Chapter 6

Chapter 6: Measuring Space: Perimeter and Area

Exercise Solutions

Exercise Set 6.1

1. The perimeter of a circle is 44 cm. What is its radius?

Perimeter (Circumference) $C = 2\pi r = 44\text{ cm}$

$2 \times \frac{22}{7} \times r = 44$

$\frac{44}{7} \times r = 44 \implies r = 7\text{ cm}$

Answer: Radius = $7\text{ cm}$

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.

Formula: Circumference $C = 2\pi r$

(i) $r = 7\text{ cm}$: $C = 2 \times \frac{22}{7} \times 7 = 44 = 44.0\text{ cm}$

(ii) $r = 10\text{ cm}$: $C = 2 \times 3.1416 \times 10 = 62.832 \approx 62.8\text{ cm}$

(iii) $r = 12\text{ cm}$: $C = 2 \times \frac{22}{7} \times 12 = \frac{528}{7} \approx 75.4\text{ cm}$

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°.

Formula: Arc Length $l = 2\pi r \times \frac{\theta}{360^\circ}$

(i) $r = 3.5\text{ cm}$, $\theta = 60^\circ$: $l = 2 \times \frac{22}{7} \times 3.5 \times \frac{60}{360} = 22 \times \frac{1}{6} = \frac{11}{3} \approx 3.67\text{ cm}$

(ii) $r = 6.3\text{ m}$, $\theta = 120^\circ$: $l = 2 \times \frac{22}{7} \times 6.3 \times \frac{120}{360} = 2 \times 22 \times 0.9 \times \frac{1}{3} = 13.2\text{ m}$

4. Find the perimeter of a sector of a circle of radius 14 cm and sector angle 75°.

Arc length $l = 2\pi r \times \frac{\theta}{360^\circ} = 2 \times \frac{22}{7} \times 14 \times \frac{75}{360} = 88 \times \frac{5}{24} = \frac{55}{3}\text{ cm}$

Perimeter of sector = $l + 2r = \frac{55}{3} + 2(14) = 18.33 + 28 = 46.33\text{ cm}$

5. Find the perimeters of the shapes in Fig. 6.14i to 6.14ix.

(i) Straight sides $= 8 + 8 = 16\text{ cm}$; Quarter circle arc $= \frac{1}{4} \times 2 \times \frac{22}{7} \times 8 = \frac{88}{7}\text{ cm}$.
Perimeter $= 16 + 12.57 = 28.57\text{ cm}$.

(ii) Rectangle with semicircular top ($r = 30\text{ m}$): Straight parts $= 80 + 80 + 60 = 220\text{ m}$; Arc $= \pi r = \frac{22}{7} \times 30 = 94.29\text{ m}$.
Perimeter $= 220 + 94.29 = 314.29\text{ m}$.

(iii) Semicircles on square sides ($r = 5\text{ cm}$, 2 semicircles = 1 circle).
Perimeter $= 2 \times (2\pi r) = 4 \times \frac{22}{7} \times 5 = 62.86\text{ cm}$.

(iv) Three-quarter circle ($r = 12\text{ cm}$): Arc $= \frac{3}{4} \times 2 \times \frac{22}{7} \times 12 = \frac{396}{7} = 56.57\text{ cm}$; Radii $= 12 + 12 = 24\text{ cm}$.
Perimeter $= 56.57 + 24 = 80.57\text{ cm}$.

(v) Quarter circle cut-outs on ends: Outer straight lengths $= 8 + 8 = 16\text{ cm}$; Arcs form 2 semicircles ($r = 4\text{ cm}$).
Perimeter $= 16 + 2 \times (\pi \times 4) = 16 + 25.14 = 41.14\text{ cm}$.

(vi) L-shape with arcs ($r = 7\text{ cm}$): Sum of straight lines and arc lengths $= 14 + 14 + 28 + \text{arcs} = 116\text{ cm}$.

(vii) Heart shape/stadium arc combinations: Sum of curved boundaries $= 2\pi r + 2\text{ straight sections} = 75.43\text{ cm}$.

(viii) Track with 4 semicircular segments ($r = 2\text{ cm}$): Total arc $= 4 \times (\pi \times 2) = 25.14\text{ cm}$.

(ix) Wavy boundary (Semicircles of diameter $10\text{ cm}$ each, $r = 5\text{ cm}$): 4 semicircles.
Perimeter $= 4 \times (\pi \times 5) = 20 \times \frac{22}{7} = 62.86\text{ cm}$.

6. If the diameter of a car tyre is 56 cm: (i) How far does the car travel in 1 revolution? (ii) How many revolutions for 10 km?

(i) Distance in 1 revolution = Circumference $= \pi d = \frac{22}{7} \times 56 = 176\text{ cm} = 1.76\text{ m}$

(ii) Total distance $= 10\text{ km} = 10,000\text{ m} = 1,000,000\text{ cm}$

Number of revolutions $= \frac{1,000,000}{176} \approx 5681.82 \approx 5682\text{ revolutions}$

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

(i) Square flower (4 petals): Side $= 14\text{ cm}$, Radius $= 7\text{ cm}$.
Each petal consists of 2 semicircular arcs. 4 petals = 8 semicircular arcs = 4 full circles.
Perimeter $= 4 \times (2\pi r) = 4 \times 2 \times \frac{22}{7} \times 7 = 176\text{ cm}$.

(ii) Hexagon flower (6 petals): Side $= 42\text{ cm}$, Radius $= 21\text{ cm}$.
6 petals = 12 circular arcs of $60^\circ$ each = 2 full circles.
Perimeter $= 2 \times (2\pi r) = 2 \times 2 \times \frac{22}{7} \times 21 = 264\text{ cm}$.

8. The ratio of the perimeters of two circles is 5:4. What is the ratio of their radii?

$\frac{C_1}{C_2} = \frac{2\pi r_1}{2\pi r_2} = \frac{r_1}{r_2}$

Since $\frac{C_1}{C_2} = \frac{5}{4}$, the ratio of their radii is $r_1 : r_2 = 5 : 4$.

Exercise Set 6.2

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

In right triangle $ABC$, $AC = 10\text{ cm}$, $BC = 8\text{ cm}$.

Height $AB = \sqrt{10^2 - 8^2} = \sqrt{36} = 6\text{ cm}$.

Base of $\Delta ADE = 10\text{ cm}$, Height $= 6\text{ cm}$.

Area $(\Delta ADE) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 6 = 30\text{ cm}^2$

2. Parallel sides of a trapezium are 40 cm and 20 cm. Non-parallel sides are both 26 cm. Find its area.

Parallel sides $a = 40$, $b = 20$. Difference in base on each side $= \frac{40 - 20}{2} = 10\text{ cm}$.

Height $h = \sqrt{26^2 - 10^2} = \sqrt{676 - 100} = \sqrt{576} = 24\text{ cm}$.

Area $= \frac{1}{2}(a + b)h = \frac{1}{2}(40 + 20) \times 24 = 30 \times 24 = 720\text{ cm}^2$

3. Find the area of a triangle with sides 8 cm, 11 cm, and perimeter 32 cm.

Third side $c = 32 - (8 + 11) = 13\text{ cm}$.

Semi-perimeter $s = \frac{32}{2} = 16\text{ cm}$.

By Heron's formula: $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{16(16-8)(16-11)(16-13)}$

$\text{Area} = \sqrt{16 \times 8 \times 5 \times 3} = \sqrt{1920} = 8\sqrt{30} \approx 43.82\text{ cm}^2$

4. Sides of a triangular plot are in ratio 3:5:7; perimeter is 300 m. Find its area.

Let sides be $3x, 5x, 7x$. Sum $= 15x = 300 \implies x = 20\text{ m}$.

Sides are $a = 60\text{ m}$, $b = 100\text{ m}$, $c = 140\text{ m}$.

$s = \frac{300}{2} = 150\text{ m}$.

$\text{Area} = \sqrt{150(150-60)(150-100)(150-140)} = \sqrt{150 \times 90 \times 50 \times 10} = 1500\sqrt{3} \approx 2598.08\text{ m}^2$

5. One diagonal of a rhombus is twice as long as the other. If area is 128 cm², find the shorter diagonal.

Let $d_1 = x$, $d_2 = 2x$.

$\text{Area} = \frac{1}{2} d_1 d_2 = \frac{1}{2}(x)(2x) = x^2 = 128$

$x = \sqrt{128} = 8\sqrt{2} \approx 11.31\text{ cm}$

Shorter diagonal $= 11.31\text{ cm}$

6. ABCD is a parallelogram. P and Q are points on AB. What is the ratio area(ΔPCD) : area(ΔQCD)?

Both triangles $\Delta PCD$ and $\Delta QCD$ share the same base $CD$ and lie between the same parallel lines $AB$ and $CD$ (having equal height $h$).

$\text{Area}(\Delta PCD) = \frac{1}{2} \times CD \times h$, $\text{Area}(\Delta QCD) = \frac{1}{2} \times CD \times h$.

Ratio: $1 : 1$

7. O is any point on diagonal PR of parallelogram PQRS. Prove area(ΔPSO) = area(ΔPQO).

Diagonal $PR$ divides parallelogram $PQRS$ into two triangles of equal area: $\text{Area}(\Delta PSR) = \text{Area}(\Delta PQR)$.

Similarly, $OR$ divides $\Delta SQR$ symmetrically, or considering altitudes from $S$ and $Q$ to $PR$ are equal ($h_1 = h_2$).

$\text{Area}(\Delta PSO) = \frac{1}{2} \times PO \times h_1 = \frac{1}{2} \times PO \times h_2 = \text{Area}(\Delta PQO)$.

8. Prove that joining the midpoints of sides of a 4-gon forms a parallelogram with half the area of the 4-gon.

Let quadrilateral be $ABCD$ with midpoints $E, F, G, H$.

By midpoint theorem, $EF \parallel AC$ and $EF = \frac{1}{2}AC$; $HG \parallel AC$ and $HG = \frac{1}{2}AC$, so $EFGH$ is a parallelogram.

Sum of areas of four corner triangles $= \frac{1}{2} \text{Area}(ABCD)$.

Hence, $\text{Area}(EFGH) = \frac{1}{2} \text{Area}(ABCD)$.

9. In ΔABC, D is midpoint of BC. P is any point on AD. Show area(ΔABP) = area(ΔACP).

Median $AD$ divides $\Delta ABC$ into equal areas: $\text{Area}(\Delta ABD) = \text{Area}(\Delta ACD)$.

In $\Delta PBC$, $PD$ is median, so $\text{Area}(\Delta PBD) = \text{Area}(\Delta PCD)$.

Subtracting the two: $\text{Area}(\Delta ABD) - \text{Area}(\Delta PBD) = \text{Area}(\Delta ACD) - \text{Area}(\Delta PCD)$.

$\implies \text{Area}(\Delta ABP) = \text{Area}(\Delta ACP)$.

10. Given square ABCD and point P inside. What is ratio of red region (ΔPAB + ΔPCD) to green region (ΔPBC + ΔPDA)?

Let side of square be $s$.

$\text{Area}(\Delta PAB) + \text{Area}(\Delta PCD) = \frac{1}{2} s h_1 + \frac{1}{2} s h_2 = \frac{1}{2} s (h_1 + h_2) = \frac{1}{2} s^2$.

Similarly, $\text{Area}(\Delta PBC) + \text{Area}(\Delta PDA) = \frac{1}{2} s^2$.

Ratio: $1 : 1$

11. Prove Area(ΔBPQ) = 1/2 Area(ΔABC) (Fig. 6.34).

Since $D$ is midpoint of $AB$, median $CD$ gives $\text{Area}(\Delta BCD) = \frac{1}{2} \text{Area}(\Delta ABC)$.

Given $CQ \parallel PD$, triangles $\Delta PDQ$ and $\Delta PDC$ lie on same base $PD$ between parallels $PD$ and $CQ$, so $\text{Area}(\Delta PDQ) = \text{Area}(\Delta PDC)$.

Adding $\text{Area}(\Delta BPD)$ to both sides: $\text{Area}(\Delta BPQ) = \text{Area}(\Delta BCD) = \frac{1}{2} \text{Area}(\Delta ABC)$.

Exercise Set 6.3

1. Find area of sector with radius 7 cm and angle 60°.

$\text{Area} = \pi r^2 \times \frac{\theta}{360^\circ} = \frac{22}{7} \times 7^2 \times \frac{60}{360} = \frac{154}{6} \approx 25.67\text{ cm}^2$

2. Find area of quadrant of circle with circumference 44 cm.

$2\pi r = 44 \implies r = 7\text{ cm}$.

$\text{Area of quadrant} = \frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 49 = 38.5\text{ cm}^2$

3. Minute hand length 7 cm. Area swept in 10 minutes?

Angle in 10 min $= 10 \times 6^\circ = 60^\circ$.

$\text{Area} = \frac{22}{7} \times 7^2 \times \frac{60}{360} = \frac{154}{6} \approx 25.67\text{ cm}^2$

4. Chord of radius 10 cm subtends 90° at centre. Find area of (i) minor sector, (ii) major sector.

(i) Minor sector: $\pi r^2 \times \frac{90}{360} = 3.14 \times 100 \times \frac{1}{4} = 78.5\text{ cm}^2$

(ii) Major sector: $3.14 \times 100 \times \frac{270}{360} = 235.5\text{ cm}^2$

5. Radius 15 cm, angle 60°. Find areas of minor and major segments.

Area of sector $= 3.14 \times 15^2 \times \frac{60}{360} = 117.75\text{ cm}^2$.

Area of equilateral triangle $= \frac{\sqrt{3}}{4} r^2 = \frac{1.73}{4} \times 225 = 97.31\text{ cm}^2$.

Minor segment: $117.75 - 97.31 = 20.44\text{ cm}^2$.

Major segment: Total area - Minor segment $= (3.14 \times 225) - 20.44 = 706.5 - 20.44 = 686.06\text{ cm}^2$.

6. Wipers with blade length 28 cm, sweep angle 120°. Total area cleaned?

Area of 2 wipers $= 2 \times \left( \frac{22}{7} \times 28^2 \times \frac{120}{360} \right) = 2 \times \left( \frac{22}{7} \times 784 \times \frac{1}{3} \right) = \frac{4928}{3} \approx 1642.67\text{ cm}^2$

7*. Show area of minor segment for angle 60° is $\pi r^2 (\frac{1}{6} - \frac{\sqrt{3}}{4})$.

$\text{Area of sector} = \pi r^2 \times \frac{60}{360} = \frac{\pi r^2}{6}$

$\text{Area of triangle} = \frac{\sqrt{3}}{4} r^2$

$\text{Area of segment} = \frac{\pi r^2}{6} - \frac{\sqrt{3}}{4} r^2 = r^2\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right)$

8*. Equilateral triangle inscribed in circle of radius r. Show area ratio is $\frac{3\sqrt{3}}{4\pi} \approx 0.413$.

Side of equilateral triangle $a = r\sqrt{3}$.

$\text{Area of triangle} = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (3r^2) = \frac{3\sqrt{3}}{4} r^2$.

$\text{Area of circle} = \pi r^2$.

$\text{Ratio} = \frac{\frac{3\sqrt{3}}{4} r^2}{\pi r^2} = \frac{3\sqrt{3}}{4\pi} \approx 0.413$.

9*. Square inscribed in circle of radius r. Show area ratio is $\frac{2}{\pi} \approx 0.637$.

Diagonal of square $= 2r \implies \text{Side } s = r\sqrt{2}$.

$\text{Area of square} = s^2 = 2r^2$.

$\text{Ratio} = \frac{2r^2}{\pi r^2} = \frac{2}{\pi} \approx 0.637$.

10*. Hexagon inscribed in circle of radius r. Show area ratio is $\frac{3\sqrt{3}}{2\pi} \approx 0.827$.

Hexagon consists of 6 equilateral triangles of side $r$.

$\text{Area of hexagon} = 6 \times \frac{\sqrt{3}}{4} r^2 = \frac{3\sqrt{3}}{2} r^2$.

$\text{Ratio} = \frac{\frac{3\sqrt{3}}{2} r^2}{\pi r^2} = \frac{3\sqrt{3}}{2\pi} \approx 0.827$.

End-of-Chapter Exercises

2. Isosceles triangle perimeter 40 cm; equal sides 15 cm each. Find area.

Base $b = 40 - 2(15) = 10\text{ cm}$.

Height $h = \sqrt{15^2 - 5^2} = \sqrt{200} = 10\sqrt{2}\text{ cm}$.

$\text{Area} = \frac{1}{2} \times 10 \times 10\sqrt{2} = 50\sqrt{2} \approx 70.71\text{ cm}^2$

3. Isosceles triangle base 10 cm, area 60 cm². Find equal sides.

$\text{Area} = \frac{1}{2} \times b \times h = 60 \implies \frac{1}{2} \times 10 \times h = 60 \implies h = 12\text{ cm}$.

Equal side $a = \sqrt{h^2 + (b/2)^2} = \sqrt{12^2 + 5^2} = \sqrt{169} = 13\text{ cm}$.

4. Right-angled triangle area 54 cm², one leg 12 cm. Find perimeter.

Leg $b = \frac{2 \times 54}{12} = 9\text{ cm}$.

Hypotenuse $c = \sqrt{12^2 + 9^2} = \sqrt{225} = 15\text{ cm}$.

Perimeter $= 12 + 9 + 15 = 36\text{ cm}$.

5. Triangle sides ratio 2:3:4, perimeter 45 cm. Find area.

$2x + 3x + 4x = 45 \implies x = 5 \implies$ sides are $10, 15, 20\text{ cm}$.

$s = 22.5\text{ cm}$.

$\text{Area} = \sqrt{22.5(12.5)(7.5)(2.5)} = \sqrt{5273.4375} \approx 72.62\text{ cm}^2$

6. Triangle sides 7 cm, 24 cm, 25 cm. Find area in two ways.

Method 1 (Right Triangle): $7^2 + 24^2 = 25^2 \implies \text{Area} = \frac{1}{2} \times 7 \times 24 = 84\text{ cm}^2$.

Method 2 (Heron's Formula): $s = \frac{56}{2} = 28 \implies \text{Area} = \sqrt{28(21)(4)(3)} = \sqrt{7056} = 84\text{ cm}^2$.

7. Bicycle wheel diameter 60 cm. Distance after 100 rotations?

Distance $= 100 \times \pi d = 100 \times \frac{22}{7} \times 60 = \frac{132000}{7} \approx 18857.14\text{ cm} \approx 188.57\text{ m}$

8. Area of quadrant with circumference 66 cm.

$2\pi r = 66 \implies r = 10.5\text{ cm}$.

$\text{Quadrant Area} = \frac{1}{4} \times \frac{22}{7} \times (10.5)^2 = 86.625\text{ cm}^2$

9. Car wheel radius 28 cm. Distance in 1 turn? Turns in 1 km?

Distance in 1 turn $= 2 \times \frac{22}{7} \times 28 = 176\text{ cm} = 1.76\text{ m}$.

Turns in 1 km ($1000\text{ m}$) $= \frac{1000}{1.76} \approx 568.18 \approx 568\text{ turns}$.

10*. Two rectangles have same area and perimeter. Are they congruent?

Yes. Let sides be $x, y$. Given $x+y = p$ and $xy = A$, $x$ and $y$ are roots of $t^2 - pt + A = 0$, which yields unique side lengths.

17–18. Fraction of rectangle covered by circles.

For $N$ identical circles packed in grid: Fraction covered $= \frac{\pi r^2}{(2r)(2r)} = \frac{\pi}{4} \approx 0.7854$ or $78.54\%$.

19*. 9 identical rectangles form large rectangle of area 72 cm². Find perimeter of one small rectangle.

Area of 1 small rectangle $= \frac{72}{9} = 8\text{ cm}^2$.

From alignment: $4 \times \text{length} = 5 \times \text{width} \implies l = 1.25w$.

$l \times w = 1.25 w^2 = 8 \implies w^2 = 6.4 \implies w \approx 2.53\text{ cm}, l \approx 3.16\text{ cm}$.

Perimeter $= 2(l + w) \approx 2(5.69) = 11.38\text{ cm}$.

22*. 4-petalled flower in square of side 2 units. Perimeter and Area?

Radius of semicircles $r = 1$.

Perimeter: 4 semicircles = 2 full circles $= 2 \times (2\pi \times 1) = 4\pi \approx 12.57\text{ units}$.

Area: Area of 4 semicircles - Area of square $= 2(\pi r^2) - 2^2 = 2\pi - 4 \approx 2.28\text{ sq. units}$.

23*. Concentric circles with chord length l tangent to inner circle. Show green area is 1/4 π l².

Let outer radius be $R$, inner radius $r$. Right triangle gives $R^2 - r^2 = (l/2)^2 = \frac{l^2}{4}$.

Ring Area $= \pi R^2 - \pi r^2 = \pi(R^2 - r^2) = \frac{1}{4} \pi l^2$.

24*. Semicircles on sides of right-angled triangle. Show Area(A) + Area(B) = Area(C).

By Pythagoras theorem $a^2 + b^2 = c^2$.

$\text{Area}(A) = \frac{1}{2} \pi (a/2)^2 = \frac{\pi a^2}{8}$, $\text{Area}(B) = \frac{\pi b^2}{8}$, $\text{Area}(C) = \frac{\pi c^2}{8}$.

$\text{Area}(A) + \text{Area}(B) = \frac{\pi}{8}(a^2 + b^2) = \frac{\pi c^2}{8} = \text{Area}(C)$.

25*. Enclosed area of two intersecting circles of radius r passing through each other's center.

Area consists of two $240^\circ$ sectors plus two equilateral triangles.

$\text{Total Area} = \left(\frac{2\pi}{3} + \frac{\sqrt{3}}{2}\right) r^2$.