RD Sharma 2020 solution class 9 chapter 17 Heron's Formula Exercise 17.1

Exercise 17.1

Page-17.8

Question 1:

Find the area of a triangle whose sides are respectively 150 cm, 120 cm and 200 cm.

Answer 1:

Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle.

If we denote area of the triangle by A, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by;

Where,

We are given:

a =150 cm

b=120 cm

c =200 cm

Here we will calculate s,

So the area of the triangle is:

Question 2:

Find the area of a triangle whose sides are 9 cm, 12 cm and 15 cm.

Answer 2:

Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle.

If we denote area of the triangle by A, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by;

Where,

We are given:

a = 9 cm, b = 12 cm, c = 15 cm

Here we will calculate s,

So the area of the triangle is:

Question 3:

Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

Answer 3:

Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle.

If we denote area of the triangle by A, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by;

Where,

We are given:

a = 18 cm

b = 10 cm, and perimeter = 42 cm

We know that perimeter = 2s, 

So 2s = 42

Therefore s = 21 cm

We know that, so

So the area of the triangle is:

Question 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.

Answer 4:

Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle.

If we denote area of the triangle by ‘Area’, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by;

Where,
We are given:

AB = 15 cm, BC = 13 cm, AC = 14 cm

Here we will calculate s,

So the area of the triangle is:

Now draw the altitude from point B on AC which intersects it at point D.BD is the required altitude. So if you draw the figure, you will see, 

Here . So,

Question 5:

The perimeter of a triangular field is 540 m and its sides are in the ratio 25 : 17 : 12. Find the area of the triangle.

Answer 5:

Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle. If we denote area of the triangle by A, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by;

Where,

We are given, and

Here,

Using these data we will find the sides of the triangle. Suppose the sides of the triangle are as follows,

Since, so

Now we know each side that is,

Now we know all the sides. So we can use Heron’s formula.

The area of the triangle is;

Question 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.

Answer 6:

Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle. If we denote area of the triangle by A, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by;

Where,

We are given, and

Here,

Using these data we will find the sides of the triangle. Suppose the sides of the triangle are as follows,

Since, so

Now we know each side that is,

Now we know all the sides. So we can use Heron’s formula.

The area of the triangle is;

$$\begin{gathered} \mathrm{A}=\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)} \\ =\sqrt{150\left(150-60\right)\left(150-100\right)\left(150-140\right)} \\ =\sqrt{150\left(90\right)\left(50\right)\left(10\right)} \\ =100\sqrt{15\times 9\times 5} \\ =100\sqrt{5\times 3\times 3\times 3\times 5} \\ =100\times 3\times 5\sqrt{3} \\ =\boxed{1500\sqrt{3} {\mathrm{m}}^2} \end{gathered}$$

Question 7:

The perimeter of a triangular field is 240 dm. If two of its sides are 78 cm and 50 dm, find the length of the perpendicular on the side of length 50 dm from the opposite vertex.

Answer 7:

Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle. If we denote area of the triangle by A, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by;

Where,  

We are given two sides of the triangle and.

That is a = 78 dm, b = 50 dm

We will find third side c and then the area of the triangle using Heron’s formula.

Now,

Use Heron’s formula to find out the area of the triangle. That is 

 

Consider the triangle ΔPQR in which 

PQ=50 dm, PR=78 dm, QR=120 dm

Where RD is the desired perpendicular length

Now from the figure we have

Question 8:

A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest of its altitudes.

Answer 8:

Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle.

If we denote area of the triangle by A, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by;

Where,

We are given: a = 35 cm; b = 54 cm; c = 61 cm

The area of the triangle is:

Suppose the triangle is ΔPQR and focus on the triangle given below,

In which PD1, QD2 and RD3 are three altitudes

Where PQ=35 cm, QR=54 cm, PR=61 cm

We will calculate each altitude one by one to find the smallest one.

Case 1 

In case of ΔPQR:

Case 2

Case 3

The smallest altitude is QD2.

The smallest altitude is the one which is drawn on the side of length 61 cm from apposite vertex.

Question 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.

Answer 9:

Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle.

If we denote area of the triangle by A, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by;

Where,

We are given, and

Here,

Using these data we will find the sides of the triangle. Suppose the sides of the triangle are as follows,

Since 2s=144, so

Now we know each side that is,

Now we know all the sides. So we can use Heron’s formula.

The area of the triangle is;

We are asked to fin out the height corresponding to the longest side of the given triangle. The longest side is c and supposes the corresponding height is H then,

Question 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.

Answer 10:

We are given that and its base is (3/2) times each of the equal sides. We are asked to find out the length of each side, area of the triangle and height of the triangle. In this case ‘height’ is the perpendicular distance drawn on the base from the apposite vertex. 

In the following triangle ΔABC 

BC = a, AC = b, AB = c and AB = AC

Let the length of each of the equal sides be x and a, b and c are the side of the triangle. So,

Since .This implies that,

Therefore all the sides of the triangle are:

All the sides of the triangle are 18 cm, 12 cm, and 12 cm.

Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle.

If we denote area of the triangle by Area, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by;

Where,

To calculate area of the triangle we need to find s:

The area of the triangle is:

Now we will find out the height, say H. See the figure, in which AD = H

So,

Question 11:

Find the area of the shaded region in the given figure.

Answer 11:

We are given the following figure with dimensions.

Figure: 

Let the point at which angle is be D.

AC = 52 cm, BC = 48 cm, AD = 12 cm, BD = 16 cm

We are asked to find out the area of the shaded region.

Area of the shaded region=Area of triangle ΔABC−area of triangle ΔABD 

In right angled triangle ABD, we have

Area of the triangle ΔABD is given by

Whenever we are given the measurement of all sides of a triangle, we basically look for Heron’s formula to find out the area of the triangle.

If we denote area of the triangle by Area, then the area of a triangle having sides a, b, c and s as semi-perimeter is given by;

Where,

Here a = 48 cm, b = 52 cm, c = 20 cm and

Therefore the area of a triangle ΔABC is given by,

Now we have all the information to calculate area of shaded region, so

Area of shaded region = Area of ΔABC − Area of ΔABD

The area of the shaded region is 384 cm².