RD Sharma Solutions for Class 9 Chapter 12 Heron's Formula Exercise 12.1

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	Exercise 12.2
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 cm²

4. In a ΔABC, AB = 15 cm, BC = 13 cm and AC = 14 cm. Find the area of ΔABC and hence its altitude on AC.

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. 

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

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

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 

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 ΔABC - Area of ΔADB 

Now in ΔADB

⇒ AB² = AD²  + BD² ...(i)

⇒ Given that AD = 12 cm,  BD = 16 cm

Substituting the values of AD and BD in the equation (i), we get