﻿ Class 9 NCERT Math Solution
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Exercise - 12.2

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

Area of ∆BCD = 1/2 x BC x CD = 1/2 x 12 x 5 = 30 m²

For ∆ABD,
s = (a + b + c)/2
s = (9 + 8 + 13)/2
s = 30/2
s = 15cm
s - a = 15 - 9 = 6cm
s - b = 15 - 8 = 7cm
s - c = 15 - 13 = 2cm
area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √15 x 6 x 7 x 2
= 6√35
= 6 x 5.916
= 35.496 m²

Area of the park = Area of ∆ABD + Area of ∆BCD
= 35.496 + 30
= 65.496
= 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.

Solution :-
```
For ∆ABC,
AB = a = 3 cm
BC = b = 4 cm
AC = c = 5 cm
s = (a + b + c)/2
s = (3 + 4 + 5)/2
s = 12/2
s = 6cm
s - a = 6 - 3 = 3cm
s - b = 6 - 4 = 2cm
s - c = 6 - 5 = 1cm
area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √6 x 3 x 2 x 1
= √36
= 6 cm²

For ∆ADC,
AD = a = 5 cm
DC = b = 4 cm
AC = c = 5 cm
s = (a + b + c)/2
s = (5 + 4 + 5)/2
s = 14/2
s = 7cm
s - a = 7 - 5 = 2cm
s - b = 7 - 4 = 3cm
s - c = 7 - 5 = 2cm
area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √7 x 2 x 3 x 2
= √84
= 2√21
= 2 x 4.583
= 9.166 cm²

Area of ABCD = Area of ∆ABC + Area of ∆ACD
= (6 + 9.166)
= 15.166
= 15.2 cm² (approximately)
```

Question-3 :-  Radha made a picture of an aeroplane with coloured paper as shown in Figure. Find the total area of the paper used.

Solution :-
```  For triangle I

This triangle is an isosceles triangle.
Perimeter = 2s = (5 + 5 + 1) cm = 11cm
2s = 11cm
s = 11/2
s = 5.5 cm
s - a = 5.5 - 5 = 0.5 cm
s - b = 5.5 - 5 = 0.5 cm
s - c = 5.5 - 1 = 4.5 cm
area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √5.5 x 0.5 x 0.5 x 4.5
= 0.75√11
= 0.75 x 3.317
= 2.488 cm² (approx)

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 = √(1)² - (0.5)²
= √0.75 cm
= 0.866 cm
Area = Area of parallelogram + Area of equilateral triangle
= 0.866 x 1 + √3/4 x (1)²
= 0.866 + 0.433
= 1.299 cm²

Area of triangle (IV) = Area of triangle in (V)
= 1/2 x 1.5 x 6
= 4.5 cm²

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

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 :-
```  For Triangle,
The sides of triangle are 26 cm, 28 cm and 30 cm,
s = (a + b + c)/2
s = (26 + 28 + 30)/2
s = 84/2
s = 42 cm
s - a = 42 - 26 = 16cm
s - b = 42 - 28 = 14cm
s - c = 42 - 30 = 12cm
area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √42 x 16 x 14 x 12
= 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.
```

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 :-
```  Let ABCD be a rhombus-shaped field.

For ∆BCD,
BD = a = 48m
BC = b = 30m
CD = c = 30m
s = (a + b + c)/2
s = (48 + 30 + 30)/2
s = 108/2
s = 54 cm
s - a = 54 - 48 = 6m
s - b = 54 - 30 = 24m
s - c = 54 - 30 = 24m
area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √54 x 6 x 24 x 24
= 432 m²

Area of field = 2 × Area of ∆BCD
= (2 × 432) m²
= 864 m²

Area for grazing for 1 cow = 864/18 =  48 m²
Each cow will get 48 m² area of grass field.
```

Question-6 :-  An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig.), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Solution :-
```  For each triangular piece,
a = 20 cm, b = 50 cm, c = 50 cm
s = (a + b + c)/2
s = (20 + 50 + 50)/2
s = 120/2
s = 60 cm
s - a = 60 - 20 = 40 cm
s - b = 60 - 50 = 10 cm
s - c = 60 - 50 = 10 cm
area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √60 x 40 x 10 x 10
= 200√6 cm²

Since there are 5 triangular pieces mad of two different  coloured  cloths ,
Area of each cloth required = 5 x 200√6 cm² = 1000√6 cm²
```

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 Fig. 12.17. How much paper of each shade has been used in it?

Solution :-
```  Area of square = 1/2 x (diagonal)²
= 1/2 x (32)²
= 512 cm²

Area of 1st shade = Area of 2nd shade
= 512/2 = 256 cm²
Therefore, the area of paper required in each shape is 256 cm².

For III rd triangle
a = 6 cm, b = 6 cm, c = 8 cm
s = (a + b + c)/2
s = (6 + 6 + 8)/2
s = 20/2
s = 10 cm
s - a = 10 - 6 = 4 cm
s - b = 10 - 6 = 4 cm
s - c = 10 - 8 = 2 cm
area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √10 x 4 x 4 x 2
= 8√5
= 8 x 2.24
= 17.92 cm²

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

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. Find the cost of polishing the tiles at the rate of 50p per cm².

Solution :-
```  a = 9cm, b = 28cm, c = 35cm
s = (a + b + c)/2
s = (9 + 28 + 35)/2
s = 72/2
s = 36 cm
s - a = 36 - 9 = 27 cm
s - b = 36 - 28 = 8 cm
s - c = 36 - 35 = 1 cm
area of the triangle = √s x (s-a) x (s-b) x (s-c)
= √36 x 27 x 8 x 1
= 36√6
= 36 x 2.45
= 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.
```

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.

Solution :-
```  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
Area of the field = 84 + 112 = 196 m²

For ∆BEC,
s = (a + b + c)/2
s = (13 + 14 + 15)/2
s = 42/2
s = 21 m
s - a = 21 - 13 = 8 m
s - b = 21 - 14 = 7 m
s - c = 21 - 15 = 6 m
area of the triangle ∆BEC = √s x (s-a) x (s-b) x (s-c)
= √21 x 8 x 7 x 6
= 84 m²

Now, Area of ∆BEC = 1/2 x CE x BF
84 = 1/2 x 15 x BF
BF = 168/15
BF = 11.2 m

Area of ABED = BF × DE = 11.2 × 10 = 112 m²
Area of the field = 84 + 112 = 196 m²
```
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