ AM + PD = 26

⇒ AM + AM = 26 [∵, AM = PD as these roads divide ABCD into 4 equal parts]

⇒ 2 AM = 26

⇒ AM = 13 m

Thus, AM = PD = 13 m

Also, AM = QE = 13 m

And, PD = TF = 13 m

Similarly, if AB = 50 m and EH = 4 m. Then,

AE + HB = AB – EH

⇒ AE + HB = 50 – 4

⇒ AE + HB = 46

⇒ AE + AE = 46 [∵, AE = HB as these roads divide ABCD into 4 equal parts]

⇒ 2 AE = 46

⇒ AE = 23 m

Thus, AE = HB = 23 m

Also, AE = MQ = 23 m

And, HB = RN = 23 m

Area of QRST is given by,

Area₁ = 4 × 4 [∵, QRST is a square with dimensions 4 m each]

⇒ Area₁ = 16 m² …(i)

Area of EHRQ is given by,

Area₂ = 4 × 13 [∵, EH = 4 m & QE = 13 m]

⇒ Area₂ = 52 m²

So,

Area of EHRQ = Area of FGST = 52 m² …(ii)

Similarly, area of ONRS is given by,

Area₃ = 4 × 23 [∵, ON = 4 m and RN = 23 m]

⇒ Area₃ = 92 m²

So,

Area of ONRS = Area of PMQT = 92 m² …(iii)

From equation (i), (ii) and (iii), we get

Area of road = Area of QRST + Area of EHRQ + Area of FGST + Area of ONRS + Area of PMQT

⇒ Area of the road = 16 + 52 + 52 + 92 + 92

⇒ Area of the road = 304 m²

Hence, area of the road is 304 m².

We have

ABCD is the rectangular land, in which

AB = 48 m

BC = 26 m

The dotted lines are 4 m width.

And EFGH is the land on which house is built.

EF = AB – (4 + 4) [One 4 from one side and the other 4 from the other side of the land]

⇒ EF = 48 – 8

⇒ EF = 40 m

And,

FG = BC – (4 + 4)

⇒ FG = 26 – 8

⇒ FG = 18 m

We need to find the area of the region in which the house is built, that is, area of EFGH.

Area of EFGH is given by,

Area of EFGH = EF × FG [∵, Area of rectangle = length × breadth]

⇒ Area of EFGH = 40 × 18

⇒ Area of EFGH = 720 m²

Hence, area of the land on which they have built their house is 720 m².

We have a rectangular sheet of

Length = 15 cm

Breadth = 8 cm

(A). According to the question,

Original dimensions of the paper are:

Length = 15 cm

Breadth = 8 cm

New dimensions of the paper are:

Length’ = 2 × Length

⇒ Length’ = 2 × 15

⇒ Length’ = 30 cm

Breadth’ = Breadth

⇒ Breadth’ = 8 cm

Let us find the original area.

Since, area of rectangle = Length × Breadth

Then,

Original area = 15 × 8

⇒ Original area = 120 cm² …(ii)

Now, let us find the new area.

New area is given by,

New area = Length’ × Breadth’

⇒ New area = 30 × 8

⇒ New area = 240 cm² …(iii)

Change in area is given by,

Hence, the area is doubled.

(B). According to the question,

Original dimensions of the paper are:

Length = 15 cm

Breadth = 8 cm

New dimensions of the paper are:

Length’ = Length

⇒ Length’ = 15 cm

Breadth’ = 2 × Breadth

⇒ Breadth’ = 2 × 8

⇒ Breadth’ = 16 cm

Let us find the original area.

Since, area of rectangle = Length × Breadth

Then,

Original area = 15 × 8

⇒ Original area = 120 cm² …(ii)

Now, let us find the new area.

New area is given by,

New area = Length’ × Breadth’

⇒ New area = 15 × 16

⇒ New area = 240 …(iii)

Change in area is given by,

Hence, the area is doubled.

(C). According to the question,

Original dimensions of the paper are:

Length = 15 cm

Breadth = 8 cm

New dimensions of the paper are:

Length’ = 2 × Length

⇒ Length’ = 2 × 15

⇒ Length’ = 30 cm

Breadth’ = 2 × Breadth

⇒ Breadth’ = 2 × 8

⇒ Breadth’ = 16 cm

We need to find the number of times the area of the paper calculated with the new dimensions will be to the area calculated in case (A).

Area of paper in case (A) = 240 cm² …(i)

Let us find the new area.

Since, area of rectangle = Length × Breadth

New area is given by,

New area = Length’ × Breadth’

⇒ New area = 30 × 16

⇒ New area = 480 cm² …(ii)

New Area = number of times × Area in case (A)

⇒ 480 = number of times × 240

⇒ Number of times = 2

Hence, the new area calculated will be twice the area calculated in case (A).

(D). According to the question,

Original dimensions of the paper are:

Length = 15 cm

Breadth = 8 cm

New dimensions of the paper are:

⇒ Breadth’ = 4 cm

Let us find the original area.

Since, area of rectangle = Length × Breadth

Then,

Original area = 15 × 8

⇒ Original area = 120 cm² …(ii)

Now, let us find the new area.

New area is given by,

New area = Length’ × Breadth’

⇒ New area = 30 cm² …(iii)

Change in area is given by,

Thus, new area is � of the original area.

Given that,

We have three square shaped paper.

Let side of the square shaped paper be x unit.

That is,

Side = x unit

Area of square is given by,

Area = (Side)²

So,

Area of one square paper is given as,

Area = (x)²

⇒ Area = x² …(i)

(a). According to the question,

Original dimensions:

Length = x

Breadth = x

New dimensions:

Length’ = 2 × Length

⇒ Length’ = 2x

New Area = (Length’)²

⇒ New Area = (2x)²

⇒ New Area = 4x² …(ii)

Then, change in area is given by

[from equation (i) and (ii), New area = 4x² and Original area = x²]

Hence, the new area is 4 times the original area of the square-shaped paper.

(b) According to the question,

Original dimensions:

Length = x

Breadth = x

New dimensions:

Breadth’ = x

New Area = Length’ × Breadth’

…(iii)

Then, change in area is given by

[from equation (i) and (iii), and original area = x²]

Hence, new area is half of the original area of the square-shaped paper.

Given:

A club house dimension,

Length = 7.2 m

Breadth = 5.5 m

Height = 4.2 m

Dimensions of door in the room,

Length = 3 m

Breadth = 1.8 m

Dimensions of each window in the room (there are 2 windows),

Length = 2.25 m

Breadth = 1.8 m

(A). Let us measure the area of the floor of the club house.

The floor comprises of 2 dimensions, that is, length and breadth.

It doesn’t have either door or any window.

So, dimensions of floor are

Length = 7.2 m

Breadth = 5.5 m

Then, area of floor is given by

Area of floor = Length × Breadth

⇒ Area of floor = 7.2 × 5.5

⇒ Area of floor = 39.6 m²

Also, let us find total expenditure of cementing the floor at the rate of Rs. 62/m².

If cost of cementing 1 m² of floor = Rs. 62

Then, cost of cementing 39.6 m² of floor = Rs. 62 × 39.6

⇒ Cost of cementing 39.6 m² of floor = Rs. 2455.2

Hence, area of the floor is 39.6 m² and cost of cementing it is Rs. 2455.2.

(B). Let us measure the area of the inner walls excluding the doors and windows.

The inner walls comprise of 4 walls.

Area of first wall = Length × Height

⇒ Area of first wall = 7.2 × 4.2

= 30.24 m²

Area of second wall = Breadth × Height

= 5.5 × 4.2

⇒ Area of second wall = 23.1 m²

Area of third wall = Length × Height

= 30.24 m²

Area of fourth wall = Breadth × Height

= 23.1 m²

Area of inner walls = Area of first wall + Area of second wall + Area of third wall + Area of fourth wall

⇒ Area of inner walls = 30.24 + 23.1 + 30.24 + 23.1

⇒ Area of inner walls = 106.68 m² …(i)

But, the walls also has 1 door and 2 windows.

So, to find the area of the walls excluding the 1 door and 2 windows, just subtract the area of the door and windows from the total area of the walls.

Area of the inner wall (excluding door and windows) = Area of inner wall – (Area of 1 door + Area of 2 windows) …(ii)

Area of 1 door is found out as,

Area of 1 door = Length of door × Breadth of door

⇒ Area of 1 door = 3 × 1.8 [given]

⇒ Area of 1 door = 5.4 m² …(iii)

Area of 1 window is found out as,

Area of 1 window = Length of window × Breadth of window

⇒ Area of 1 window = 2.25 × 1.8 [given]

⇒ Area of 1 window = 4.05 m²

If area of 1 window = 4.05 m²

Then, area of 2 windows = 2 × 4.05

⇒ Area of 2 windows = 8.1 m² …(iv)

Substituting equations (i), (iii) and (iv) in equation (ii), we get

Area of inner walls (excluding door and windows) = 106.68 – (5.4 + 8.1)

⇒ Area of inner walls (excluding door and windows) = 106.68 – 13.5

⇒ Area of inner walls (excluding door and windows) = 93.18 m²

Hence, the area of inner walls excluding door and windows is 93.18 m².

(C). To find the area of the ceiling of the room, we need to know the length and breadth of the ceiling.

Length = 7.2 m

Breadth = 5.5 m

So, area of the ceiling is given by

Area = Length of the ceiling × Breadth of the ceiling

⇒ Area = 7.2 × 5.5

⇒ Area = 39.6 m²

Note that, area of the ceiling is equal to the area of the floor.

Hence, area of the ceiling is 39.6 m².

(D). Now, we need to find the expenditure in whitewashing inner walls and ceiling without door and windows.

Let us recall that,

Cost of whitewashing 1 m² = Rs. 12

We have,

Area of inner walls (excluding door and windows) = 93.18 m²

Area of ceiling = 39.6 m²

Total area for whitewashing = Area of inner walls (excluding door and windows) + Area of ceiling

⇒ Total area for whitewashing = 93.18 + 39.6

⇒ Total area for whitewashing = 132.78 m²

If cost of whitewashing 1 m² = Rs. 12

Then, cost of whitewashing 132.78 m² = Rs. 12 × 132.78

⇒ Cost of whitewashing 132.78 m² = Rs. 1593.36

Hence, total amount for whitewashing inner walls and ceiling except door and windows is Rs. 1593.36.

As per the solution,

My arrangement

Sticks in first arrangement (a) = 5

Sticks in second arrangement = 8

Common Difference (d) = 8 – 5 = 3

Sticks in n-th position = a + (n – 1) d

= 5 + (n – 1) (3)

= 5 + 3n – 3

= 3n + 2

Ayan’s arrangement,

Sticks in first arrangement (a) = 3

Sticks in second arrangement = 6

Common Difference (d) = 6 – 3 = 3

Sticks in n-th position = a + (n – 1) d

= 3 + (n – 1) (3)

= 3 + 3n – 3

= 3n

Amita’s arrangement,

Sticks in first arrangement (a) = 3

Sticks in second arrangement = 5

Common Difference (d) = 5 – 3 = 2

Sticks in n-th position = a + (n – 1) d

= 3 + (n – 1) (2)

= 3 + 2n – 2

= 2n + 1

Hence calculated the number of sticks required at n-th position in each of the arrangement.

The new arrangement would be

Sticks in first arrangement (a) = 2

Sticks in second arrangement = 5

Common Difference (d) = 5 – 2 = 3

Sticks in n-th position = a + (n – 1) d

= 2 + (n – 1) (3)

= 2 + 3n – 3

= 3n – 1

the sticks required in n-th position are “3n – 1”

i. Length = Breadth = 3 – x m

Area = Length Breadth

Area =

Area =

Area of the rectangular land =

ii. Length = Breadth = 5x + 2 m

Area = Length Breadth

Area =

Area =

Area of the rectangular land =

iii. Area = Breadth = 4x m

Area = Length Breadth

Length =

Length =

Length of the rectangular land = m

iv. Area = Length = m

Area = Length Breadth

Breadth =

Breadth =

Breadth of the rectangular land = m

v. Length = Breadth = 2 – 5x m

Area = Length Breadth

Area =

Area =

Area of the rectangular land =

vi. Area = Breadth = (4 + 10p) m

Area = Length Breadth

Length =

Length = (-10p + 4) m

Length of the rectangular land = (-10p + 4) m

vii. Area = Length = (11m - 13n) m

Area = Length Breadth

Breadth =

Breadth = 11m + 13n

Breadth of the rectangular land = (11m + 13n) m

viii. Length = 9x - y Breadth = 9x + y m

Area = Length Breadth

Area = (9x - y) × (9x + y )

Area =

Area of the rectangular land =

i)In first position, there are 6 sticks after that 5 sticks increase in each position.

Therefore,

In second position, there are (6 + 5) sticks

In third position, there are (6 + 5 + 5 = 6 + 5(2)) sticks

Similarly, in nth position there are (6 + (n – 1)5) = (6+5n-5)

= (1+5n) sticks

ii) In first position, there are 7 sticks after that 5 sticks increase in each position.

Therefore,

In second position, there are (7 + 5) sticks

In third position, there are (7 + 5 + 5 = 7 + 5(2)) sticks

Similarly, in nth position there are (7 + (n – 1)5) = (7 + 5n – 5)

= (2 + 5n) sticks

iii) In first position, there are 5 sticks after that 4 sticks increase in each position.

Therefore,

In second position, there are (5 + 4) sticks

In third position, there are (5 + 4 + 4 = 5 + 4(2)) sticks

Similarly, in nth position there are (5 + (n – 1)4) = (5 + 4n – 4)

= (1 + 4n) sticks

Given,

Side of equilateral triangle = (4y + 2) cm

We know,

Perimeter of equilateral triangle = 3 × side

= 3(4y + 2) cm

= (12y + 6) cm

Given,

Length of rectangle = (8x + 3y) cm

Breadth of rectangle = (8x – 3y) cm

Area of rectangle = length × breadth

= (8x + 3y) (8x – 3y)

= (64x² – 24xy + 24xy – 9y²) cm²

= 64x²– 9y² cm²

We know, area of square = (side)²

Given, side = (3m – 4) meters

⇒ Area of square = (3m – 4)² m²

Also, Perimeter of square = 4 × side

Given, perimeter = 8 meters

⇒ 8 = 4 × (3m – 4)

⇒ 2 = 3m – 4

⇒ 3m = 4 + 2

⇒ 3m = 6

⇒ m = 2

(b)

Given,

(i) 6a² + 2

(ii) -3a² + 3a

(iii) -2a +3

Now, (i) + (ii) + (iii)

= 6a² + 2 – 3a² + 3a – 2a + 3

Add the like terms,

= 6a² – 3a² + 3a – 2a + 2+ 3

= 3a² + a + 5

Also, (ii) – (i)

= -3a² + 3a – (6a² + 2)

Opening the brackets,

= -3a² + 3a – 6a² – 2

Subtracting the like terms,

= -9a² + 3a – 2

and (iii) – (i)

= -2a + 3 – (6a² + 2)

Opening the brackets,

= -2a + 3 – 6a² – 2

Subtracting the like terms,

= -6a² – 2a + 1

(c)

Given,

(i) 9m² – 2mn + n²

(ii) m² + n²

(iii) m² – 3mn – 2n²

Now, (i) + (ii) + (iii)

= 9m² – 2mn + n² + m² + n² + m² – 3mn – 2n²

= 9m² – 5mn + 2n² + 2m² – 2n²

Add the like terms,

= 11m²– 5mn

Also, (i) – (ii)

= 9m² – 2mn + n² – (m² + n²)

Opening the brackets,

= 9m² – 2mn + n² – m² – n²

Subtracting the like terms,

= 8m² – 2mn

and (ii) – (iii)

= m² + n² – (m² – 3mn – 2n²)

Opening the brackets,

= m² + n² – m² + 3mn + 2n²

Subtracting the like terms,

= 3n² + 3mn

(b)

First Part,

Given, (i) x⁴ – 4x³ + 6x²

(ii) x²

(i) × (ii)

= (x⁴ – 4x³ + 6x²) × x²

= x⁴x² – 4x³x² + 6x²x²

We know (x^ax^b= x^a+b)

= x⁴⁺²– 4x³⁺² + 6x²⁺²

= x⁶ – 4x⁵ + 6x⁴

(i) ÷ (ii)

We know

= x^4-2 – 4x^3-2 + 6x^2-2

= x² – 4x + 6

Second Part

Given, (i) 3m²n³ + 40m³n⁴ – 5m⁴n⁵

(ii) 10m²n²

(i) × (ii)

= (3m²n³ + 40m³n⁴ – 5m⁴n⁵) × 10m²n²

We know (x^ax^b= x^a+b)

= 30m²m²n³n² + 400m³m²n⁴n² – 50m⁴m²n⁵n²

= 30m²⁺²n³⁺² + 400m³⁺²n⁴⁺² – 50m⁴⁺²n⁵⁺²

= 30m⁴n⁵ + 400m⁵n⁶ – 50m⁶n⁷

(i) ÷ (ii)

We know

(c)

Given,

(i) 49l² – 100m²

(ii) (7l + 10m)

(i) × (ii)

= (49l² – 100m²)(7l + 10m)

We know (x^ax^b= x^a+b)

= 343l²l+ 490l²m – 700m²l – 1000m²m

= 343l²⁺¹ + 490l²m – 700lm² – 1000m²⁺¹

= 343l³ + 490l²m – 700lm² – 1000m³

(i) ÷ (ii)

∵ a² – b² = (a – b)(a + b), taking a = 7l, b = 10m we have

Cancelling out the common terms from numerator and denominator, we get

= 7l – 10m

(d)

Given,

(i) 625a⁴ – 81b⁴

(ii) 5a + 3b

(i) × (ii)

= (625a⁴ – 81b⁴)(5a + 3b)

= 625a⁴a + 1875a⁴b – 405ab⁴ – 243b⁴b

We know (x^ax^b= x^a+b)

= 625a⁴⁺¹ + 1875a⁴b – 405ab⁴ – 243b⁴⁺¹

= 3125a⁵ + 1875a⁴b – 405ab⁴ – 243b⁵

(i) ÷ (ii)

∵ a² – b² = (a – b)(a + b), taking a = 25a², b = 9b², we have

∵ a² – b² = (a – b)(a + b), taking a = 5a, b = 3b we have

=

Cancelling out the common terms from numerator and denominator, we get

= (5a – 3b)(25a² + 9b²)

= 125aa² + 45ab² – 75a²b – 27bb²

We know (x^ax^b= x^a+b)

= 125a²⁺¹ + 45ab² – 75a²b – 27b²⁺¹

= 125a³ + 45ab² – 75a²b – 27b³

_{i. (a – b) + (b – c) + (c – a)}

= a – b + b – c + c – a

= a – a + b – b + c – c

= 0

ii. (a + b)(a – b) + (b – c)(b + c) + (c + a)(c –a)

We know, (a + b)(a – b) = a² – b², therefore

= a² – b² + b² – c² + c² – a²

= 0

iii.

Cancelling out common terms from numerator and denominator,

= (x² – y²)(x² + y²)

= x⁴ – y⁴ [∵ a² – b² = (a – b)(a + b), Here a = x², b = y²]

iv. a(b – c) + b(c – a) + c(a – b)

= ab – ac + bc – ab + ac – bc

= ab – ab + bc – bc + ac – ac

= 0

v. x²(y² – z²) + y²(z² – x²) + z²(x² – y²)

= x²y² – x²z² + y²z² – x²y² + x²z² – y²z²

= x²y² – x²y² + y²z² – y²z² + x²z² – x²z²

= 0

vi. (x³ + y³)(x³ – y³) + (y³ + z³)(y³ – z³) + (z³ – x³)(z³ + x³)

We know, (a + b)(a – b) = a² – b², therefore

= (x³)² –(y³)² + (y³)² – (z³)² + (z³)² – (x³)²

= x⁶ – y⁶ + y⁶ – z⁶ + z⁶ – x⁶

= 0

_{i. (5x – 2y)}²

_{Apply the formula (a-b)}² = a² + b² -2ab

_{Here a = 5x and b = 2y}

= (5x)² – 2(5x)(2y) + (2y)²

= 25x² – 20xy + 4y²

ii. (7 + 2m)²

Using, (a + b)² = a² + 2ab + b²

Here, a = 7, b = 2m

= (7)² + 2(7)(2m) + (2m)²

= 49 + 28m + 4m²

iii. (x + y + z)²

= (x + (y + z))²

Using, (a + b)² = a² + 2ab + b²

Here, a = x, b = y + z

= x² + 2(x)(y + z) + (y + z)²

Again, using, (a + b)² = a² + 2ab + b²

Here, a = y, b = z

= x² + 2xy + 2xz + y² + z² + 2yz

= x² + y² + z² + 2xy + 2yz + 2xz

iv. (a + b – c – d)²

= (a + b – (c + d))²

Using, (A - B)² = A² - 2AB + B²

Here, A = a + b, B = c + d

= (a + b)² – 2(a + b)(c + d) + (c + d)²

Using, (a + b)² = a² + 2ab + b²

= (a² + b² + 2ab) – 2(ac + ad + bc + bd) + (c² + d² + 2cd)

= a² + b² + c² + d² + 2ab – 2bc + 2cd – 2ad–2bd – 2ac

i.

Comparing with a² + b² – 2ab = (a – b)²,

Here a = 3x,

ii. 25m² – 70mn + 49n²

= (5m)² – 2(5n)(7n) + (7n)²

Comparing with a² + b² – 2ab = (a – b)²,

Here a = 5m, b = 7n

= (5m – 7n)²

iii. (2a – b)² + (4a – 2b)(a + b) + (a + b)²

= (2a – b)² + 2(2a – b)(a + b) + (a + b)²

Comparing with A² + B² – 2AB = (A – B)²,

Here A = 2a – b and B = a + b

= (2a – b + a + b)²

= a²

iv.

Comparing with a² + b² – 2ab = (a – b)²,

Here

i. 391 × 409

= (400 – 9)(400 + 9)

[∵ a² – b² = (a – b)(a + b), here taking a = 400, b = 9]

= (400)² – 9²

ii. (4x + 3y)(2x – 3y)

= (3x + x + 3y)(3x – x – 3y)

= (3x + x + 3y)(3x – (x + 3y))

[∵ a² – b² = (a – b)(a + b), Here taking a = 3x, b = x + 3y]

= (3x)² – (x + 3y)²

iii. x

= x.1

[∵ a² – b² = (a – b)(a + b), here taking ]

_{i. 225m}² – 100n²

= (15m)² – (10n)²

[∵ a² – b² = (a – b)(a + b)]

Here, a = 15m, b = 10n

= (15m – 10n)(15m + 10n)

ii.

[∵ a² – b² = (a – b)(a + b)]

Here, a = 5x,

iii. 7ax² + 14ax + 7a

= 7a(x² + 2x + 1)

Since, a² + 2ab + b²= (a + b)²

Here, a = x, b = 1

= 7a(x + 1)²

iv. 3x⁴ – 6x²a² + 3a⁴

= 3(x⁴ – 2x²a² + a⁴)

= 3[(x²)² – 2(x²)(a²) + (a²)²]

Since, a² + 2ab + b²= (a + b)²

Here, a = x², b = a²

= 3(x² – a²)²

[∵ a² – b² = (a – b)(a + b)]

Here, a = x, b = a

= 3[(x – a)(x + a)]²

= 3(x – a)²(x + a)²

v. 4b²c² – (b² + c² – a²)²

= (2bc)² – (b² + c² – a²)²

= (2bc + b² + c² – a²) (2bc – (b² + c² – a²)) [∵ a² – b² = (a – b)(a + b)]

= (b² + c² + 2bc – a²) (2bc – b² – c² + a²)

[∵ (a + b)² = a² + 2ab + b²]

So, if a = b, b = c in b² + c² + 2bc then after applying the identity we get,

b² + c² + 2bc = (b + c)²

= [(b + c)² – a²] [a² – (b² + c² – 2bc)]

= (b + c – a) (b + c + a)[a²– (b – c)²]

[∵ (a – b)² = a² – 2ab + b²_, Here a = b, b = c]

= (b + c – a) (b + c + a)(a – (b – c))(a + b – c)

= (b + c – a) (b + c + a)(a – b + c)(a + b – c)

vi. 64ax² – 49a(x – 2y)²

= a(64x² – 49(x – 2y)²)

= a[(8x)² – (7(x – 2y))²]

[∵ a² – b² = (a – b)(a + b)]

Here, a = 8x, b = 7(x – 2y)

= a[8x + 7(x – 2y)][8x – 7(x – 2y)]

= a(8x + 7x – 14y)(8x - 7x + 14y)

= a(15x – 2y)(x + 14y)

vii. x² – 9 – 4xy + 4y²

= x² – 4xy + 4y² – 9

= x² – 2(x)(2y) + (2y)² – 9

Since, a² - 2ab + b²= (a - b)²

Here, a = x, b = 2y

= (x – 2y)² - 3²

[∵ a² – b² = (a – b)(a + b)]

Here, a = x – 2y, b = 3

= (x – 2y + 3)(x – 2y – 3) [∵ a² – b² = (a – b)(a + b)]

viii. x² – 2x – y² + 2y

= x² – y² – 2x + 2y

= (x + y)(x – y) – 2(x – y)

[∵ a² – b² = (a – b)(a + b), here a = x, b = y]

= (x – y)(x + y – 2) [Taking (x – y) as common]

ix. 3 + 2a – a²

= 2 + 1 + 2a – a²

= 2 + 2a + 1² – a²

= 2(1 + a) + (1 – a)(1 + a)

[∵ a² – b² = (a – b)(a + b), here a = 1, b = a]

= (1 + a)(2 + 1 – a) [Taking (1 + a) as common]

= (1 + a)(3 – a)

x. x⁴ – 1

= (x²)² - 1²

= (x² – 1)(x² + 1)

[∵ a² – b² = (a – b)(a + b), here a = x², b = 1]

= (x – 1)(x + 1)(x² + 1)

[∵ a² – b² = (a – b)(a + b), here a = x, b = 1]

xi. a²– b² – c² + 2bc

= a² – (b² + c² – 2bc)

= a² – (b - c)²

[∵ (a – b)² = a² – 2ab + b², here a = b and b = c]

= (a + b – c)(a – (b – c))

[∵ a² – b² = (a – b)(a + b), here a = a, b = b - c]

= (a + b – c)(a – b +c)

xii. ac + bc + a + b

= ac + a + bc + b

= a(c + 1) + b(c + 1)

= (c + 1)(a + b) [Taking (c + 1) as common]

xiii. x⁴ + x²y² + y⁴

= x⁴ + 2x²y² – x²y² + y⁴

= x⁴ + 2x²y² + y⁴ – x²y²

= (x²)² + 2(x²)(y²) + (y²)² – (xy)²

= (x² + y²)² – (xy)²

[∵ (a + b)² = a² + 2ab + b², here a = x², b = y²]

= (x² + y² + xy)(x² + y² – xy)

[∵ a² – b² = (a – b)(a + b), here a = x² + y² and b = xy]

_{Formula used:}

_{(a + b)(a – b) = a}² – b²

_{i. (xy + pq)(xy – pq)}

Taking a = xy and b = pq, then from above identity,

= (xy)² – (pq)²

= x²y² – p²q²

ii. 49 × 51

= (50 – 1)(50 + 1)

Taking a = 50 and b = 1, then from above identity,

= 50² - 1²

= 2500 – 1

= 2499

iii. (2x – y + 3z)(2x + y + 3z)

Taking a = 2x and b = y + 3z, the from the above identity we have

= (2x)² – (y + 3z)²

= 4x² – (y² + 2(y)(3z) + (3z)²) [∵ (a + b)² = a² + 2ab + b²]

= 4x² – (y² + 6yz + 9z²)

= 4x² – y² – 6yz – 9z²

iv. 1511 × 1489

= (1500 + 11)(1500 – 11)

Taking a = 1500 and b = 11, then from above identity,

= (1500)² - 11²

= 2250000 – 121

= 2249879

v. (a – 2)(a + 2)(a² + 4)

= (a² – 2²)(a² + 4) [Taking a = a, and b = 2 in above identity]

= (a² – 4)(a² + 4)

= (a²)² - 4² [Taking a = a² and b = 4 in above identity]

= a⁴ – 16

_{Formulas used:}

_{(a + b)}² = a² + 2ab + b² (1)

_{(a – b)}² = a² – 2ab + b² (2)

_{(a) Given,}

Squaring both side, we get

⇒

Taking a = x, b , in identity (1)
⇒
⇒

Again squaring both side, we get

Squaring both side, we get

Taking a = x², b , in identity (1)
⇒
⇒
⇒

(b) Given,

Squaring both sides, we get

Taking a = m, b , in identity (1)

⇒
⇒
⇒

(c) i)

Given,

Squaring both sides, we get

Taking a = p, b , in identity (2)

⇒
⇒

ii)

From Part (i), we have

Adding 2 both side,

⇒

(d) Given,

a + b = 5

Squaring both sides, we get

(a + b)² = 25

⇒ a² + 2ab + b² = 25 [1]

Also, a – b = 1

Squaring both sides, we get

(a – b)² = 1

⇒ a² – 2ab + b² = 1 [2]

Adding [1] and [2], we get

a² + 2ab + b² + a² – 2ab + b² = 25 + 1

⇒ 2a² + 2b² = 26

⇒ 2(a² + b²) = 26 [3]

Subtracting [2] from [1], we get

a² + 2ab + b² – (a²– 2ab + b²) = 25 – 1

⇒ a² + 2ab + b² – a² + 2ab – b² = 24

⇒ 4ab = 24 [4]

Multiplying [3] and [4], we get

2(a² + b)² (4ab) = 26 × 24

⇒ 8ab(a² + b²) = 624

Hence, Proved!

(e) Given,

x – y = 3

xy = 28

To find: x² + y²

We know,

(x – y)² = x² + y² – 2xy

Putting values,

⇒ 3² = x² + y² -2(28)

⇒ 9 = x² + y² – 56

⇒ x² + y² = 56 + 9

⇒ x² + y² = 65

_{Formulas used:}

_{(a + b)}² = a² + 2ab + b² [1]

_{(a – b)}² = a² – 2ab + b² [2]

_{i. 2(a}² + b)²

_{= 2a}² + 2b²

_{= a}² + b² + a² + b²

_{= a}² + 2ab + b² + a² – 2ab + b²

_{Using identities [1] and [2],}

_{= (a + b)}² + (a – b)²

_{ii. 50x}² + 18y²

_{= 25x}² + 25x² + 9y² + 9y²

_{= 25x}² + 9y² + 25x² + 9y²

_{= (5x)}² + (3y)² + (5x)² + (3y)²

_{Using identities [1] and [2],}

_{Here, a = 5x, b = 3y}

_{= (5x)}² + 2(5x)(3y) + (3y)² + (5x)² – 2(5x)(3x) + (3x)²

_{= (5x + 3y)}² + (5x – 3y)²

_{iii. a}² + b² + c² + d² + 2(ac – bd)

_{= a}² + b² + c² + d² + 2ac – 2bd

_{Using identities [1] and [2],}

_{= a}² + 2ac + c² + b² – 2bd + d²

_{= (a + c)}² + (b – d)²

i. For making a complete square, we need to reduce the expression in x²± 2xy + y²form, so if we put t = ±1, then we get expression as

Hence, for t = ±1, the given equation will be a whole square form.

ii. For making a complete square, we need to reduce the expression in x² + 2xy + y²form, so if we add ±4x to given equation, we have

a² + 4 ± 4x

= a²± 4x + 4

= a²± 2(2)x + 2²

= (a ± 2)²

Hence, if we add 4x to the given expression, the expression becomes a whole square.

iii. Given,

a² – b² = 9 × 11

we know, a² – b² = (a - b)(a + b)

⇒ (a – b)(a + b) = 9 × 11

On comparison, we get

a – b = 9 [1]

a + b = 11 [2]

Adding equation [1] and [2]

a + b + a – b = 9 + 11

⇒ 2a = 20

⇒ a = 10

Putting value of ‘a’ in equation [1], we get

10 – b = 9

⇒ b = 10 – 9

⇒ b = 1

iv. Identity, as

Taking LHS

Now, we know

(x + y)² = x² + 2xy + y²

and (x – y)² = x² – 2xy + y²

Applying above identities in LHS, we get

= (x + y)² – (x – y)²

= x² + 2xy + y² – (x² – 2xy + y²)

= x² + 2xy + y² – x² + 2xy – y²

= 4xy

= RHS

As, LHS = RHS the given expression is an identity!

v. Square of any number is always positive

[Explanation: Suppose -a (a > 0) is a negative number, then

(-a)² = -a × (-a) = a²]

Therefore, a²and b² both are positive numbers

⇒ a² + b² is a positive number, as sum of two positive numbers is positive!

i. 6x = 72

Dividing both side by 6, we get

x = 12

ii. 9x + 2 = 20

⇒ 9x + 2 – 2 = 20 – 2

⇒ 9x = 18

Dividing both side by 9, we get

⇒ x = 2

iii. 4x – 2x + 3 = 9 – 4x

Simplifying, we get

2x + 3 = 9 – 4x

⇒ 2x + 4x = 9 – 3

⇒ 6x = 6

Dividing both side by 6, we get

⇒ x = 1

iv.

Taking LCM, we get

Cross Multiplying, we get

⇒ -6x = 4(21 – 2x)

⇒ -6x = 82 – 8x

⇒ 8x – 6x = 82

⇒ 2x = 82

Dividing both side by 2

⇒ x = 41

v. 2x – 5{7 – (x – 6) + 3x} – 28 = 39

⇒ 2x – 5(7 – x + 6 + 3x) – 28 = 39

⇒ 2x – 5(13 + 2x) – 28 = 39

⇒ 2x – 65 – 10x – 28 = 39

⇒ 2x – 10x = 39 + 28 + 65

⇒ -8x = 132

Dividing both side by -8, we get

vi.

Taking LCM,

⇒

Cross-Multiplying we get,

⇒ 5(7x + 1) = 12(x + 79)

⇒ 35x + 5 = 12x + 948

⇒ 35x – 12x = 948 – 5

⇒ 23x = 943

Dividing both side by 23, we get

⇒ x = 41

(i). This is a simple activity, which can be done through some facts and handy items.

Materials Required: Paper, Scissors, Scale, Pencil

Process: Let us take a piece of paper and scissors. Cut this piece of paper into a square using scale and scissors.

Now, we have a square sheet of paper.

We can also mark the vertices for identification of angles.

A fact: We know that, the angles of a square are right angles.

That is, ∠ABC = ∠BCD = ∠CDA = ∠DAB = 90°

So, it is easy to construct a 90°.

Since, we know that each angle of a square is a right angle, so we can say that

∠ABC = 90°

Hence, we have constructed a 90° angle.

Now, let us try to construct 45°.

Mathematically,

⇒ Half of 90° = 45°

So, just like we did mathematically we need to do practically, that is, fold the paper into halves from the vertex of ∠ABC.

Then, unfold it to get 45° angle.

Hence, we have constructed a 45° angle.

Now, let us construct .

By looking at the right side, we can say that we need to half the angle 45°.

Practically,

We need to fold the paper by BD. We get 45°.

Then, fold the paper again by the same vertex.

We get half of 45°.

Then, unfold the paper.

The angle obtained is the required angle.

Hence, we have constructedangle.

(ii). For this activity,

Materials required: a sheet of paper, scale, scissors, pencil and protractor

Theory:

Let us understand what a parallelogram is.

A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.

Process:

Let us draw a parallelogram on a sheet of paper first and cut it along the edges.

Name it ABCD.

Here, AD ∥ BC & AB ∥ DC.

Also, AD = BC & AB = DC.

Here, the angles are ∠ABC, ∠BCD, ∠CDA and ∠DAB.

Let us draw the angles.

Now, along BD fold the paper such that there appears a visible crease.

And then unfold it.

Similarly, along AC fold the paper such that there appears a visible crease.

And then unfold it.

We have got the intersection of the four angles.

We need to analyze this intersection. Let this angle be O.

Using a protractor, measure the angles ∠AOB, ∠BOC, ∠COD and ∠DOA.

Thus, the angles formed by the intersecting diagonals AC and BD are not right-angles.

Thus, the diagonals are bisecting the other diagonal.

Hence, we have analyzed that the diagonals when intersected bisect each other.

(iii). For this activity,

Materials Required: sheet of paper, pencil, scale, scissors and protractor.

We need to verify the properties of diagonals of square.

Process:

Let us understand what square is.

The square is a geometric shape that belongs to the quadrilateral family because it has 4 sides. The 4 sides are the same length and are parallel to each other.

Properties of diagonals of square:

(a). The two diagonals are of equal length.

(b). Each diagonal bisects the other.

A diagonal is a line segment joining two vertices of a polygon or polyhedron, when those vertices are not on the same edge.

With the help of scale and pencil, draw diagonals connecting AC and BD.

Name the intersection point of AC and BD as O.

Let us verify:

(a). Let us measure the length of the diagonals AC and BD.

Thus, the diagonals of square are of equal length.

(b). Measure the length OA, OB, OC and OD.

Thus, the diagonal bisects the other diagonal.

We need to verify the properties of diagonals of rectangle.

Process:

Let us understand what rectangle is.

A rectangle is a quadrilateral with four right angles. It can also be defined as an equiangular quadrilateral, since equiangular means that all of its angles are equal. It can also be defined as a parallelogram containing a right angle.

Properties of diagonals of rectangle:

(a). The two diagonals are of equal length.

(b). Each diagonal bisects the other.

(c). Opposite central angles are equal.

A diagonal is a line segment joining two vertices of a polygon or polyhedron, when those vertices are not on the same edge.

With the help of scale and pencil, draw diagonals connecting AC and BD.

Name the intersection point of AC and BD as O.

Let us verify:

(a). Let us measure the length of the diagonals AC and BD.

Thus, the diagonals of a rectangle are of equal length.

(b). Measure OA, OB, OC and OD.

Thus, the diagonal bisects the other diagonal.

(c). Measure the central angles, that is, ∠AOB, ∠BOC, ∠COD and ∠DOA.

⇒ ∠AOB = ∠COD

⇒ ∠BOC = ∠DOA

Thus, opposite central angles are equal.

We need to verify the properties of diagonals of rhombus.

Process:

Let us understand what rhombus is.

A rhombus is a simple quadrilateral whose four sides all have the same length. Another name is equilateral quadrilateral, since equilateral means that all of its sides are equal in length.

Properties of diagonals of rhombus:

(a). The two diagonals bisect each other.

(b). The angle at which the diagonals cross is always 90°.

A diagonal is a line segment joining two vertices of a polygon or polyhedron, when those vertices are not on the same edge.

With the help of scale and pencil, draw diagonals connecting AC and BD.

Name the intersection point of AC and BD as O.

Let us verify:

(a). Let us measure the length of OA, OB, OC and OD.

Thus, the diagonals of a rhombus bisect each other.

(b). Now, measure angles ∠AOB, ∠BOC, ∠COD and ∠DOA.

Thus, the angles at which diagonals cross is 90°.

(iv). We need to construct a rhombus by 4 set squares.

First, let us understand what a set square is.

A set square or triangle is an object used in engineering and technical drawing, with the aim of providing a straightedge at a right angle or other particular planar angle to a baseline. The simplest form of set square is a triangular piece of transparent plastic with the centre removed.

And we know what a rhombus is.

A rhombus is a simple quadrilateral whose four sides all have the same length. Another name is equilateral quadrilateral, since equilateral means that all of its sides are equal in length.

Materials required: A sheet of paper, pencil, set squares, protractor and ruler

Process:

(a). Draw a line of length 4 cm using a ruler.

(b). Using a protractor, take any angle (say, 70°), draw a line AD with ∠BAD = 70°.

(c). Using a ruler and a set square, draw a free line. Using ruler and set square, we can ensure to draw a line parallel to AD. Measure BC = 4 cm using a ruler.

(d). Now, just join CD.

We can also measure CD, which should come out to be 4 cm.

Hence, we have got the rhombus using set protractor.

(v). We need to draw angles 30°, 45°, 60°, 90°, 105° and 120° using set square.

We know that,

A set square or triangle is an object used in engineering and technical drawing, with the aim of providing a straight edge at a right angle or other particular planar angle to a baseline. The simplest form of set square is a triangular piece of transparent plastic with the centre removed.

Basically,

Set squares are used in conjunction with T-squares to draw accurate angles. There are two main types of set square. One has an angle of 45 degrees and the other 30/60 degree angles.

Materials Required: A sheet of paper, set protractor, ruler and pencil.

Note: Set squares are only accurate if they are used along with a T-square. The set square must rest on the T-square which should be pushed against the edge of the board.

Steps of construction of 30°:

1. We need to construct an angle of 30°. Let ∠AOB = 30°.

2. To draw an angle of 30°, place the set square with the 30° angle, and then draw a ray OB starting from the vertex measuring 30°.

3. Now, join OA for the other edge, so that we get ∠AOB = 30°.

Thus, we have got the angle 30°.

Steps of construction of 45°:

1. We need to construct an angle of 45°. Let ∠AOB = 45°.

2. To draw an angle of 45°, place the set square with the 45° angle, and then draw a ray OB starting from the vertex measuring 45°.

3. Join OA for the other edge, so that we get ∠AOB = 45°.

Thus, we have got the angle 45°.

Steps of construction of 60°:

To construct an angle of 60°, use Set Square of 30° that we used earlier.

1. In order to draw 60°, place the set square with 60° angle, and then draw a ray OB starting from the vertex measuring 60°.

2. Join OA to form the other edge of the angle, so that we get ∠AOB = 60°.

Thus, we have got the angle 60°.

Steps of construction of 90°:

Both the set squares have 90°, so either of them can be used to construct an angle of 90°.

1. Place the 30/60 degree set square or 45 degree set square at 90 degree, and draw a ray OB starting from the vertex measuring 90°.

2. Join line OA for the other edge, so that we can get ∠AOB = 90°.

Thus, we have got the angle 90°.

Steps of construction of 105°:

In order to construct an angle of 105°, we need to use both the set squares (30/60 degree and 45 degree), because

60° + 45° = 105°

1. Place the 45 degree set square at 45°, and then draw a ray OC starting from the vertex measuring 45°.

2. But as we want 105° angle, so place the 60 degree set square just adjacent to 45 degree set square such that its vertex coincides with O, draw a ray OB starting from O vertex.

3. We can rub out OC and join OA. We get ∠AOB = 105°.

Thus, we have got the angle 105°.

Steps of construction of 120°:

In order to construct an angle of 120°, we need to use two 60 degree set squares, because

60° + 60° = 120°

1. Place the 60 degree set square at 60°, and then draw a ray OC starting from the vertex measuring 60°.

2. But as we want a 120° angle, so place the 60 degree set square again just adjacent to the 60 degree set square just used such that its vertex coincides with O, draw a ray OB starting from O vertex.

3. We can rub out OC and join OA. We get ∠AOB = 120°.

Thus, we have got the angle 120°.

We need to draw angles 90°, 45°, , 60°, 30°, 120°, 75°, 105°, 135° and 150°.

Materials Required: Sheet of papers, pencil, compass and a scale.

Steps of construction of 90°:

1. Draw a line OA of any length using a scale.

2. Place the point of a compass at vertex O and draw an arc or circle of any radius smaller than length of OA, cutting the line OA at Q.

3. Place the point of the compass at Q, draw an arc or circle of radius equals to the length of OQ, cutting the arc drawn in step 2 at R.

4. Now, taking R as center, place the point of the compass at R and draw an arc or circle of radius equals to the length of OQ. Cut the arc drawn in step 2 at S.

5. Now, take S as center and draw an arc of radius equals to the length of OQ, cutting the circle/arc drawn in step 4 at B.

6. Join OB.

Thus, the angle obtained ∠AOB = 90°.

Steps of construction of 45°:

1. We have already constructed 90° angle.

2. Now, taking O as centre again draw an arc or circle of any radius cutting OA at X and OB at Y.

3. Taking X as centre, place the point of compass at X and draw an arc or circle of radius OX.

4. Now, taking Y as centre, draw an arc with the same radius OX cutting the previous arc at Z.

5. Join OZ and extend the line to C.

Thus, the angle obtained ∠AOC = 45°.

Steps of construction of :

In order to construct , we can simplify it first.

1. We have already constructed 45°.

2. Now, taking O as centre again draw an arc or circle of any radius cutting OA at X’ and OC at Y’. We can rub out other lines in order to avoid confusion.

3. Taking X’ as centre, place the point of compass at X’ and draw an arc or circle of radius OX’.

4. Now, taking Y’ as centre, draw an arc with the same radius OX’ cutting the previous arc at Z’.

5. Join OZ’ and extend the line to D.

Thus, the angle obtained .

Steps of construction of 60°:

1. Draw a line OA of any radius using a scale.

2. Fix one end of a compass at point O, and draw an arc or circle of radius equals to approximately half of length OA, cutting the line OA at Q.

3. From the point Q, draw an arc or circle of same radius cutting the arc formed in step 2 at R.

4. Join OR and extend it to B.

Thus, we have got ∠AOB = 60°.

Steps of construction of 30°:

1. Draw a line OA of any radius using a scale.

2. Place one end of compass at point O, and draw an arc or circle of any radius convenient, cutting the line OA at Q.

3. Place the compass at point Q, and draw an arc or radius of the same radius cutting the previous arc at R.

4. Now, place the compass at point R, and draw an arc or circle of the same radius cutting the arc drawn in step 3 at S.

5. Join OS, and extend the line to B.

Thus, we have got ∠AOB = 30°.

Steps of construction of 120°:

1. Draw a line OA of any radius using a scale.

2. Place one end of compass at point O, and draw an arc or circle of any radius convenient, cutting the line OA at Q.

3. Place one end of compass at point Q, and draw an arc or circle of the same radius, cutting the previous arc at R.

4. Place one end of the compass at point R, and draw an arc or circle of the same radius cutting the previous arc at S.

5. Join OS, and also extend it to B.

Thus, we have got the ∠AOB = 60°.

Steps of construction of 75°:

1. Draw a line OA of any radius using a scale.

2. Place one end of compass at point O, and draw an arc or circle of any radius convenient, cutting the line OA at Q.

3. Place one end of compass at point Q, and draw an arc or circle of the same radius, cutting the previous arc at R.

4. Place one end of the compass at point R, and draw an arc or circle of the same radius cutting the previous arc at S.

5. Now, take S as centre and draw an arc or circle of same radius cutting the previous arc at C.

6. Join OC, and name the intersection with the arc drawn at step 1 as Z.

7. Now, set the compass at Z with a radius just more than half of ZR and draw an arc or radius.

8. With the same radius, place the compass at R and draw an arc or radius cutting the previous arc at Y.

9. Join OY, and extend the line to B.

Thus, we have got ∠AOB = 75°.

Steps of construction of 105°:

1. Draw a line OA of any radius using a scale.

2. Place one end of compass at point O, and draw an arc or circle of any radius convenient, cutting the line OA at Q.

3. Place one end of compass at point Q, and draw an arc or circle of the same radius, cutting the previous arc at R.

4. Place one end of the compass at point R, and draw an arc or circle of the same radius cutting the previous arc at S.

5. Now, take S as centre and draw an arc or circle of same radius cutting the previous arc at C.

6. Join OC, and name the intersection with the arc drawn at step 1 as Z.

7. Now, set the compass at Z with a radius just more than half of ZS and draw an arc or radius.

8. Now, with S as centre, draw an arc or circle with the same radius cutting the previous arc at Y.

9. Join OY, and then extend the line to B.

Thus, we have got ∠AOB = 105°.

Steps of construction of 135°:

1. Draw a line OA of any radius using a scale.

2. Place one end of compass at point O, and draw an arc or circle of any radius convenient, cutting the line OA at Q.

3. Place the compass at Q and draw an arc or circle of the same radius, cutting the arc drawn in step 2 at R.

4. Setting the compass at R, with the same radius draw an arc or circle cutting the previous arc at S.

5. Again, setting the compass at S, draw an arc or circle with the same radius cutting the previous arc at P and T. Also, join OT so that we get ∠AOT = 180°.

6. Also, join OP so that ∠AOP = 90° cutting the arc drawn in step 2 at Z.

7. Taking Z and T as centers, with the same radius draw an arc or circle cutting each other at intersection Y.

8. Join OY and extend the line to B.

Thus, we have got ∠AOB = 135°.

Steps of construction of 150°:

1. Draw a line OA of any radius using a scale.

2. Place one end of compass at point O, and draw an arc or circle of any radius convenient, cutting the line OA at Q.

3. Place the compass at Q and draw an arc or circle of the same radius, cutting the arc drawn in step 2 at R.

4. Setting the compass at R, with the same radius draw an arc or circle cutting the previous arc at S.

5. Again, setting the compass at S, draw an arc or circle with the same radius cutting the previous arc at P and T. Also, join OT so that we get ∠AOT = 180°.

6. Also, join OP so that ∠AOP = 90° cutting the arc drawn in step 2 at Z.

7. Taking T as center, with the same radius draw an arc or circle cutting the arc drawn in step 5 at Y such that, ∠POY = 60°.

8. Extend the line OY to B.

Thus, we have got ∠AOB = 150°.

Given: A quadrilateral PLAN with

PL = AN = 6 cm

PN = LA = 5 cm

To Do: Draw three types of quadrilateral PLAN, and then assess the criteria as to when it will be rectangle.

Process:

Let us understand what a quadrilateral is.

A quadrilateral is a polygon with four edges and four vertices or corners. There are basic 6 types of quadrilaterals:

- Rectangle

- Square

- Parallelogram

- Rhombus

- Trapezium

- Kite

Let’s draw a type of quadrilateral: Square

A square is a regular quadrilateral, which means that it has four equal sides and four equal angles.

Here, PL = LA = AN = NP = 5 cm

∠PLA = ∠LAN = ∠ANP = ∠NPL = 90°

⇒ PLAN is a square.

Let’s draw another type of quadrilateral: Parallelogram.

A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.

In the quadrilateral PLAN above,

PL ∥ AN and PN ∥ LA

PL = AN and PN = LA

Also, ∠PNA = ∠PLA and ∠LAN = ∠LPN

⇒ PLAN is a parallelogram.

Let’s draw third type of quadrilateral: Rectangle.

A rectangle is a quadrilateral with four right angles. It can also be defined as an equiangular quadrilateral, since equiangular means that all of its angles are equal.

Here, PL = AN = 6 cm and LA = NP = 5 cm

∠PLA = ∠LAN = ∠ANP = ∠NPL = 90°

Hence, PLAN is a rectangle.

So, the criteria required for a quadrilateral to become a rectangle are,

- Opposite sides are parallel and equal

- All angles are right angle

Here, we are supposed to write the conditions required to draw a fixed square, that is, a square with a fixed radius.

Let us understand what a condition is.

Condition is a requirement necessary for a given statement or theorem to hold. Also called a criterion.

Let us take radius to be r = 4 cm.

The conditions required to draw a fixed square of radius 4 cm are:

1. Sides are equal – There are four sides in a square, and all the four sides are equal. We need to fix the length of the sides of the square to be 4 cm as fixed.

2. Angles are right-angled – There are four angles in a square. All the four angles are equal and are necessarily 90° angled.

A fixed square can be formed using these two simple conditions.

We are supposed to write the conditions required to draw a fixed parallelogram, that is, a quadrilateral.

Let us understand what a condition is.

Condition is a requirement necessary for a given statement or theorem to hold. Also called a criterion.

Let us fix the base at the length of 6 cm.

The conditions required to draw a fixed parallelogram are:

1. Sides are equal and parallel: There are four sides in a parallelogram, and the opposite sides are equal and parallel. So, once the base is fixed at length 6 cm, the opposite side also measures 6 cm. Also, these two sides are parallel.

While take the other opposite sides to be of same length, say, of 5 cm and parallel to each other.

2. Opposite angles are equal: There are four angles in a parallelogram. Check for opposite angles to be equal.

A fixed parallelogram can be formed using these two simple conditions.

It is given that,

DE = 5.6 cm

To do: Draw a square DEAR.

Required: Paper, pencil, and ruler and a compass.

Steps:

Note that, a square is a type of quadrilateral that has 4 equal sides and each angle making 90°.

1. Draw a line DE = 5.6 cm on a paper using a ruler.

2. Now, we need to construct 90° angle at vertex D as well as E (since, this is a square). So, keep one end of the compass at D, and draw an arc of any radius smaller than DE, say, 2 cm, cutting the line DE at Q.

3. Keeping one end of compass at Q, draw an arc or circle of the same radius cutting the arc drawn in step 2 at S.

4. Taking S as centre, draw an arc or circle of same radius cutting the previous arc at T.

5. Now, taking T as centre draw an arc or circle of same radius cutting the previous arc at Z.

6. Join DZ and extend it to R, such that, DR = 5.6 cm. (since, DEAR is a square)

Thus, we have got ∠EDR = 90°.

7. Similarly, taking E as centre and repeat the steps.

Thus, we have got ∠DEA = 90°.

8. Now, join AR.

Now, measure the angles that should be 90° each.

Thus, we have got the square DEAR.

Given that,

BE = 6 cm and ES = 4.8 cm

To do: We need to draw a rectangle BEST.

Materials required: Scale, compass, pencil and a sheet of paper.

Steps:

Note that, a rectangle is a type of quadrilateral that has equal and parallel opposite sides with each angle at 90°.

1. Draw a line on a sheet of paper using ruler, say, BE = 6 cm.

2. Now, we need to construct 90° angle at B as well as E (since, this is a rectangle). So, keep one end of compass at B and draw an arc or circle of radius 2 cm (say), cutting the line BE at P.

3. Keep the compass at P, and draw an arc or circle of the same radius cutting the previous arc at Q.

4. Keep one end of the compass at Q and draw an arc or circle of the same radius cutting the arc drawn in step 2 at R.

5. Taking R as centre, draw an arc or circle of the same radius cutting the previous arc at Z.

6. Join BZ, and extend the line to T.

We have got ∠EBT = 90°.

7. Similarly take E as centre and repeat the steps.

We have got ∠BES = 90°.

8. Join TS.

Now, measure all the angles that should come out to be 90°.

Thus, we have got the rectangle BEST.

Given that,

∠HOM = 60° and HO = 6 cm

To do: Draw a rhombus HOME.

Materials required: ruler, compass, pencil and a sheet of paper.

Steps:

Note that, a rhombus is a simple quadrilateral whose four sides all have the same length.

1. Draw a line HO = 6 cm using a ruler and a pencil on s sheet of paper.

2. Now, let us draw angle 90° at vertex O, so that ∠HOM = 60°. Keep one end of the compass at O and draw an arc or circle of the radius, say, 2 cm. Let it cut the line HO at P.

3. Keep one end of the compass at point P and draw an arc or circle of the same radius cutting the previous arc at Q.

4. Join OQ and extend the line to M, such that OM = 6 cm.

5. Now, using a ruler draw a straight line ME = 6 cm.

6. Now, join EH.

Note that, the opposite angles are equal in a rhombus.

Thus, we have got the rhombus HOME.

Given that,

RA = 8 cm and OD = 6 cm

To do: Draw a rhombus ROAD.

Materials required: ruler, compass, pencil and a sheet of paper.

Steps:

Note that, a rhombus is a simple quadrilateral whose four sides all have the same length.

1. Draw a line RA = 8 cm using a ruler on a sheet of paper.

2. Draw perpendicular bisector XY of AC. For this, keep one end of the compass at R and draw an arc or circle of radius, say, 4.5 cm above as well as below the line RA.

3. Similarly, taking A as centre draw an arc or circle above and below the line RA intersecting the previous arc at X and Y.

4. Join XY cutting the line RA at O’. XY is the perpendicular bisector of the line RA.

5. From O’, extend the line to O’D on line O’X, where O’D = 3 cm (half of 6 cm) using a ruler.

6. Similarly, from O’ extend the line to O’O on line O’Y, where O’O = 3 cm (half of 6 cm).

7. Join RO, OA, AD and DR.

Here, O’D = 3 cm and O’O = 3 cm.

And, OD = O’D + O’O

⇒ OD = 3 + 3

⇒ OD = 6 cm

Thus, we have formed a rhombus, where RA = 8 cm and OD = 6 cm.

We are given with,

GO = 7 cm

OL = 5.8 cm

GL = 5.8 cm

To do: Construct a parallelogram GOLD.

Recall: a parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.

So, we need to keep in mind, that

- The opposite sides are equal and parallel.

- The opposite angles are equal.

- The diagonals bisect each other.

- No angle can be of 90° or else it will be a square.

Materials required: ruler, pencil, compass, and a sheet of paper

Steps:

1. Firstly, draw a line of 7 cm on a paper using ruler.

2. Now, set the compass at 5.8 cm using a ruler then keeping one of the compass at G, draw an arc or circle at that fixed 5.8 cm radius.

3. Similarly, without changing the radius draw an arc or circle keeping O at centre now. Note that this arc cuts the previous arc, name it L. (Since, we wanted OL = GL = 5.8 cm)

4. We can join GL and OL.

Here, GL is a diagonal and OL is one of the sides of the parallelogram.

5. Now, since opposite sides of parallelogram are equal and parallel, draw a straight line from L measuring 7 cm, call it LD.

6. Then, join DG as well.

Measure the angles in the end.

Thus, we have got the parallelogram GOLD.

i. We are given that,

There is a rectangle named ABCD, where

AC = 5 cm.

We need to find the length of the other diagonal, BD.

We know that,

A rectangle is a quadrilateral with four right angles having equal opposite sides.

This means,

AD = BC

AB = DC

And

∠ABC = ∠BCD = ∠CDA = ∠DAB = 90°

Take ∆ADC and ∆BAD,

AD = AD [∵, common sides]

DC = AB [∵, opposite sides are equal in rectangle]

∠ADC = ∠BAD [∵, each angle measure 90° in a rectangle]

⇒ ∆ADC ≅ ∆BAD by SAS congruency.

So, by cpct (corresponding parts of congruent triangles),

AC = BD

So, if AC = 5 cm, then BD = 5 cm.

ii. We are given that,

There is a square PQRS.

PR and QS are two diagonals intersecting at O.

PR = 5 cm.

We need to find the length of QO.

We know that,

The square is a geometric shape that belongs to the quadrilateral family because it has 4 sides. The 4 sides are the same length and are parallel to each other.

This implies,

PQ = QR = RS = SP

And,

∠PQR = ∠QRS = ∠RSP = ∠SPQ = 90°

Take ∆PQR and ∆QRS,

PQ = QR [∵, every side is equal in a square]

QR = RS [∵, every side is equal in a square]

∠PQR = ∠QRS [∵, every angle is equal to 90° in a square]

⇒ ∆PQR ≅ ∆QRS by SAS congruency.

So, by cpct (corresponding parts of congruent triangles),

PR = QS

So, if PR = 5 cm, then QS = 5 cm.

Also, we know the property of the square that the diagonals bisects each other.

So, QS = QO + OS

Where, QO = OS.

⇒ QS = QO + QO

⇒ QS = 2 QO

And QS = 5 cm

⇒ QO = 2.5 cm

Thus, QO = 2.5 cm.

iii. We are given that,

There is a parallelogram ABCD.

∠ABC = 60°

We need to find the ∠ADC.

We know that,

A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.

This means,

AB = DC & AB ∥ DC

BC = AD & BC ∥ AD

Also,

∠ABC = ∠ADC

∠BCD = ∠DAB

So, if ∠ABC = 60°, then ∠ADC = 60°.

iv. We are given that,

There is a rhombus ABCD.

AC and BD are diagonals of the rhombus intersecting at O.

We need to find ∠AOB.

We know that,

A rhombus is a simple quadrilateral whose four sides all have the same length, with opposite angles being equal. The diagonals bisect each other at right angles.

This means,

AB = BC = CD = DA

Also,

∠AOB = ∠BOC = ∠COD = ∠DOA = 90°

Thus, ∠AOB = 90°.

v. A square is always a rhombus but a rhombus is not always a square.

Explanation – Recall, square and rhombus both have 4 sides that is why they belong to quadrilateral family.

A square has 4 equal sides with interior angles of 90° each.

While, a rhombus has 4 equal sides but opposite interior angles are equal.

So, we can say that, a square can also be called a rhombus since it satisfies rhombus’s properties. But a rhombus cannot be called a square always since it doesn’t have interior angles as 90°.

iv. A square is always a rectangle but a rectangle is not always a square.

Explanation – Recall, square and rectangle both have 4 sides that is why they belong to quadrilateral family.

A square has 4 equal sides with interior angles of 90° each.

While, rectangle has equal opposite sides with interior angles of 90° each. The adjacent sides are not equal.

Since, square is a quadrilateral with all interior angles of 90°, it can be called a rectangle. (Since, it satisfies every property of a rectangle)

But, a rectangle cannot be called a square since every side of the rectangle is not equal.