RD Sharma Solutions Class 9 Herons Formula Exercise 12.1

RD Sharma Class 9 Solutions Chapter 12 Ex 12.1 Free Download

RD Sharma Solutions Class 9 Chapter 12 Ex 12.1

Q1 . Find  the area of a triangle whose sides are respectively 150 cm, 120 cm and 200 cm.

Solution:

Say, the sides of the given triangle be a, b, c respectively.

Given here,

a = 150 cm

b = 120 cm

c = 200 cm

From Heron’s Formula;

The area of the triangle = \(\sqrt{s\times (s-a)\times (s- b)\times (s-c)}\)

Semi perimeter of a triangle = s

2s = a + b + c

s = \(\frac{(a+b+c)}{2}\)

s = \(\frac{(150+200+120)}{2}\)

s = 235 cm

Hence, area of the triangle = \(\sqrt{s\times (s-a)\times (s- b)\times (s-c)}\)

= \(\sqrt{235\times (235-150)\times (235-200)\times (235-120)}\)

= 8966.56 sq.cm

Q2. Find  the area of a triangle whose sides are respectively 9 cm, 12 cm and 15 cm.

 Solution:

Say, the sides of the given triangle be a, b, c respectively.

So given,

a = 9 cm

b = 12 cm

c = 15 cm

From Heron’s Formula;

The area of the triangle = \(\sqrt{s\times (s-a)\times (s- b)\times (s-c)}\)

Semi perimeter of a triangle = s

2s = a + b + c

s = \(\frac{(a+b+c)}{2}\)

s = \(\frac{(9+12+15)}{2}\)

s = 18 cm

Hence, area of the triangle = \(\sqrt{s\times (s-a)\times (s- b)\times (s-c)}\)

= \(\sqrt{18\times (18-9)\times (18-12)\times (18-15)}\)

= \(54cm^{2}\)

 

Q3. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42cm.

Solution:

If we are given with measurements of all sides of a triangle, we apply Heron’s formula to find out the area of the triangle.

Let us denote area of the triangle by A, and the sides of the triangle as a, b, c and s as its semi-perimeter. Thus;

A = \(\sqrt{s\times (s-a)\times (s- b)\times (s-c)}\)

Where, s = \(\frac{(a+b+c)}{2}\)

Already given;

a = 18 cm

b = 10 cm, and perimeter = 42 cm

We know, perimeter = 2s,

So, 2s = 42

So, s = 21 cm

We know , s = \(\frac{(a+b+c)}{2}\)

21 = \(\frac{(18+10+c)}{2}\)

42 = 28 + c

c = 14 cm

Therefore, the area of the triangle is:

A = \(\sqrt{21\times (21-18)\times (21-10)\times (21-14)}\)

A = \(\sqrt{21\times (3)\times (11)\times (7)}\)

A = \(\sqrt{4851}\)

A = \(21\sqrt{11}cm^{0}\)

Q4 . In a triangle ABC, AB = 15cm, BC = 13cm and AC = 14cm. Find the area of triangle ABC and hence its altitude on AC.

Solution:

Say, the sides of the given triangle be AB = a, BC = b, AC =  c respectively.

Given in the question;

a = 15 cm

b = 13 cm

c = 14 cm

From Heron’s Formula;

The area of the triangle = \(\sqrt{s\times (s-a)\times (s- b)\times (s-c)}\)

Semi perimeter of a triangle= 2s

2s = a + b + c

s = \(\frac{(a+b+c)}{2}\)

s = \(\frac{(15+13+14)}{2}\)

s = 21 cm

Hence, area of the triangle = \(\sqrt{s\times (s-a)\times (s- b)\times (s-c)}\)

= \(\sqrt{21\times (21-15)\times (21-13)\times (21-14)}\)

= \(84cm^{2}\)

Let, BE is a perpendicular on AC

Now, area of triangle = \(84cm^{2}\)

\(\frac{1}{2}\times BE \times AC\) = \(84cm^{2}\)

BE = 12cm

Hence, the length of BE is 12 cm

Q5 . The perimeter of a triangular field is 540 m and its sides are in the ratio 25:17:12. Find the area of triangle.

Solution:

Say, the sides of the given triangle be a = 25x, b = 17x, c = 12x respectively,

So,

a = 25x cm

b = 17x cm

c = 12x cm

Given, Perimeter of triangle = 540 cm

2s = a + b + c

a + b + c = 540 cm

25x + 17x + 12x = 540 cm

54x = 540 cm

x = 10 cm

So, the sides of a triangle are

a = 250 cm

b = 170 cm

c = 120 cm

Since, Semi perimeter s = \(\frac{(a+b+c)}{2}\)

= \(\frac{540}{2}\)

= 270 cm

From Heron’s Formula;

The area of the triangle = \(\sqrt{s\times (s-a)\times (s-b)\times (s-c)}\)

= \(\sqrt{270\times (270-250)\times (270-170)\times (270-120)}\)

= \(9000cm^{2}\)

Hence, the area of the triangle is \(9000cm^{2}\)

Q6. The perimeter of a triangle is 300 m. If its sides are in the ratio of 3: 5: 7. Find the area of the triangle.

Solution:

Given, the perimeter of a triangle is 300 m and the sides are in a ratio of 3: 5: 7

Say, the sides a, b, c of a triangle be 3x, 5x, 7x respectively

So, the perimeter = 2s = a + b + c

200 = a + b + c

300 = 3x + 5x + 7x

300 = 15x

So, x = 20 m

Now, the respective sides are;

a = 60 m

b = 100 m

c = 140 m

Since, semi perimeter s = \(\frac{a+b+c}{2}\)

= \(\frac{60+100+140}{2}\)

= 150 m

From Heron’s Formula;

The area of a triangle = \(\sqrt{s\times (s-a)\times(s-b) \times(s-c) }\)

= \(\sqrt{150\times (150-60)\times(150-100) \times(150-140) }\)

= \(1500\sqrt{3}m^{2}\)

Hence, the area of a triangle is \(1500\sqrt{3}m^{2}\)

Q7. The perimeter of a triangular field is 240 dm. If two of its sides are 78 dm and 50 dm, find the length of the perpendicular on the side of length 50 dm from the opposite vertex.

Solution:

Given here,

In a △ABC, a = 78 dm = AB, b = 50 dm = BC

Now, Perimeter of triangle = 240 dm

As, AB + BC + AC = 240 dm

78 + 50 + AC = 240

AC = 240-(78+50)

AC = 112 dm = c

Since, 2s = a + b + c

2s = 78 + 50 + 112

s = 120 dm

Area of a △ABC = \(\sqrt{s\times (s-a)\times(s-b) \times(s-c) }\)

= \(\sqrt{120\times (120-78)\times(120-50) \times(120-112) }\)

= \(1680dm^{2}\)

Assume, AD be a perpendicular on BC

Area of the △ABC = \(\frac{1}{2}\times AD\times BC\)

\(\frac{1}{2}\times AD\times BC\)  = \(1680dm^{2}\)

therefore, AD = 67.2 dm

Q8. A triangle has sides 35 cm, 54 cm, 61 cm long. Find its area. Also, find the smallest of its altitudes?

Solution:

Given here;

The sides of the triangle are;

a = 35 cm

b = 54 cm

c = 61 cm

Since, Perimeter of triangle, 2s = a + b + c

2s = 35 + 54 + 61 cm

So, Semi perimeter s = 75 cm

From Heron’s Formula;

Area of the triangle = \(\sqrt{s\times (s-a)\times(s-b) \times(s-c) }\)

= \(\sqrt{75\times (75-35)\times(75-54) \times(75-61) }\)

= \(939.14cm^{2}\)

As we know, the altitude will be smallest, provided the side corresponding to this altitude is longest.

Thus, the longest side = 61 cm

Area of the triangle = \(\frac{1}{2}\times h\times 61\)

\(\frac{1}{2}\times h\times 61\)  = \(939.14cm^{2}\)

h = 30.79cm

Therefore, the length of the smallest altitude is 30.79cm

Q9.The lengths of the sides of a triangle are in a ratio of 3 : 4 : 5 and its perimeter is 144 cm. Find the area of the triangle and the height corresponding to the longest side?

Solution:

Given, the perimeter of a triangle is 160m and the sides are in a ratio of  3 : 4 : 5

Say, the sides a, b, c of a triangle be 3x, 4x, 5x respectively

Since, the perimeter = 2s = a + b + c

144 = a + b + c

144 = 3x+ 4x+ 5x

So, x = 12cm

Hence, the respective sides are;

a = 36cm

b = 48cm

c = 60cm

Since, semi perimeter s = \(\frac{a+b+c}{2}\)

= \(\frac{36+48+60}{2}\)

= 72 cm

From Heron’s Formula;

The area of a triangle = \(\sqrt{s\times (s-a)\times(s-b) \times(s-c) }\)

= \(\sqrt{72\times (72-36)\times(72-48) \times(72-60) }\)

= \(864cm^{2}\)

Hence, the area of a triangle is \(864cm^{2}\)

The altitude will be smallest, provided the side corresponding to this altitude is longest.

The longest side = 60 cm

Area of the triangle = \(\frac{1}{2}\times h\times 60\)

\(\frac{1}{2}\times h\times 60\)  = \(864cm^{2}\)

h = 28.8 cm

Therefore, the length of the smallest altitude is 28.8 cm

Q10. The perimeter of an isosceles triangle is 42 cm and its base is \(\frac{3}{2}\) times each of the equal side. Find the length of each of the triangle, area of the triangle and the height of the triangle.

Solution:

Assume ‘x’ be the length of two equal sides,

Therefore the base = \(\frac{1}{2}\times x\)

Say, the sides a, b, c of a triangle be \(\frac{1}{2}\times x\), x and x respectively

Since, the perimeter = 2s = a + b + c

42 = a + b + c

42 = \(\frac{3}{2}\times x\) + x+ x

Therefore, x = 12 cm

Hence, the respective sides are;

a = 12 cm

b = 12 cm

c = 18 cm

Since, semi perimeter s = \(\frac{a+b+c}{2}\)

= \(\frac{12+12+18}{2}\)

= 21 cm

From Heron’s Formula;

The area of a triangle = \(\sqrt{s\times (s-a)\times(s-b) \times(s-c) }\)

= \(\sqrt{21\times (21-12)\times(21-12) \times(21-18) }\)

= \(71.42cm^{2}\)

So, the area of a triangle is \(71.42cm^{2}\)

The altitude will be smallest provided the side corresponding to this altitude is longest.

The longest side = 18 cm

Area of the triangle = \(\frac{1}{2}\times h\times 18\)

\(\frac{1}{2}\times h\times 18\)  = \(71.42cm^{2}\)

h = 7.93 cm

Therefore, the length of the smallest altitude is 7.93 cm

Q11. Find the area of the shaded region in fig. below

Find the area of shaded region

Solution:

From the figure,

Area of the shaded region = \(Area\;of\;\Delta ABC-Area\;of\;\Delta ADB\)

In △ADB,

\(AB^{2} = AD^{2}+BD^{2}\)………(i)

Given, AD = 12 cm, BD =16 cm

Putting the value of AD and BD in eq (i), we get;

\(AB^{2} = 12^{2}+16^{2}\)

= 400\(cm^{2}\)

AB = 20 cm

As, area of a triangle = \(\frac{1}{2}\times AD\times BD\)

= \(96 cm^{2}\)

In △ABC,

s = \(\frac{1}{2}\times (AB+BC+CA)\)

= \(\frac{1}{2}\times (52+48+20)\)

= 60 cm

From Heron’s Formula;

The area of a triangle = \(\sqrt{s\times (s-a)\times(s-b) \times(s-c) }\)

= \(\sqrt{60\times (60-20)\times(60-48) \times(60-52) }\)

= \(480cm^{2}\)

Thus, the area of a triangle is 480 sq. cm.

Since, Area of shaded region = Area of triangle ABC – Area of triangle ADB

= (480 – 96) cm2

= 384 cm2

Therefore, Area of shaded region = 384 cm2

2 Comments

  1. useful, awesome, fantastic ,mind blowing ,topnotch, superb ,marvelous ,excellent.

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