Exercise 12.2 Class 9 Maths

Heron’s Formula Exercise 12.2 Class 9 Maths

Exercise 12.2 Class 9 Maths Question 1.

A park, in the shape of a quadrilateral ABCD, has ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

Solution.

The following figure shows:- AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. ∠C = 90°

IN Δ DCB ∠C = 90º

BD² = DC² + BC² $\left (By Pythagoras Theorem \right )$

BD² = 5² + 12²

BD² = 25 + 144

BD² = 169

BD = $\sqrt{169}$

BD = 13 m

IN Δ BCD

Area of Δ BCD = $\frac{1}{2 } \times Base\times Height$

= $\frac{1}{2}$ × 12 × 5

= 30 m²

IN Δ ADB

a = 9 m b = 8 m c = 13 m

S = $\frac{a+b+c}{2}$ (Semi-Perimeter)

$S = \frac{9+8+13}{2}$

S = $\frac{30}{2}$

S = 15m

The area of the triangle ABD = $\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$ (Heron’s Formula)

$=\sqrt{15\left ( 15-9 \right )\left ( 15-8 \right )\left ( 15-13 \right )}$

$= \sqrt{15\times 6\times 7\times 2}$

$= \sqrt{3\times 5\times 3\times 2\times 7\times 2}$ $= 3\times 2\sqrt{35}$

Thus, The area of the triangle ABD $= 6\sqrt{35}m^{2}$

Area of quadrilateral ABCD = Area of BCD + Area of ABD

= 30 + $6\sqrt{35}$

= 30 + 6 × 5.91

= 30 + 35.46

Area of quadrilateral ABCD = 65.46m² 〈 APPROX 〉

Exercise12.2 Class 9 Maths Question 2.

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.

Solution.

The following figure shows:- AB = 3 cm, BC = 4 cm, CD = 4 cm , DA = 5 cm and AC = 5 cm.

IN Δ ADC

a = 5 cm b = 5 cm c = 4 cm

S = $\frac{a+b+c}{2}$ (Semi-Perimeter)

$S = \frac{5+5+4}{2}$

S = $\frac{14}{2}$

S = 7 cm

The area of the triangle ADC = $\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$ (Heron’s Formula)

$=\sqrt{7\left ( 7-5\right )\left ( 7-5 \right )\left ( 7-4 \right )}$

$= \sqrt{7\times 2\times 2\times 3}$

$= 2\sqrt{21}cm^{2}$

IN ΔABC

a = 3 cm b = 4 cm c = 5 cm

S = $\frac{a+b+c}{2}$ (Semi-Perimeter)

$S = \frac{3+4+5}{2}$

S = $\frac{12}{2}$

S = 6 cm

The area of the triangle ABC = $\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$ (Heron’s Formula)

$=\sqrt{6\left ( 6-3 \right )\left ( 6-4 \right )\left ( 6-5\right )}$

$= \sqrt{6\times 3\times 2\times 1}$

$= {\sqrt{6\times 6}$

= 6 cm²

AREA OF Quadrilateral ABCD = AREA Δ ADC + AREA Δ ABC

$= 2\sqrt{21} + 6$

= 2×4.58 + 6

= 9.16 + 6

= 15.16 cm² OR 15.2 cm²

Exercise 12.2 Class 9 Maths Question 3.

Radha made a picture of an Aeroplan with coloured paper as shown in Fig . Find
the total area of the paper used.

Solution.

Area of first part :-

a= 5 cm b = 5 cm c = 1 cm

S = $\frac{a+b+c}{2}$ (Semi-Perimeter)

$S = \frac{5+5+1}{2}$

S = $\frac{11}{2}$ cm

The area of first part = $\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$ (Heron’s Formula)

$=\sqrt{\frac{11}{2}\left ( \frac{11}{2}-5\right )\left ( \frac{11}{2}-5 \right )\left ( \frac{11}{2}-1\right )}$

$= \sqrt{\frac{11}{2}\left ( \frac{11-10}{2} \right )\left ( \frac{11-10}{2} \right )\left ( \frac{11-2}{2} \right )}$

$= \sqrt{\frac{11}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{9}{2}}$

$= \sqrt{\frac{11}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{3}{2}\times 3}$

$= \sqrt{11\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\times 3\times 3}$

$= \frac{1}{2}\times \frac{1}{2}\times 3\sqrt{11}$

$= \frac{3}{4}\sqrt{11} cm^{2}$ = 0.75 × 3.3 = 2.475 cm²

Area of second part = L × B

= 6.5 × 1 = 6.5 cm²

Area of third part :-

Area of BDC = $\frac{\sqrt{3}}{4}\left ( 1 \right ) = \frac{\sqrt{3}}{4}$

$\frac{1}{2}\times 1\times h = \frac{\sqrt{3}}{4}$

$h = \frac{\sqrt{3}}{2}$

Area of third part = $\frac{1}{2}\left [ 2 + 1 \right ]\times \frac{\sqrt{3}}{2}$

$= \frac{1}{2}\times 3\times \frac{\sqrt{3}}{2}$

$= \frac{3}{4}\sqrt{3 } cm^{2}$ = 0.75 x 1.73 = 1.2975 cm ² or 1.30 cm²

Area of forth part = $\frac{1}{2}\times base \times height$

= $\frac{1}{2}\times 1.5 \times 6$

= 4.5 cm²

Area of forth part = $\frac{1}{2}\times base \times height$

= $\frac{1}{2}\times 1.5 \times 6$

= 4.5 cm²

The total area of the paper used = 2.5 + 6.5 + 1.3 + 4.5 + 4.5 = 19.3 cm²

Exercise 12.2 Class 9 Maths Question 4.

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

Solution.

IN Δ ABC

a = 26 cm b = 30 cm c = 28 cm

S = $\frac{a+b+c}{2}$ (Semi-Perimeter)

$S = \frac{26+30+28}{2}$

S = $\frac{84}{2}$

S = 42 cm

The area of the triangle ABC = $\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$ (Heron’s Formula)

$=\sqrt{42\left ( 42-26 \right )\left ( 42-30\right )\left ( 42-28\right )}$

$= \sqrt{42\times 16\times 12\times 14}$

$= \sqrt{14\times 3\times 4\times 4\times 3\times 2\times 2\times 14}$

= 14 × 4 × 3 × 2

∴ AREA OF BEDC = AREA OF triangle ABC

BASE × HEIGHT = 14 × 4 × 3 × 2

28 × HEIGHT = 14 × 4 × 3 × 2

HEIGHT = $\frac{14 \times 4 \times 3 \times 2}{28}$

HEIGHT = 12 cm

Exercise12.2 Class 9 Maths Question 5.

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?

Solution.

IN Δ ACD

a = 30 m b = 30 m c = 48 m

S = $\frac{a+b+c}{2}$ (Semi-Perimeter)

$S = \frac{30+30+48}{2}$

S = $\frac{108}{2}$

S = 54 m

The area of the triangle ACD = $\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$ (Heron’s Formula)

$=\sqrt{54\left ( 54-30 \right )\left ( 54-30\right )\left ( 54-48\right )}$

$= \sqrt{54\times 24\times 24\times 6}$

$= \sqrt{3\times 3\times 6\times 24\times 24\times 6}$

= 3 × 24 × 6 m²

TOTAL COWS = 18

Area will be getting by each Cow = $\frac{2\times 3\times 24\times 6}{18}$

Area of rhombus ABCD = 48 m²

Exercise 12.2 Class 9 Maths Question 6.

An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Solution.

IN Δ ABC

a = 20 cm b = 50 cm c = 50 cm

S = $\frac{a+b+c}{2}$ (Semi-Perimeter)

$S = \frac{20+50+50}{2}$

S = $\frac{120}{2}$

S = 60 cm

The area of the triangle ABC = $\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$ (Heron’s Formula)

$=\sqrt{60\left (60-20 \right )\left ( 60-50\right )\left ( 60-50\right )}$

$= \sqrt{60\times 40\times 10\times 10}$

$= \sqrt{10\times 6\times 4\times 10\times 10\times 10}$

$= \sqrt{10\times 10\times 10\times 10\times 3\times 2\times 2\times 2}$

$= 10 \times 10 \times 2\sqrt{6}$

Area of each colours cloth :- $\frac{10\times 200\sqrt{6}}{2}$ cm²

$= 1000\sqrt{6}$ cm²

FIRST AREA = SECOND AREA $= 1000\sqrt{6}$ cm²

Ex.12.2 class 9 Maths Question 7.

A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it ?

Solution.

IN Δ ABD ∠ A = 90°

a² + a² = 32²

2a² = 32 × 32

a² = 16 × 32

Area of first part = $\frac{a^{2}}{2}$ = Area of square

= $\frac{16 \times 32}{2}$

Area of first part = 256 cm²

Area of second part = 256 cm²

Area of third part :-

a = 6 cm b = 6 cm c = 8 cm

S = $\frac{a+b+c}{2}$ (Semi-Perimeter)

$S = \frac{6+6+8}{2}$

S = $\frac{20}{2}$

S = 10 cm

The area of the third part = $\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$ (Heron’s Formula)

$=\sqrt{10\left (10-6 \right )\left ( 10-6\right )\left ( 10-8\right )}$

$= \sqrt{10\times 4\times 4\times 2}$

$= \sqrt{2\times 5\times 4\times 4\times 2}$

$= 2\times 4\sqrt{5}$

$= 8\sqrt{5}$

= 8 × 2.24 = 17.92

The area of the third part = 17.92 cm²

Exercise12.2 Class 9 Maths Question 8.

A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle
being 9 cm, 28 cm and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate
of 50p per cm2.