Area And Perimeter Of Triangle And Quadrilateral Shaped Region

The distance to be covered to go one round along perimeter of this pentagon is 16.2 cm + 10 cm + 12 cm + 13 cm + 14.8 cm

The distance to be covered to go one round along perimeter of this quadrilateral is

10 cm + 7 cm + 19.4 cm + 21 cm

The distance to be covered to go one round along perimeter of this quadrilateral is

12 cm + 3 cm + 5.6 cm + 19 cm

The distance to be covered to go one round along perimeter of this hexagon is

9 cm + 4 cm + 8 cm + 19 cm + 6 cm + 15 cm

The distance to be covered to go one round along perimeter of this quadrilateral is

12 cm + 9 cm + 16 cm + 26 cm

Do it yourself.

The length of diagonal of square land meter

Where a = side of square

meter

Perimeter of square land

Where a = side of square

The length in meter required for fencing a wall, perimeter meter

Hence, 80 meter length required for fencing a wall surrounding of it.

The length of rectangular land meter

The width of rectangular land meter

The distance covered to go one round along outer path ABCD = perimeter of rectangle

Where a = length and b = breadth

meter

meter

Cost required for fencing around the outer side of path = Perimeter × Rs. per meter

Hence, cost required for fencing around the outer side of path is Rs. 2,052.

Note: Perimeter of equilateral triangle with side ‘a’ is

(i)Perimeter of rectangle

Where a = length and b = breadth of rectangle respectively

cm

cm

Now, the perimeter of Equilateral triangle cm

cm

cm

(ii)Perimeter of square

Where a = side of square

cm

cm

Now, the perimeter of equilateral triangle = 36 cm

cm

cm

(iii)Perimeter of quadrilateral = The distance covered to go one round along perimeter

cm

cm

Now, the perimeter of equilateral triangle = 39 cm

cm

cm

(iv)Perimeter of parallelogram = 2 (a + b)

Where a = length and b = breadth

cm

cm

Now, the perimeter of equilateral triangle = 66 cm

cm

cm

(v)Perimeter of triangle = (a + b + c)

Where a, b and c are sides of triangle respectively

cm

cm

Now, the perimeter of equilateral triangle = 30 cm

cm

cm

(vi)Perimeter of triangle = (2a + b)

Where a and b are sides of isosceles triangle

cm

cm

cm

Now, the perimeter of equilateral triangle = 45 cm

cm

cm

(i) Given triangle is a right angle triangle whose area would be

Area of triangle

In a right angled triangle, according to the Pythagoras theorem

Hypothenuse² = Base² + height² ………. eq 1

In the given triangle base = 5cm

Hypotenuse = 13cm

Height = ?

By eq 1 we have

⇒ 13² = 5² + height²

⇒ 169 - 25 = height²

⇒ height = √144 cm

⇒ height = 12cm

Area of the given triangle

= 30cm²

(ii) Given triangle has all the sides equal as 6 cm so it`s a equilateral triangle

Area of a equilateral triangle

Here side = 6cm

So area of triangle

= 9√3 cm²

(iii) Two sides of the given triangle are equal = 6cm and base = 8 cm

So it’s an isosceles triangle

Area of an isosceles triangle

(∵ half of base = 4cm)

=

= 4× √20

= 8 √5 cm²

(iv) Given figure is a trapezium with one angle ∠ ABC = 90°

In right triangle ABC

CB = 5cm, is the base

AB = 12 cm, is the height of the triangle

AC is the hypothenuse which will be given by Pythagoros theorem

Hypothenuse² = Base² + height² ………………….eq1

In the given figure

AC² = CB² + AB²

⇒ AC² = 5² + 12²

⇒ AC^{2 =} 25 + 144

⇒ AC = √169

⇒ AC = 13cm

Area of right - angled triangle ABC

= 30cm²

Now in triangle ACD

AC = 13cm

DC = 10cm given

AD = 7 cm given

Area of triangle with given three sides

Where a, b, and c are the sides of the triangle

and s (semi perimeter of triangle )

⇒

⇒ s = 15 cm

Area of ADC triangle

=

=

= 30√2 cm²

Area of given figure = area of triangle ABC + area of triangle ACD

Area of given figure = 30cm² + 30√2 cm²

= 60√2 cm²

Let the depth of the water be BD = x cm

According to the given condition

AD = CD = (x + 2) cm because the straight lotus submerged into lake covering distance BC = 15cm

Now in right angled triangle DBC

CD = x + 2, is hypotenuse

BC = 15, is base and

BD = x, is the height which is also the depth of water

According to the Pythagoras theorem

Hypothenuse² = Base² + height²

CD² = BD² + BC²

(x + 2)² = x² + 15²

⇒ x² + 4 + 4x = x² + 225 (∵ (a + b)² = a² + b² + 2ab)

⇒ 4x = 221

⇒

⇒ x = 55.25 cm

The depth of water = 55.25cm

Given

In an isosceles right angled triangle

Hypotenuse = 12√2 cm

Since it is the isosceles triangle

⇒ the length of the remaining two sides would be same

Let the length of remaining two sides be x cm

According to the Pythagoras theorem,

Hypothenuse² = Base² + height²

⇒ (12√2)² = x² + x²

⇒ 144 × 2 = 2x²

⇒

⇒ 144 = x²

⇒ x = ± 12 cm

Since length cannot be negative

⇒ Length (x) of remaining two sides = 12cm

Area of an isosceles triangle

= 72cm²

Given,

In a triangular park length of three sides are AB = 65 m, AC = 70 m and BC = 75 m and AD = x m, is the height of the perpendicular drawn to BC

AD is the perpendicular drawn on the longest side BC

Since all the three sides are different of the triangular park it forms a scalene triangle

Area of a scalene triangle with given three sides

Where a, b, and c are the sides of the triangle

and s (semi perimeter of triangle )

Here,

⇒

⇒ s = 105 m

Area of triangular park m²

m²

= m²

= 2100 m²

But, Area of triangle

So area of triangle ABC

⇒

⇒

⇒ x = 56 m

Therefore, height of the perpendicular drawn from the vertex opposite to longest side is 56m

Area of a triangle

Given there are two triangles

And ratio of their area

Also ratio of their height

Where, b₁is the base and h₁is the height and A₁is the area of suja`s triangle

And b₂is the base and h₂is the height and A₂is the area of my triangle

Ratio of area of triangle

⇒

⇒

⇒

⇒

The ratio of the bases of two triangles is 16:9

From the figure, we see that the garden measures 20m x 20m

Area = L × B = 20m × 20m = 400m²

(ii) Varandah measures 10m x 5m.

Total Area of Varandah = L × B = 10m × 5m = 50m²

Cost per square meter = Rs. 30

If 1 m² costs Rs. 30

50m² costs Rs. 50 × 30 = Rs. 1500

(iii) Length of the reading room = 5m

Breadth of the reading room = 6m

As,

Area of rectangle = length × breadth

Area of floor of his room = 5 × 6 = 30 m²

Area of one tile = length × breadth

= 25 cm × 25 cm

= 0.25 m × 0.25m

= 0.0625 m²

No of tiles required

We know, area of rectangle = length × breadth

(i) Area of colored region(green) = area of outer rectangle – area of inner rectangle

Now, dimensions of outer rectangle = 12 × 8

⇒ area of outer rectangle = 96 m²

Width of each strip = 4 m

Therefore, dimension of inner rectangle = (12 – 3) × (8 – 3)

⇒ area of inner rectangle = 45 m²

⇒ area of colored region = 96 – 45 = 51 m²

(ii) Area of colored region = area of whole rectangle – area of 4 small rectangles

Now, dimensions of whole rectangle = 26 × 14

⇒ area of outer rectangle = 364 m²

Width of each strip = 3 m

Let length of rectangle be ‘l’ and width be ‘b’

Now,

Length of 2 small rectangles + width of strip = 26 m

⇒ 2l + 3 = 26

⇒

breadth of 2 small rectangles + width of strip = 14 m

⇒ 2b + 3 = 14

⇒

⇒ area of one small rectangle = lb

⇒ area of four small rectangles = 231 m²

⇒ area of colored region = 364 – 231 = 133 m²

(iii) Area of colored region(violet) = area of outer rectangle – area of inner rectangle

Now, dimensions of inner rectangle = 16 × 9

⇒ area of outer rectangle = 144 m²

Width of each strip = 4 m

Therefore, dimension of outer rectangle = (16 + 2(4)) × (9 + 2(4)) = 24 × 17

⇒ area of inner rectangle = 408 m²

⇒ area of colored region = 408 – 144 = 264 m²

(iv) Area of colored region(orange) = area of outer rectangle – area of inner rectangle

Now, dimensions of outer rectangle = 28 × 20

⇒ area of outer rectangle = 560 m²

Width of each strip = 3 m

Therefore, dimension of outer rectangle = (28 - 2(3)) × (20 - 2(3)) = 22 × 14

⇒ area of inner rectangle = 308 m²

⇒ area of colored region = 560 – 308 = 252 m²

(v)

Area of colored region = area of strip I + 4 × area of strip II

Now, dimension of strip I = 3 cm × 120 cm

Hence, area of strip I = 360 cm²

Now, width of strip II = 3 cm

Also, 2 × length of strip II + width of strip I = 90 cm

⇒ 2 × length of strip II + 3 = 90

⇒ length of strip II = 43.5 cm

Hence, area of II strip = 43.5 × 3 = 130.5 cm²

Hence, area of required region = 360 + 4(130.5) = 882 cm²

Perimeter = 336m

Let length of the field be L, and breadth be B.

Perimeter of a rectangle = 2(L + B) = 336

Or L + B = 168

Given

3L = 4B

Or L =

Substituting the above result in L + B = 168,

On solving, we get B = 72

And since 3L = 4B, L = 96

Area = L × B = 96 × 72 = 6912m²

1 m² - - - Rs. 3.50

Area - - - - - Rs. 1400

So area = = 400m²

Let the side of square land be L.

Since it is a square land, Area = L²

Or L = = 20m

Perimeter of a square = 4L = 4 × 20 = 80m

1m - - - Rs. 8.5

80m - - - - - Rs. 80 × 8.5 = Rs. 680

Let original length be L and original breadth be B.

L × B = 500.

Since length decreased by 3m and Breadth increased by 2m gives rise to a square,

L - 3 = B + 2 (Because in a square, L = B)

Or L = B + 5

(B + 5) × B = 500

B² + 5B - 500 = 0

On solving the quadratic equation, we have B = 20 or B = - 25

Since the measurement cannot be negative,

B = 20

So L = B + 5 = 25

Area of wall = Area of outer square – Area of inner square

Length of outer square = 300 + 0.3m (3dcm = 0.3m)

= 300.3m

We know, area of square = (side)²

Area of outer square = 300.3² = 90180.09m²

Area of inner square = 300² = 90000m²

Area of wall = 90180.09 – 90000 = 180.09m²

Rate per square metre of wall = = Rs. 50/sq m

So for 1 m² - - - - Rs. 50

180.09m² - - - Rs. 50 × 180.09 = Rs. 9004.5

We know, area of rectangle = length × breadth

Area of outer rectangle = 14m × 12m = 168m²

Let the width of path be ‘x’ m.

Dimensions of inner rectangle will be (14 – 2x) × (12 – 2x)

And area of inner rectangular ground = (14 – 2x)(12 – 2x)

Now, area of path = area of whole garden – area of garden without path

Area of path

⇒ 69 = 168 – (14 – 2x)(12 – 2x)

⇒ 69 = 168 – 168 + 28x + 24x – 4x²

⇒ 4x² – 52x + 69 = 0

⇒ 4x² - 46x – 6x + 69 = 0

⇒ 2x(2x – 23) – 3(2x – 23) = 0

⇒ (2x – 3)(2x – 23) = 0

⇒ 2x – 3 = 0 or 2x – 23 = 0

⇒ x = 1.5 m or x = 11.5 m

Now, logically 11.5 m is not possible, because in that case the length of inner rectangular ground will be negative,

Hence, width of path = x = 1.5 m

Let the length of rectangular garden be L and breadth be B

We know, area of rectangle = length × breadth

Given, L = 40cm, A = 1200cm²

A = L × B

⇒ 1200 = 40 × B

⇒ B = 30cm

We know, diagonal of a rectangle

Where, L = length of rectangle and B = breadth of rectangle

Diagonal of rectangle = side of square

Also, we know area of square = (side)²

= (50)² = 2500 cm²

We know, area of rectangle = length × breadth

Area of square = (side)² = side × side

There are 4 walls, 1 floor, 3 doors and 4 windows in the hall.

Area of Wall 1 = 4m × 4m = 16m²

Area of Wall 2 = 6m × 4m = 24m²

Area of Wall 3 = 4m × 4m = 16m²

Area of Wall 4 = 6m × 4m = 24m²

Area of Floor = 4m × 6m = 24m²

Area of each door = 1.5m × 1m = 1.5m²

Total area for 3 doors = 4.5m²

Area of each windows = 1.2m × 1m = 1.2m²

Total area for 4 windows = 4.8m²

Net area = Area of floor + Area of walls – Area of doors – Area of windows

= 24 + 16 + 24 + 16 + 24 - 4.5 - 4.8 = 94.7m²

Total cost at the rate of Rs. 70 per sq m

= Rs. 94.7 × 70 = Rs. 6629

Let length of the room be L = 4m

Let breadth of the room be B = ?

We know, area of rectangle = length × breadth

Area of floor = L × B = 12m² = 4 × B

So B = 3m

Let the height of room is H.

Area of four walls = 2(L × H) + 2(B × H) = 48m²

So (L × H) + (B × H) = 24m²

Substituting L and B,

3H + 4H = 24m²

7H = 24

Or H = 3.24m

Let length be L, breadth be B.

We know, area of rectangle = length × breadth

Given, Area of rectangular picture = 84 cm²

⇒ L × B = 84cm²

Also, Given the difference of length and breadth is 5 cm.

⇒ L - B = 5cm

Or L = B + 5

Substituting, (B + 5) × B = 84

B² + 5B - 84 = 0

B² + 12B - 7B - 84 = 0

B(B + 12) - 7(B + 12) = 0

(B - 7)(B + 12) = 0

B cannot be negative, so B = 7cm.

L = 12cm

Perimeter of a rectangle = 2(L + B)

= 2(12 + 5) = 34cm

Let L and B be the total length and total breadth of the rectangle covering the path and the garden respectively.

Length of the garden

= L - 2 × width of path

= L – 2(2.5) = (L – 5)

Breadth of the garden

= B - 2 × width of path

= B – 5

We know, area of square = length × breadth

Area of the garden = (L - 5)(B - 5)

Total Area = L × B

Area of path = Total Area – Area of garden = 165

LB - (L - 5)(B - 5) = 165

⇒ LB – LB + 5L + 5B - 25

⇒ 5B + 5L - 25 = 165

⇒ 5(L + B) = 190

⇒ L + B = 38m

Let L = 20m

B = 18m

Area of the garden = (20 - 5)(18 - 5) = 195m²

We know, diagonal of a rectangle

Where, L = length of rectangle and B = breadth of rectangle

Length of the Diagonal = = 27m

We know, that the length of diagonal of square = L√2 units, where L is the side of square

Diagonal of the square = 20√2m

On comparing,

side of the square land = 20m

Also, we know

Area of square = (side)²

Area of the square land = 20² = 400m²

Cost for 1 m² to plant grass = Rs. 20

Cost for 400m² to plant grass = Rs. 20 × 400 = Rs. 8000

We know, diagonal of a rectangle

Where, L = length of rectangle and B = breadth of rectangle

L = 12m, B = 7m

Diagonal = = 13m

Since there are two diagonals, the length of fence is 13 × 2 = 26m

Or 5L = 9B

Perimeter of Rectangle = 2(L + B) = 140m

We know that L = B

B + B = 70

B = 70

B = 25m

L = 45m

Area of rectangle = 25m × 45m = 1125m²

Area of each tile = 25cm × 20cm = 0.25m × 0.2m = 0.05m²

Number of tiles required = = 22500

Rate for 100 tiles = Rs. 500

Rate for 1 tile = Rs. 5

Rate for 22500 tiles = Rs. 22500 × 5 = Rs. 1,12,500

Let original length be L and breadth be B.

L = 18m. B = ?

Area = L × B = 18B

New length = 18m, new breadth = B - 4 m

New Area = 18(B - 4)

Let the rate be X per sqm.

Now, 18B × X = 2160

18(B - 4) × X = 1620

Dividing two equations,

1620B = 2160(B - 4)

B = 16m

Perimeter = 2(L + B) = 68m

Area = L × B = 16 × 18m² = 288m²

Diagonal = 15m, Length - Breadth = 3m

From Pythagorean theorem,

Length² + Breadth² = Diagonal²

As, Length = Breadth + 3

Let Length = B + 3, Breadth = B and Diagonal = 15 m

(B + 3)² + B² = 15²

2B² + 6B - 216 = 0

Or, B² + 3B - 108 = 0

B² + 12B - 9B - 108 = 0

B(B + 12) - 9(B + 12) = 0

Breadth cannot be negative, so B = 9m

Length = 12m

Perimeter = 2(L + B) = 42m

Area = L × B = 108m²

Let the side of each tile is ‘a’, Let number of tiles required be N.

We know area of rectangle = length × breadth

Area of courtyard = 385 × 60 = 22100 m²

Now, we have to calculate the longest size of square tile that can be used for paving the courtyard,

As, 22100 = 2 × 2 × 5 × 5 × 17 × 13

Clearly, we have to make 17 and 13 eliminate to make it perfect square, therefore we take square tiles of (2 × 5 = 10 cm) side.

And no of tiles

We know that, diagonal of a square is a√2, where a is the side of the square.

If the length of the diagonal of the square is 12√2cm, then,

a√2 = 12√2

⇒ a = 12 cm

We know,

Area of square = a²

= 12² = 144sqcm.

Let the side of square A1 be ‘L’

We know,

Area of square = (side)²

A1 Area = L²

Also, we know Diagonal of square = L√2

Area of square drawn on diagonal A2 = (a√2)

= 2L²

Ratio of A1:A2 is L²:2L² = 1:2

We know,

Area of rectangle = Length × breadth

Area of the place = 6m × 4m = 24sqm = 2400sqcm

Also,

Area of square = (side)²

Area of each tile = (2cm)² = 4sqcm

Number of tiles = = 600

Let the rectangle have dimensions L and B

We know,

Perimeter = 2(L + B), Area = L × B

⇒ R = LB

Let the square have side as ‘a’

We know,

Perimeter = 4a, Area = a²

Given,

Perimeter of square = Perimeter of rectangle

⇒ 2(L + B) = 4a

⇒ L + B = 2a

Now, Area of square

Adding and subtracting from RHS

Since, , [Being a square]

⇒ S > R

Let sides of rectangle be L and B.

We know, Diagonal =

Or Diagonal² = L² + B²

⇒ 10² = L² + B²

⇒ 100 sq. cm = L² + B²

This can be rewritten as

100 = (L + B)² – 2LB

Area of rectangle = L × B = 62.5 sqcm

Hence,

100 = (L + B)² - 2 × 62.5

⇒ (L + B)² = 100 + 125

Or L + B = = 15cm

Let original Length be L.

10% increase is 0.1L

New Length = L + 0.1L = 1.1L.

Change in area = (1.1L)² –L² = 0.21L²

%change in area =

= 21%

Let original length and breadth be L and B respectively.

Original Area = L × B

10% increase in length corresponds to 0.1L.

10% decrease in length corresponds to 0.1B.

New length = 1.1L

New breadth = 0.9B

New area = (1.1L)(0.9B) = 0.99LB

%change in area =

= = - 1%

So 1% area will be decreased.

We see that it is not possible for the perpendicular to be of 2cm when the length of the rectangle is 5cm. (It should be 2.5cm)

No solution exists.

Let us consider a square ABCD with diagonals intersecting at ‘O’.

OP is perpendicular to any of its sides such that

OP = 2√2 cm

Let the side of square be ‘L’

Then diagonal of square, BD = L√2

Also,

Diagonals of square bisect each other, therefore

Also, By symmetry

Now, In ΔOPD, By Pythagoras theorem

Hypotenuse² = Perpendicular² + Base²

⇒ OD² = OP² + DP²

⇒ L² = 32

⇒ L² = 4√2 cm

Perimeter = 2(L + B) = 34cm

Area = LB = 60sqcm

So L + B = 34, squaring on both sides

(L + B)² = 1156

L² + B² + 2LB = 1156

LB = 60

L² + B² + 120 = 1156

Or L² + B² = 1036

We know that Diagonal² = L² + B²

So D² = 1036

D = 32.18cm

(i) Since all the three sides of a triangle are given as equal it forms a equilateral triangle with sides measure = 10 cm

Area of a equilateral triangle

= where a is the side of the triangle

Here, Area of equilateral triangle =

=

= 25 √3 cm²

(ii) In the triangle two sides AB and AC (both a) are equal and

The base BC (b) = 8 cm

Area of the isosceles triangle with the given equal length sides and base =

=

= 4× √84

= 4× 9.17

= 36.66cm²

(iii) In the given trapezium ABCD, AD ∥ BC and DC is transversal (both are at 90°)

Area of trapezium

= 1/2 × sum of the parallel two sides of a trapezium

× the distance between the parallel sides

Here, parallel sides AD= 5cm and BC = 4 cm and distance between them, DC= 3cm

Area of trapezium = 1/2 × (5+4) × 3

= 1/2 × 9 × 3

= 13.5cm²

(iv) Given in the trapezium ABCD parallel sides are DC and AB and distance between parallel sides AD = 9 cm ( arrow means parallel sides)

AD= 9 cm

DC= 40 cm

AB = 15cm

And ∠ ADC= 90°

Area of trapezium

= 1/2 × sum of the parallel two sides of a trapezium

× the distance between the parallel sides

= 1/2 × (15 +40) × 9

= 1/2 × 55 × 9

= 247.5 cm²

(v) Arrows in the figure indicates side DC = AB and AD = BC hence it is a rectangle with both the pair of opposite sides parallel and angle between adjacent sides is 90°

⇒ DC = AB = 38

∠ ADC = 90° (given)

AC = 42cm

In Δ ADC

AC= 42

DC = 38

∠ ADC = 90

By using the Pythagoras theorem

AC² = AD² +DC²

⇒ 42² = AD² + 38²

⇒ 1764 – 1444 = AD²

⇒ AD = √320

⇒ AD = 17.89cm

Area of rectangle = l × b, where l and b are length and breadth of rectangle

Area of rectangle ABCD = 17.89 × 38

= 679.76cm²

Given,

The perimeter of equilateral triangle = 48 cm

But perimeter of equilateral triangle = 3a

Where ‘a’ is side of equilateral triangle.

⇒ 48 = 3a

⇒ a = 16 cm

Also, we know

Area of an equilateral triangle

Area of required triangle

= 64√3 cm²

Given, in equilateral triangle height (h)= 5 √ 3 cm

Area of equilateral triangle =

And height of equilateral triangle = , where a is side of equilateral triangle

⇒

⇒ 10 = a

Area of equilateral triangle =

=

= √3 × 25cm²

= 25√3 cm²

Given, equal side of isosceles triangle (a)= 10cm

Length of base (b) = 4cm

Area of isosceles triangle =

where ‘b’ is base and ‘a’ is length of two equal sides

=

= 2√96 cm²

Given, length of base of isosceles triangle (b) = 12cm

Length of equal side (a) = 10cm

Area of isosceles triangle =

=

=6 √64

= 6 × 8

= 48cm²

Given, perimeter of isosceles triangle = 544cm

Let length of base side be b

Then according to the given condition

Length of equal side =

Perimeter of isosceles triangle = 2a +b, where a is equal side and b is base

⇒

⇒

⇒ 544 × 6 = 16b

⇒ 3264 = 16 b

⇒ b = 204 cm

⇒ equal side (a) =

⇒ a =170 cm

Area of =

=

=

= 102 × √ 18496

= 102 × 136

= 13872 cm²

Given, length of hypotenuse of an isosceles triangle = 12√ 2 cm

Let the equal sides of the isosceles right-angled triangle be x cm

But according to the Pythagoras theorem

⇒ x² +x² = (12√ 2)²

⇒ 2x² = 144 × 2

⇒ x = √ 144

⇒ x = 12 cm

Area of isosceles right-angled triangle

= where x is the equal side of isosceles triangle

Area =

=

=72 cm²

Given, AC ⊥ BD and ABCD is the parallelogram so its diagonals bisects each other at O

AO = OC = 4cm

BO = OD = 3cm

Now in Δ AOB

AO = 4cm

BO = 3cm

And ∠ AOB = 90° so its right angled triangle

And AB is the hypotenuse

AB² = AO² +OB²

⇒ AB =

⇒ AB =

⇒ AB = √ 25

⇒ AB =5 cm

Similarly, BC = DC = AD = 5cm

Since all the sides are equal and the diagonals bisects each other at 90°

⇒ the given parallelogram is a rhombus

Let the length of sides of a triangular park be x cm

Given, length of sides = 2x, 3x and 4x

Also perimeter of triangular park = 2x +3x+4x = 216(given)

⇒ 9x = 216

⇒ x = 24 cm

⇒ three sides of triangle are 2× 24 = 48cm, 3× 24 = 72cm and 4× 24 = 96cm

i) Area of Triangle =

And s = semi perimeter of the triangle

⇒

⇒

=168 cm

Area of triangular park =

=

=

= 11804.49cm²

ii) Distance is to be walked from opposite vertex of longest side to that side straightly

Here longest side is 96cm

The distance walked from the opposite vertex of longest side to that side straightly is perpendicular distance from the vertex to the longest side

This is the height of the triangular park (h)

Area of triangular park =

⇒

⇒

⇒

⇒ h = 245.93 cm

Given sides of triangle a= 26 m, b = 28m, c = 30m

(i) Since the grass has to be planted within the area of triangular field we will calculate area of the triangular field

Area =

Where s is the semi perimeter of triangle and a, b, c are the sides of the triangle

⇒

⇒ s = 42 cm

Area =

=

=

=336 m²

Cost of planting grass = area of triangular field × rate per sq meter

= 336 × 5

= Rs1680

(ii) Since the fencing is to done on the edges, we will calculate the perimeter of triangular field

Perimeter of triangular field = 26 + 28+30

= 84cm

Since 5 meter space has to be left for entrance gate construction

So the total perimeter to be fenced = 84-5 = 79cm

Cost of fencing the triangular field = perimeter of the field to be fenced × rate of fencing per meter

Cost of fencing = 79 × 18

= Rs 1422

Given, PQR is an equilateral triangle

Let PQ = QR= PR = x cm and AC = 10cm, AB = 12 cm and AD = 8cm

Area of the equilateral triangle =

⇒ area of equilateral triangle PQR =

Here,

Area of triangle PQR = area of triangle PAR

+ area of triangle RAQ

+ area of triangle PAQ

= (∵ area of triangle= 1/2 ×base× height)

In this case the perpendicular is height of the that particular triangle

= 4x + 6x + 5x

According to the given condition

15x =

= x

Now

Area of equilateral triangle PQR =

= cm²

= cm²

= 300√3cm²

Given, length of equal sides of triangle = 20m

Angle between the equal sides = 45°

Area of a isosceles triangle with two equal sides and angle between them

= , where s is the length of equal sides and θ is the angle between them

Here

Area of isosceles triangle =

= (∵)

Given, length of equal sides of triangle = 20m

Angle between the equal sides = 30°

Area of a isosceles triangle with two equal sides and angle between them

= , where s is the length of equal sides and θ is the angle between them

Here

Area of isosceles triangle =

= (∵ sin 30° = 1/2 )

= 100 cm²

Given, Perimeter of isosceles right angled triangle = (√ 2+1)cm

Let the equal sides of the isosceles triangle = a cm

Then the hypotenuse = a√ 2 cm

According to the given condition

a+ a + √ 2 a = (√ 2+1)

⇒ 2a+ √ 2 a = (√ 2+1)

⇒ a (2+√ 2) = (√ 2+1)

⇒

By rationalizing the denominator

Here, In denominator a = 2 and b = √2

= cm

Now hypotenuse = a√ 2

⇒ Hypotenuse

⇒ Hypotenuse = 1cm

Area = 1/2 base × height

Here hypotenuse is the height of the right isosceles triangle

Area =

= cm²

Given, the field is equilateral triangle

Let the side of triangle be x cm then each side of triangle = x km

She travels at speed of 18km/hour

⇒ She travels at the speed of 3 km in 10 minutes (

⇒

⇒ Distance she covers in 10 minutes = 3 × 10 = 30 km

Now according to the given condition

3x = 30

⇒ x = 10

Now each side of equilateral triangle = 10km

Midpoint of the side of the field directly from the opposite vertex means the perpendicular distance to that side which is also the height of the triangle

Area of triangle = 1/2 × base × height

⇒

⇒

⇒ h = km

=

= 0.3464 km

Time taken to cover this distance

Time

= 0.0192 hours

Let side of equilateral triangle be x m

Then area of equilateral triangle =

According to the given condition

If sides of equilateral triangle = x+1

Then area of this triangle =

⇒

⇒

⇒

⇒ 2x = 3

⇒ x =

Side of equilateral triangle =

Area of equilateral triangle = ,

Where, x is the equal side of triangle

Area of square = side × side

Let side of square here be x cm

In a square the diagonal is hypotenuse

Diagonal = hypotenuse = √ 2 a

According to the given condition

60 = √ 2a

⇒

⇒ a = 30√2

⇒ Area of square = (30√2 )² = 3600 ………….eq1

Given

From eq 1

⇒ Area of equilateral triangle = 1800√ 3cm

But Area of equilateral triangle =

⇒ 1800√ 3 =

⇒

⇒ 84.85 = x cm

Perimeter of the equilateral triangle = 3x

= 3 × 84.85

= 254.56 cm

Let the base, perpendicular and Hypotenuse of right-angled triangle be b, p and h respectively.

We know, Perimeter of right-angled triangle = b + p + h

Given, perimeter = 30 cm

⇒ b + p + h = 30

⇒ b + p + 13 = 30 [∵ h, hypotenuse = 13 cm]

⇒ b = 17 – p [1]

Also, we know By Pythagoras theorem in right-angled triangle

(Hypotenuse)² = (Base)² + (Perpendicular)²

⇒ h² = p² + b²

⇒ (13)² = p² + (17 – p)²

⇒ 169 = p² + p² + 289 – 34p

⇒ 2p² – 34p + 120 = 0

⇒ p² – 17p + 60 = 0

⇒ p² – 12p - 5p + 60 = 0

⇒ (p – 12)(p - 5) = 0

⇒ p = 12 cm or p = 5 cm

Case I: p = 12 cm

⇒ b = 17 – 12 = 5 cm

and we know, area of triangle

Case II: p = 5 cm

⇒ b = 17 – 5 = 12 cm

and we know, area of triangle

Therefore, Area of triangle is 30 cm².

Given, Length of the height of the right- triangle = 12cm

Length of base = 5cm

According to the Pythagoras theorem

⇒ AC² + CB ² = AB²

⇒ 12² + 5² = AB²

⇒ 144 + 25 = AB²

⇒ AB = √ 169

⇒ AB = 13 cm

Now, Area of triangle ABC = ……. Eq 1

⇒ Area of right Triangle ABC =

⇒ Area of right-angle triangle = 30 cm²

Now according to given condition CD is perpendicular to AB

⇒ Area of triangle ABC = AB × h (from eq 1)

⇒

⇒

⇒ h = 4.62cm

Given, sides of the right angled triangle are 3 cm, 4cm and 5 cm

Let BFDE is the largest square that can be inscribed in the right triangle ABC right angled at B

Also let BF = x cm so AF = 4-x cm

In Δ ABC and Δ AFD

∠ A= ∠ A common

∠ AFD = ∠ ABC (each 90°)

Δ ABC ∼ Δ AFD (by AA similarity)

So

⇒

⇒ 3 (4 - x) = 4x

⇒ 12 - 3x = 4x

⇒ 12 = 7x

⇒ cm

Length of square =

Given each side of equilateral triangle 4cm

Height of the equilateral triangle =

=

= 2√3 cm

Hence the height of equilateral triangle = 2√3

Given the length of equal side of isosceles right-triangle = a

Then hypotenuse =

⇒ Hypotenuse = √2a units

Perimeter of the triangle = 2a + √2a

= (2 +√2)a units

Hence the solution is option (b)

_{Given Area of equilateral triangle = a}

_{Perimeter = s}

_{And height = h}

But area of equilateral triangle = = a

Height of equilateral triangle = = h

Perimeter of equilateral triangle = 3 × side = s

Now,

=

_And

=

Putting these values in

=

=

Hence the solution is option C

Given length of equal sides of isosceles triangle = 5cm

Base = 6cm

Area of isosceles triangle =

=

=

= 3 × 4

= 12 cm²

Hence the solution is option B

Given

⇒ 2AD = 3DC

But AC = AD +CD

⇒

⇒

Area of triangle ABC = 40 cm² (given)

But

⇒ DC × BD = 32 ………eq 1

Also Area of triangle BDC =

Area of triangle BDC =

= 16 cm2

Hence the solution is option A

Let the semi perimeter of the triangle be x cm

Then the length of three sides of the triangle is

x - 8, x - 7 and x - 5 cm

Semi-perimeter of the triangle =

Where a, b and c are the sides of the triangle.

⇒

⇒ 2x = 3x – 20

⇒ x= 20

Hence the semi perimeter = 20 cm

Sides of triangle are 12 cm, 13cm and 15cm

Area of triangle =

=

=

=

= 20√14 cm²

Hence the correct option is C

Since the numerical values of area and height of an equilateral triangle are equal

⇒ , where a is side of the triangle

Length of the side of the triangle a=

= 2 units

The area of an equilateral triangle of side a =

A=

Here A₁ =

Now, if we double the side it becomes 2a

the new area is

A=

A=

A =

⇒ A = 4A₁

The increase in percentage of the area of the equilateral triangle

=

=

= 3 × 100

=300

Hence the % increase in area of equilateral triangle is 300%

The area of an equilateral triangle of side a =

A=

Here A₁ =

Now, if the side is tripled, the side it becomes 3a

the new area is

A=

A=

A =

⇒ A = 27 A₁

The increase in percentage of the area of the equilateral triangle

=

=

= 26 × 100

= 2600

Hence the % increase in area of equilateral triangle is 2600 %

In the right angle triangle, hypotenuse is the longest side

In this case, (x+ 2) would be the longest side of the triangle

That would be given by Pythagoras theorem

⇒ x² + (x-2)² = (x+2)²

⇒ x² + x² + 4 - 4x = x² +4 +2x

⇒ x²- 6x = 0

⇒ x( x-6) = 0

⇒ x = 0 or x= 6

⇒ Since the side cannot be zero

⇒ x =6

Hence, the hypotenuse = x= 8cm

Height of equilateral triangle =

Given the height of the triangle is the side of square

Area of square = side²

⇒ Area of square=

⇒ Area of square =

Area of equilateral triangle =

Now

=

=

Hence the ratio of the two areas = √3:3

Given, a parallelogram with base =5cm

And height = 4cm

Formula used:

area of parallelogram = base × height

hence, area of parallelogram drawn by Ratul is,

= 5 × 4

= 20cm

Given, The base of a parallelogram is twice its height the area of shape of parallelogram is 98 sq.cm.

Consider length of base = b

And height = h

According to given question, b = 2h ..(i)

Area of parallelogram = base × height

98 = b × h

From eq. (i),

⟹ 98 = 2h ×h

98 = 2h²

h²=

⟹ h² = 49

⟹ h =

⟹ h = 7cm

and base, b = 2h

⟹ b = 2 × 7

⟹ b = 14cm

Given, a parallelogram land,

Consider a parallelogram ABCD, adjacent sides are given, AB=13m

BC=15m

And diagonal AC=14m

Semi perimeter of ∆ABC, S =

S= 21m

Hence area of ∆ABC = sq. m

Area ∆ABC = 84 sq. m

Area of parallelogram ABCD = 2×area of ∆ABC

Area of parallelogram ABCD = 2×84

= 168 sq. m

Given, a parallelogram of which adjacent sides are 25 cm and 15 cm and length of one diagonal is 20 cm.

Consider a parallelogram ABCD, adjacent sides are given, AB=25cm

BC=15cm

And diagonal AC=20cm

Semi perimeter of ∆ABC, S =

S = 30cm

Hence area of ∆ABC = sq. cm

Area ∆ABC = 150 sq. cm

Area of parallelogram ABCD = 2×area of ∆ABC

Area of parallelogram ABCD = 2×150

= 300sq. cm

Let, Height AE = h

Area of parallelogram ABCD= base × height

⟹ 300 = 25 × h

⟹ h =

⟹ h = 12cm

Given, The length of adjacent two sides are 15 cm. and 12 cm. of a parallelogram and distance between two smaller sides is 7.5 cm.

According to question, area of the parallelogram

= smaller side (base) × distance between smaller sides

= 12 × 7.5

= 90 sq. cm

Area of the parallelogram = longer side (base) × distance between longer sides

⟹ 90 = 15 × h

⟹ h =

⟹ h = 6cm

Given, the measure of two diagonals of a rhombus are 15 meter and 20 meter.

Formula used:

Area of a rhombus = ×product of length of diagonals of Rhombus.

Area of a rhombus = × 15 × 20

= 150 sq. cm

Given, perimeter of a rhombus is 440 meter and distance between two parallel sides are 22meter.

Perimeter of Rhombus = 4×side of square(a)

Hence, 440 = 4a

⟹ a =

⟹ side, a = 11m

Area of rhombus = side of square (a)× height (distance between two parallel sides)

Hence, Area of rhombus = 11 × 22

= 242 sq. m

Given, perimeter of a rhombus is 20 cm. and length of its one diagonal is 6 cm

Side AB =

⟹ AB =

⟹ AB = 5cm

The diagonal AC =6cm

Hence, AO = 3cm

∠AOB = 90 ( the two diagonals of rhombus perpendicular bisected to each other)

Hence, OB²=AB²−AO²

⟹ OB²=5²−3²

⟹ OB = 4cm

∴ BD = 4×2

⟹ BD = 8cm

∴ Area of the Rhombus = ×8×6

= 24 sq.cm

Given, The area of field shaped in trapezium is 1400 sq. dcm and the perpendicular distance between two parallel sides are 20 dcm and the length of two parallel sides are in the ratio 3 : 4

Formula used:

Area of a Trapezium = × sum of parallel sides × distance between two parallel sides

If the length of two parallel sides are in the ratio 3 : 4. Let length of one side is 3x and the length of the other side is 4x.

By using the formula,

⟹ 1400 = × (3x+4x) × 20

⟹ 1400 = 10×(3x+4x)

⟹ 7x = 140

⟹ x = 20dcm

Hence, length of one side = 3x

= 3×20

= 60 dcm.

And the length of the other side = 4x

= 4×20

= 80 dcm.

Given, a regular hexagon field of which length of sides is 8 cm

We know that the regular hexagon consists of six equilateral triangles.

Hence,

Area of hexagonal field = 6 × area of a equilateral triangle.

Area of hexagonal field = 6 ×a²

= 6××8²

= 166.3 sq. cm.

Given, a quadrilateral ABCD in which,

AB = 5m

BC = 12m

CD = 14m

And DA = 15m

∠ABC = 90

In ∆ABC,

⟹ AC²=AB²+BC²

⟹ AC²=5²+12²

⟹ AC²=169

⟹ AC=13m

∴ area of ∆ABC = ×5×12

= 30 sq. m

And diagonal AC=13m

Semi perimeter of ∆ADC, S =

S = 21m

Hence area of ∆ADC = sq. m

Area ∆ABC = 84 sq. m

∴ total area of the quadrilateral = area of (∆ABC +∆ADC)

= 30 + 84

= 114 sq. m

Given, a trapezium ABCD of which length of diagonal BD is 11 cm

And two perpendicular of which length are 5 cm and 11 cm respectively from the points A and C on the diagonal BD.

According to the figure the area of the trapezium = area of ∆ABD + area ∆BCD

= ×BD×AM + ×BD×CN

= ×11×5 + ×11×11

= 27.5 + 60.5

= 88 sq. cm.

Given: ABCDE is a pentagon of which side BC is parallel to diagonal AD, EP is perpendicular on BC and EP intersects AD at the point Q. BC = 7 cm, AD = 13 cm, PE = 9 cm

Also,

and PQ + EQ = PE

⇒ 4 + EQ = 9

⇒ EQ = 5 cm

Now,

Area of pentagon ABCDE = area(ΔAED) + area(trapezium ABCD)

Now, we know

Area of trapezium sum of parallel sides × Height

⇒ area(trapezium ABCD)

Also,

Area of triangle

⇒ area(ABCDE) = 40 + 32.5 = 72.5 cm²

Given, The length of a rhombus is equal to length of a square and the length of diagonal of square is 10√2 cm and the length of diagonals of a rhombus are in the ratio 3 : 4

The diagonal of a square = a

10 = a

a = 10cm

i.e. the side of the square = 10 cm

according to given question side of the rhombus is equal to the side of the square

∴ side of the rhombus = 10 cm

The diagonals of a rhombus bisect each other. Let the diagonals are 3x and 4x.

Hence, 10² = (3x)² + (4x)²

⟹ 10² = 25x²

⟹ X² = 4

⟹ X = 2

∴ the diagonals are, 3x = 3×2

= 6cm

And, 4x = 4×2

= 8cm

The area of the rhombus = ×6×8

= 24 sq. cm.

Given, In a trapezium, the length of each slant sides is 10 cm. and the length of parallel sides are 5 cm. and 17 cm. respectively.

According to the figure two right angled triangles are formed.

In ∆AEC,

⟹ AC²=AE²+CE² -------(i)

⟹ CE = FD

And 17 = CE +FD +EF

⟹ 17 = 2CE + 5

⟹ 2CE = 12

⟹ CE = 6cm

∴ 10² = AE² + 6²

⟹ AE²= 100 – 36

⟹ AE²= 64

⟹ AE = 8cm

Hence, area of the trapezium =×(5+17)×8

= 88 sq. cm.

Given, The lengths of parallel sides of a trapezium are 19 cm. and 9 cm. and the length of slant sides are 8 cm. and 6 cm.

In ∆AEC,

⟹ AC²=AE²+CE² -------(i)

In ∆BFD,

⟹ BD²=BF²+FD² -------(ii)

From eq(i),

⟹ 6² = AE² +CE²⟹ AE²=36−CE²

From eq(ii),

⟹ 8²= AE² + FD² { AE = BF }

⟹ AE²=64−FD²

From figure, 19 = CE+EF+FD

⟹ CE + FD = 19−EF

⟹ CE + FD = 19−9

⟹ CE + FD = 10 ----(iii)

⟹ 36−CE²=64−FD²

⟹ 36−(10−FD)²=64−FD²

⟹ 36−100−FD²+20FD = 64−FD²

⟹ 20FD = 128

⟹ FD = 6.4cm and then CE = 10 – FD

CE = 10 – 6.4

CE = 3.6

⟹ AE² = 36−CE²

⟹ AE² = 36−3.6²

⟹ AE² = 36−12.96

⟹ AE = 4.8cm

Area of trapezium = ×(9+19)×4.8

= 67.2 sq. cm

Formula used:

area of parallelogram = base × height

⟹ 192 = base ×base

⟹ 192 = b²

⟹ b² = 192×3

⟹ b² = 576

⟹ b = 24cm

then height = base

= ×24

= 8cm

∆ABD and ∆CBD are equilateral triangles.

You may know that the height of an equilateral triangle is .side

∴ AO = OC = .AB

=.6

∴ AC (diagonal) = AO+OC

= 6

In ∆AOB, AB²=BO²+AO²

⟹ 36 = BO² +( .6)²

⟹ BO²=36−27

⟹ BO² = 9

⟹ BO = 3cm

The diagonal AC = 2×3

= 6cm

Hence, the area of the rhombus=×6×6

=18 sq. cm

Given, The length of one diagonal of rhombus is three times of another diagonal and the area of field in the shape of rhombus is 96 cm².

Let diagonal one and two be D1 and D2 respectively.

Area of the rhombus = ×product of diagonals

(D₂)² = 64

D₂ =

D₂ = 8cm

Hence the longer diagonal D₁ = 3D₂

= 3×8

= 24cm.

The height h of the rhombus will always be less than x i.e.

h<x

multiplying by x both sides

hx<x² and in this case hx = y

∴ y<x²

Formula used:

Area of a Trapezium = × sum of parallel sides × distance between two parallel sides

⟹ 162 = × (23 +b) ×6

⟹ 162 ×2 = (23+b) × 6

⟹ b = 54 −23

⟹ b = 31cm

Let us consider a parallelogram ABCD such that

Area(ABCD) = 96 cm² and BD = 12 cm

To find: length of perpendicular length drawn on diagonal BD from the point A i.e. AE.

Now, In ΔABD and ΔBCD,

AB = CD [Opposite sides of parallelogram are equal]

AD = BC [Opposite sides of parallelogram are equal]

BD = BD [Common]

ΔABD ≅ ΔBCD [By SSS congruency criterion]

[Congruent triangles have equal areas]

⇒

Now, we know

Area of a triangle

⇒ 6AE = 48

∴ AE = 8 cm

Given, The length of adjacent sides of a parallelogram are 5 cm. and 3 cm and the distance between the longer side is 2 cm.

Area of the parallelogram = base × height

Area = 5 × 2

= 10 cm²

Also, area = base(shorter side) × distance between shorter sides (height)

10 = 3 × smaller side

Smaller side =

Smaller side = 3.33 cm

Given, Length of height of rhombus is 4 cm. and length of side is 5 cm.

Hence, DB= 4cm and then BO = 2cm

By Pythagoras theorem,

BO² + AO² = AB²

2² + AO² = 5²

AO² = 25 – 4

AO =

AO = 4.6cm

Hence, AC = 2.AO

AC = 2×4.6

AC = 9.2 cm

Area of rhombus = ×4×9.2

= 18.4 sq. cm

Let us consider a trapezium ABCD, such that AB || CD and BC be the slant height such that BC = 62 cm, and ∠ABC = 45°

To find: Distance between two parallel lines i.e. CP.

In ΔACP

We know,

⇒ CP = 31√2 cm

Consider a parallelogram ABCD, such that AB = 4 cm, BC = 6 cm and ∠ABC = 30°, Draw AE ⊥ BC

We know, area of parallelogram = Base × Height

⇒ area(ABCD) = AE × BC

In ΔAEB

We know,

⇒ AE = 2 cm

Hence, area of ABCD = 2 × 6 = 12 cm²