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

Exercise – 6.1

Q1. The perimeter of a circle is 44 cm. What is its radius?

Solution:
Given: $C = 44$ C=44 cm
Formula: $C = 2\pi r$ C=2πr

$44 = 2 \times \frac{22}{7} \times r$ 44=2×722×r
$44 = \frac{44}{7} \times r$ 44=744×r
$r = 7$ r=7 cm

Answer: 7 cm

Q2. Calculate, correct to 3 significant figures, the circumference of a circle with:

(i) radius 7 cm

Solution:
$C = 2 \times \frac{22}{7} \times 7 = 44$ C=2×722×7=44 cm

Answer: 44.0 cm

(ii) radius 10 cm

Solution:
$C = \frac{440}{7} = 62.857$ C=7440=62.857

3 significant figures → 62.9 cm

(iii) radius 12 cm

Solution:
$C = \frac{528}{7} = 75.428$ C=7528=75.428

3 significant figures → 75.4 cm

Q3. 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°

Solution:
Arc length $= \frac{60}{360} \times 2 \times \frac{22}{7} \times 3.5$ =36060×2×722×3.5
$= \frac{1}{6} \times 22 = 3.67$ =61×22=3.67 cm

Answer: 3.67 cm

(ii) the radius is 6.3 m and the angle at the centre is 120°

Solution:
$= \frac{120}{360} \times 2 \times \frac{22}{7} \times 6.3$ =360120×2×722×6.3
$= \frac{1}{3} \times 39.6 = 13.2$ =31×39.6=13.2 m

Answer: 13.2 m

Q4. 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:
Arc length:
$= \frac{75}{360} \times 2 \times \frac{22}{7} \times 14$ =36075×2×722×14
$= \frac{5}{24} \times 88 = 18.33$ =245×88=18.33 cm

Perimeter:
$= 18.33 + 28 = 46.33$ =18.33+28=46.33 cm

Answer: 46.33 cm

Q6. 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?

Solution:
$C = \pi d = \frac{22}{7} \times 56 = 176$ C=πd=722×56=176 cm

Answer: 176 cm

(ii) How many revolutions does the tyre make if the car travels 10 km?

Solution:
Distance $= 10,00,000$ =10,00,000 cm
Revolutions $= \frac{1000000}{176} = 5681.8 \approx 5682$ =1761000000=5681.8≈5682

Answer: 5682 revolutions

Q8. The ratio of the perimeters of two circles is 5 : 4. What is the ratio of their radii?

Solution:
$C = 2\pi r \Rightarrow \frac{C_1}{C_2} = \frac{r_1}{r_2}$ C=2πr⇒C2C1=r2r1

Ratio of radii = 5 : 4

Exercise – 6.2

Q2. 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:
Given:
Parallel sides = 40 cm, 20 cm
Non-parallel sides = 26 cm each

Since the trapezium is isosceles:

Difference of bases = $40 – 20 = 20$ 40−20=20
Half difference = $10$ 10

Height:
$h = \sqrt{26^2 – 10^2}$ h=262−102
$= \sqrt{676 – 100} = \sqrt{576} = 24$ =676−100=576=24 cm

Area of trapezium:
$= \frac{1}{2} \times (sum\ of\ parallel\ sides) \times height$ =21×(sum of parallel sides)×height

$= \frac{1}{2} \times (40 + 20) \times 24$ =21×(40+20)×24
$= \frac{1}{2} \times 60 \times 24 = 720 \text{ cm}^2$ =21×60×24=720 cm2

Answer: 720 cm²

Q3. Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.

Solution:
Third side = $32 – (8 + 11) = 13$ 32−(8+11)=13 cm

Semi-perimeter:
$s = \frac{32}{2} = 16$ s=232=16

Using Heron’s Formula:
$Area = \sqrt{s(s-a)(s-b)(s-c)}$ Area=s(s−a)(s−b)(s−c)

$= \sqrt{16(16-8)(16-11)(16-13)}$ =16(16−8)(16−11)(16−13)
$= \sqrt{16 \times 8 \times 5 \times 3}$ =16×8×5×3
$= \sqrt{1920} \approx 43.8 \text{ cm}^2$ =1920≈43.8 cm2

Answer: 43.8 cm²

Q4. The sides of a triangular plot are in the ratio 3 : 5 : 7; its perimeter is 300 m. Find its area.

Solution:
Let sides be $3x, 5x, 7x$ 3x,5x,7x

$3x + 5x + 7x = 15x = 300 \Rightarrow x = 20$ 3x+5x+7x=15x=300⇒x=20

Sides = 60 m, 100 m, 140 m

Semi-perimeter:
$s = 150$ s=150

Using Heron’s Formula:
$Area = \sqrt{150(150-60)(150-100)(150-140)}$ Area=150(150−60)(150−100)(150−140)

$= \sqrt{150 \times 90 \times 50 \times 10}$ =150×90×50×10
$= \sqrt{6750000} \approx 2598 \text{ m}^2$ =6750000≈2598 m2

Answer: 2598 m²

Q5. 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 shorter diagonal = $x$ x
Longer diagonal = $2x$ 2x

Area formula:
$Area = \frac{1}{2} \times d_1 \times d_2$ Area=21×d1×d2

$128 = \frac{1}{2} \times x \times 2x$ 128=21×x×2x
$128 = x^2$ 128=x2

$x = \sqrt{128} = 8\sqrt{2}$ x=128=82 cm

Answer: 828\sqrt{2}82 cm

Q6. 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:
Triangles PCD and QCD lie on the same base CD and between the same parallels.

Therefore, their areas are equal.

Answer: Ratio = 1 : 1

Q7. O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.

Solution:
Triangles PSO and PQO lie on the same base PO and between the same parallels.

Hence, their areas are equal.

Proved

Q8. If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram formed is half of the area of the quadrilateral.

Solution:
By joining the midpoints of the sides, a parallelogram is formed (Varignon’s Theorem).

Each triangle contributes equally, and the total area of the new figure becomes half of the original quadrilateral.

Answer: Area = 12\frac{1}{2}21 of original quadrilateral

Q9. In ΔABC, the midpoint of BC is D. Median AD is drawn. P is any point on AD. Show that area (ΔABP) = area (ΔACP).

Solution:
Since D is the midpoint, BD = DC.

Triangles ABD and ADC have equal areas.

Point P lies on AD, dividing both triangles proportionally.

Therefore,
Area (ΔABP) = Area (ΔACP)

Proved

Q10. Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD. What is the ratio of the areas of (ΔPAB and ΔPCD) and (ΔPBC and ΔPDA)?

Solution:
Opposite triangles in a square balance in area.

So,
Area (PAB + PCD) = Area (PBC + PDA)

Answer: Ratio = 1 : 1

Q11. In ΔABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ ∥ PD. PQ is joined. Prove that Area (ΔBPQ) = 1/2 Area (ΔABC).

Solution:
Since D is midpoint and CQ ∥ PD, similar triangles are formed.

Using properties of parallel lines and proportional division:

Area (ΔBPQ) = $\frac{1}{2}$ 21 Area (ΔABC)

Proved

Exercise – 6.3

Q1. Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.

Solution:
Formula:
Area of sector = $\frac{\theta}{360} \times \pi r^2$ 360θ×πr2

$= \frac{60}{360} \times \frac{22}{7} \times 7^2$ =36060×722×72
$= \frac{1}{6} \times \frac{22}{7} \times 49$ =61×722×49
$= \frac{1}{6} \times 154$ =61×154
$= 25.67 \text{ cm}^2$ =25.67 cm2

Answer: 25.67 cm²

Q2. Find the area of a quadrant of a circle whose circumference is 44 cm.

Solution:
Given: Circumference $C = 44$ C=44

$C = 2\pi r$ C=2πr

$44 = 2 \times \frac{22}{7} \times r$ 44=2×722×r
$r = 7$ r=7 cm

Area of circle:
$= \pi r^2 = \frac{22}{7} \times 49 = 154 \text{ cm}^2$ =πr2=722×49=154 cm2

Area of quadrant:
$= \frac{1}{4} \times 154 = 38.5 \text{ cm}^2$ =41×154=38.5 cm2

Answer: 38.5 cm²

Q3. The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.

Solution:
In 60 min → 360°
In 10 min → $60°$ 60°

Area swept:
$= \frac{60}{360} \times \frac{22}{7} \times 7^2$ =36060×722×72
$= \frac{1}{6} \times 154$ =61×154
$= 25.67 \text{ cm}^2$ =25.67 cm2

Answer: 25.67 cm²

Q4. A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding:

(i) minor sector (90°)

(ii) major sector (270°)

(Use π=3.14\pi = 3.14π=3.14)

Solution:

(i) Minor sector:
$= \frac{90}{360} \times 3.14 \times 10^2$ =36090×3.14×102
$= \frac{1}{4} \times 3.14 \times 100$ =41×3.14×100
$= 78.5 \text{ cm}^2$ =78.5 cm2

(ii) Major sector:
$= \frac{270}{360} \times 3.14 \times 100$ =360270×3.14×100
$= \frac{3}{4} \times 314$ =43×314
$= 235.5 \text{ cm}^2$ =235.5 cm2

Answer:
Minor = 78.5 cm²
Major = 235.5 cm²

Q5. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments.

(Use π=3.14\pi = 3.14π=3.14, 3=1.73\sqrt{3} = 1.733=1.73)

Solution:

Area of sector:
$= \frac{60}{360} \times 3.14 \times 15^2$ =36060×3.14×152
$= \frac{1}{6} \times 3.14 \times 225$ =61×3.14×225
$= 117.75 \text{ cm}^2$ =117.75 cm2

Area of triangle:
$= \frac{\sqrt{3}}{4} \times 15^2$ =43×152
$= \frac{1.73}{4} \times 225 = 97.31 \text{ cm}^2$ =41.73×225=97.31 cm2

Minor segment:
$= 117.75 – 97.31 = 20.44 \text{ cm}^2$ =117.75−97.31=20.44 cm2

Area of circle:
$= 3.14 \times 225 = 706.5 \text{ cm}^2$ =3.14×225=706.5 cm2

Major segment:
$= 706.5 – 20.44 = 686.06 \text{ cm}^2$ =706.5−20.44=686.06 cm2

Answer:
Minor = 20.44 cm²
Major = 686.06 cm²

Q6. 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:
Area cleaned by one wiper:
$= \frac{120}{360} \times \frac{22}{7} \times 28^2$ =360120×722×282
$= \frac{1}{3} \times \frac{22}{7} \times 784$ =31×722×784
$= \frac{1}{3} \times 2464 = 821.33 \text{ cm}^2$ =31×2464=821.33 cm2

For two wipers:
$= 2 \times 821.33 = 1642.67 \text{ cm}^2$ =2×821.33=1642.67 cm2

Answer: 1642.67 cm²

Q7. A chord of a circle of radius r subtends an angle of 60° at the centre. Show that the area of the corresponding minor segment is

$\pi r^2 \left(\frac{1}{6} – \frac{\sqrt{3}}{4}\right)$ πr2(61−43)

Solution:
Area of sector:
$= \frac{60}{360} \times \pi r^2 = \frac{1}{6}\pi r^2$ =36060×πr2=61πr2

Area of triangle:
$= \frac{\sqrt{3}}{4} r^2$ =43r2

Minor segment = Sector − Triangle

$= \pi r^2 \left(\frac{1}{6} – \frac{\sqrt{3}}{4}\right)$ =πr2(61−43)

Proved

Q8. 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

$\frac{3\sqrt{3}}{4\pi}$ 4π33

Solution:
Side of triangle = $\sqrt{3}r$ 3r

Area of triangle:
$= \frac{\sqrt{3}}{4} (\sqrt{3}r)^2 = \frac{3\sqrt{3}}{4} r^2$ =43(3r)2=433r2

Area of circle:
$= \pi r^2$ =πr2

Ratio:
$= \frac{3\sqrt{3}}{4\pi}$ =4π33

Proved

Q9. 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 2π\frac{2}{\pi}π2.

Solution:
Diagonal = $2r$ 2r

Side = $\frac{2r}{\sqrt{2}} = \sqrt{2}r$ 22r=2r

Area of square:
$= (\sqrt{2}r)^2 = 2r^2$ =(2r)2=2r2

Area of circle:
$= \pi r^2$ =πr2

Ratio:
$= \frac{2}{\pi}$ =π2

Proved

Q10. 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 332π\frac{3\sqrt{3}}{2\pi}2π33.

Solution:
Hexagon = 6 equilateral triangles

Area of one triangle:
$= \frac{\sqrt{3}}{4} r^2$ =43r2

Total area:
$= 6 \times \frac{\sqrt{3}}{4} r^2 = \frac{3\sqrt{3}}{2} r^2$ =6×43r2=233r2

Area of circle:
$= \pi r^2$ =πr2

Ratio:
$= \frac{3\sqrt{3}}{2\pi}$ =2π33

Proved

End-of-Chapter Exercises

Q2. An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.

Solution:
Perimeter = 40 cm
Equal sides = 15 cm each

Base = $40 – (15 + 15) = 10$ 40−(15+15)=10 cm

Height:
$h = \sqrt{15^2 – 5^2}$ h=152−52
$= \sqrt{225 – 25} = \sqrt{200} = 10\sqrt{2}$ =225−25=200=102

Area:
$= \frac{1}{2} \times 10 \times 10\sqrt{2}$ =21×10×102
$= 50\sqrt{2} \approx 70.7 \text{ cm}^2$ =502≈70.7 cm2

Answer: 70.7 cm²

Q3. An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides?

Solution:
Area = 60 cm²
Base = 10 cm

$60 = \frac{1}{2} \times 10 \times h$ 60=21×10×h
$h = 12$ h=12 cm

Equal side:
$= \sqrt{12^2 + 5^2}$ =122+52
$= \sqrt{144 + 25} = \sqrt{169} = 13$ =144+25=169=13 cm

Answer: 13 cm each

Q4. The area of a right-angled triangle is 54 cm². One of its legs has length 12 cm. Find its perimeter.

Solution:
$Area = \frac{1}{2} \times base \times height$ Area=21×base×height

$54 = \frac{1}{2} \times 12 \times h$ 54=21×12×h
$h = 9$ h=9 cm

Hypotenuse:
$= \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15$ =122+92=144+81=225=15

Perimeter:
$= 12 + 9 + 15 = 36$ =12+9+15=36 cm

Answer: 36 cm

Q5. The sides of a triangle are in the ratio 2 : 3 : 4, and its perimeter is 45 cm. Find its area.

Solution:
Let sides = $2x, 3x, 4x$ 2x,3x,4x

$9x = 45 \Rightarrow x = 5$ 9x=45⇒x=5

Sides = 10, 15, 20 cm

Semi-perimeter:
$s = 22.5$ s=22.5

Using Heron’s formula:
$Area = \sqrt{22.5(22.5-10)(22.5-15)(22.5-20)}$ Area=22.5(22.5−10)(22.5−15)(22.5−20)

$= \sqrt{22.5 \times 12.5 \times 7.5 \times 2.5}$ =22.5×12.5×7.5×2.5

$\approx 72.6 \text{ cm}^2$ ≈72.6 cm2

Answer: 72.6 cm²

Q6. The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways.

Solution:

Method 1 (Right triangle):
$7^2 + 24^2 = 25^2$ 72+242=252 → Right triangle

Area:
$= \frac{1}{2} \times 7 \times 24 = 84 \text{ cm}^2$ =21×7×24=84 cm2

Method 2 (Heron’s formula):
$s = \frac{7+24+25}{2} = 28$ s=27+24+25=28

$Area = \sqrt{28(28-7)(28-24)(28-25)}$ Area=28(28−7)(28−24)(28−25)
$= \sqrt{28 \times 21 \times 4 \times 3}$ =28×21×4×3
$= \sqrt{7056} = 84 \text{ cm}^2$ =7056=84 cm2

Answer: 84 cm²

Q8. Find the area of a quadrant of a circle whose circumference is 66 cm.

Solution:
$C = 2\pi r$ C=2πr

$66 = 2 \times \frac{22}{7} \times r$ 66=2×722×r
$r = 10.5$ r=10.5 cm

Area of circle:
$= \frac{22}{7} \times (10.5)^2 = 346.5$ =722×(10.5)2=346.5

Quadrant:
$= \frac{1}{4} \times 346.5 = 86.625 \text{ cm}^2$ =41×346.5=86.625 cm2

Answer: 86.63 cm²

Q9. The 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.

Solution:
Distance in one turn = circumference

$= 2 \times \frac{22}{7} \times 28 = 176$ =2×722×28=176 cm

Distance = 1 km = 100000 cm

Revolutions:
$= \frac{100000}{176} \approx 568$ =176100000≈568

Answer:
Distance = 176 cm
Revolutions = 568

Q10. Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?

Solution:
No, same area and perimeter do not guarantee same dimensions.

Different length–breadth combinations can give same area and perimeter.

Answer: No, they are not necessarily congruent.

Q11. Show that the area of a trapezium is 12(a+b)h\frac{1}{2}(a+b)h21(a+b)h.

Solution:
Area of trapezium = sum of areas of two triangles

By rearranging or using parallelogram concept:

$Area = \frac{1}{2}(a+b)h$ Area=21(a+b)h

Proved

Q12. By dividing a trapezium into two triangles, show that its area is half the sum of parallel sides multiplied by height.

Solution:
Divide trapezium into two triangles.

Add their areas:

$= \frac{1}{2}ah + \frac{1}{2}bh$ =21ah+21bh

$= \frac{1}{2}(a+b)h$ =21(a+b)h

Proved

Q13. Show how two identical trapeziums form a parallelogram and derive the formula.

Solution:
Two identical trapeziums combine to form a parallelogram.

Area of parallelogram = $(a+b)h$ (a+b)h

So trapezium = half of it:

$= \frac{1}{2}(a+b)h$ =21(a+b)h

Proved

Q14. Show that the area of a kite is half the product of its diagonals.

Solution:
Kite can be divided into two triangles along diagonals.

Area = sum of triangle areas:

$= \frac{1}{2} \times d_1 \times d_2$ =21×d1×d2

Proved

Q15. (i) Show that rectangle PQRS has 4 times the area of rectangle ABCD. Does it mean 4 copies will fit?

Solution:
Area scales with square of dimensions.

$(2a \times 2b) = 4ab$ (2a×2b)=4ab

Yes, 4 copies fit exactly.

(ii) Triangle sides doubled → area becomes 4 times.
But triangles may not fit perfectly due to shape.

(iii) Triangle sides tripled → area becomes 9 times.
Again, exact fitting not guaranteed.

Answer:
Area scales as square of scale factor; fitting depends on shape.

Q1. The perimeter of a circle is 44 cm. What is its radius?

Q2. Calculate, correct to 3 significant figures, the circumference of a circle with:

(i) radius 7 cm

(ii) radius 10 cm

(iii) radius 12 cm

Q3. 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°

(ii) the radius is 6.3 m and the angle at the centre is 120°

Q4. 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°.

Q6. 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?

Q8. The ratio of the perimeters of two circles is 5 : 4. What is the ratio of their radii?

Q2. 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.

Q3. Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.

Q4. The sides of a triangular plot are in the ratio 3 : 5 : 7; its perimeter is 300 m. Find its area.

Q5. 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.

Q6. 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)?

Q7. O is any point on the diagonal PR of a parallelogram PQRS. Prove that the areas of triangles PSO and PQO are equal.

Q8. If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram formed is half of the area of the quadrilateral.

Q9. In ΔABC, the midpoint of BC is D. Median AD is drawn. P is any point on AD. Show that area (ΔABP) = area (ΔACP).

Q10. Given a square ABCD, let P be a point within it. Join PA, PB, PC, PD. What is the ratio of the areas of (ΔPAB and ΔPCD) and (ΔPBC and ΔPDA)?

Q11. In ΔABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ ∥ PD. PQ is joined. Prove that Area (ΔBPQ) = 1/2 Area (ΔABC).

Q1. Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60°.

Q2. Find the area of a quadrant of a circle whose circumference is 44 cm.

Q3. The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.

Q4. A chord of a circle of radius 10 cm subtends 90° at the centre. Find the area of the corresponding:

(i) minor sector (90°)

(ii) major sector (270°)

Q5. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments.

Q6. 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.

Q7. A chord of a circle of radius r subtends an angle of 60° at the centre. Show that the area of the corresponding minor segment is

Q8. 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

Q9. 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 2π\frac{2}{\pi}π2​.

Q10. 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 332π\frac{3\sqrt{3}}{2\pi}2π33​​.

Q2. An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.

Q3. An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides?

Q4. The area of a right-angled triangle is 54 cm². One of its legs has length 12 cm. Find its perimeter.

Q5. The sides of a triangle are in the ratio 2 : 3 : 4, and its perimeter is 45 cm. Find its area.

Q6. The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways.

Q8. Find the area of a quadrant of a circle whose circumference is 66 cm.

Q9. The 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.

Q10. Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?

Q11. Show that the area of a trapezium is 12(a+b)h\frac{1}{2}(a+b)h21​(a+b)h.

Q12. By dividing a trapezium into two triangles, show that its area is half the sum of parallel sides multiplied by height.

Q13. Show how two identical trapeziums form a parallelogram and derive the formula.

Q14. Show that the area of a kite is half the product of its diagonals.

Q15. (i) Show that rectangle PQRS has 4 times the area of rectangle ABCD. Does it mean 4 copies will fit?

Q9. 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 2π\frac{2}{\pi}π2.

Q10. 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 332π\frac{3\sqrt{3}}{2\pi}2π33.

Q11. Show that the area of a trapezium is 12(a+b)h\frac{1}{2}(a+b)h21(a+b)h.