Measuring Space: Perimeter and Area Class 9 Maths Ganita Manjari Part 1 Chapter 6 NCERT Solutions

Measuring Space: Perimeter and Area Class 9 Maths Ganita Manjari Part 1 Chapter 6 NCERT Solutions Looking for the Measuring Space: Perimeter and Area Class 9 Maths Ganita Manjari Part 1 Chapter 6 NCERT Solutions? You’ve come to the right place. This chapter helps students understand important concepts of perimeter, area, and practical geometry, making it essential for exams and building a strong mathematical foundation.

Exercise Set 6.1

Unless stated otherwise, use the approximation $\frac{22}{7}$ for π.
1. The perimeter of a circle is 44 cm. What is its radius?
Solution:
Let the radius of the circle be ‘x’ cm.
Circumference = 44 cm = $2\pi r$
r = $\frac{44}{2\pi }$ = $\frac{22}{\pi }$
Using $\pi =\frac{22}{7}$,
r = $\frac{22\times 7}{22}$ = 7 cm
Therefore, the radius is 7 cm.

2. Calculate, correct to 3 significant figures, the circumference of a circle with:
(i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm.
Solution:
(i) Radius (r) = 7 cm
Circumference = $2\pi r$
= 2 × 3.142 × 7 = 43.988 = 44.0 cm

(ii) Radius (r) = 10 cm
Circumference = $2\pi r$
= 2 × 3.142 × 10 = 62.84 = 62.8 cm

(iii) Radius (r) = 12 cm
Circumference = $2\pi r$
= 2 × 3.142 × 12 = 75.408 = 75.4 cm

3. Calculate the length of the arc of a circle if:
(i) the radius is 3.5 cm, and the angle at the centre is 60°, and
(ii) the radius is 6.3 m, and the angle at the centre is 120°.
Solution:
(i) Radius = 3.5 cm, angle = 60°
Length of the arc = $\frac{\theta }{360°}$ × $2\pi r$
= $\frac{60°}{360°}$ × 2 × $\frac{22}{7}$ × 3.5
= $\frac{1}{6}$ × 2 × $\frac{22}{7}$ × $\frac{35}{10}$
= $\frac{11}{3}$ = 3.67 cm

(ii) Radius = 6.3 m, angle = 120°
Length of the arc = $\frac{\theta }{360°}$ × $2\pi r$
= $\frac{120°}{360°}$ × 2 × $\frac{22}{7}$ × 6.3
= $\frac{1}{3}$ × 2 × $\frac{22}{7}$ × $\frac{63}{10}$
= $\frac{66}{5}$ = 13.2 cm

4. Find the perimeter of a sector (i.e., the curved portion as well as the two straight portions) of a circle of radius 14 cm and sector angle 75°.
Solution:

Radius of the circle r = 14 cm
Angle of the sector θ = 75°
Perimeter of a sector = length of arc + 2r
= $\frac{\theta }{360°}$ × $2\pi r$ + 2r
= 2r $(\frac{\theta }{360°}\times \pi +1)$
= 2 × 14 $(\frac{75°}{360°}\times \frac{22}{7}+1)$
= 28 $(\frac{15}{72}\times \frac{22}{7}+1)$
= 28 $(\frac{5}{24}\times \frac{22}{7}+1)$
= 28 $(\frac{55}{84}+1)$
= 28 $\left(\frac{55+84}{84}\right)$
= 28 $\left(\frac{139}{84}\right)$
= $\frac{139}{3}$ cm = 46$\frac{1}{3}$ cm
Therefore, the perimeter of the sector is 46$\frac{1}{3}$ cm.

5. Find the perimeters of the following shapes (taking the arcs to be quarter or half or three-quarters of a circle, as appropriate) (Fig. 6.14i to 6.14ix):

Solution:

(i) Radius of semicircles a and b = 30 cm
Length of lines c and d = 80 + 80 = 160 cm
Perimeter = Length of semicircles a and b + Length of lines c and d
= $2\pi r$ + 160
= 2 × $\frac{22}{7}$ × 30 + 160
= $\frac{1320}{7}$ +160
= $\frac{1120+1320}{7}$ = $\frac{2440}{7}$ = 348$\frac{4}{7}$ m

(ii) Radius of outer semicircle a = 6 cm
Raidus of inner semicircle b = 4 cm
Length of lines c and d = $\frac{12\text{–}8}{2}$ = $\frac{4}{2}$ = 2 + 2 = 4 cm
Perimeter = Length of semicircle a + Length of semicircle b + Length of lines c and d
= $\pi \times 6$ + $\pi$ × 4 + 4
= 6$\pi$ + 4$\pi$ + 4
= 10$\pi$ + 4
= 10 × $\frac{22}{7}$ + 4
= $\frac{220}{7}$ + 4
= $\frac{220+28}{7}$ = $\frac{248}{7}$ = 35$\frac{3}{7}$ cm

(iii) Radius of semicircles a, b, c and d = 5 cm
Perimeter = Length of 4 semicircles a, b, c and d
= 4 × $\pi r$
= 4$\pi r$
= 4 × $\frac{22}{7}$ × 5 = $\frac{440}{7}$ = 62$\frac{6}{7}$ cm

(iv) Radius of semicircles a, b and c = 6 cm
Perimeter = Length of 3 semicircle a, b and c
= 3 × $\pi r$
= 18$\pi$
= 18 × $\frac{22}{7}$
= $\frac{396}{7}$ = 56$\frac{4}{7}$ cm

(vi) Radius of semicircle a = 14 cm
Radius of semicircles b, c, d and e = $\frac{28}{4\times 2}$ = 3.5 cm
Perimeter = Length of semicircle a + Length of 4 semicircles b, c, d and e
= $\pi \times 14$ + 4 × $\pi \times 3.5$
= $14\pi$ + 14$\pi$
= 28$\pi$
= $28\times \frac{22}{7}$ = 88 cm

(vii) Using Pythagoras theorem,
a² = 6² + 8²
a² = 36 + 64
a² = 100
a = 10 cm
Thus, diameter of semicircle b = 10 cm
Radius of semicircle b = 5 cm
Radius of semicircle c = 4 cm
Radius of semicircle d = 3 cm
Perimeter = Length of semicircle b + Length of semicircle c + Length of semicircle d
= $\pi \times 5$ + $\pi \times 4$ + $\pi \times 3$
= $5\pi$ + $4\pi$ + $3\pi$
= $12\pi$
= $12\times \frac{22}{7}$ = $\frac{264}{7}$ = 35$\frac{5}{7}$ cm

(viii) Radius of semicircle a = 6 cm
Radius of semicircles b, c and d = 2 cm
Perimeter = Length of semicircle a + Length semicircles b, c and d
= $\pi \times 6$ + 3 × $\pi \times 2$
= $6\pi$ + $6\pi$
= $12\pi$
= $12\times \frac{22}{7}$ = $\frac{264}{7}$ = 35$\frac{5}{7}$ cm

(ix) Radius of semicircle a = 10 cm
Radius of semicircles b, and c = 5 cm
Perimeter = Length of semicircle a + Length of semicircles b and c
= $\pi \times 10$ + 2 × $\pi \times 5$
= $10\pi$ + $10\pi$
= $20\pi$
= $20\times \frac{22}{7}$ = $\frac{440}{7}$ = 62$\frac{6}{7}$ cm

6. If the diameter of a car tyre is 56 cm, then:
(i) How far does the car need to travel for the tyre to complete one revolution?
(ii) How many revolutions does the tyre make if the car travels 10 km?
Solution:

Diameter of the tyre = 56 cm
(i) Distance covered in one revolution = Circumference of the tyre = πd
= $\frac{22}{7}$ × 56 = 176 cm
Therefore, the car needs to travel 176 cm to complete one revolution.
(ii) Distance travelled by the car = 10 km
= 10 × 1000 × 100
= 1000000 cm
Number of revolutions made by the tyre = $\frac{1000000}{176}$ = 5681.81
Therefore, the tyre makes 5681 revolutions.

7. Find the total perimeter of all the petals in each of the given flowers.

Solution:
(i) Side of the given square = 14 cm
Radius of each arc = $\frac{14}{2}$ = 7 cm
Each petal consists of two quarter-circles.
∴ Length of one quarter-circle = $\frac{1}{4}$ × $2\pi r$ = $\frac{1}{4}$ × 2 × $\frac{22}{7}$ × 7 = 11 cm
So, perimeter of one petal = 11 + 11 = 22 cm
There are 4 petals.
Hence, the total perimeter of all petals = 4 × 22 = 88 cm

(ii) Side of the regular hexagon = 42 cm
Radius of each arc = 42 cm
Angle subtended by each arc = $\frac{360°}{6}$ = 60°
Length of one arc = $\frac{\theta }{360°}$ × $2\pi r$
= $\frac{60°}{360°}$ × $2\times \frac{22}{7}$ × 42
= $\frac{1}{6}$ × 44 × 6 = 44 cm
Each petal consists of two such arcs.
So, perimeter of one petal = 44 + 44 = 88 cm
There are 6 petals.
Therefore, total perimeter of all petals = 6 × 88 = 528 cm

8. The ratio of the perimeters of two circles is 5 : 4. What is the ratio of their radii?
Solution:
Let the radii of the two circles be r_1​ and r₂​.
Then,
$2\pi$r₁ : $2\pi$r₂ = 5 : 4
r₁ : r₂ = 5 : 4
Hence, the ratio of their radii is 5 : 4.

Exercise Set 6.2

1. Find the area of triangle ADE in Fig. 6.31.

Solution:
Height of △ADE = Length of rectangle ABCD = 10 cm
Base of △ADE = Breadth of rectangle ABCD = 8 cm
Therefore,
Area of △ADE = $\frac{1}{2}$ × Base × Height
= $\frac{1}{2}$ × 8 × 10
= 4 × 10 = 40 cm²

2. The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.
Solution:

Let ABCD be the given trapezium in which AB = 40 cm, DC = 20 cm, and AD = BC = 26 cm.
Draw CE $\parallel$ AD.
Now, ADCE is a parallelogram in which AD $\parallel$ CE and AE $\parallel$ CD.
AE = DC = 20 cm, CE = 26 cm, and BE = AB – AE = 40 – 20 = 20 cm.
In △BCE,
s = $\frac{26+26+20}{2}$ = $\frac{72}{2}$ = 36
Area of △BCE = $\sqrt{s(s-a)(s-b)(s-c)}$
= $\sqrt{36(36-26)(36-26)(36-20)}$
= $\sqrt{36(10)(10)(16)}$= 240 cm² …………. (1)
Let h be the height of △BCE, then
Area of △BCE = $\frac{1}{2}$ × Base × Height
= $\frac{1}{2}$ × 20 × Height = 10 × h = 10h ………… (2)
From (i) and (ii), we get
10h = 240 ⇒ h = 24 cm
Since the height of trapezium ABCD is same as that of △BCE.
∴ Area of trapezium = $\frac{1}{2}$ (AB + DC) × h = $\frac{1}{2}$ (40 + 20) × 24 = $\frac{1}{2}$ × 60 × 24 = 720 cm²

3. Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.
Solution:
Sides of the triangle a = 8 cm, b = 11 cm.
Perimeter = 32 cm
32 = a + b + c
c = 32 – (a + b)
c = 32 – (8 + 11)
c = 32 – 19 = 13 cm
Now, semi-perimeter = s = $\frac{32}{2}$ = 16 cm
Using Heron’s formula,
Area = $\sqrt{s(s-a)(s-b)(s-c)}$
= $\sqrt{16(16-8)(16-11)(16-13)}$
= $\sqrt{16(8)(5)(3)}$
= $\sqrt{1920}$= $\sqrt{64\times 30}$= 8$\sqrt{30}$cm²
Therefore, the area of the triangle is 8$\sqrt{30}$cm².

4. The sides of a triangular plot are in the ratio 3: 5: 7; its perimeter is 300 m. Find its area.
Solution:
Let the sides of the triangular plot be 3x, 5x, and 7x.
Perimeter = 300 m
3x + 5x + 7x = 300
15x = 300
x = 20 m
Therefore, the sides are:
3x = 3(20) = 60 cm
5x = 5(20) = 100 cm
7x = 7(20) = 140 cm
Now, s = $\frac{60+100+140}{2}$ = $\frac{300}{2}$ = 150
Using Heron’s formula.,
Area = $\sqrt{s(s-a)(s-b)(s-c)}$
= $\sqrt{150(150-60)(150-100)(150-140)}$
= $\sqrt{150(90)(50)(10)}$
= $\sqrt{6750000}$= 1500$\sqrt{3}$m²
Hence, the area of the triangular plot is 1500$\sqrt{3}$m².

5. One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area 128 cm², find the length of the shorter diagonal.
Solution:
Let the shorter diagonal of the rhombus be x cm.
Then, the longer diagonal = 2x cm.
Area of the rhombus = $\frac{1}{2}$ × d₁ × d₂
Given,
128 = $\frac{1}{2}$ × x × 2x
128 = x²
x = $\sqrt{128}$
x = $\sqrt{64\times 2}$
x = 8$\sqrt{2}$cm
Therefore, the length of the shorter diagonal is $8\sqrt{2}$cm.

6. ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area (ΔPCD): area (ΔQCD)?
Solution:

ΔPCD and ΔQCD are on the same base DC and lie between the same parallels AB and DC.
Therefore,
$\frac{Area\mathrm{\Delta }PCD}{Area\mathrm{\Delta }QCD}$ = $\frac{\frac{1}{2}\times DC\times h}{\frac{1}{2}\times DC\times h}$ = $1$
Hence
Area (△PCD) : Area (△QCD) = 1 : 1.

7. O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.
Solution:

Join SQ.
Let the diagonals PR and SQ intersect at A.
Since the diagonals of a parallelogram bisect each other. Therefore, A is the midpoint of AC and BD.
Since a median of a triangle divides it into two triangles of equal area.
In ΔPQS, PA is median,
area (ΔPSA) = area (ΔPQA) ………. (1)
In ΔOSQ, OA is median,
area (ΔOSA) = area (ΔOQA) …………. (2)
Adding (1) and (2), we get
area (ΔPSA) + area (ΔOSA) = area (ΔPQA) + area (ΔOQA)
area (ΔPSO) = area (ΔPQO).
Hence, verified.

8. If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram, and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.)

9. In ∆ABC, the midpoint of BC is D (Fig. 6.32). Median AD is drawn. P is any point on AD. Show that area (∆ABP) = area (∆ACP).

Solution:
In ∆ABC, AD is the median, and median divides a triangle into two triangles of equal area.
∴ area (△ABD) = area (△ACD) ……………. (1)
Again, in ∆PBC, PD is the median
∴ area (△PBD) = area (△PCD) ……………… (2)
Subtracting (2) from (1), we get
area (△ABD) – area (△PBD) = area (△ACD) – area (△PCD)
Hence, area (∆ABP) = area (∆ACP)

10. Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD (Fig. 6.33). What is the ratio of the areas of the red region (∆PAB and ∆PCD) and the green region (∆PBC and ∆PDA)?

Solution:

Let h₁​ = perpendicular distance from P to AB, h₂​ = perpendicular distance from P to CD, h₃​ = perpendicular distance from P to AD, and h₄​ = perpendicular distance from P to BC.
Since AB $\parallel$ CD and both are sides of the square,
h₁ ​+ h₂​ = a …………… (1)
Since AD $\parallel$ BC and both are sides of the square,
h₃​ + h₄​ = a …………… (2)
Area of red region:
area (△PAB) + area (△PCD) = $\frac{1}{2}$​⋅a⋅h₁​ + $\frac{1}{2}$​⋅a⋅h₂​ = $\frac{1}{2}$​a(h₁​ + h₂​) = $\frac{1}{2}$​a⋅a = $\frac{a^2}{2}$
Area of green region:
area (△PBC) + area (△PDA) = $\frac{1}{2}$​⋅a⋅h₄ + $\frac{1}{2}$​⋅a⋅h₃​ = $\frac{1}{2}$​a(h₄​ + h₃​) = $\frac{1}{2}$​a⋅a = $\frac{a^2}{2}$
Therefore,
area (red) : area (green) = $\frac{a^2}{2}$ : $\frac{a^2}{2}$ = 1 : 1.

11. In ∆ABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ $\parallel$ PD. PQ is joined (Fig. 6.34). Prove that Area (∆BPQ) = $\frac{1}{2}$ Area (∆ABC).

Solution:

Join CD.
Since ∆PDQ and ∆PDC are on the same base PD and between the same parallels PD and QC.
∴ Area (∆PDQ) = Area (∆PDQ) …………… (1)
Since in △ABC, D is the midpoint of AB, and CD is the median.
∴ Area (∆BCD) = $\frac{1}{2}$ Area (∆ABC)
Area (∆BPD) + Area (∆PDC) = $\frac{1}{2}$ Area (∆ABC)
Area (∆BPD) + Area (∆PDQ) = $\frac{1}{2}$ Area (∆ABC)
Area (∆BPQ) = $\frac{1}{2}$ Area (∆ABC).

Exercise Set 6.3

Unless stated otherwise, use the approximation $\frac{22}{7}$ for π.
1. Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.
Solution:
Radius = 7 cm
Angle of the sector = 60°
Area of the sector = $\pi r^2$ × $\frac{\theta }{360°}$
= $\frac{22}{7}$ × 7 × 7 × $\frac{60°}{360°}$
= 22 × 7 × $\frac{1}{6}$
= 11 × 7 × $\frac{1}{3}$ = $\frac{77}{3}$ cm²
Therefore, area of the sector is $\frac{77}{3}$ cm².

2. Find the area of a quadrant of a circle whose circumference is 44 cm.
Solution:
Circumference = 44 cm
∴ 2$\pi r$ = 44
2 × $\frac{22}{7}$ × r = 44
$\frac{44}{7}$ × r = 44
r = 44 × $\frac{7}{44}$ = 7 cm
Since a quadrant divides a circle into 4 equal sectors, and the angle of each sector is 90°.
∴ Area of the quadrant = $\pi r^2$ × $\frac{\theta }{360°}$
= $\frac{22}{7}$ × 7 × 7 × $\frac{90°}{360°}$
= 22 × 7 × $\frac{1}{4}$
= 11 × 7 × $\frac{1}{2}$ = $\frac{77}{2}$ cm² = 38.5 cm²

3. The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.
Solution:
Length of minute hand = 7 cm
In 60 minutes, the minute hand sweeps an angle = 360 °.
Therefore, in 10 minutes, the angle swept = $\frac{360°}{60°}$ = 60°
Area swept by the minute hand = $\pi r^2$ × $\frac{\theta }{360°}$
= $\frac{22}{7}$ × 7 × 7 × $\frac{60°}{360°}$
= 22 × 7 × $\frac{1}{6}$
= 11 × 7 × $\frac{1}{3}$ = $\frac{77}{3}$ cm²
Hence, the area swept by the minute hand in 10 minutes is $\frac{77}{3}$ cm².

4. A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding:
(i) minor sector (that subtends 90° at the centre), and
(ii) major sector (that subtends 270° at the centre). (Use π ≈ 3.14.)
Solution:
Radius of the circle (r) = 10 cm
(i) Angle of the minor sector = 90°
Area of the minor sector = = $\pi r^2$ × $\frac{\theta }{360°}$
= 3.14 × 10 × 10 × $\frac{90°}{360°}$
= 3.14 × 100 × $\frac{1}{4}$
= $\frac{314}{100}$ × 100 × $\frac{1}{4}$
= $\frac{314}{4}$ = 78.5 cm²

(ii) Angle of the major sector = 270°
Area of the minor sector = $\pi r^2$ × $\frac{\theta }{360°}$
= 3.14 × 10 × 10 × $\frac{270°}{360°}$
= 3.14 × 100 × $\frac{1}{4}$
= $\frac{314}{100}$ × 100 × $\frac{3}{4}$
= $\frac{942}{4}$ = 235.5 cm²

5. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre of the circle. Find the areas of the corresponding minor and major segments of the circle. (Use π ≈ 3.14 and $\sqrt{3}$≈ 1.73.)
Solution:
Radius of the circle (r) = 15 cm
Angle subtended by the chord at the centre = 60°
Area of the minor sector = $\pi r^2$ × $\frac{\theta }{360°}$
= 3.14 × 15 × 15 × $\frac{60°}{360°}$
= 3.14 × 225 × $\frac{1}{6}$
= 706.5 × $\frac{1}{6}$
= 117.75 cm²
Since OA = OB = 15 cm and ∠AOB = 60°,
△OAB is an equilateral triangle of side 15 cm.
Area of triangle = $\frac{\sqrt{3}}{4}{a}^2$
= $\frac{1.73}{4}$ × 15 × 15
= 97.31 cm²
Area of minor segment = Area of sector − Area of triangle
= 117.75 − 97.31 = 20.44 cm²
Area of the circle = 3.14 × 15 × 15 = 706.5 cm²
Area of major segment = 706.5 − 20.44 = 686.06 cm²

6. A car has two wipers which do not overlap. Each wiper has a blade of length 28 cm and sweeps through an angle of 120°. Find the total area cleaned at each sweep of the blades.
Solution:
Length of each wiper blade (r) = 28 cm
Angle swept by each wiper = 120°
Area cleaned by two blades = 2 × $\pi r^2$ × $\frac{\theta }{360°}$
= 2 × $\frac{22}{7}$ × 28 × 28 × $\frac{120°}{360°}$
= 44 × 4 × 28 × $\frac{1}{3}$
= 4928 × $\frac{1}{6}$
= 1642 $\frac{2}{3}$ cm²
Therefore, the total area cleaned at each sweep of the blades is 1642 $\frac{2}{3}$ cm².

7. A chord of a circle of radius r subtends an angle of 60° at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to r² $(\frac{\pi }{6}-\frac{\sqrt{3}}{4})$.
Solution:
Let AB be the chord, and O be the centre of the circle.
Radius of the circle = r
∠AOB = 60°
Area of sector AOB = $\pi r^2$ × $\frac{\theta }{360°}$
= $\pi r^2$ × $\frac{60°}{360°}$ = $\frac{\pi r^2}{6}$
Since OA = OB = r and ∠AOB = 60°, △AOB is an equilateral triangle.
Area of △AOB = $\frac{\sqrt{3}}{4}{r}^2$
Therefore,
Area of the minor segment = $\frac{\pi r^2}{6}$ – $\frac{\sqrt{3}}{4}{r}^2$
= $r^2$ $(\frac{\pi }{6}-\frac{\sqrt{3}}{4})$.

8. An equilateral triangle is inscribed in a circle of radius r. Show that the ratio of the area of the triangle to the area of the circle is equal to $\frac{3\sqrt{3}}{4\pi }$ ≈ 0.413.
Solution:
Let the side of the equilateral triangle be a.
Since the triangle is inscribed in a circle of radius r,
r = $\frac{a}{\sqrt{3}}$ ⇒ a = $\sqrt{3}r$
Area of equilateral triangle = $\frac{\sqrt{3}}{4}{a}^2$
= $\frac{\sqrt{3}}{4}(\sqrt{3}r{)}^2$
= $\frac{3\sqrt{3}}{4}{r}^2$
Area of the circle = $\pi r^2$
Therefore,
$\frac{Areaofequilateraltriangle}{Areaofcircle}$ = $\frac{\frac{3\sqrt{3}}{4}{r}^2}{\pi r^2}$ = $\frac{3\sqrt{3}}{4\pi }$ ≈ 0.413.

9. A square is inscribed in a circle of radius r. Show that the ratio of the area of the square to the area of the circle is equal to $\frac{2}{\pi }$ ≈ 0.637.
Solution:
Let the side of the square be a.
Since the square is inscribed in the circle,
Diagonal of square = 2r
But, Diagonal = a$\sqrt{2}$
Therefore, ​
a$\sqrt{2}$= 2 r
a = $\sqrt{2}r$
Area of square = a² = ${\left(\sqrt{2}r\right)}^2$ = 2r²
Area of circle = $\pi r^2$
Therefore,
$\frac{Areaofsqaure}{Areaofcircle}$ = $\frac{2r2}{\pi r^2}$ = $\frac{2}{\pi }$ ≈ 0.637.

10. A hexagon is inscribed in a circle of radius r. Show that the ratio of the area of the hexagon to the area of the circle is equal to $\frac{3\sqrt{3}}{2\pi }$ ≈ 0.827. Can you see why the answer is exactly twice the answer to Question 8?
Solution:
A regular hexagon can be divided into 6 equilateral triangles.
Each triangle has side r.
Area of one equilateral triangle = $\frac{\sqrt{3}}{4}{r}^2$
Area of hexagon = 6 × $\frac{\sqrt{3}}{4}{r}^2$ = $\frac{3\sqrt{3}}{2}{r}^2$
Area of circle = $\pi r^2$
Therefore,
$\frac{Areaofhexagon}{Areaofcircle}$ = $\frac{\frac{3\sqrt{3}}{2}{r}^2}{\pi r^2}$ = $\frac{3\sqrt{3}}{2\pi }$ ≈ 0.827
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