CBSE · NCERT · Class 9 Maths · Chapter 6

NCERT Solutions: Class 9 Maths Chapter 6 - Measuring Space: Perimeter and Area

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Chapter-wise NCERT intext questions and exercise answers for Measuring Space: Perimeter and Area, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
Sections in this chapter
Exercise 6.1 6Exercise 6.2 8Exercise 6.3 8Exercise 6.4 14
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1Exercise 6.16 questions
Q.1The perimeter of a circle is 44 cm. What is its radius?v
Solution

Using $\pi=\dfrac{22}{7}$, circumference $=2\pi r$. So $44=2\times\dfrac{22}{7}\times r=\dfrac{44}{7}r$. Hence $r=7$ cm.

Answer:

The radius is $7$ cm.

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Q.2Calculate, correct to 3 significant figures, the circumference of a circle with: (i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm.v
Solution

Circumference $=2\pi r$. With $\pi=\dfrac{22}{7}$: for $r=7$, $C=44.0$ cm; for $r=10$, $C=\dfrac{440}{7}=62.857\ldots\approx 62.9$ cm; for $r=12$, $C=\dfrac{528}{7}=75.428\ldots\approx 75.4$ cm.

Answer:

(i) $44.0$ cm
(ii) $62.9$ cm
(iii) $75.4$ cm

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Q.3Calculate 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°.v
Solution

Arc length $=\dfrac{\theta}{360°}\times 2\pi r$. For (i), $\dfrac{60}{360}\times2\times\dfrac{22}{7}\times3.5=\dfrac{11}{3}$ cm. For (ii), $\dfrac{120}{360}\times2\times\dfrac{22}{7}\times6.3=13.2$ m.

Answer:

(i) $\dfrac{11}{3}$ cm, i.e. about $3.67$ cm
(ii) $13.2$ m

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Q.4Find 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°.v
Solution

Arc length $=\dfrac{75}{360}\times2\times\dfrac{22}{7}\times14=\dfrac{55}{3}$ cm. The sector perimeter includes two radii, so perimeter $=14+14+\dfrac{55}{3}=\dfrac{139}{3}$ cm.

Answer:

$\dfrac{139}{3}$ cm, i.e. about $46.3$ cm.

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Q.6If 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?v
Solution

One revolution covers one circumference: $\pi d=\dfrac{22}{7}\times56=176$ cm. Now $10$ km $=1,000,000$ cm. Number of revolutions $=\dfrac{1000000}{176}=\dfrac{62500}{11}\approx5681.8$, about $5682$ revolutions.

Answer:

(i) $176$ cm
(ii) $\dfrac{62500}{11}\approx 5682$ revolutions

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

For a circle, perimeter $=2\pi r$. Since $2\pi$ is the same constant for both circles, the ratio of perimeters equals the ratio of radii.

Answer:

The ratio of their radii is $5:4$.

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2Exercise 6.28 questions
Q.2The 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.v
Solution

The trapezium is isosceles. The difference of the parallel sides is $40-20=20$ cm, so each right triangle at the ends has horizontal leg $10$ cm. Height $=\sqrt{26^2-10^2}=\sqrt{676-100}=24$ cm. Area $=\dfrac{1}{2}(40+20)\times24=720\text{ cm}^2$.

Answer:

The area is $720\text{ cm}^2$.

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Q.3Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.v
Solution

The third side is $32-8-11=13$ cm. With sides $8,11,13$, the semiperimeter is $s=16$. By Heron's formula, area $=\sqrt{16(16-8)(16-11)(16-13)}=\sqrt{16\times8\times5\times3}=8\sqrt{30}\text{ cm}^2$.

Answer:

$8\sqrt{30}\text{ cm}^2$.

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Q.4The sides of a triangular plot are in the ratio $3:5:7$; its perimeter is 300 m. Find its area.v
Solution

Let the sides be $3x,5x,7x$. Then $15x=300$, so $x=20$ and the sides are $60,100,140$ m. The semiperimeter is $150$ m. Area $=\sqrt{150(150-60)(150-100)(150-140)}=\sqrt{150\times90\times50\times10}=1500\sqrt{3}\text{ m}^2$.

Answer:

$1500\sqrt{3}\text{ m}^2$.

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Q.5One diagonal of a rhombus is twice as long as the other diagonal. If the rhombus has area $128\text{ cm}^2$, find the length of the shorter diagonal.v
Solution

Let the shorter diagonal be d cm. Then the longer diagonal is $2d$ cm. Area of a rhombus $=\dfrac{1}{2}\times d\times2d=d^2$. Since the area is $128$, $d^2=128$, so $d=8\sqrt2$ cm.

Answer:

$8\sqrt{2}$ cm.

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Q.6ABCD is a parallelogram. P and Q are any two points on side AB. What can you say about the ratio area $(\triangle PCD)$ : area $(\triangle QCD)$?v
Solution

Both triangles have the same base CD. Since P and Q lie on AB and AB is parallel to CD, the perpendicular distance from P to CD equals the perpendicular distance from Q to CD. Thus the triangles have the same base and height, so their areas are equal.

Answer:

The ratio is $1:1$.

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

Equivalently, take PO as the common base direction. The perpendicular distances from Q and S to the diagonal PR are equal because Q and S are opposite vertices of a parallelogram on the two sides of PR. Therefore triangles PQO and PSO, which have the same base PO and equal heights, have equal areas.

Answer:

Let PR be a diagonal and O lie on PR. The diagonal PR divides parallelogram PQRS into two equal-area triangles, $\triangle PQR$ and $\triangle PSR$. For the same fraction of the way from P to R, point O cuts off the same fraction of each of these two equal-area triangles. Hence $\text{area}(\triangle PQO)=\text{area}(\triangle PSO)$.

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Q.9In $\triangle ABC$, the midpoint of BC is D. Median AD is drawn. P is any point on AD. Show that area $(\triangle ABP) =$ area $(\triangle ACP)$.v
Answer:

Since D is the midpoint of BC, $BD=DC$. Triangles ABD and ACD have equal bases BD and DC and the same height from A, so their areas are equal. For any point P on AD, triangles PBD and PCD also have equal bases BD and DC and the same height from P, so their areas are equal. Subtracting equal areas from equal areas gives $\text{area}(\triangle ABP)=\text{area}(\triangle ACP)$.

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Q.10Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD. What is the ratio of the areas of the red region $(\triangle PAB$ and $\triangle PCD)$ and the green region $(\triangle PBC$ and $\triangle PDA)$?v
Solution

Let the side of the square be s. If the perpendicular distances from P to AB and CD are $h_1$ and $h_2$, then $h_1+h_2=s$, so $\text{area}(\triangle PAB)+\text{area}(\triangle PCD)=\dfrac{1}{2}s(h_1+h_2)=\dfrac{s^2}{2}$. Similarly, the sum of the areas of $\triangle PBC$ and $\triangle PDA$ is also $\dfrac{s^2}{2}$. Hence the two regions have equal area.

Answer:

The ratio is $1:1$.

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3Exercise 6.38 questions
Q.1Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.v
Solution

Sector area $=\dfrac{60}{360}\times\pi r^2=\dfrac{1}{6}\times\dfrac{22}{7}\times49=\dfrac{77}{3}\text{ cm}^2$.

Answer:

$\dfrac{77}{3}\text{ cm}^2$, i.e. about $25.7\text{ cm}^2$.

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Q.2Find the area of a quadrant of a circle whose circumference is 44 cm.v
Solution

From $2\pi r=44$ and $\pi=\dfrac{22}{7}$, $r=7$ cm. Area of a quadrant $=\dfrac{1}{4}\pi r^2=\dfrac{1}{4}\times\dfrac{22}{7}\times49=38.5\text{ cm}^2$.

Answer:

$38.5\text{ cm}^2$.

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Q.3The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.v
Solution

In 10 minutes, the minute hand covers $\dfrac{10}{60}$ of a full circle, i.e. $60°$. Area swept $=\dfrac{60}{360}\times\dfrac{22}{7}\times7^2=\dfrac{77}{3}\text{ cm}^2$.

Answer:

$\dfrac{77}{3}\text{ cm}^2$, i.e. about $25.7\text{ cm}^2$.

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Q.4A 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 $\pi \approx 3.14$.)v
Solution

Total area of the circle is $3.14\times10^2=314\text{ cm}^2$. The minor sector is $\dfrac{90}{360}=\dfrac14$ of the circle, so its area is $78.5\text{ cm}^2$. The major sector is $\dfrac{270}{360}=\dfrac34$ of the circle, so its area is $235.5\text{ cm}^2$.

Answer:

(i) $78.5\text{ cm}^2$
(ii) $235.5\text{ cm}^2$

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Q.5A 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 $\pi \approx 3.14$ and $\sqrt{3}\approx 1.73$.)v
Solution

The minor sector area is $\dfrac{60}{360}\times3.14\times15^2=117.75\text{ cm}^2$. The triangle formed by the two radii and chord is equilateral with side 15 cm, so its area is $\dfrac{\sqrt3}{4}\times15^2\approx\dfrac{1.73}{4}\times225=97.3125\text{ cm}^2$. Minor segment $=117.75-97.3125=20.4375\text{ cm}^2$. Circle area $=3.14\times225=706.5\text{ cm}^2$, so major segment $=706.5-20.4375=686.0625\text{ cm}^2$.

Answer:

Minor segment area $\approx 20.44\text{ cm}^2$; major segment area $\approx 686.06\text{ cm}^2$.

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Q.6A 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.v
Solution

Each wiper sweeps a sector of radius 28 cm and angle $120°$. Area of one sector $=\dfrac{120}{360}\times\dfrac{22}{7}\times28^2=\dfrac{2464}{3}\text{ cm}^2$. There are two non-overlapping wipers, so total area $=2\times\dfrac{2464}{3}=\dfrac{4928}{3}\text{ cm}^2$.

Answer:

$\dfrac{4928}{3}\text{ cm}^2$, i.e. about $1643\text{ cm}^2$.

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Q.7A 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 $r^2\left(\dfrac{\pi}{6}-\dfrac{\sqrt3}{4}\right)$.v
Solution

The minor sector area is $\dfrac{60}{360}\pi r^2=\dfrac{\pi r^2}{6}$. The triangle formed by the two radii and the chord is equilateral with side r, so its area is $\dfrac{\sqrt3}{4}r^2$. Therefore minor segment area $=$ sector area $-$ triangle area $=\dfrac{\pi r^2}{6}-\dfrac{\sqrt3}{4}r^2=r^2\left(\dfrac{\pi}{6}-\dfrac{\sqrt3}{4}\right)$.

Answer:

The minor segment area is $r^2\left(\dfrac{\pi}{6}-\dfrac{\sqrt3}{4}\right)$.

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Q.8An equilateral triangle is inscribed in a circle of radius r. Show the area of the triangle in terms of r.v
Solution

For an equilateral triangle of side a, the circumradius is $\dfrac{a}{\sqrt3}$. Since the circumradius is r, $a=\sqrt3r$. Hence area $=\dfrac{\sqrt3}{4}a^2=\dfrac{\sqrt3}{4}(\sqrt3r)^2=\dfrac{3\sqrt3}{4}r^2$.

Answer:

The area of the equilateral triangle is $\dfrac{3\sqrt3}{4}r^2$.

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4Exercise 6.414 questions
Q.2An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.v
Solution

The base is $40-15-15=10$ cm. The altitude bisects the base, so half-base is 5 cm. Height $=\sqrt{15^2-5^2}=\sqrt{200}=10\sqrt2$ cm. Area $=\dfrac12\times10\times10\sqrt2=50\sqrt2\text{ cm}^2$.

Answer:

$50\sqrt2\text{ cm}^2$.

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Q.3An isosceles triangle has base 10 cm, and its area is $60\text{ cm}^2$. What are the lengths of the equal sides?v
Solution

Area $=\dfrac12\times10\times h=60$, so $h=12$ cm. The altitude bisects the base, giving half-base 5 cm. Each equal side is $\sqrt{12^2+5^2}=\sqrt{169}=13$ cm.

Answer:

Each equal side is $13$ cm.

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Q.4The area of a right-angled triangle is 54 sq. cm. One of its legs has length 12 cm. Find its perimeter.v
Solution

Let the other leg be x cm. Area $=\dfrac12\times12\times x=54$, so $6x=54$ and $x=9$. The hypotenuse is $\sqrt{12^2+9^2}=15$ cm. Perimeter $=12+9+15=36$ cm.

Answer:

The perimeter is $36$ cm.

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Q.5The sides of a triangle are in the ratio $2:3:4$, and its perimeter is 45 cm. Find its area.v
Solution

Let the sides be $2x,3x,4x$. Then $9x=45$, so $x=5$ and the sides are $10,15,20$ cm. The semiperimeter is $22.5$ cm. By Heron's formula, area $=\sqrt{22.5\times12.5\times7.5\times2.5}=\dfrac{75\sqrt{15}}{4}\text{ cm}^2$.

Answer:

$\dfrac{75\sqrt{15}}{4}\text{ cm}^2$, i.e. about $72.6\text{ cm}^2$.

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Q.6The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways.v
Solution

Method 1: Since $7^2+24^2=49+576=625=25^2$, the triangle is right-angled. Area $=\dfrac12\times7\times24=84\text{ cm}^2$. Method 2: By Heron's formula, $s=\dfrac{7+24+25}{2}=28$, so area $=\sqrt{28(21)(4)(3)}=\sqrt{7056}=84\text{ cm}^2$.

Answer:

The area is $84\text{ cm}^2$.

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Q.7If the wheel of a bicycle has a diameter of 60 cm, find how far a cyclist will have travelled after the wheel has rotated 100 times.v
Solution

One rotation covers the circumference: $\pi d=\dfrac{22}{7}\times60=\dfrac{1320}{7}$ cm. For 100 rotations, distance $=100\times\dfrac{1320}{7}=\dfrac{132000}{7}$ cm $\approx188.6$ m.

Answer:

$\dfrac{132000}{7}$ cm, i.e. about $188.6$ m.

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Q.8Find the area of a quadrant of a circle whose circumference is 66 cm.v
Solution

From $2\pi r=66$ and $\pi=\dfrac{22}{7}$, $r=10.5$ cm. Area of a quadrant $=\dfrac14\pi r^2=\dfrac14\times\dfrac{22}{7}\times(10.5)^2=86.625\text{ cm}^2$.

Answer:

$86.625\text{ cm}^2$.

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Q.9The wheel of a car has an outer radius of 28 cm. Calculate how far the car travels after one complete turn of the wheel, and how many times the wheel turns during a journey of 1 km.v
Solution

Circumference $=2\pi r=2\times\dfrac{22}{7}\times28=176$ cm. Since 1 km $=100000$ cm, the number of turns is $\dfrac{100000}{176}=\dfrac{6250}{11}\approx568.2$.

Answer:

One complete turn covers $176$ cm. In 1 km the wheel turns $\dfrac{6250}{11}\approx568.2$ times.

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Q.10Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?v
Solution

Let the side lengths of one rectangle be a and b, and the other be c and d. Same perimeter gives $a+b=c+d$, and same area gives $ab=cd$. The side lengths are therefore two numbers with the same sum and product. Such two numbers are the roots of the same quadratic equation, so the pair $\{a,b\}$ equals the pair $\{c,d\}$. Thus the rectangles are congruent.

Answer:

Yes. They must have the same side lengths, possibly interchanged, so they are congruent.

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Q.11You know that the area of a parallelogram is base × height. Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides × height, i.e., $\dfrac{1}{2}(a+b)h$.v
Answer:

Two congruent copies of the trapezium can be arranged to form a parallelogram with base $a+b$ and height h. The parallelogram area is $(a+b)h$, so one trapezium has half this area: $\dfrac12(a+b)h$.

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Q.12By dividing a trapezium into two triangles show that its area is, half the sum of the parallel sides multiplied by the height.v
Answer:

Draw a diagonal to divide the trapezium into two triangles. If the parallel sides are a and b and the height is h, the two triangles have bases a and b with the same height h. Their total area is $\dfrac12ah+\dfrac12bh=\dfrac12(a+b)h$.

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Q.13Show how we can use two identical copies of a trapezium to make a parallelogram. How will this give us the formula for the area of a trapezium?v
Answer:

Place one copy of the trapezium upside down next to the first so that the non-parallel sides match. The combined shape is a parallelogram with base $a+b$ and height h. Its area is $(a+b)h$, so the area of one trapezium is $\dfrac12(a+b)h$.

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Q.14Show that the area of a kite is half the product of its diagonals. Show this: (i) using algebra, and (ii) using geometry.v
Solution

In a kite, one diagonal is the perpendicular bisector of the other. Algebraically, if one diagonal is split into parts x and y and the other diagonal has length d, the kite is two triangles with areas $\dfrac12dx$ and $\dfrac12dy$. Total area $=\dfrac12d(x+y)=\dfrac12d_1d_2$. Geometrically, the kite can be seen as two triangles sharing the same base diagonal; adding their areas gives half the product of the diagonals.

Answer:

If the diagonals of a kite have lengths $d_1$ and $d_2$, then its area is $\dfrac12d_1d_2$.

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Q.15Three problems about fitting congruent shapes together: (i) Rectangle ABCD has sides a, b, and rectangle PQRS has sides $2a, 2b$. Show that PQRS has 4 times the area of ABCD. Does this mean that 4 copies of rectangle ABCD will fit into rectangle PQRS? (ii) $\triangle ABC$ has sides a, b, c, and $\triangle PQR$ has sides $2a, 2b, 2c$. Show that $\triangle PQR$ has 4 times the area of $\triangle ABC$. Does this mean that 4 copies of $\triangle ABC$ will fit into $\triangle PQR$? (iii) $\triangle ABC$ has sides a, b, c, and $\triangle PQR$ has sides $3a, 3b, 3c$. Show that $\triangle PQR$ has 9 times the area of $\triangle ABC$. Does this mean that 9 copies of $\triangle ABC$ will fit into $\triangle PQR$?v
Solution

When all lengths in a shape are multiplied by k, its area is multiplied by $k^2$. For the rectangle and the triangle with doubled side lengths, $k=2$, so area becomes 4 times. For the triangle with tripled side lengths, $k=3$, so area becomes 9 times. Standard subdivision of the sides into equal parts gives the corresponding tilings.

Answer:

(i) Area scales by $2\times2=4$, and 4 copies of the smaller rectangle fit in a $2$ by $2$ arrangement.
(ii) The triangles are similar with scale factor 2, so area scale factor is $2^2=4$; 4 congruent copies can tile the larger similar triangle.
(iii) The triangles are similar with scale factor 3, so area scale factor is $3^2=9$; 9 congruent copies can tile the larger similar triangle.

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