NCERT solution class 9 chapter 12 Heron's Formula exercise 12.2 mathematics ~ evidya

EXERCISE 12.2

Q 1, Ex. 12.2 - Heron's Formula (NCERT Maths Class 9th)

Page No 206:

Question 1:

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?

Answer:

Let us join BD.
In ΔBCD, applying Pythagoras theorem,
BD² = BC² + CD²
= (12)² + (5)²
= 144 + 25
BD² = 169
BD = 13 m

Area of ΔBCD$=\frac{1}{2} \times \mathrm{BC} \times \mathrm{CD}$
$=\left(\frac{1}{2} \times 12 \times 5\right) \mathrm{m}^{2}=30 \mathrm{m}^{2}$

For ΔABD,
$s=\frac{\text { Perimeter }}{2}$

$=\frac{(9+8+13) m}{2}$

=15m

By Heron’s formula,
Area of triangle$=\sqrt{s(s-a)(s-b)(s-c)}$
Area of ΔABD $=[\sqrt{15(15-9)(15-8)(15-13)}] \mathrm{m}^{2}$
$=(\sqrt{15 \times 6 \times 7 \times 2}) \mathrm{m}^{2}$
$=6 \sqrt{35} \mathrm{m}^{2}$
$=(6 \times 5.916) \mathrm{m}^{2}$
$=35.496 \mathrm{m}^{2}$

Area of the park = Area of ΔABD + Area of ΔBCD
= 35.496 + 30 m²= 65.496 m² = 65.5 m² (approximately)

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.

Answer:

For ΔABC,
AC² = AB² + BC²
(5)² = (3)² + (4)²
Therefore, ΔABC is a right-angled triangle, right-angled at point B.
Area of ΔABC$=\frac{1}{2} \times \mathrm{AB} \times \mathrm{BC}$
$=\frac{1}{2} \times 3 \times 4=6 \mathrm{cm}^{2}$

For ΔADC,
Perimeter = 2s = AC + CD + DA = (5 + 4 + 5) cm = 14 cm
s = 7 cm
By Heron’s formula,
Area of triangle$=\sqrt{s(s-a)(s-b)(s-c)}$

Area of $\Delta \mathrm{ADC}=[\sqrt{7(7-5)(7-5)(7-4)}] \mathrm{cm}^{2}$

$=(\sqrt{7 \times 2 \times 2 \times 3}) \mathrm{cm}^{2}$

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

$=(2 \times 4.583) \mathrm{cm}^{2}$

=9.166cm²

Area of ABCD = Area of ΔABC + Area of ΔACD
= (6 + 9.166) cm² = 15.166 cm² = 15.2 cm² (approximately)

Q 3, Ex. 12.2, Page No 206 - Heron's Formula (NCERT Maths Class 9th)

Question 3:

Radha made a picture of an aeroplane with coloured papers as shown in the given figure. Find the total area of the paper used.

Answer:

For triangle I
This triangle is an isosceles triangle.
Perimeter = 2s = (5 + 5 + 1) cm = 11cm
$s=\frac{11 \mathrm{cm}}{2}=5.5 \mathrm{cm}$

Area of the triangle$=\sqrt{s(s-a)(s-b)(s-c)}$
$=[\sqrt{5.5(5.5-5)(5.5-5)(5.5-1)}] \mathrm{cm}^{2}$
$=[\sqrt{(5.5)(0.5)(0.5)(4.5)}] \mathrm{cm}^{2}$
$=0.75 \sqrt{11} \mathrm{cm}^{2}$
$=(0.75 \times 3.317) \mathrm{cm}^{2}$
$=2.488 \mathrm{cm}^{2}$ (approximately)

For quadrilateral II
This quadrilateral is a rectangle.
Area = l × b = (6.5 × 1) cm²= 6.5 cm²
For quadrilateral III
This quadrilateral is a trapezium.
Perpendicular height of parallelogram $=(\sqrt{1^{2}-(0.5)^{2}}) \mathrm{cm}$

$=\sqrt{0.75} \mathrm{cm}$
=0.866cm

Area = Area of parallelogram + Area of equilateral triangle
$=(0.866) 1+\frac{\sqrt{3}}{4}(1)^{2}$
= 0.866 + 0.433 = 1.299 cm²

Area of triangle (IV) = Area of triangle in (V)
$=\left(\frac{1}{2} \times 1.5 \times 6\right) \mathrm{cm}^{2}=4.5 \mathrm{cm}^{2}$

Total area of the paper used = 2.488 + 6.5 + 1.299 + 4.5 × 2
= 19.287 cm²

Q 4, Ex. 12.2, Page No 206 - Heron's Formula (NCERT Maths Class 9th)

Question 4:

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.

Answer:

For triangle
Perimeter of triangle = (26 + 28 + 30) cm = 84 cm
2s = 84 cm
s = 42 cm
By Heron’s formula,
Area of triangle$=\sqrt{s(s-a)(s-b)(s-c)}$
Area of triangle$=[\sqrt{42(42-26)(42-28)(42-30)}] \mathrm{cm}^{2}$
$=[\sqrt{42(16)(14)(12)}] \mathrm{cm}^{2}$
= 336 cm²

Let the height of the parallelogram be h.
Area of parallelogram = Area of triangle
h × 28 cm = 336 cm²
h = 12 cm
Therefore, the height of the parallelogram is 12 cm.

Q 5, Ex. 12.2, Page No 207 - Heron's Formula (NCERT Maths Class 9th)

Page No 207:

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?

Answer:

Let ABCD be a rhombus-shaped field.
For ΔBCD,
Semi-perimeter, $s=\frac{(48+30+30) \mathrm{cm}}{2}$
= 54 m

By Heron’s formula,
Area of triangle$=\sqrt{s(s-a)(s-b)(s-c)}$

Therefore, area of ΔBCD$=[\sqrt{54(54-48)(54-30)(54-30)}] \mathrm{m}^{2}$
$=\sqrt{54(6)(24)(24)}$

=3×6×24

=432m²

Area of field = 2 × Area of ΔBCD
= (2 × 432) m² = 864 m²
Area for grazing for 1 cow $=\frac{864}{18}$= 48 m²
Each cow will get 48 m² area of grass field.

Q 6, Ex. 12.2, Page No 207 - Heron's Formula (NCERT Maths Class 9th)

Question 6:

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?

Answer:

For each triangular piece,
Semi-perimeter, $s=\frac{(20+50+50) \mathrm{cm}}{2}$
=60cm

By Heron’s formula,
Area of triangle$=\sqrt{s(s-a)(s-b)(s-c)}$

Area of each triangular piece $=[\sqrt{60(60-50)(60-50)(60-20)}] \mathrm{cm}^{2}$
$=[\sqrt{60(10)(10)(40)}] \mathrm{cm}^{2}=200 \sqrt{6} \mathrm{cm}^{2}$

Since there are 5 triangular pieces made of two different coloured cloths,
Area of each cloth required$=(5 \times 200 \sqrt{6}) \mathrm{cm}^{2}$
$=1000 \sqrt{6} \mathrm{cm}^{2}$

Q 7, Ex. 12.2, Page No 207 - Heron's Formula (NCERT Maths Class 9th)

Question 7:

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

Answer:

We know that
Area of square $=\frac{1}{2}$(diagonal)²
Area of the given kite $=\frac{1}{2}(32 \mathrm{cm})^{2}=512 \mathrm{cm}^{2}$
Area of 1^st shade = Area of 2^nd shade
$=\frac{512 \mathrm{cm}^{2}}{2}=256 \mathrm{cm}^{2}$

Therefore, the area of paper required in each shape is 256 cm².
For III^rd triangle
Semi-perimeter, $s=\frac{(6+6+8) \mathrm{cm}}{2}=10 \mathrm{cm}$
By Heron’s formula,
Area of triangle$=\sqrt{s(s-a)(s-b)(s-c)}$
Area of III^rd triangle$=\sqrt{10(10-6)(10-6)(10-8)}$
$=(\sqrt{10 \times 4 \times 4 \times 2}) \mathrm{cm}^{2}$

$=(4 \times 2 \sqrt{5}) \mathrm{cm}^{2}$

$=8 \sqrt{5} \mathrm{cm}^{2}$

$=(8 \times 2.24) \mathrm{cm}^{2}$

$=17.92 \mathrm{cm}^{2}$

Area of paper required for III^rd shade = 17.92 cm²

Q-8, Ex - 12.2 - Heron's Formula, Chapter 12, NCERT Class 9th 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 the given figure). Find the cost of polishing the tiles at the rate of 50p per cm².

Answer:

It can be observed that
Semi-perimeter of each triangular-shaped tile, $s=\frac{(35+28+9) \mathrm{cm}}{2}=36 \mathrm{cm}$
By Heron’s formula,
Area of triangle$=\sqrt{s(s-a)(s-b)(s-c)}$
Area of each tile $=[\sqrt{36(36-35)(36-28)(36-9)}] \mathrm{cm}^{2}$
$=[\sqrt{36 \times 1 \times 8 \times 27}] \mathrm{cm}^{2}$
$=36 \sqrt{6} \mathrm{cm}^{2}$
= (36 × 2.45) cm²
= 88.2 cm²
Area of 16 tiles = (16 × 88.2) cm²= 1411.2 cm²
Cost of polishing per cm² area = 50 p
Cost of polishing 1411.2 cm² area = Rs (1411.2 × 0.50) = Rs 705.60
Therefore, it will cost Rs 705.60 while polishing all the tiles.

Q 9, Ex. 12.2, Page No 207 - Heron's Formula (NCERT Maths Class 9th)

Question 9:

A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Answer:

Draw a line BE parallel to AD and draw a perpendicular BF on CD.
It can be observed that ABED is a parallelogram.
BE = AD = 13 m
ED = AB = 10 m
EC = 25 − ED = 15 m
For ΔBEC,
Semi-perimeter, $s=\frac{(13+14+15) \mathrm{m}}{2}$
=21m

By Heron’s formula,
Area of triangle$=\sqrt{s(s-a)(s-b)(s-c)}$
Area of ΔBEC $=[\sqrt{21(21-13)(21-14)(21-15)}] \mathrm{m}^{2}$
$=[\sqrt{21(8)(7)(6)}]$m²
= 84 m²

Area of ΔBEC $=\frac{1}{2} \times C E \times B F$
⇒ 84 = 12 × 15 × BF⇒ BF = 16815 = 11.2 m
Area of ABED = BF × DE = 11.2 × 10 = 112 m²
Area of the field = 84 + 112 = 196 m²

Sunday, May 31, 2020