# Area

##### Class 8th Mathematics (new) MHB Solution

##### Class 8^{th} Mathematics (new) MHB Solution

**Practice Set 15.1**- If the base of a parallelogram is 18 cm and its height is 11 cm, find its area.…
- If the area of a parallelogram is 29.6 sq cm and its base is 8 cm, find its height.…
- Area of a parallelogram is 83.2 sq cm. If its height is 6.4 cm, find the length of its…

**Practice Set 15.2**- Lengths of the diagonals of a rhombus are 15cm and 24 cm, find its area.…
- Length of the diagonals of a rhombus are 16.5 cm and 14.2 cm, find its area.…
- If the perimeter of a rhombus is 100 cm and length of one diagonal is 48 cm, what is…
- If the length of a diagonal of a rhombus is 30 cm and its area is 240 sq cm, find its…

**Practice Set 15.3**- In ABCD, l (AB) = 13 cm, l (DC) = 9 cm, l (AD) = 8 cm, find the area of ABCD. .5…
- Length of the two parallel sides of a trapezium is 8.5 cm and 11.5 cm respectively and…
- PQRS is an isosceles trapezium l (PQ) = 7 cm. seg PM ⊥ seg SR, l(SM) = 3 cm,Distance…

**Practice Set 15.4**- Sides of a triangle are cm 45 cm, 39 cm, and 42 cm, find its area.…
- Look at the measures shown in the adjacent figure and find the area of PQRS.…
- Some measures are given in the adjacent figure, find the area of ABCD.…

**Practice Set 15.5****Practice Set 15.6**- Radii of the circles are given below, find their areas.(1) 28 cm(2) 10.5 cm(3) 17.5 cm…
- Areas of some circles are given below find their diameters.(1) 176 sq cm(2) 394.24 sq…
- The diameter of the circular garden is 42 m. There is a 3.5 m wide road around the…
- Find the area of the circle if its circumference is 88 cm.

**Practice Set 15.1**

- If the base of a parallelogram is 18 cm and its height is 11 cm, find its area.…
- If the area of a parallelogram is 29.6 sq cm and its base is 8 cm, find its height.…
- Area of a parallelogram is 83.2 sq cm. If its height is 6.4 cm, find the length of its…

**Practice Set 15.2**

- Lengths of the diagonals of a rhombus are 15cm and 24 cm, find its area.…
- Length of the diagonals of a rhombus are 16.5 cm and 14.2 cm, find its area.…
- If the perimeter of a rhombus is 100 cm and length of one diagonal is 48 cm, what is…
- If the length of a diagonal of a rhombus is 30 cm and its area is 240 sq cm, find its…

**Practice Set 15.3**

- In ABCD, l (AB) = 13 cm, l (DC) = 9 cm, l (AD) = 8 cm, find the area of ABCD. .5…
- Length of the two parallel sides of a trapezium is 8.5 cm and 11.5 cm respectively and…
- PQRS is an isosceles trapezium l (PQ) = 7 cm. seg PM ⊥ seg SR, l(SM) = 3 cm,Distance…

**Practice Set 15.4**

- Sides of a triangle are cm 45 cm, 39 cm, and 42 cm, find its area.…
- Look at the measures shown in the adjacent figure and find the area of PQRS.…
- Some measures are given in the adjacent figure, find the area of ABCD.…

**Practice Set 15.5**

**Practice Set 15.6**

- Radii of the circles are given below, find their areas.(1) 28 cm(2) 10.5 cm(3) 17.5 cm…
- Areas of some circles are given below find their diameters.(1) 176 sq cm(2) 394.24 sq…
- The diameter of the circular garden is 42 m. There is a 3.5 m wide road around the…
- Find the area of the circle if its circumference is 88 cm.

###### Practice Set 15.1

**Question 1.**If the base of a parallelogram is 18 cm and its height is 11 cm, find its area.

**Answer:**we know that,

Area of parallelogram = base × height

Given that base of parallelogram = 18cm

And, the height of parallelogram = 11cm

Area of parallelogram = 18 × 11

= **198 sq cm**

**Question 2.**If the area of a parallelogram is 29.6 sq cm and its base is 8 cm, find its height.

**Answer:**We know that,

Area of parallelogram = base × height

Given that area of parallelogram = 29.6cm

And, the base of parallelogram = 8cm

**= 3.7 cm**

**Question 3.**Area of a parallelogram is 83.2 sq cm. If its height is 6.4 cm, find the length of its base.

**Answer:**We know that,

Area of parallelogram = base × height

Given that area of parallelogram = 83.2cm

And, the height of parallelogram = 6.4cm

**= 13 cm**

**Question 1.**

If the base of a parallelogram is 18 cm and its height is 11 cm, find its area.

**Answer:**

we know that,

Area of parallelogram = base × height

Given that base of parallelogram = 18cm

And, the height of parallelogram = 11cm

Area of parallelogram = 18 × 11

= **198 sq cm**

**Question 2.**

If the area of a parallelogram is 29.6 sq cm and its base is 8 cm, find its height.

**Answer:**

We know that,

Area of parallelogram = base × height

Given that area of parallelogram = 29.6cm

And, the base of parallelogram = 8cm

**= 3.7 cm**

**Question 3.**

Area of a parallelogram is 83.2 sq cm. If its height is 6.4 cm, find the length of its base.

**Answer:**

We know that,

Area of parallelogram = base × height

Given that area of parallelogram = 83.2cm

And, the height of parallelogram = 6.4cm

**= 13 cm**

###### Practice Set 15.2

**Question 1.**Lengths of the diagonals of a rhombus are 15cm and 24 cm, find its area.

**Answer:**We know that,

Area of rhombus = ×product of the length of diagonals

Given that length of one of the diagonals is 15cm

And the other is 24cm

⇒ Area of rhombus = 1/2×15×24

**= 180 sq cm**

**Question 2.**Length of the diagonals of a rhombus are 16.5 cm and 14.2 cm, find its area.

**Answer:**We know that,

Area of rhombus = ×product of the length of diagonals

Given that length of one of the diagonals is 16.5cm

And the other is 14.2cm

**= 117.5 sq cm**

**Question 3.**If the perimeter of a rhombus is 100 cm and length of one diagonal is 48 cm, what is the area of the quadrilateral?

**Answer:**

We know that perimeter of rhombus = 4 × side of the rhombus

Given perimeter of rhombus = 100cm

Side AB of rhombus = 100/4 = 25cm

Let BD be the diagonal given = 48cm

We know that diagonals of a rhombus bisect each other

E is the midpoint of BD

BE = 24 cm

Now, ∆ABE is the right angle triangle at E

∴using Pythagoras theorem,

AE^{2} + BE^{2}= AB^{2}

AE = 7cm

Area of rhombus = 4×area of ∆ABE

= 2 × 24 × 7

**= 336 sq cm**

**Question 4.**If the length of a diagonal of a rhombus is 30 cm and its area is 240 sq cm, find its perimeter.

**Answer:**

we know that,

Given that area of rhombus = 240 sq cm

And diagonal BD = 30cm

⇒other diagonal, AC = 240 × 2 ÷ 30

AC = 16cm

We know that diagonals of a rhombus bisect each other,

So let E be the midpoint of their point of intersection.

Now, AE = 16/2 = 8cm

And BE = 30/2 = 15cm

Now, ∆ABE is right angle triangle

∴ using Pythagoras theorem,

AE^{2}+ BE^{2} = AB^{2}

⇒AB = 17cm

We know that perimeter of rhombus = 4 × side of rhombus

= 4 × 17

**= 68 cm**

**Question 1.**

Lengths of the diagonals of a rhombus are 15cm and 24 cm, find its area.

**Answer:**

We know that,

Area of rhombus = ×product of the length of diagonals

Given that length of one of the diagonals is 15cm

And the other is 24cm

⇒ Area of rhombus = 1/2×15×24

**= 180 sq cm**

**Question 2.**

Length of the diagonals of a rhombus are 16.5 cm and 14.2 cm, find its area.

**Answer:**

We know that,

Area of rhombus = ×product of the length of diagonals

Given that length of one of the diagonals is 16.5cm

And the other is 14.2cm

**= 117.5 sq cm**

**Question 3.**

If the perimeter of a rhombus is 100 cm and length of one diagonal is 48 cm, what is the area of the quadrilateral?

**Answer:**

We know that perimeter of rhombus = 4 × side of the rhombus

Given perimeter of rhombus = 100cm

Side AB of rhombus = 100/4 = 25cm

Let BD be the diagonal given = 48cm

We know that diagonals of a rhombus bisect each other

E is the midpoint of BD

BE = 24 cm

Now, ∆ABE is the right angle triangle at E

∴using Pythagoras theorem,

AE^{2} + BE^{2}= AB^{2}

AE = 7cm

Area of rhombus = 4×area of ∆ABE

= 2 × 24 × 7

**= 336 sq cm**

**Question 4.**

If the length of a diagonal of a rhombus is 30 cm and its area is 240 sq cm, find its perimeter.

**Answer:**

we know that,

Given that area of rhombus = 240 sq cm

And diagonal BD = 30cm

⇒other diagonal, AC = 240 × 2 ÷ 30

AC = 16cm

We know that diagonals of a rhombus bisect each other,

So let E be the midpoint of their point of intersection.

Now, AE = 16/2 = 8cm

And BE = 30/2 = 15cm

Now, ∆ABE is right angle triangle

∴ using Pythagoras theorem,

AE^{2}+ BE^{2} = AB^{2}

⇒AB = 17cm

We know that perimeter of rhombus = 4 × side of rhombus

= 4 × 17

**= 68 cm**

###### Practice Set 15.3

**Question 1.**In ABCD, l (AB) = 13 cm, l (DC) = 9 cm, l (AD) = 8 cm, find the area of ABCD.

**Answer:**We know that,

From the fig. it is clear that AB and CD are the 2 parallel sides

Given that AB = 13cm, CD = 9cm and AD = 8cm

Here sum of parallel sides, i.e., AB+CD = 13+9 = 22

Hence,

**= 88 sq cm**

**Question 2.**Length of the two parallel sides of a trapezium is 8.5 cm and 11.5 cm respectively and its height is 4.2 cm, find its area.

**Answer:**We know that,

Given that length of 2 parallel sides = 8.5cm and 11.5cm

⇒sum of parallel sides = 8.5 + 11.5 = 20

And, distance between them = 4.2cm

**= 42 sq cm**

**Question 3.**PQRS is an isosceles trapezium l (PQ) = 7 cm. seg PM ⊥ seg SR, l(SM) = 3 cm,

Distance between two parallel sides is 4 cm, find the area of PQRS

**Answer:**

Given that the trapezium is isosceles. Therefore from the fig. it is clear that SM = NR = 3cm

Also, PQ = MN = 7cm

Now, length of side SR = 3 + 7 + 3 = 13cm

Therefore, the sum of parallel sides of trapezium = 7 + 13 = 20

And the distance between them = 4 cm

**= 40 sq cm**

**Question 1.**

In ABCD, l (AB) = 13 cm, l (DC) = 9 cm, l (AD) = 8 cm, find the area of ABCD.

**Answer:**

We know that,

From the fig. it is clear that AB and CD are the 2 parallel sides

Given that AB = 13cm, CD = 9cm and AD = 8cm

Here sum of parallel sides, i.e., AB+CD = 13+9 = 22

Hence,

**= 88 sq cm**

**Question 2.**

Length of the two parallel sides of a trapezium is 8.5 cm and 11.5 cm respectively and its height is 4.2 cm, find its area.

**Answer:**

We know that,

Given that length of 2 parallel sides = 8.5cm and 11.5cm

⇒sum of parallel sides = 8.5 + 11.5 = 20

And, distance between them = 4.2cm

**= 42 sq cm**

**Question 3.**

PQRS is an isosceles trapezium l (PQ) = 7 cm. seg PM ⊥ seg SR, l(SM) = 3 cm,

Distance between two parallel sides is 4 cm, find the area of PQRS

**Answer:**

Given that the trapezium is isosceles. Therefore from the fig. it is clear that SM = NR = 3cm

Also, PQ = MN = 7cm

Now, length of side SR = 3 + 7 + 3 = 13cm

Therefore, the sum of parallel sides of trapezium = 7 + 13 = 20

And the distance between them = 4 cm

**= 40 sq cm**

###### Practice Set 15.4

**Question 1.**Sides of a triangle are cm 45 cm, 39 cm, and 42 cm, find its area.

**Answer:**To find the area of a triangle whose three sides are given we have the Heron’s formula

Where, ∆ is an area of a triangle.

And a, b, c are the three sides of the triangle

In this question, we have the three sides of the triangle which are 45cm, 39cm, and 42cm

= 63m

S - a = 63 - 45 = 18

S - b = 63 - 39 = 24

S - c = 63 - 42 = 21

Hence

**= 756 sq m**

**Question 2.**Look at the measures shown in the adjacent figure and find the area of PQRS.

**Answer:**In the given fig. ∆ PRS is right angle triangle at S

Therefore, using Pythagoras theorem,

PS^{2} + SR^{2} = PR^{2}

= 39m

Now,

**Area of ∆PRS =****× base × height**

= ×PS×SR

= ×36×15

= 270 sq m

Now the area of triangle PQR, using heron’s formula

Here, sides are 56 cm, 25 cm, and 39 cm

Therefore,

S = 60

S - a = 60 - 56 = 4

S - b = 60 - 25 = 35

S - c = 60 - 39 = 21

= 420 sq m

Hence, the area of the quadrilateral PQRS = area of ∆PQR + ∆PSR

= 420 + 270

**= 690 sq m**

**Question 3.**Some measures are given in the adjacent figure, find the area of ABCD.

**Answer:**In the given fig. ABD is right angled triangle at A,

Given that AB = 40cm, and AD = 9cm

Therefore, the area of triangle ABD

= 180 sqm

Now, the area of triangle, ∆BCD

Now area of quadrilateral ABCD,

= 180 + 390

**= 570 sq m**

**Question 1.**

Sides of a triangle are cm 45 cm, 39 cm, and 42 cm, find its area.

**Answer:**

To find the area of a triangle whose three sides are given we have the Heron’s formula

Where, ∆ is an area of a triangle.

And a, b, c are the three sides of the triangle

In this question, we have the three sides of the triangle which are 45cm, 39cm, and 42cm

= 63m

S - a = 63 - 45 = 18

S - b = 63 - 39 = 24

S - c = 63 - 42 = 21

Hence

**= 756 sq m**

**Question 2.**

Look at the measures shown in the adjacent figure and find the area of PQRS.

**Answer:**

In the given fig. ∆ PRS is right angle triangle at S

Therefore, using Pythagoras theorem,

PS^{2} + SR^{2} = PR^{2}

= 39m

Now,

**Area of ∆PRS =****× base × height**

= ×PS×SR

= ×36×15

= 270 sq m

Now the area of triangle PQR, using heron’s formula

Here, sides are 56 cm, 25 cm, and 39 cm

Therefore,

S = 60

S - a = 60 - 56 = 4

S - b = 60 - 25 = 35

S - c = 60 - 39 = 21

= 420 sq m

Hence, the area of the quadrilateral PQRS = area of ∆PQR + ∆PSR

= 420 + 270

**= 690 sq m**

**Question 3.**

Some measures are given in the adjacent figure, find the area of ABCD.

**Answer:**

In the given fig. ABD is right angled triangle at A,

Given that AB = 40cm, and AD = 9cm

Therefore, the area of triangle ABD

= 180 sqm

Now, the area of triangle, ∆BCD

Now area of quadrilateral ABCD,

= 180 + 390

**= 570 sq m**

###### Practice Set 15.5

**Question 1.**Find the areas of given plots. (All measures are in meters.)

(1)

(2)

**Answer:**(1)

Given that,

PA = 30m, AC = 30m, and CT = 30m

PC = PA + AC = 30 + 30 = 60m

∆PCT is right angled triangle at C

Area of ∆PCT = 1/2 × PC × CT

= 900m…………(1)

In, ∆SCT is right angled triangle at C

SB = 60m, BC = 30m, and CT = 30m

Area of ∆SCT = 1/2× base × height

= ×30×90

= 1350m…………….(2)

In ∆SBR is right angled triangle at B

SB = 60m, BR = 25m

Area of ∆SBR = 1/2 × base × height

= × SB × BR

= × 60 × 25

= 750m…………..(3)

In ∆APQ is right angled triangle at A

AP = 30m, AQ = 50m

Area of ∆APQ = × base × height

= ×AP×AQ

= ×50×30

= 750m…………(4)

Now, in trapezium ABRQ

AQ and RB are the 2 parallel sides

Also, AQ = 50m and BR = 25m

⇒AQ + BR = 75m

The distance between AQ and BR = 60m

Hence,

= 2250 sq m………….(5)

Now area of quadrilateral PQRST = (1)+(2)+(3)+(4)+(5)

= 900+1350+750+750+2250

= 6000 sq m

(2) the data for this question is inadequate.

**Question 1.**

Find the areas of given plots. (All measures are in meters.)

(1)

(2)

**Answer:**

(1)

Given that,

PA = 30m, AC = 30m, and CT = 30m

PC = PA + AC = 30 + 30 = 60m

∆PCT is right angled triangle at C

Area of ∆PCT = 1/2 × PC × CT

= 900m…………(1)

In, ∆SCT is right angled triangle at C

SB = 60m, BC = 30m, and CT = 30m

Area of ∆SCT = 1/2× base × height

= ×30×90

= 1350m…………….(2)

In ∆SBR is right angled triangle at B

SB = 60m, BR = 25m

Area of ∆SBR = 1/2 × base × height

= × SB × BR

= × 60 × 25

= 750m…………..(3)

In ∆APQ is right angled triangle at A

AP = 30m, AQ = 50m

Area of ∆APQ = × base × height

= ×AP×AQ

= ×50×30

= 750m…………(4)

Now, in trapezium ABRQ

AQ and RB are the 2 parallel sides

Also, AQ = 50m and BR = 25m

⇒AQ + BR = 75m

The distance between AQ and BR = 60m

Hence,

= 2250 sq m………….(5)

Now area of quadrilateral PQRST = (1)+(2)+(3)+(4)+(5)

= 900+1350+750+750+2250

= 6000 sq m

(2) the data for this question is inadequate.

###### Practice Set 15.6

**Question 1.**Radii of the circles are given below, find their areas.

(1) 28 cm

(2) 10.5 cm

(3) 17.5 cm

**Answer:**(1) We know that

area of circle = πr^{2}

Here given that radius of the circle is 28cm

= 784π sq cm

= 2464 sq cm

(2) Here the radius of the circle = 10.5 cm

= 110.25π sq cm

= 346.5 sq cm

(3) Here the radius of the circle is 17.5cm

= 306.25π sq cm

= 961.625 sq cm

**Question 2.**Areas of some circles are given below find their diameters.

(1) 176 sq cm

(2) 394.24 sq cm

(3) 12474 sq cm

**Answer:**(1) We know that area of circle = πr^{2}

Here area of circle = 176cm

(2) Here area of circle = 394.24 sq cm

r = 11.2 cm

D = 2r = 2(11.20) = 22.4 cm

(3) Here area of circle = 12474 sq cm

D = 2r = 2(63) = 126cm

**Question 3.**The diameter of the circular garden is 42 m. There is a 3.5 m wide road around the garden. Find the area of the road.

**Answer:**Given that the diameter of the garden (inner circle) = 42m

Therefore, inner radius, r = 21m

Also, given that road surrounds the garden and is 3.5 m wide.

Therefore, the diameter of the road (outer circle) will be = 42+2(3.5) = 49m

And then outer radius, R = 24.5m

Now, the area of road = area of the outer circle – area of the inner circle

area of outer circle = πR^{2}

= π(24.5)^{2}

= 1885 sq m

area of inner circle = πr^{2}

= π(21)^{2}

= 1385 sq m

Hence, area of road = 1885-1385 = 500 sq m

**Question 4.**Find the area of the circle if its circumference is 88 cm.

**Answer:**we know that,

The Circumference of a circle = 2πr

Given circumference = 88cm

⇒2πr = 88

r = 14cm

Now, area of circle = πr^{2}

= π(14)^{2}

= 615.75 sq cm

**Question 1.**

Radii of the circles are given below, find their areas.

(1) 28 cm

(2) 10.5 cm

(3) 17.5 cm

**Answer:**

(1) We know that

area of circle = πr^{2}

Here given that radius of the circle is 28cm

= 784π sq cm

= 2464 sq cm

(2) Here the radius of the circle = 10.5 cm

= 110.25π sq cm

= 346.5 sq cm

(3) Here the radius of the circle is 17.5cm

= 306.25π sq cm

= 961.625 sq cm

**Question 2.**

Areas of some circles are given below find their diameters.

(1) 176 sq cm

(2) 394.24 sq cm

(3) 12474 sq cm

**Answer:**

(1) We know that area of circle = πr^{2}

Here area of circle = 176cm

(2) Here area of circle = 394.24 sq cm

r = 11.2 cm

D = 2r = 2(11.20) = 22.4 cm

(3) Here area of circle = 12474 sq cm

D = 2r = 2(63) = 126cm

**Question 3.**

The diameter of the circular garden is 42 m. There is a 3.5 m wide road around the garden. Find the area of the road.

**Answer:**

Given that the diameter of the garden (inner circle) = 42m

Therefore, inner radius, r = 21m

Also, given that road surrounds the garden and is 3.5 m wide.

Therefore, the diameter of the road (outer circle) will be = 42+2(3.5) = 49m

And then outer radius, R = 24.5m

Now, the area of road = area of the outer circle – area of the inner circle

area of outer circle = πR^{2}

= π(24.5)^{2}

= 1885 sq m

area of inner circle = πr^{2}

= π(21)^{2}

= 1385 sq m

Hence, area of road = 1885-1385 = 500 sq m

**Question 4.**

Find the area of the circle if its circumference is 88 cm.

**Answer:**

we know that,

The Circumference of a circle = 2πr

Given circumference = 88cm

⇒2πr = 88

r = 14cm

Now, area of circle = πr^{2}

= π(14)^{2}

= 615.75 sq cm