## Find the capacity in litres of a conical vessel with

[question] Question. Find the capacity in litres of a conical vessel with (i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: (i) Radius (r) of cone = 7 cm Slant height (l) of cone = 25 cm Height $(h)$ of cone $=\sqrt{l^{2}-r^{2}}$ $=\left(\sqrt{25^{2}-7^{2}}\right) \mathrm{cm}$ $=24 \mathrm{~cm}$ Volume of cone $=\frac{1}{3} \pi r^{2} h$ $=\left(\frac{1}{3} \times \frac{22}{7} \times(7)...

## An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see the given figure),

[question] Question. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see the given figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella? [/question] [solution] Solution: For each triangular piece, Semi-perimeter, $s=\frac{(20+50+50) \mathrm{cm}}{2}=60 \mathrm{~cm}$ By Heron's formula, Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ Area of each triangular piece $=[\sqrt{60(60-50)(60-50)(60-20)}] \mat...

## A rhombus shaped field has green grass for 18 cows to graze.

[question] Question. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting? [/question] [solution] Solution: Let ABCD be a rhombus-shaped field. For $\triangle B C D$, Semi-perimeter, $s=\frac{(48+30+30) \mathrm{cm}}{2}=54 \mathrm{~m}$ By Heron’s formula, Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ Therefore, area of $\triangle B C D=[\sqrt{54(54-48)(54-30)(54-30)}] \...

## A triangle and a parallelogram have the same base and the same area

[question] Question A triangle and a parallelogram have the same base and the same area. If the sides of triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram. [/question] [solution] Solution: For triangle Perimeter of triangle $=(26+28+30) \mathrm{cm}=84 \mathrm{~cm}$ $2 s=84 \mathrm{~cm}$ $s=42 \mathrm{~cm}$ By Heron's formula, Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ Area of triangle $=[\sqrt{42(42-26)(42-28)(42-30)}] \mat...

## Radha made a picture of an aeroplane with coloured papers as shown in the given figure.

[question] Question. Radha made a picture of an aeroplane with coloured papers as shown in the given figure. Find the total area of the paper used. [solution] Solution: For triangle I This triangle is an isosceles triangle. Perimeter $=2 s=(5+5+1) \mathrm{cm}=11 \mathrm{~cm}$ $s=\frac{11 \mathrm{~cm}}{2}=5.5 \mathrm{~cm}$ Area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ $=[\sqrt{5.5(5.5-5)(5.5-5)(5.5-1)}] \mathrm{cm}^{2}$ $=[\sqrt{(5.5)(0.5)(0.5)(4.5)}] \mathrm{cm}^{2}$ $=0.75 \sqrt{11} \mathrm{~...

## Find the area of a quadrilateral ABCD in which AB = 3 cm

[question] Question. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. [/question] [solution] Solution: For $\triangle \mathrm{ABC}$, $A C^{2}=A B^{2}+B C^{2}$ $(5)^{2}=(3)^{2}+(4)^{2}$ Therefore, $\triangle \mathrm{ABC}$ is a right-angled triangle, right-angled at point $\mathrm{B}$. Area of $\triangle \mathrm{ABC}=\frac{1}{2} \times \mathrm{AB} \times \mathrm{BC}=\frac{1}{2} \times 3 \times 4=6 \mathrm{~cm}^{2}$ For $\triangle \mathrm{ADC}...

## A park, in the shape of a quadrilateral ABCD,

[question] Question. A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy? [/question] [solution] Solution: Let us join BD. In $\triangle B C D$, applying Pythagoras theorem, $B D^{2}=B C^{2}+C D^{2}$ $=(12)^{2}+(5)^{2}$ $=144+25$ $B D^{2}=169$ $B D=13 \mathrm{~m}$ Area of $\triangle B C D=\frac{1}{2} \times B C \times C D=\left(\frac{1}{2} \times 12 \times 5\right) \mathrm{m}^{2}=30 \mathrm{~m}^{2}$ For $\triangle \...

## An isosceles triangle has perimeter 30 cm

[question] Question. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle. [/question] [solution] Solution: Let the third side of this triangle be x. Perimeter of triangle = 30 cm $12 \mathrm{~cm}+12 \mathrm{~cm}+x=30 \mathrm{~cm}$ $x=6 \mathrm{~cm}$ $s=\frac{\text { Perimeter of triangle }}{2}=\frac{30 \mathrm{~cm}}{2}=15 \mathrm{~cm}$ By Heron’s formula, Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ $=[\sqrt{15(15-12)(15-12)(15-6)}] \mathr...

## Sides of a triangle are in the ratio of 12: 17: 25

[question] Question. Sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540 cm. Find its area. [/question] [solution] Solution: Let the common ratio between the sides of the given triangle be x. Therefore, the side of the triangle will be $12 x, 17 x$, and $25 x$. Perimeter of this triangle $=540 \mathrm{~cm}$ $12 x+17 x+25 x=540 \mathrm{~cm}$ $54 x=540 \mathrm{~cm}$ $x=10 \mathrm{~cm}$ Sides of the triangle will be $120 \mathrm{~cm}, 170 \mathrm{~cm}$, and $250 \mathrm{~cm}...

## Find the area of a triangle two sides of which are 18 cm

[question] Question. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm. [/question] [solution] Solution: Let the third side of the triangle be x. Perimeter of the given triangle = 42 cm $18 \mathrm{~cm}+10 \mathrm{~cm}+x=42$ $x=14 \mathrm{~cm}$ $s=\frac{\text { Perimeter }}{2}=\frac{42 \mathrm{~cm}}{2}=21 \mathrm{~cm}$ By Heron's formula, Area of a triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ Area of the given triangle $=(\sqrt{21(21-18)(21-10)(21-14)}) \mathrm...

## There is a slide in the park.

[question] Question. There is a slide in the park. One of its side walls has been painted in the same colour with a message “KEEP THE PARK GREEN AND CLEAN” (see the given figure). If the sides of the wall are 15m, 11m, and 6m, find the area painted in colour. [/question] [solution] Solution: t can be observed that the area to be painted in colour is a triangle, having its sides as 11 m, 6 m, and 15 m. Perimeter of such a triangle $=(11+6+15) \mathrm{m}$ $2 s=32 \mathrm{~m}$ $s=16 \mathrm{~m}$ By...

## A traffic signal board, indicating ‘SCHOOL AHEAD’

[question] Question. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board? [/question] [solution] Solution: Side of traffic signal board = a Perimeter of traffic signal board $=3 \times a$ $2 s=3 a \Rightarrow s=\frac{3}{2} a$ By Heron’s formula, Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ Area of given triangle $=\sqrt{\frac{3}{2...