Chapter 12 Heron's Formula RD Sharma Solutions Exercise 12.1 Class 9 Maths
Chapter Name  RD Sharma Chapter 12 Heron's Formula Exercise 12.1 
Book Name  RD Sharma Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 12.1 Solutions
1. Find the area of a triangle whose sides are respectively 150 cm, 120 cm and 200 cm.
Solution
The triangle whose sides are
a = 150 cm
b = 120 cm
c = 200 cm
2. Find the area of a triangle whose sides are 9 cm, 12 cm and 15cm.
Solution
The triangle whose sides are a = 9 cm, b = 12cm and c = 15 cm
3. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Solution
21√11 cm^{2}
Solution
The triangle sides are
5. The perimeter of a triangular field is 540m and its sides are in the ratio 25 : 17 :12. Find the area of the triangle.
Solution
The sides of a triangle are in the ratio 25: 17 : 12
Let the sides of a triangle are a = 25x, b = 17x and c = 12x say.
Let the sides of a triangle are a = 25x, b = 17x and c = 12x say.
Perimeter = 25 = a + b + c = 540 cm
6. The perimeter of a triangle is 300 m. If its sides are in the ratio 3 : 5 : 7. Find the area of the triangle.
Solution
Given that
The perimeter of a triangle = 300 m
The sides of a triangle in the ratio 3 : 5 : 7
Let 3x, 5x, 7x be the sides of the triangle
Let 3x, 5x, 7x be the sides of the triangle
Perimeter ⇒ 2x = a + b + c
⇒ 3x + 5x +7x = 300
⇒ 15x = 300
⇒ x =20m
The triangle sides are a = 3x
= 3(20)m = 60 m
b = 5x = 5(20) m = 100 m
c = 7x = 140 m
Semi perimeter s = (a + b + c)/2
= 300/2 = 150m
⇒ 3x + 5x +7x = 300
⇒ 15x = 300
⇒ x =20m
The triangle sides are a = 3x
= 3(20)m = 60 m
b = 5x = 5(20) m = 100 m
c = 7x = 140 m
Semi perimeter s = (a + b + c)/2
= 300/2 = 150m
7. The perimeter of a triangular field is 240 dm. If two of its sides are 78 dm and 50dm, find the length of the perpendicular on the side of length 50 dm from the opposite vertex.
Solution
ABC be the triangle, Here a = 78 dm = AB,
BC = b = 50 dm
8. A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest of its altitudes.
Solution
The sides of a triangle are a = 35 cm, b = 54 cm and c = 61 cm
Now, perimeter a + b + c = 25
9. The lengths of the sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 144 cm. Find the area of the triangle and the height corresponding to the longest side.
Solution
Let the sides of a triangle are 3x, 4x and 5x.
Now, a = 3x, b = 4x and c = 5x
The perimeter 2s = 144
The perimeter 2s = 144
10. The perimeter of an isosceles triangle is 42 cm and its base is (3/2) times each of the equal sides. Find the length of each side of the triangle, area of the triangle and the height of the triangle.
Solution
Let 'x' be the measure of each equal sides
11. Find the area of the shaded region in Fig. Below.
Solution
Area of shaded region = Area of Î”ABC  Area of Î”ADB
Now in Î”ADB
⇒ AB^{2} = AD^{2} + BD^{2} ...(i)
⇒ Given that AD = 12 cm, BD = 16 cm
Substituting the values of AD and BD in the equation (i), we get