S Chand ICSE Mathematics Class 9 Solution Sixteenth Chapter Area of Plane Figures Exercise 16B (O.P. Malhotra, S.K. Gupta, Anubhuti Gangal)

S Chand ICSE Mathematics Class 9 Solution Sixteenth Chapter Area of Plane Figures Exercise 16B

Exercise 16B

(1) For any triangle, complete the following table

  (i) (ii) (iii) (iv) (v) (vi) (vii)
Base 6 8 5 X 6 ? ?
Height 10 14 20 2x ? 5 11
area ? ? ? ? 30 50 110

 Solution:

(2) The area of a triangle is 6 cm2 and its base is 4 cm. Find its height.

Solution: Area = 6 cm2

Base = 4 cm

6 = ½ x 4 x height

Or, height = 3 cm.

(3) Find the base of a triangle is its

(i) area is 25 ares and height 20 m.

(ii) area is 16 hectares and height 40 decametres

1 are = 100 m2

1 hectare = 10000 m2

1 decameter = 10 m

Solution:

(4) Find the area of the triangle whose sides are

(i) 26 cm, 28 cm, 30 cm

(ii) 48 cm, 73 cm, 55 cm

(iii) 21 cm, 20 cm, 13 cm

(iv( 7.5 cm, 18 cm, 19.5 cm

Solution:  

(i) a = 26 cm, b = 28 cm, c = 30 cm

S = a + b + c / 2

= 26 + 28 + 30 / 2

= 84 / 2

= 42

Area = √42 (42 – 36) (42 – 28) (42 – 30)

= √42 x 16 x 14 x 12

= √112896

= √336

(ii) a = 48 cm, b = 73 cm, c = 55 cm

S = a + b + c / 2

= 48 + 73 + 55 / 2

= 176 /2

= 88

∴ Area = √88(88 – 48) (88 – 73) (88 – 55)

= √88 x 40 x 15 x 33 cm2

=√1742400

= 1320 cm2

(iii) a = 21 cm, b = 20 cm, c = 13 cm

S = a +_ b + c / 2

= 21 + 20 + 13 / 2

= 54 / 2

= 27

∴ Area =-√27 (27 – 21) (27 – 20) (27 – 13)

= √27 x 6 x 7 x 14

= √15876

= 126

(iv)

(5) The perimeter of a triangle is 540 m and its sides are in the ration 25 : 17 : 12. Find the area of the triangle.

Solution: Hence, the sides of the triangular = 25x, 17x, 12x

Perimeter = (25x + 17x + 12x)

= 54x

∴ 54x = 540

x = 10

∴ Sides are = 250, 170, 120

(6) The given figure, ABCD represents a square of side  6 cm. F is a point on DC such that the area of the triangle ADF is one third of the area of the square. Find the length of FD.

Solution:

Area of square = 6 x 6 = 36 cm2

Area of ΔADF = 1/3 x 36 = 12 cm2

∴ ½ x DF x 6 = 12

DF = 4 cm

(7) Find the area of a triangle with base 5 cm and whose height is equal to that of a rectangle with base 5 cm and area 20 cm2

Solution:

Let, l be the length of the rectangle,

∴ l x 5 = 20

l = 4

∴ Area of triangle = ½ x 5 x 4

= 10 cm2

(8) Answer true or false:

(i) The area of a triangle with base 4 cm and perp. Height 6 cm is 24 cm2

(ii) The area of a triangle with sides measuring, a, b, c, is given by √s(s – a) (s – b) (s – c) where s is the perimeter of the triangle.

(iii) The area of a rectangle and a triangle are equal if they stand on the same base and are between the same parallels.

Solution: (i) base = 4 cm, height = 6 cm

Area = ½ x 4 x 6

= 12 cm2

False

(ii) False

S is semi-perimeter not perimeter.

(iii) False,

Area of triangle = ½ x Area of rectangle.

(9) ABC is triangle in which AB = AC = 4 cm and ∠A = 90o, calculate the area of ΔABC.

Solution:

∴ Area of ΔABC = 1/2 x 4 x 4

= 8 Cm2

(11) The sides of a triangular field are 975 m, 1050 m and 1125 m . If this field is sold at the rate of Rs. 1000 per hectare, find its selling price.

Solution:

(12) Find the area of the equilateral triangle whose each side is (i) 12 cm  (ii) 5 cm

Solution:

(13) The perimeter of an equilateral triangle is 24 cm. Find its area.

Solution:

Let ‘a’ be the side of the equilateral triangle,

3a = 24

a = 8

Area = √3/4 x 82

= √3/4 x 64

= 16√3

= 27.712 cm2

(14) The perimeter of an equilateral triangle is √3 times its area. Find the length of each side.

Solution: Let, a be the length of each side,

Perimeter = √3 x area

3a = √3 x √3/4 a2

Or, a = 4

∴ length = 4 unit.

(15) The area of an equilateral triangle is 173.2 m2. Find its perimeter.

Solution: Let, a be the side

√3/4 a= 173.2

Or, a2 = 173.2 x 4 / √3

a =20 m

Perimeter = 3 x 20 = 60 m

(16) Find the area of the isosceles triangle whose

(i) each of the equal sides is 8 cm  and the base is 9 cm

(ii) each of the equal sides is 10 cm and the base is 12 cm

(iii) each of the equal sides is 7.4  cm and the base is 6.2 cm;

(iv) Perimeter is 11 cm and te base is 4 cm.

Solution:

∴ Area = 1/2 x 4 x √33/2

= √33 cm2

= 5.74 cm2

(17) (i) The base of an isosceles triangle is 24 cm and its area is 192 sq. cm. Find its perimeter.

(ii) Find the base of an isosceles triangle whose area is 12 cm2 and one of the equal sides is 5 cm.

Solution: 

(18) The perimeter of an isosceles triangle is 42 cm. Its base is 2/3 times the sum of equal sides. Find the length of each side and the area of the triangle.

Solution:

(19) PQR is an isosceles triangle whose equal sides PQ and PR are 13 cm each, and the base QR measures 10 cm. PS is the perpendicular from P to QR and O is a point on PS such that ∠QOR = 900. Find the area of shaded region.

Solution: 

(20) The perimeter of a right triangle is 50 cm and the hypotenuse is 18 cm. Find its area.

Solution: 


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