Maharashtra Board Textbook Solutions for Standard Eight

Chapter 15 - Area

Practice set 15.1

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

IMG 20230308 220856 Chapter 15 – Area

Given:

Base of the parallelogram = 18 cm

Height of the parallelogram = 11 cm

 

To find:

Area of parallelogram

 

Solution:

Area of parallelogram = Base × Height

∴ Area of parallelogram = 18 × 11

∴ Area of parallelogram = 198 sq.cm

 

Ans: Area of parallelogram is 198 sq.cm

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

IMG 20230308 221213 Chapter 15 – Area

Given:

Area of the parallelogram = 29.6 sq cm

Base of the parallelogram = 8 cm

 

To find:

Height of the parallelogram

 

Solution:

Area of parallelogram = Base × Height

∴ 29.6 = 8 × Height

∴ \(\large \frac {29.6}{8}\) = Height

∴ Height = 3.7 cm

 

Ans: Height of the parallelogram is 3.7 cm

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

IMG 20230308 221055 Chapter 15 – Area

Given:

Area of the parallelogram = 83.2 sq cm

Height of the parallelogram = 6.4 cm

 

To find:

Base of the parallelogram

 

Solution:

Area of parallelogram = Base × Height

∴ 83.2 = Base × 6.4

∴ \(\large \frac {83.2}{6.4}\) = Base

∴ Base = 13 cm

 

Ans: Base of the parallelogram is 13 cm

Practice set 15.2

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

IMG 20230309 020909 Chapter 15 – Area

Given:

Length of 1st diagonal = 15 cm

Length of 2nd diagonal = 24 cm

 

To find:

Area of rhombus

 

Solution:

Area of rhombus = \(\large \frac {1}{2}\) × Product of diagonals

∴ Area of rhombus = \(\large \frac {1}{2}\) × 15 × 24

∴ Area of rhombus = 15 × 12

∴ Area of rhombus = 180 sq.cm

 

Ans: Area of rhombus is 180 sq.cm

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

IMG 20230308 221114 Chapter 15 – Area

Given:

Length of 1st diagonal = 16.5 cm

Length of 2nd diagonal = 14.2 cm

 

To find:

Area of rhombus

 

Solution:

Area of rhombus = \(\large \frac {1}{2}\) × Product of diagonals

∴ Area of rhombus = \(\large \frac {1}{2}\) × 16.5 × 14.2

∴ Area of rhombus = 16.5 × 7.1

∴ Area of rhombus = 117.15 sq.cm

 

Ans: Area of rhombus is 117.15 sq.cm

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

IMG 20230308 221348 Chapter 15 – Area

Given:

Perimeter of rhombus = 100 cm

Length of one diagonal = 48 cm

 

To find:

Area of rhombus 

 

Solution:

Let □ ABCD be the rhombus with perimeter 100 cm.

Diagonal AC and BD intersect at point M.

 

We know that,

Perimeter of a rhombus = 4 × side

∴ 100 = 4 × side

∴ \(\large \frac {100}{4}\) = Side

∴ Side = 25 cm

i.e. AB = 25 cm

 

Also,

Diagonals of a rhombus are perpendicular bisectors of each other

 

BD = 48 cm …[Given]

∴ BM = \(\large \frac {1}{2}\) × BD

∴ BM = \(\large \frac {1}{2}\) × 25

∴ BM = \(\large \frac {1}{2}\) × 25

 

In ∆ AMB, ∠ AMB = 90⁰

∴ ∆ AMB is a right angled triangle

∴ AB² = AM² + BM² …[By applying Pythagoras Theorem]

∴ 25² = AM² + 24²

∴ 625 = AM² + 576

∴ 625 – 576 = AM²

∴ AM² = 49

∴ \( \sqrt {AM²}\) = \( \sqrt {49}\) …[By Taking square root of both sides] 

∴ AM = 7 cm

 

Now, 

AC = 2 × AM

∴ AC = 2 × 7

∴ AC = 14 cm

Area of rhombus = \(\large \frac {1}{2}\) × Product of diagonals

∴ Area of rhombus = \(\large \frac {1}{2}\) × AC × BD

∴ Area of rhombus = \(\large \frac {1}{2}\) × 14 × 48

∴ Area of rhombus = 14 × 24

∴ Area of rhombus = 336 sq.cm

 

Ans: Area of rhombus is 336 sq.cm

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

IMG 20230308 221334 Chapter 15 – Area

Given:

Length of one diagonal = 30 cm

Area of rhombus = 240 sq.cm

 

To find:

Perimeter of rhombus 

 

Solution:

Let □ PQRS be the rhombus.

Diagonal PR and QS intersect at point M.

 

Area of rhombus = \(\large \frac {1}{2}\) × Product of diagonals

∴ 240 = \(\large \frac {1}{2}\) × PR × QS

∴ 240 = \(\large \frac {1}{2}\) × 30 × QS

∴ 240 = 15 × QS

∴ \(\large \frac {240}{15}\) = QS

∴ QS = 16 sq.cm

 

We know that,

Diagonals of a rhombus are perpendicular bisectors of each other

 

∴ PM = \(\large \frac {1}{2}\) × RP

∴ PM = \(\large \frac {1}{2}\) × 30

∴ PM = 15 cm

 

and QM = \(\large \frac {1}{2}\) × QS

∴ QM = \(\large \frac {1}{2}\) × 16

∴ QM = 8 cm

 

In ∆ PMQ, ∠ PMQ = 90⁰

∴ ∆ PMQ is a right angled triangle

∴ PQ² = PM² + QM² …[By applying Pythagoras Theorem]

∴ PQ² = 15² + 8²

∴ PQ² = 225 + 64

∴ PQ² = 289

∴ \( \sqrt {PQ²}\) = \( \sqrt {289}\) …[By Taking square root of both sides] 

∴ PQ = 17 cm

 

Now,

Perimeter of a rhombus = 4 × PQ

∴ Perimeter of a rhombus = 4 × 17

∴ Perimeter of a rhombus = 68 cm

 

Ans: Perimeter of a rhombus is 68 cm

Practice set 15.3

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

IMG 20230308 221432 Chapter 15 – Area

Given:

In □ABCD, 

l(AB) = 13 cm

l(DC) = 9 cm

l(AD) = 8 cm

 

To find:

Area of □ABCD

 

Solution:

Area of □ABCD = \(\large \frac {1}{2}\) × Sum of parallel sides × Height

∴ Area of □ABCD = \(\large \frac {1}{2}\) × (AB + CD) × AD

∴ Area of □ABCD = \(\large \frac {1}{2}\) × (13 + 9) × 8

∴ Area of □ABCD = 22 × 4

∴ Area of □ABCD = 88 sq.cm

 

Ans: Area of □ABCD is 88 sq.cm

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

IMG 20230308 220940 Chapter 15 – Area

Given:

Parallel side 1 = 8.5 cm

Parallel side 2 = 11.5 cm

Height of the trapezium = 4.2 cm

 

To find:

Area of trapezium 

 

Solution:

Area of trapezium = \(\large \frac {1}{2}\) × Sum of parallel sides × Height

∴ Area of trapezium = \(\large \frac {1}{2}\) × (8.5 + 11.5) × 4.2

∴ Area of trapezium = 20 × 2.1

∴ Area of trapezium = 42 sq.cm

 

Ans: Area of trapezium is 42 sq.cm 

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.

IMG 20230308 221410 Chapter 15 – Area

Given:

□PQRS is an isosceles trapezium,

l(PQ) = 7 cm

seg PM ⊥ seg SR

l(SM) = 3 cm

Distance between two parallel sides = 4 cm

 

To find:

Area of □PQRS

 

Solution:

In ∆ PMS, ∠ PMS = 90⁰

∴ ∆ AMB is a right angled triangle

∴ PS² = PM² + SM² …[By applying Pythagoras Theorem]

∴ PS² = 3² + 4²

∴ PS² = 9 + 16

∴ PS² = 25

∴ \( \sqrt {PS²}\) = \( \sqrt {25}\) …[By Taking square root of both sides] 

∴ PS = 5 cm

 

PS = QR [∵ □PQRS is an isosceles trapezium]

∴ QR = 5 cm

 

Draw seg QM ⊥ seg RS such that point N lies on seg RS

∴ QN = PM [Distance between two parallel lines are same.]

∴ QN = 4 cm

 

In ∆ QNR, ∠ QNR = 90⁰

∴ ∆ QNR is a right angled triangle

∴ QR² = QN² + RN² …[By applying Pythagoras Theorem]

∴ 5² = 4² + RN²

∴ 25 = 16 + RN²

∴ 25 – 16 = RN²

∴ RN² = 9

∴ \( \sqrt {RN²}\) = \( \sqrt {9}\) …[By Taking square root of both sides] 

∴ RN = 3 cm

 

In □PMNQ,

PM || QN [∵ They are ⊥ to the same line]

∴ PQ || MN [Parts of parallel sides]

 

∴  □PMNQ is a parallelogram.

∴ MN = PQ [Opposite sides of parallelogram are congruent]

∴ MN = 7 cm

 

Now,

SR = SM + MN + NR

∴ SR = 3 + 7 + 3

∴ SR = 13 cm

 

Area of □PQRS = \(\large \frac {1}{2}\) × Sum of parallel sides × Height

∴ Area of □PQRS = \(\large \frac {1}{2}\) × (PQ + RS) × PM

∴ Area of □PQRS = \(\large \frac {1}{2}\) × (7 + 13) × 4

∴ Area of □PQRS = 20 × 2

∴ Area of □PQRS = 40 sq.cm

 

Ans: Area of trapezium is 40 sq.cm 

Practice set 15.4

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

IMG 20230308 220812 Chapter 15 – Area

Given:

a = 45 cm

b = 39 cm

c = 42 cm

 

To find:

Area of triangle

 

Solution:

We know that,

semiperimeter (s) = \(\large \frac {a\, +\, b\, +\, c}{2}\)

∴ s = \(\large \frac {45\, +\, 39\, +\, 42}{2}\)

∴ s = \(\large \frac {126}{2}\)

∴ s = 63

 

Area of triangle = \( \sqrt {s\, (s\, –\, a)(s\, –\, b)(s\, –\, c) }\)

∴ Area of triangle = \( \sqrt {63\, (63\, –\, 45)(63\, –\, 39)(63\, –\, 42) }\)

∴ Area of triangle = \( \sqrt {63\, ×\, 18\, ×\, 24\, ×\, 21) }\)

∴ Area of triangle = \( \sqrt {9\, ×\, 7\, ×\, 2\, ×\, 9\, ×\, 2\, ×\, 4\, ×\, 3\, ×\, 3\, ×\, 7) }\)

∴ Area of triangle = \( \sqrt {9²\, ×\, 7²\, ×\, 2²\, ×\, 2²\, ×\, 3²) }\)

∴ Area of triangle = 9 × 7 × 2 × 2 ×3

∴ Area of triangle = 756 sq.cm

 

Ans: Area of triangle is 756 sq.cm

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

IMG 20230308 221500 Chapter 15 – Area

Given:

PS = 36 m

SR = 15 m

PQ = 56 m

QR = 25 m

∠ PSR = 90⁰

 

To find:

Area of □PQRS

 

Solution:

In ∆ PSR, ∠ PSR = 90⁰

∴ ∆ PSR is a right angled triangle

∴ PR² = PS² + SR² …[By applying Pythagoras Theorem]

∴ PR² = 36² + 15²

∴ PR² = 1296 + 225

∴ PR² = 1521

∴ \( \sqrt {PR²}\) = \( \sqrt {1521}\) …[By Taking square root of both sides] 

∴ PR = 39 m

 

Area of ∆PSR = \(\large \frac {1}{2}\) × Base × Height

∴ Area of ∆PSR = \(\large \frac {1}{2}\) × SR × PS

∴ Area of ∆PSR = \(\large \frac {1}{2}\) × 15 × 36

∴ Area of ∆PSR = 15 × 18

∴ Area of ∆PSR = 270 sq.m …(i)

 

For ∆PQR,

semiperimeter (s) = \(\large \frac {a\, +\, b\, +\, c}{2}\)

∴ s = \(\large \frac {39\, +\, 25\, +\, 56}{2}\)

∴ s = \(\large \frac {120}{2}\)

∴ s = 60

 

Area of ∆PQR = \( \sqrt {s\, (s\, –\, a)(s\, –\, b)(s\, –\, c) }\)

∴ Area of ∆PQR = \( \sqrt {60\, (60\, –\, 39)(60\, –\, 25)(60\, –\, 56) }\)

∴ Area of ∆PQR = \( \sqrt {60\, ×\, 21\, ×\, 35\, ×\, 4) }\)

∴ Area of ∆PQR = \( \sqrt {4\, ×\, 5\, ×\, 3\, ×\, 3\, ×\, 7\, ×\, 7\, ×\, 5\, ×\, 4) }\)

∴ Area of ∆PQR = \( \sqrt {4²\, ×\, 5²\, ×\, 3²\, ×\, 7²) }\)

∴ Area of ∆PQR = 4 × 5 × 3 × 7

∴ Area of ∆PQR = 420 sq.m …(ii)

 

Now,

Area of □PQRS = Area of ∆PSR + Area of ∆PQR

∴ Area of □PQRS = 270 + 420 …[From (i) and (ii)]

∴ Area of □PQRS = 690 sq.m

 

Ans: Area of □PQRS is 690 sq.m

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

IMG 20230308 221324 Chapter 15 – Area

Given:

AD = 9 m

AB = 40 m

BE = 13 m

DC = 60 m

∠ BAD = 90⁰

∠ BEC = 90⁰

 

To find:

Area of □ABCD

 

Solution:

Area of ∆BAD = \(\large \frac {1}{2}\) × Base × Height

∴ Area of ∆BAD = \(\large \frac {1}{2}\) × AD × AB

∴ Area of ∆BAD = \(\large \frac {1}{2}\) × 9 × 40

∴ Area of ∆BAD = 9 × 20

∴ Area of ∆BAD = 180 sq.m …(i)

 

Area of ∆BCD = \(\large \frac {1}{2}\) × Base × Height

∴ Area of ∆BCD = \(\large \frac {1}{2}\) × DC × BE

∴ Area of ∆BCD = \(\large \frac {1}{2}\) × 60 × 13

∴ Area of ∆BCD = 30 × 13

∴ Area of ∆BCD = 390 sq.m …(ii)

 

Now,

Area of □ABCD = Area of ∆BAD + Area of ∆BCD

∴ Area of □ABCD = 180 + 390 …[From (i) and (ii)]

∴ Area of □ABCD = 570 sq.m

 

Ans: Area of □ABCD is 570 sq.m

Practice set 15.5

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

1.

IMG 20230308 221521 Chapter 15 – Area

Given:

AP = 30

QA = 50

AB = 30

BC = 30

RC = 25

CS = 60

BT = 30

 

To find:

Area of the given plot

 

Solution:

Area of ∆PAQ = \(\large \frac {1}{2}\) × Base × Height

∴ Area of ∆PAQ = \(\large \frac {1}{2}\) × PA × QA

∴ Area of ∆PAQ = \(\large \frac {1}{2}\) × 30 × 50

∴ Area of ∆PAQ = 30 × 25

∴ Area of ∆PAQ = 750 sq.m …(i)

 

Area of ∆RCS = \(\large \frac {1}{2}\) × Base × Height

∴ Area of ∆RCS = \(\large \frac {1}{2}\) × CS × RC

∴ Area of ∆RCS = \(\large \frac {1}{2}\) × 60 × 25

∴ Area of ∆RCS = 30 × 25

∴ Area of ∆RCS = 750 sq.m …(ii)

 

Area of ∆PTS = \(\large \frac {1}{2}\) × Base × Height

∴ Area of ∆PTS = \(\large \frac {1}{2}\) × PS × TB

∴ Area of ∆PTS = \(\large \frac {1}{2}\) × 150 × 30

∴ Area of ∆PTS = 75 × 30

∴ Area of ∆PTS = 2250 sq.m …(iii)

 

□QRCA is a trapezium 

Area of □QRCA = \(\large \frac {1}{2}\) × Sum of parallel sides × Height

∴ Area of □QRCA = \(\large \frac {1}{2}\) × (AQ + CR) × AC

∴ Area of □QRCA = \(\large \frac {1}{2}\) × (50 + 25) × 60

∴ Area of □QRCA = 75 × 30

∴ Area of □QRCA = 2250 sq.cm …(iv)

 

Now,

Area of the given plot = Area of ∆PAQ + Area of ∆RCS + Area of ∆PTS + Area of □QRCA

∴ Area of the given plot = 750 + 750 + 2250 + 2250

∴ Area of the given plot = 6000 sq.m

 

Ans: Area of the given plot is 6000 sq.m

2.

IMG 20230308 221545 Chapter 15 – Area

Given:

AB = 24

BC = 26

BE = 30

CE = 28

FD = 16

∠ BAE = 90⁰

 

To find: 

Area of the given plot

 

Solution:

In ∆ BAE, ∠ BAE = 90⁰

∴ ∆ BAE is a right angled triangle

∴ BE² = BA² + AE² …[By applying Pythagoras Theorem]

∴ 30² = 24² + AE²

∴ 900 = 576 + AE²

∴ 900 – 576 = AE²

∴ AE² = 324

∴ \( \sqrt {AE²}\) = \( \sqrt {324}\) …[By Taking square root of both sides] 

∴ AE = 18 m 

 

Area of ∆ABE = \(\large \frac {1}{2}\) × Base × Height

∴ Area of ∆ABE = \(\large \frac {1}{2}\) × AB × AE

∴ Area of ∆ABE = \(\large \frac {1}{2}\) × 24 × 18

∴ Area of ∆ABE = 24 × 9

∴ Area of ∆ABE = 216 sq.m …(i)

 

Area of ∆CDE = \(\large \frac {1}{2}\) × Base × Height

∴ Area of ∆CDE = \(\large \frac {1}{2}\) × CE × DF

∴ Area of ∆CDE = \(\large \frac {1}{2}\) × 28 × 16

∴ Area of ∆CDE = 28 × 8

∴ Area of ∆CDE = 224 sq.m …(ii)

 

For ∆BCE,

semiperimeter (s) = \(\large \frac {a\, +\, b\, +\, c}{2}\)

∴ s = \(\large \frac {26\, +\, 30\, +\, 28}{2}\)

∴ s = \(\large \frac {84}{2}\)

∴ s = 42

 

Area of ∆BCE = \( \sqrt {s\, (s\, –\, a)(s\, –\, b)(s\, –\, c) }\)

∴ Area of ∆BCE = \( \sqrt {42\, (42\, –\, 26)(42\, –\, 30)(42\, –\, 28) }\)

∴ Area of ∆BCE = \( \sqrt {42\, ×\, 16\, ×\, 12\, ×\, 14) }\)

∴ Area of ∆BCE = \( \sqrt {7\, ×\, 6\, ×\, 4\, ×\, 4\, ×\, 6\, ×\, 2\, ×\, 2\, ×\, 7) }\)

∴ Area of ∆BCE = \( \sqrt {7²\, ×\, 6²\, ×\, 4²\, ×\, 2²) }\)

∴ Area of ∆BCE = 7 × 6 × 4 × 2

∴ Area of ∆BCE = 336 sq.m …(iii)

 

Now,

Area of the given plot = Area of ∆ABE + Area of ∆CDE + Area of ∆BCE

∴ Area of the given plot = 216 + 224 + 336

∴ Area of the given plot = 776 sq.m

 

Ans: Area of the given plot is 776 sq.m

Practice set 15.6

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

(1) 28 cm 

IMG 20230308 221300 Chapter 15 – Area

Given:

Radius of the circle = 28 cm

 

To find:

Area of the circle

 

Solution:

Area of the circle = πr²

∴ Area of the circle = \(\large \frac {22}{7}\) × 28 × 28

∴ Area of the circle = 22 × 4 × 28

∴ Area of the circle = 2464 sq.cm

 

Ans: Area of the circle is 2464 sq.cm

(2) 10.5 cm 

IMG 20230308 221234 Chapter 15 – Area

Given:

Radius of the circle = 10.5 cm

 

To find:

Area of the circle

 

Solution:

Area of the circle = πr²

∴ Area of the circle = \(\large \frac {22}{7}\) × 10.5 × 10.5

∴ Area of the circle = 22 × 1.5 × 10.5

∴ Area of the circle = 346.5 sq.cm

 

Ans: Area of the circle is 346.5 sq.cm

(3) 17.5 cm

IMG 20230308 221032 Chapter 15 – Area

Given:

Radius of the circle = 17.5 cm

 

To find:

Area of the circle

 

Solution:

Area of the circle = πr²

∴ Area of the circle = \(\large \frac {22}{7}\) × 17.5 × 17.5

∴ Area of the circle = 22 × 2.5 × 17.5

∴ Area of the circle = 962.5 sq.cm

 

Ans: Area of the circle = 962.5 sq.cm

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

(1) 176 sq cm 

IMG 20230308 221010 Chapter 15 – Area

Given:

Area of the circle = 176 sq cm

 

To find:

Diameter of the circle

 

Solution:

Area of the circle = πr²

∴ 176 = \(\large \frac {22}{7}\) × r²

∴ \(\large \frac {176\, ×\, 7}{22}\) = r²

∴ r² = 56 

∴ \( \sqrt {r²}\) = \( \sqrt {56}\) …[By Taking square root of both sides] 

∴ r = \( \sqrt {56}\) m 

 

Diameter of the circle = 2 × radius 

∴ Diameter of the circle = 2 × \( \sqrt {56}\) 

∴ Diameter of the circle = 2\( \sqrt {56}\) m

 

Ans: Diameter of the circle = 2\( \sqrt {56}\) m

(2) 394.24 sq cm 

IMG 20230308 220957 Chapter 15 – Area

Given:

Area of the circle = 394.24 sq cm 

 

To find:

Diameter of the circle

 

Solution:

Area of the circle = πr²

∴ 394.24 = \(\large \frac {22}{7}\) × r²

∴ \(\large \frac {394.24\, ×\, 7}{22}\) = r²

∴ r² = 17.92 × 7

∴ r² = 125.44

∴ \( \sqrt {r²}\) = \( \sqrt {125.44}\) …[By Taking square root of both sides] 

∴ r = 11.2 m 

 

Diameter of the circle = 2 × radius 

∴ Diameter of the circle = 2 × 11.2

∴ Diameter of the circle = 22.4 m

 

Ans: Diameter of the circle is 22.4 m

(3) 12474 sq cm

IMG 20230309 021915 Chapter 15 – Area

Given:

Area of the circle = 12474 sq cm

 

To find:

Diameter of the circle

 

Solution:

Area of the circle = πr²

∴ 12474 = \(\large \frac {22}{7}\) × r²

∴ \(\large \frac {12474\, ×\, 7}{22}\) = r²

∴ r² = 567 × 7

∴ r² = 3969

∴ \( \sqrt {r²}\) = \( \sqrt {3969}\) …[By Taking square root of both sides] 

∴ r = 63 m 

 

Diameter of the circle = 2 × radius 

∴ Diameter of the circle = 2 × 63

∴ Diameter of the circle = 126 m

 

Ans: Diameter of the circle = 126 m

3. 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.

IMG 20230308 220800 Chapter 15 – Area

Given:

Diameter of the circular garden = 42 m

Width of the road = 3.5 m

 

To find:

Area of the road

 

Solution:

Diameter of the circular garden = 42 m

∴ Radius of the circular garden = \(\large \frac {42}{2}\)

∴ Radius of the circular garden = 21 m

 

Area of the circular garden = πr²

∴ Area of the circular garden = \(\large \frac {22}{7}\) × 21 × 21

∴ Area of the circular garden = 22 × 21 × 3

∴ Area of the circular garden = 1386 sq.cm

 

Now,

Width of the road = 3.5 m

Outer Radius = Radius of the circular garden + Width of the road

∴ Outer Radius = 21 + 3.5

∴ Outer Radius = 24.5

 

Area of the outer circle = πr²

∴ Area of the outer circle = \(\large \frac {22}{7}\) × 24.5 × 24.5

∴ Area of the outer circle = 22 × 24.5 × 3.5

∴ Area of the outer circle = 1886.5 sq.cm

 

Area of the road = Area of the outer circle – Area of the circular garden

∴ Area of the road = 1886.5 – 1386

∴ Area of the road = 500.5 sq.m

 

Ans: Area of the road is 500.5 sq.m

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

IMG 20230308 221250 Chapter 15 – Area

Given:

Circumference of the circle = 88 cm

 

To find:

Area of the circle

 

Solution:

We know that,

Circumference of the circle = 2πr

∴ 88 = 2 × \(\large \frac {22}{7}\) × r

∴ 88 = \(\large \frac {44}{7}\) × r

∴ \(\large \frac {88\, ×\, 7}{44}\) = r

∴ r = 2 × 7

∴ r = 14 cm

 

Now,

Area of the circle = πr²

∴ Area of the circle = \(\large \frac {22}{7}\) × 14 × 14

∴ Area of the circle = 22 × 14 × 2

∴ Area of the circle = 616 sq.cm


Ans: Area of the circle is 616 sq.cm