Perimeter And Area

Shape 1

Perimeter = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

= 22 units

Area = 18×1

= 18 square units

Shape 2

Perimeter = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

= 18 units

Area = 18×1

= 18 square units

Shape 3

Perimeter =

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

= 22 units

Area = 16×1

= 16 square units

Shape 4

Perimeter =

1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1

= 22 units

Area = 18×1

= 18 square units

So option A. is not correct

Dimensions of bigger rectangle are 6 cm × 5 cm

So area of bigger rectangle

= 6 cm × 5 cm

= 30cm²

Now,

Dimension of smaller rectangle = 3 cm × 2 cm

Area of smaller rectangle = 3 cm × 2 cm

= 6 cm²

Area of remaining sheet = area of bigger rectangle - Area of smaller rectangle

⇒ 30 – 6

⇒ 24 cm²

Area of rectangle formed

= 36 units

We have, 36 = 6×6

= 2

=

=

So the sides of the rectangle are 4cm and 9cm.

So perimeter = 2(l + b)

= 2(4 + 9)

= 2(13)

= 2

= 26 units

Given, side of square = 22cm

Perimeter of square and circumference of the circle are equal, because the wire has same length.

Perimeter of square = circumference of circle

⇒ So radius is 14cm

Area of circle =

⇒

⇒

Given dimensions of rectangle

L = 14cm B = 11cm

Given area of rectangle = area of circle

⇒ l×b = πr²

From the given figure,

Length of rectangle = 5cm and breadth of rectangle = 3 + 1 = 4cm

Area of shaded region =

=

= 10 cm²

Area of parallelogram = base × corresponding height

So area of parallelogram ABCD

⇒ AD×BE = BC×BE (Since AD = BC)

⇒ So area of parallelogram ABCD = DC×BF

So answer = A.

DE × DC is not equal to area of parallelogram ABCD

Area of triangle =

= ×OQ

Area of MNO = ×OQ

Area of parallelogram MNOP = base × corresponding height

Area of parallelogram MNOP = MP×OQ

= NO×OQ

Since MP = NO

Ratio =

⇒ Ratio =

⇒ Ratio = 1:2

Area of MNO = ×MO

=

= 10 cm²

Area of MOP = ×OP

= 5

= 5 cm²

Area of MPQ = ×PQ

= 5

= 15 cm²

Ratio = 10:5:15

= 2:1:3

In parallelogram EFGH,

EF = HG = 10 cm [Given]

Area of parallelogram EFGH = Base × Corresponding height = 10 × 4 = 40 cm²

_{Circumference of a circle}

Circumference =

=

=

We know that,
Circumference of a circle = 2πr
∴ Circumference = 2×3.14×r [∵ π = 3.14]
⇒ Circumference = 3.14×d [∵ d = 2r]
So, circumference of circle is always more than three times of its diameter.

Area of triangle PQR = ×PR

Given Area of triangle PQR is 100 cm²

100 = ×PR

100 = ×PR

100 = 5 PR

= PR

So PR = 20cm

Given that, PR = 12cm, QR = 6cm and PL = 8cm

Now in right angled PLR, using Pythagoras theorem,

= -

=

= 144- 64

= 80

4

LQ + QR

LQ = LR-QR

LQ = 4

Area of PLR = ×PL

= 4×8

= 16 cm²

Area of PLQ = ×PL

= 4×8

= 4(4

= 16 cm²

Area of PLR = Area of PLQ + Area of PQR

16 = (16 + Area of PQR

Area of PQR = 24 cm²

×QM = 24

×QM = 24

4cm

Given ∆MNO is a right-angled triangle

According to Pythagoras theorem

=

= +

= 36 + 64

= 100

MO = 10cm

Area of ∆MNO =

⇒ ×NO = ×NP

⇒ ×8 = ×NP

⇒ NP =

⇒ NP = 4.8cm

Given, Area of a right-angled triangle is 30 cm²

And smallest side i.e. base = 5cm

Area of triangle = ×height

30 = ×height

Height =

Height = 12cm

According to Pythagoras

⇒ =

⇒

⇒

⇒

⇒ Hypotenuse = 13cm

Circumference of a circle = 2

2r = diameter = 5cm

Circumference of a circle = 5

Circumference = 15.7cm

Given Circumference of a circle disc = 88 cm

Circumference of a circle = 2

88 = 2

r =

r = 14cm

In order to find the length of tape required to cover the edges of a semi-circular disc, we have to find the perimeter of semi-circle

perimeter of semicircle = Circumference of semicircle + diameter

Circumference of semi circle =

=

=

= 31.4cm

Total tape required

= 31.4 + 10 + 10

= 51.4 cm

Area of circular garden =

Given diameter = 8m

So radius = = 4m

area =

= 50.24 m²

Given, diameter = m

radius = n and circumference = p

∴ Area of circle = π

given r = n so π

Diameter = 2.8 cm

So radius = = = 1.4cm

Area of table top = area of semicircle =

=

= 3.08 m²

Given 1m² = × mm²

⇒ (1000mm)² = × mm²

⇒ 1000000mm² = × mm²

⇒ × = 1000000

Area of square = side×side

Area of 1 square of side 1mm = 1×1 mm²

= 1mm²

Area of square of side 1cm = 1×1cm²

= 1 cm²

Area of square of side 1mm = area of square 1cm

⇒ p × 1mm² = 1cm²

⇒ pmm² = (10 mm)²

⇒ pmm² = 100 mm²

For option A,

Area of square with side 12cm = 12×12

= 144 m²

For option B,

Area of 12 squares with side 1m each = 12×1×1

= 12 m²

For option C,

Area of 3 squares with side 4 m each = 3×4×4

= 48 m²

For option D,

Area of 4 squares with side 3 m each = 4×3×3

= 36 m²

So option B is correct

Let b be the side and h be the height of a rhombus.
∴ Area of rhombus = b × h

[∵ area of rhombus = base × corresponding height]

If each side of rhombus is doubled, then side of rhombus = 2b
Now, area of rhombus = 2b × h = 2(b × h)

= 2 times of original

Hence, the area of rhombus will be increased by 2 times.

Let the length and breadth of the parallelogram be l and b, respectively.

⇒ Then, perimeter = 2(l + b) [∵ perimeter of parallelogram = 2 × (length + breadth)]

⇒ If both sides are increased twice, then new length and breadth will be 2l and 2b, respectively.

⇒ Now, new perimeter = 2(2l + 2b)

⇒ 2 × 2(l + b)

⇒ 2 times of original perimeter.

Hence, the perimeter of parallelogram will be increased 2 times

Let r be the radius of the circle.

∴ Area of circle = π2

If radius is increased to twice its original length, then radius will be 2r.
Now, area of new circle = π(2r)2 = 4π2 = 4 times of original area
Hence, the area of circle will be increased by 4 times

Given, radius of circle = 10 cm

The largest square that can be cut-out of a circle of radius 10 cm will have its diagonal equal to the diameter of the circle.
Let the side of a square be ×

Incomplete

Area of square = ײ

Now in right triangle DAB

(BD) ² = (AD) ² + (AB) ²

(20) ² = ײ + ײ

400 = 2ײ

200 = side²

area of the largest square = 200 square cm

Largest circle will have diameter equals smaller side i.e. 8cm

So diameter is 8cm

Radius =

So radius = 4cm

Perimeter = sum of sides

So, AJ + IL + IH + HG + GF + FE + DE + CD + BC + AB

⇒ (AJ + IH + GF + BC) + 3 + 5 + 2 + 20 + 4 + 6

⇒ DE + 40 [AJ + IH + GF + BC = DE]

⇒ 20 + 40

⇒ 60cm

Let the radius of the circle be R

Area of circle = πR²

81 πr² = πR²

R = r

R = 9r

Circumference = 2 πr

⇒ 2πR

⇒ 2 π(9r)

⇒ 18 πr

Let the side of square be a cm

area of a square = 100 cm²

area of square = a²

⇒ a² = 100cm²

⇒ a = 10cm

Now for the largest circle in the square, diameter of the circle must be equal to the side of the square

Diameter = side of square = 10cm

⇒ 2r = 10cm

⇒ r = 5cm

Circumference of the circle = 2πr

⇒ 2 π×5

⇒ 10 π

Let r be the radius of a circle.

∴ Area of circle = πr²

If radius is tripled, then new radius will be 3r.

∴ Area of new circle = π(3r)²

= 9π²

9 times of original

Hence, the area of a circle becomes 9 times to the original area.

Given radius of a semicircle = 4r

Area of semicircle =

=

=

= 8πr²

Perimeter of regular polygon = Length of one side × Number of sides

When we change the shape, then the perimeter remains same as the length of wire is fixed, but area changes as shape changes.

Given, area of square MNOP

= 144 cm²
Since, there are 8 identical triangles in the given square MNOP.
Hence, area of each triangle = 1/8 × Area of square MNOP = 1/8 × 144 = 18 cm²

We know,

Area of parallelogram = Base × Height

Clearly, Height of parallelogram BCEF = CD

Area of parallelogram BCEF = EF × CD

Since BCEF is a parallelogram,

BC = EF [Opposite sides of a parallelogram are equal]… [Eq 1]

Also, ACDF is a rectangle

AC = FD [Opposite sides of a rectangle are equal]… [Eq 2]

Subtracting Eq 1 from Eq 2, we get

AC – BC = FD – EF

⇒ AB = ED

⇒ ED = 3 cm [∵ AB = 3 cm]

Also,

EF = FD – ED

⇒ EF = 10 cm – 3 cm = 7 cm [∵ FD = 10 cm]

Therefore,

Area of parallelogram BCEF

= EF × CD

= 7 cm × 5 cm [∵ CD = 5 cm]

= 35 cm²

While calculating the area of the parallelogram, we can choose any side as base

Perpendicular dropped on the base of a parallelogram from the opposite vertex is known as the corresponding height/altitude of the base

The distance around a circle is its circumference.
In case of circle, perimeter is known as circumference.

Circumference = 2πr [2r = d where d is diameter]

⇒ C = πd

⇒

⇒

Area of triangle = base × height

90 = × 20 × height

Height = 9cm

We know that π =

= 3.14

Circumference = 2πr

Since, diameter (d) = 2r

So,

C = π×d

Hence, π is the answer

Circumference = 2π × r

Hence, r is the answer.

We know that, 1m = 100 cm

1m² = (100)²cm².

1m². = 10000 cm².

We know that, 1cm = 10mm

∴ 1 cm² = (10)² mm²

1 cm² = 100 mm²

1 hectare = 10000 m²

Area of triangle = base × height

We know that 1km = 1000m

So 1 km² = (1000m)²

= 1000000 m²

Let number of squares having side 1 cm = a

According to the question,

Area of side 6 m square = Area of side 1 cm square

[∵ area of square = (side) ² ]

(6m)² = a(1cm)²

[∵1m = 100cm]

⇒ (600 cm)² = a (1 cm)²

⇒ 360000 cm² = a cm²

⇒ a = 360000

⇒ 10 cm² = 10² m²

⇒ 10 cm² = m²

⇒ 10 cm² = 0.001 m²

Perimeter is the sum of sides of any polygon and area is space that the polygon required. So, by observing the figures we can say that, perimeter of (ii) is greater than (i) and area is less than that of (i).

(a) True
Area of both figures is same, because in both number of blocks are same.

(b) False
Because 2 new sides are added in (ii). So, the perimeter of (ii) is greater than (i).

(c) False

∴ Area of 1 square = 1 × 1

= 1 unit squares

∵ Number of squares = 12

So, total area = 12 × 1 = 12 unit squares

(d) True

∵ Perimeter is the sum of all sides. So, it is 18 units.

Their corresponding sides and height may be different. So, area cannot be equal

True
Congruent triangles have equal shape and size. Hence, their areas are also equal.

False
It is not necessary that all parallelograms having equal areas have same perimeters as their base and height may be different.

True
It is clear from the figure that all triangles have same base AB and all the vertices lie on the same line, so the distance between vertex and base of triangle (i.e. length of altitude) are equal

False
As shown in the figure,

Any triangle can have any value of base, vertex etc. Hence, not all triangles are congruent.

True
Because the triangles on same base and between same parallel lines have equal in area.

True
It is clear from the figure that all triangles may not have the same perimeter.

Different figures have different perimeter.

True

Using the formula: Area of triangle = × b × h

Area of ΔABC = × BC × AC

Area of ΔACD = × CD × AC

⇒ Ratio of the area of ΔABC to that of the area of ΔACD is the same as the ratio of base BC of ΔABC to the base CD of ΔACD.

False

We know that, Area of triangle = × b × h

i.e. Area of a triangle depends directly on base as well as the height of the triangle.

So, if triangles have same base but different heights, their area would be unequal.

True

We know that, the circumference C of a circle of radius r, is C = 2πr.

On rearranging the terms, we obtain,

⇒ Ratio of circumference of a circle to its radius is always 2π : 1.

False

We know that, 1 hectare = 10,000 m²

So, 5 hectare = 5 × 10,000 = 50,000 m²

False

An increase in perimeter does not necessarily mean an increase in area of the figure.

For example:

Perimeter= 4 × 4 = 16 cm Perimeter =18 cm

Area = 16 cm² Area = 14 cm²

The perimeter gets increased but there is a reduction in area.

True

Consider the following example:

Perimeter= 4 × 4 = 16 cm Perimeter =20 cm

Area = 16 cm² Area = 16 cm²

So, two figures can have the same area but different perimeters.

True

Consider the following example:

Perimeter= 4 × 4 = 16 cm Perimeter =18 cm

Area = 16 cm² Area = 14 cm²

Though the first figure has the larger area, its perimeter is less than the second one.

⇒ Out of two figures if one has larger area, then its perimeter need not to be larger than the other figure.

Given, length of the lawn = 72 m

Breadth of the lawn = 18 m

We know that, perimeter of rectangle = 2 × (l + b)

So, perimeter of the lawn = 2 × (72 + 18) = 2 × 90 = 180 m

i.e. Length of hedge boundary = 180 m

No. of shrubs to be planted in 1 m of hedge = 3

No. of shrubs to be planted in 18 0m of hedge = 3 × 180 = 540

Using the formula: Area of a rectangle = l × b

Total area of land= 15 × 10=150 m²

Area kept for camel = 5 × 3 = 15 m²

Area kept for plants = 9 × 1 = 9 m²

Now, diameter of the circular field kept for ox = 2.8 m

Radius = =1.4 m

Using the formula: Area of a circle = ∏r²

Area kept for ox = × (1.4)² = 6.16 m²

Area kept for animals = 15 + 9 + 6.16 = 30.16 m²

Living area = Total area of land – Area kept for animals = 150 – 30.16 = 119.84 m²

Ratio of the area kept for animals and plants to that of the living area = 377:1498

Let breadth of the rectangle be x.

Then, length of the rectangle = 5x – 4

We know that perimeter of rectangle = 2 × (l + b)

40 = 2 × (x + 5x - 4)

40 = 2 × (6x - 4)

12x - 8 = 40

12x = 48

x = = 4

So, breadth of the rectangle = x = 4 m

Length of the rectangle = 5x – 4 = 5 × 4 – 4 = 20 – 4 = 16 m

Area of rectangle = l × b = 16 × 4 = 64 m²

Hence, area of rectangle = 64 m²

Given, a wall of a room is of dimensions 1.5 m × 1m.

Area of the wall = l × b = 5 × 4 = 20 m²

Area of the window = l × b = 1.5 × 1 = 1.5 m²

Area of the door = l × b = 2.25 × 1 = 2.25 m²

Area of the wall to be painted = Area of the wall – (Area of the window + Area of the door)

∴ Area of the wall to be painted = 20 – (1.5 + 2.25) = 20 – 3.75 = 16.25m²

Area of the wall which is to be painted= 16.25 m²

Area of one rectangle = 8 m²

And breadth = 2 m

We know that,

Area of a rectangle = l × b

⇒ l × 2 = 8

⇒ l = 4 m

MNOP is made up of 4 congruent rectangles, i.e. they have same dimensions (equal length and breadth).

∴ Perimeter of MNOP = PF + FA + AM + MN + NC + CD + DO + OP

= 2 + 4 + 2 + 4 + 2 + 4 + 2 + 4

= 24 m

The perimeter of MNOP = 24 m

We know that, area of a triangle = × b × h

and area of a parallelogram = b × h (corresponding height)

Given, Area of ∆AFB = Area of parallelogram ABCD

× AB × EF = DC × h

× 10 × 16 = 10 × h

10h = 80

h = 8 m

O is the mid point of DC, hence DO = × DC = × 10 = 5 m

Area of ∆DAO = × DO × h = × 5 × 8 = 20 m²

We know that, area of a triangle = × b × h

Given, Area of ∆WXZ = 56 cm²

× XZ × 8 = 56

XZ = = 14 cm

∵

4XY = 42 – 3XY

7XY = 42

XY = 6 cm

YZ = XZ – XY = 14 – 6 = 8 cm

So, XY = 6 cm and YZ = 8 cm

Given, side of the square = 70 m

Diameter of the semicircular lawn = side of the square = 70 m

∴ Radius = × 70 = 35 m

We know that circumference C of a circle of radius r, is C = 2πr

Perimeter of a semicircle = × 2πr + 2r = πr + 2r

i) So, perimeter of the lawn = × 35 + 2 × 35 = 110 + 70 = 180 m

ii) Area of square = (side)² = (70)² = 4900 m²

Area of semicircular lawn = × πr² = × × (35)² = 11 × 5 × 35 = 1925 m²

Area of the square field excluding the lawn = Area of square - Area of semicircular lawn

= 4900- 1925 = 2975 m²

Given, Area of shaded triangle = 9 cm²

× BE × AE = 9

× 3 × AE = 9

AE = 6cm

Area of a parallelogram = b × h (corresponding height) (corresponding height)

Area of parallelogram ABCD = 7 × 6 = 42 cm²

Given, side of square pizza = 45 cm

∴ Area of square pizza = (45)² = 2025 cm²

Price of 2025 cm² of square pizza = Rs. 150

Price of 1 cm² of square pizza = = Rs. 13.5

Diameter of circular pizza = 50 cm

i.e. Radius = =25 cm

∴ Area of circular pizza = πr² = × (25)² =1964.28 cm²

Price of 1964.28 cm² of square pizza = Rs. 160

Price of 1 cm² of circular pizza = = Rs. 12.27

Since price of 1 cm² of circular pizza is lesser, hence circular pizza is a better deal.

Side of first square = 6m

Side of second square = = 3m

Side of third square = = 1.5m

Perimeter of the completed figure= 6 + 6 + 3 + 3 + 1.5 + 1.5 + 1.5 + 1.5 + 3 + 6

Perimeter of the completed figure= 33m

Given, side of square ABCD = 15cm

We know that diagonal of a square = √2 × side

∴ BD = √2 × 15 = 15√2 cm

Side of square BDFE = 15√2 cm

∴ Area of square BDFE = (15√2)² =15√2 × 15√2 = 450 cm²

Given, perimeter of ∆ABC = perimeter of ∆PQR

AB + BC + CA = PQ + QR + RP

AB + 5 + 13 = 6 + 10 + 14

AB + 18 = 30

∴ AB = 30 – 18 = 12 cm

We know that, Area of triangle = × b × h

∴ Area of ∆ABC = × BC × AB = × 5 × 12 = 30 cm²

We know that, Area of a parallelogram = b × h (corresponding height)

Area of MGHK with GH as base = Area of MGHK with HK as base

GH × MN = HK × MO

6 × 8 = HK × 4

∴ HK = 12 cm

Since opposite sides of a parallelogram are equal to each other,

MK = GH = 6 cm

GM = HK = 8 cm

∴ Perimeter of parallelogram MGHK = GM + MK + KH + HG = 8 + 6 + 8 + 6 = 36 cm

We know that, Area of triangle = × b × h

Area of ∆PQR = 20 cm²

× QR × PQ = 20

× 5 × PQ = 20

∴ PQ = 8 cm

Area of ∆PQS = 44 cm²

× QS × PQ = 44

× QS × 8 = 44

∴ QS = 11 cm

RS = QS – QR = 11 – 5 = 6 cm

Length of RS = 6 cm

Consider an isosceles ∆ABC with base BC, equal sides AB and BC.

We know that, Area of triangle = × b × h

48 = × BC × 8

∴ BC = 12 cm

In an isosceles triangle, the altitude divides base into half.

So, DC = = 6 cm

Applying Pythagoras theorem in ∆ADC,

(AD)² + (DC)² = (AC)²

(8)² + (6)² = (AC)²

(AC)² = 64 + 36 = 100

AC =√100 = 10 cm

Now, AB = AC = 10 cm

Perimeter of ∆ABC = 10 + 10 + 12 = 32 cm

AB = DC =18 m

Perimeter of parallelogram ABCD = AB + BC + CD + DA

96 = 18 + BC + 18 + BC

(opposite sides of a parallelogram are equal)

2 BC = 96 – 36 = 60

BC = AD = 30 m

Given, area of parallelogram ABCD = 270 m²

We know that,

Area of a parallelogram = b × h (corresponding height)

Taking AB as the base,

AB × l = 270

18 × l = 270

l = 15 m

Taking AD as the base,

AD × m = 270

30 × m = 270

m = 9 m

Length of the other side = 30 m

Length of altitude l = 15 m

Length of altitude m = 9 m

Given, area of ∆PQR = 60 cm² with PQ = 8 cm

We know that, Area of triangle = × b × h

Area of ∆PQR = 60 cm²

× PQ × QR = 60

× 8 × QR = 60

∴ QR = 15 cm

Applying Pythagoras theorem in ∆PQR,

(PQ)² + (QR)² = (PR)²

(8)² + (15)² = (PR)²

(PR)² = 64 + 225 = 289

AC =√289 = 17 cm

Hence, the length of two sides are 15 cm and 17 cm.

Since the rectangles are congruent, their length and breadth are equal.

Given, Perimeter of the rectangle = 264 cm

It can be observed from the figure that

4l + 5b = 264

And 2l = 3b

On solving the above equations:

11b = 264

⇒ b = 24 cm

∴ l = × 24 = 36 cm

Hence, perimeter of smaller rectangles = 2 × (l + b) = 2 × (36 + 24) = 120 cm

Given a square ABCD inscribed in a circle of radius 7 cm.

For ∆AOB, OA = OB = 7 cm

We know that, Area of triangle = × b × h

Area of ∆AOB = × OB × OA = × 7 × 7 = cm²

Since the square is made up of four congruent triangles, area of square ABCD = 4 × = 98 cm²

Area of the inscribed square = 98 cm²

We know that, area of a circle = ∏r²

Area of given circle = ∏ × (7)² = × 7 × 7 = 154 cm²

Shaded area = Area of circle – Area of square = 154 cm² - 98 cm² = 56 cm²

∴ Shaded area = 56 cm²

We know that, area of rectangle = l × b

And area of semicircle = × ∏r²

Area of rectangle = 10.2 × 1.5 = 15.3 cm²

Now, Diameter of the semicircle = 10.2 – 3.9 = 6.3 cm

So, radius = = 3.15 cm

Area of semicircle = × × 3.15 × 3.15 = 15.5925 cm²

Total area = Area of rectangle + Area of semicircle = 15.3 + 15.5925 = 30.8925 cm²

∴ Total area = 30.8925 cm²

We know that, area of rectangle = l × b

And area of triangle = × b × h

Area of rectangle = 13 × 4 = 52 cm²

Now, base of triangle = 13 – 8 = 5 cm

And height of triangle = 16 – 4 = 12 cm

Area of triangle = × 5 × 12 = 30 cm²

Total area = Area of rectangle + Area of triangle = 52 + 30 = 82 cm²

∴ Total area = 82 cm²

We know that, area of rectangle = l × b

And area of triangle = × b × h

Area of rectangle = 15 × 3 = 45 cm²

Now, base of triangle = 15 – 10 = 5 cm

Given, height of triangle = 4 cm

Area of triangle = × 5 × 4 = 10 cm²

Total area = Area of rectangle + Area of triangle =45 + 10 = 55 cm²

∴ Total area = 55 cm²

We know that, area of triangle = × b × h

And area of semicircle = × ∏r²

Radius of the semicircle = 10 cm

Area of semicircle = × × 10 × 10 = cm² = 157.14 cm²

Base of triangle = diameter of the semicircle = 2 × 10 = 20 cm

Height of the triangle = 17 – r = 17 – 10 = 7 cm

Area of triangle = × 20 × 7 = 70 cm²

Total area = Area of semicircle + Area of triangle = 157.14 + 70 = 227.14 cm²

∴ Total area = 227.14 cm²

Radius of the smaller circle, r = cm

Radius of the larger circle, R = + 7 = cm

Area of shaded region = ∏R² - ∏r² =∏(R² - r²)

= = 303 cm²

∴ Area of shaded region = 303 cm²

Diameter of the small circles = cm

Radius of the small circles = cm

Area of two small circles = 2 × = cm² = 4.8 cm²

Diameter of the large circle =14 cm

Radius of the large circle = × 7 × 7 = 154 cm²

Area of shaded region = Radius of the large circle - Area of two small circles = 154 - 4.8 = 149.2 cm²

∴ Area of shaded region = 149.2 cm²

Perimeter of the circle in first fig includes 4 arcs of equal length.

While in the second fig, perimeter includes 4 arcs as well as radius twice.

So the perimeter is increased by 2r = 2 × 16 = 32 cm

Perimeter of rectangle = 2 × (l + b)

16 = 2 × (l + b)

⇒ l + b = 8 cm

Side of the square = l + b = 8 cm

Hence area = (side)² = 8² = 64 cm²

Given, area of ∆AED = 6 cm²

× DE × AE = 6

× 4 × AE = 6

AE = 3 cm

In ∆AEC, AE = 3 cm, AC = 5 cm

Applying Pythagoras theorem in ∆AEC,

(AE)² + (EC)² = (AC)²

(3)² + (EC)² = (5)²

(EC)² = 25 - 9 = 16

EC =√16 = 4 cm

DC = DE + EC = 4 + 4 = 8 cm

Area of ∆ADC = × DC × AE = × 8 × 3 = 12 cm²

Since the diagonal divides the parallelogram into two congruent triangles,

Area of parallelogram ABCD = 2 × Area of ∆ADC = 2 × 12 = 24 cm²

Applying Pythagoras theorem in ∆AED,

(AE)² + (DE)² = (AD)²

(3)² + (4)² = (AD)²

(AD)² = 9 + 16 = 25

AD =√25 = 5 cm

Perimeter of parallelogram ABCD = 2 (l + b) = 2(DC + AD) = 2(8 + 5) = 26 cm

Area of parallelogram ABCD = 24 cm²

Perimeter of parallelogram ABCD = 26 cm

Total distance around the track = Length of two parallel strips + Length of two semicircles (r =16 cm)

= 2 × 52 + 2 × 3.14 × 16

= 104 + 100.5

= 205 cm (approx.)

Given, Length of the table cover = 3 m 25 cm = 3.25 m

and Breadth of the table cover = 2 m 30 cm = 2.3 m

Area of the table cover = 3.25 × 2.3 = 7.475 m²

The table cover hangs 30 cm around the table, and so:

Length of the table top = Length of the table cover – 30 cm= 3.25 m - 2 × 0.3 m= 2.65 m

Breadth of the table top = Breadth of the table cover – 30 cm = 2.3 m – 2 × 0.3 m = 1.7 m

Area of the table top = 2.65 × 1.7 = 4.505 m²

Area of the hanging cover = Area of the table cover - Area of the table top

= 7.475 – 4.505 = 2.97 m²

Cost of polishing 1 m² of table top = Rs. 16

∴ Cost of polishing 4.505 m² of table top = Rs. 4.505 × 16 = Rs. 72.08

Area of the hanging cover = 2.97 m²

Cost of polishing 4.505 m² of table top = Rs. 4.505 × 16 = Rs. 72.08

Area of the plot = 200 × 150 = 30,000 m²

Area of roads = 3 × (200 × 3) = 1,800 m²

Area available for building the houses = Area of the plot - Area of roads

= 30,000 – 1,800 = 2,8200 m²

∴ Area available for building the houses = 2,8200 m²

Area of the floor = 4.5 × 4 = 18 m² = 18,0000 cm² (1 m = 100 cm)

Area of one tile = 15 × 10= 150 cm²

Number of tiles required to cover the floor = = 1200

∴ Cost of covering the floor with tiles at the rate of Rs. 4.50 per tile = 4.5 × 1200 = Rs. 5400

Circumference of a circle = ∏d

So, length of wooden fencing required = × 28 = 88 m

Cost of fencing at the rate of Rs 300 per metre = 88 × 300 = Rs. 26,400

Given, radius of the circle = 14 cm

and length of the rectangle = 24 cm

Since the same wire is rebent to form different shapes,

Perimeter of rectangle = Circumference of circle

2 × (l + b) = 2∏r

2 × (24 + b) = 2 × × 14

24 + b = 44

⇒ b = 44 - 24 = 20 cm

Area of rectangle = l × b = 24 × 20 = 480 cm²

And area of circle = ∏r² = × 14 × 14 = 616 cm²

Hence, circle encloses more area than the rectangle.

A wheel covers a distance equal to its circumference in one rotation.

Radius of wheel = 25 cm= m =

So, distance covered in one rotation = 2∏r = 2 × = m

Hence, distance covered in 350 rotations = × 350 = 550 m

Let R and r be the radius of the outer and inner circle respectively.

We know that the circumference of a circle =2∏r

Given, circumference of outer circle = 44 m

2∏R = 44

2 × × R = 44

R = 7 m

R = R – 2 = 7 – 2 = 5 m

Inner circumference = 2 × 3.14 × 5 = 3.14 m

Area of the path = Area of outer circle - area of inner circle

=∏R² - ∏r² =∏(R² - r²)

= 3.14 (7² – 5²) = 3.14 × (49 – 25) = 3.14 × 24 = 75.36 m²

Length of the carpet = 5 m

Breadth of the carpet = 2m

∴ Area of the carpet =5 × 2 = 10 m²

Length of border = 25 cm = 0.25 m

Length of inner blue portion = Length of the carpet - 2 × length of border = 5 - 2 × 0.25 = 4.5 m

Breadth of inner blue portion = Breadth of the carpet - 2 × breadth of border = 2 - 2 × 0.25 = 1.5 m

Area of blue portion = 4.5 × 1.5 = 6.75 m²

Area of red portion = Area of the carpet - Area of blue portion= 10 – 6.75 = 3.25 m²

Ratio of areas of red portion to that of blue portion = 13:27

A. Area of land used to grow hay = 17.8 × 10.6 = 188.68 m²

B. Area of vegetable garden = 49 × 15.2 = 744.8 m²

Cost of fertilizing 1 m² of vegetable garden = Rs. 91

∴ Cost of fertilizing 744.8 m² of vegetable garden = 91 × 744.8 = Rs. 67,776.8

C. Perimeter of the house = 2 × (l + b) = 2 × (18.7 + 12.6) = 2 × 31.3 = 62.6 m

D. Area of the orchard = 20 × 15.7 = 314 m²

No. of banana trees that can be planted in 1.25 m² of ground space =1

Hence, No. of banana trees that can be planted in 314 m² of ground space = = 251.25

i.e. ,251 banana trees can be planted in the orchard.

A. Total area covered by both the bedrooms and the kitchen

= 2 × area of one bedroom + area of kitchen

= 2 × ( x × 5) + (15-(x + 2)) × 5

= 10x + (75 – 5x – 10)

= (5x + 65) m²

B. Perimeter of the living room = 15 + 2 + 5 + (15 – x) + 5 + x + 2 = 44 m

C. Total area of the bedrooms and the living room

= area of bedroom1 + (area of bedroom 2 + area of living room)

= 5x + 15 × 7

= (5x + 105) m²

Cost of carpeting (5x + 105) m² at the cost of Rs. 50/ m² = 50 × (5x + 105) = Rs. 250(x + 21)

D. Total area of bathroom and kitchen = (15 – x) × 5 m²

Total cost of carpeting bathroom and kitchen floors at the cost of Rs. 30/ m²

= 30 × (15 – x) × 5

= Rs. 150(15 - x)

E. Given, area of floor of each bedroom = 35 m²

Area of one bedroom = 5x m²

⇒ 5x= 35

⇒ x = 7 m

Width of flower bed = 50 cm = 0.5 m

Length of the remaining portion = 10 – (0.5 × 2) = 9 m

Breadth of the remaining portion = 4 – 0.5 = 3.5 m

Area of remaining portion = 9 × 3.5 = 31.5 m²

Total area = 120 × 80 = 9,600 m²

Area of the vertical road lying above the horizontal road = 77 × 2 = 154 m²

Area of the horizontal road = 120 × 3 = 360 m²

Area of the remaining playground = 9,600 – (154 + 360)

= 9086 m²

Area of the flower bed = Area of the park – Area of the park other than the flower bed

= 20 × 15 – (15-1) × (20–1)

= 300 – 266 = 34 m²

Total cost of manuring for the flower bed at the rate of Rs 45 per m² = 34 × 45 = Rs. 1530

Length of outer rectangle = 60 + 6 + 6 = 72 cm

Breadth of outer rectangle = 38 + 6 + 6 = 50 cm

Area of wooden frame = Area of outer rectangle – area on inner rectangle

= 72 × 50 – 60 × 38

= 3,600 – 2,280 = 1,320 m²

Area of one triangle = × 10 × 30 = 150 cm²

Area of 4 such triangles = 4 × 150 = 600 cm²

Since the triangles are congruent, it can be concluded that the side of the square = 10 + 30 = 40 cm

Area of square = 40² = 1600 cm²

Area of shaded portion = 1600 – 600 = 1000 cm²

(i) We know that area of a square = (side)²

and area of a circle = ∏r²

Area of quarter circle = × ∏r² =

Area of the shaded portion of first tile = Area of square - 4 × Area of one quarter circle

= (20)² - 4 ×

= 400 - 4 × 3.14 × × 10 × 10

= 400 - 314

= 86 cm²

(ii) In the second tile, diameter of the circle = Side of the tile = 20 cm

i.e. radius = 10 cm

Area of the shaded portion of second tile = Area of square - Area of circle

= (20)² – 3.14 × 10 × 10

= 400 - 314

= 86 cm²

Shaded area in both the tiles is same.

Total area of the field = 48 × 12 m²

Area of one triangular bed = × 2 × 4 = 4 m²

Number of flower beds that can be laid in the field = = 144

Let the increase in length be x m.

Give, increase in area of the field = 650 m²

Area of expanded wheat field – Area of original wheat field= 650 m²

(32 + x ) × 26 – 32 × 26 = 650

26 (32 + x – 32) = 650

x = = 25 m

Length of expanded wheat field = 32 + x = 32 + 25 = 57 m

In ∆AEC, AE = 15 cm, AC = 25 cm

Applying Pythagoras theorem in ∆AEC,

(AE)² + (EC)² = (AC)²

(15)² + (EC)² = (25)²

(EC)² = 625 - 225 = 400

EC =√400 = 20 cm

EB = EC – BC = 20 – 7 = 13 cm

Using the formula: Area of triangle = × b × h

Area of ΔAEC = × AE × EC = × 15 × 20 = 150 cm²

Area of ΔAEB = × AE × EB = × 15 × 13 = 97.5 cm²

Area of ΔABC = Area of ΔAEC - Area of ΔAEB

= 150 – 97.5 = 52.5 cm²

Also, area of ΔABC = × DB × AC

52.5 = × DB × 25

DB = = 4.2 cm

Hence, Area of ΔABC =52.5 cm² and DB = 4.2 cm

Number of pieces that can be cut from the sheet of chocolate

= = = 108

As there are 9 short line segments along length and breadth of the large square,

Length of the large square = 9 × 1 = 9 cm

Breadth of the large square = 1 × 9 = 9 cm

There are 41 small squares of 1 cm × 1 cm each.

∴ Area of shaded region = Area of large square – Area of 41 small squares

= 9 × 9 – 41 × 1 × 1 = 81 – 41 = 40 cm²

(i) Length of the inner rectangle (house, excluding path) = 4 + 2.5 + 4 =10.5 m

Breadth of the inner rectangle (house, excluding path) = 3 + 3 =6 m

Area of the inner rectangle = 10.5 × 6 = 63 m²

Length of the outer rectangle = Length of the inner rectangle + 2 × 1 =10.5 + 2 =12.5 m

Breadth of the outer rectangle = Breadth of the inner rectangle + 2 × 1 = 6 + 2 = 8 m

Area of the outer rectangle = 12.5 × 8 = 100 m²

Area of the path = Area of the outer rectangle – Area of the inner rectangle

= 100 – 63 = 37 m²

(ii) Area of the bathroom = 2.5 × 2 = 5 m²

Area of the floor to be covered = 63 – 5 = 58 m²

Cost of flooring = 120 × 58 = Rs. 6960

(iii) The living room can be divided into two rectangles of dimensions

6.5 m × 3 m and 2.5 m × 1 m

Area of living room = 19.5 + 2.5 = 22 m²

A. (i) Length of family room = 5.48 m

Breadth of family room = 4.57 m

Perimeter of family room = 2 (l + b) = 2 (5.48 + 4.57) = 2 × 10.05 = 20.1 m²

Moulding required for family room = 20.1 m²

(ii) Length of living room = 7.53 m

Breadth of living room = 3.81 m

Perimeter of living room = 2 (l + b) = 2 (7.53 + 3.81) = 2 × 11.84 = 22.68 m²

Moulding required for living room = 22.68 m²

(iii) Length of dining room = 5.48 m

Breadth of dining room = 5.41 m

Perimeter of dining room = 2 (l + b) = 2 (5.48 + 5.41) = 2 × 10.89 = 21.78 m²

Moulding required for dining room = 20.1 m²

(iv) Side of bedroom 1 = 3.04 m

Perimeter of bedroom 1 = 4 × side = 4 × 3.04 = 12.16 m²

Moulding required for bedroom 1 = 12.16 m²

(v) Length of bedroom 2 = 3.04 m

Breadth of bedroom 1 = 2.43 m

Perimeter of dining room = 2 (l + b) = 2 (3.04 + 2.43) = 2 × 5.47 = 10.94 m²

Moulding required for dining room = 20.1 m²

B. Area of family room = 5.48 × 4.57 = 25.0436 m²

Cost of carpeting family room at the rate of Rs 200/m² = 200 × 25.0436 =Rs. 5008.72

Area of living room = 7.53 × 3.81= 28.6893 m²

Cost of carpeting living room at the rate of Rs 200/m² = 200 × 28.6893 =Rs. 5737.86

Area of dining room = 5.48 × 5.41= 29.6468 m²

Cost of carpeting dining room at the rate of Rs 200/m² = 200 × 29.6468 =Rs. 5929.36

Area of bedroom1 = 3.04 × 3.04 = 9.2416 m²

Cost of carpeting bedroom1 at the rate of Rs 200/m² = 200 × 9.2416 =Rs. 1848.32

Area of bedroom2 = 3.04 × 2.43 = 7.3872 m²

Cost of carpeting bedroom 2 at the rate of Rs 200/m² = 200 × 7.3872 =Rs. 1477.44

C. Total perimeter of all five rooms = 20.1 + 22.68 + 21.78 + 12.16 + 10.94 = 87.66 m

Cost of moulding 1 m = Rs 500

Total cost of moulding all five rooms = 500 × 87.66 = Rs. 43,830

Given, P is the mid point of AB, so AP = 40 cm

And S is the mid point of AD, so AS = 30 cm

Area of ABCD = 80 × 60 = 4,800 cm²

Area of ∆SAP = × AP × AS = × 40 × 30 = 600 cm²

Area of 4 unshaded congruent triangles =4 × 600 cm² = 2,400 cm²

Area of unshaded circle = ∏r² = 3.14 × 10 × 10 = 314 cm²

∴ Area of shaded area = 4,800 – (2,400 + 314) = 2086 cm²

Total area of rectangular sheet = 100 × 80 = 8,000 cm²

Area of each square = (10)² = 100 cm²

Area of triangle = × 10 × 10 = 50 cm²

Area of the remaining part of the paper = 8,000 – (100 + 150) = 7,550 cm²

Let the radius of inner and outer circle be r and R, respectively.

Given, inner circumference = 352 mm

⇒ 2∏r = 352

⇒ 2 × × r = 352

⇒ r = = 56 mm

and outer circumference = 396 mm

⇒ 2∏R = 396

⇒ 2 × × R = 396

⇒ r = = 63 mm

∴ Width of circular design = R – r = 63 -56 = 7 mm

Length of path described by moon in one complete revolution

= 2∏r

= 2 × 3.14 × 3,84,000

= 2,411,520 km

Since the photograph has dimensions as th of its actual size, hence

Actual length = 25 × 10 = 250 cm

Actual breadth = 10 × 10 = 100 cm

Actual area of the table = 250 × 100 = 25,000 cm²

Diameter of 1 hole = 0.5 × 10 = 5 cm

Radius of 1 hole = 2.5 cm

Area of 6 holes = 6 × ∏r² = 6 × × 2.5 × 2.5 = 118 cm²

Area to be polished = Area of table – Area of 6 holes = 25,000 – 118 = 24,882 cm² = 2.4882 m²

∴ Cost of polishing at the rate of Rs 200/ m² = 2.4882 × 200 = Rs 497