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

MCQS

Page-17.24

Question 1:

Mark the correct alternative in each of the following:

The sides of a triangle are 16 cm, 30 cm, 34 cm. Its area is

(a) 225 cm²

(b) 240 cm²

(c) $225\sqrt{2}$ cm²

(d) 450 cm²

Answer 1:

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
, where

Therefore the area of a triangle say A, having sides 16 cm, 30 cm and 34 cm is given by

a = 16 cm ; b = 30 cm ; c = 34 cm






Therefore the area of the triangle is 

Hence, the correct option is (b).

Question 2:

The base of an isosceles right triangle is 30 cm. Its area is

(a) 225 cm²

(b) 225 $\sqrt{3}$ cm²

(c) 225 $\sqrt{2}$ cm²

(d) 450 cm²

Answer 2:

$$\begin{gathered} \mathrm{Let} \mathrm{ABC} \mathrm{be} \mathrm{the} \mathrm{right} \mathrm{triangle} \mathrm{in} \mathrm{which} \angle \mathrm{B}=90°. \\ \mathrm{Now}, \mathrm{base}=\mathrm{BC}; \mathrm{perpendicular}=\mathrm{AB}; \mathrm{Hypotenuse}=\mathrm{AC} \\ \mathrm{Now}, \mathrm{BC}=30 \mathrm{cm} \left(\mathrm{given}\right) \\ \mathrm{Now}, ∆\mathrm{ABC} \mathrm{is} \mathrm{an} \mathrm{isosceles} \mathrm{right} \mathrm{angled} ∆ \mathrm{and} \mathrm{we} \mathrm{know} \mathrm{that} \mathrm{hypotenuse} \mathrm{is} \mathrm{the} \mathrm{longest} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{right} ∆. \\ \mathrm{So}, \mathrm{AB}=\mathrm{BC}=30 \mathrm{cm} \\ \mathrm{area} \mathrm{of} ∆\mathrm{ABC}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height} \\ =\frac{1}{2}\times \mathrm{BC}\times \mathrm{AB} \\ =\frac{1}{2}\times 30\times 30 \\ =450 {\mathrm{cm}}^2 \end{gathered}$$


Hence, the correct option is (d).

Question 3:

The sides of a triangle are 7 cm, 9 cm and 14 cm. Its area is

(a) $12\sqrt{5} c{m}^2$

(b) $12\sqrt{3} c{m}^2$

(c) $24\sqrt{5} c{m}^2$

(d) $63 c{m}^2$

Answer 3:

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
, where

Therefore the area of a triangle having sides 7 cm, 9 cm and 14 cm is given by
a = 7 cm ; b = 9 cm ; c = 14 cm


Therefore the answer is (a).

Question 4:

The sides of a triangular field are 325 m, 300 m and 125 m. Its area is

(a) 18750 m²

(b) 37500 m²

(c) 97500 m²

(d) 48750 m²

Answer 4:

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
, where


Therefore the area of a triangular field, say A having sides 325 m, 300 m and 125 m is given by
a = 325 m ; b = 300 m ; c = 125 m


Therefore, the correct answer is (a).

Question 5:

The sides of a triangle are 50 cm, 78 cm and 112 cm. The smallest altitude is

(a) 20 cm

(b) 30 cm

(c) 40 cm

(d) 50 cm

Answer 5:

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
, where

Therefore the area of a triangle, say A having sides 50 cm, 78 cm and 112 cm is given by



The area of a triangle, having p as the altitude will be,
$\mathrm{Area} = \frac{1}{2}\times \mathrm{base}\times \mathrm{height}$
Where, A = 1680
We have to find the smallest altitude, so will substitute the value of the base AC with the length of each side one by one and find the smallest altitude distance i.e. p
Case 1 

Case 2

Case 3

Therefore, the answer is (b).

Question 6:

The sides of a triangle are 11 m, 60 m and 61 m. The altitude to the smallest side is

(a) 11 m

(b) 66 m

(c) 50 m

(d) 60 m

Answer 6:

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
, where

We need to find the altitude to the smallest side

Therefore the area of a triangle having sides 11 m, 60 m and 61 m is given by
a = 11 m ; b = 60 m ; c = 61 m



The area of a triangle having base AC and height p is given by

We have to find the height p corresponding to the smallest side of the triangle. Here smallest side is 11 m 
AC = 11 m


Therefore, the answer is (d).

Question 7:

The sides of a triangle are 11 cm, 15 cm and 16 cm. The altitude to the largest side is

(a) $30\sqrt{7} cm$

(b) $\frac{15\sqrt{7}}{2}cm$

(c) $\frac{15\sqrt{7}}{4}cm$

(d) 30 cm

Answer 7:

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
, where

We need to find the altitude corresponding to the longest side

Therefore the area of a triangle having sides 11 cm, 15 cm and 16 cm is given by
a = 11 m ; b = 15 cm ; c = 16 cm



The area of a triangle having base AC and height p is given by

We have to find the height p corresponding to the longest side of the triangle.Here longest side is 16 cm, that is AC=16 cm 


Therefore, the answer is (c).

Question 8:

The base and hypotenuse of a right triangle are respectively 5 cm and 13 cm long. Its area is

(a) 25 cm²

(b) 28 cm²

(c) 30 cm²

(d) 40 cm²

Answer 8:

In right angled triangle ABC having base 5 cm and hypotenuse 13 cm we are asked to find its area
Using Pythagorean Theorem

Where, AB = hypotenuse = 13 cm, AC = Base = 5 cm, BC = Height

Area of a triangle, say A having base 5 cm and altitude 12 cm is given by

Where, Base = 5 cm; Height = 12 cm

Therefore, the answer is (c).

Question 9:

The length of each side of an equilateral triangle of area $4\sqrt{3} c{m}^2$, is

(a) 4 cm

(b) $\frac{4}{\sqrt{3}} cm$

(c) $\frac{\sqrt{3}}{4} cm$

(d) 3 cm

Answer 9:

Area of an equilateral triangle say A, having each side a cm is given by 

We are asked to find the side of the triangle
Therefore, the side of the equilateral triangle says a, having area is given by

Therefore, the correct answer is (a). 

Question 10:

If an isosceles right triangle has area 8 cm²,then the length of its hypotenuse is
(a) $\sqrt{32} \mathrm{cm}$
(b) $\sqrt{48} \mathrm{cm}$
(c) $\sqrt{24} \mathrm{cm}$
(d) 4 cm

Answer 10:

Given: Area of an isosceles right triangle is 8 cm²

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{an} \mathrm{isosceles} \mathrm{right} \mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ =\frac{1}{2}\times \mathrm{Base}\times \mathrm{Base} \left(\because \mathrm{Base}=\mathrm{Height}\right) \\ =\frac{1}{2}\times {\left(\mathrm{Base}\right)}^2 \\ \Rightarrow 8=\frac{1}{2}\times {\left(\mathrm{Base}\right)}^2 \\ \Rightarrow 16={\left(\mathrm{Base}\right)}^2 \\ \Rightarrow \mathrm{Base}=4 \mathrm{cm} =\mathrm{Perpendicular} \\ \mathrm{In} \mathrm{a} \mathrm{right} \mathrm{angled} \mathrm{triangle}, \mathrm{using} \mathrm{pythagoras} \mathrm{theorem} \\ {\left(\mathrm{Hypotenuse}\right)}^2={\left(\mathrm{Base}\right)}^2+{\left(\mathrm{Perpendicular}\right)}^2 \\ ={\left(4\right)}^2+{\left(4\right)}^2 \\ =16+16 \\ =32 \\ \Rightarrow \mathrm{Hypotenuse}=\sqrt{32} \mathrm{cm} \end{gathered}$$

Hence, the correct option is (a).

Question 11:

The perimeter of an equilateral triangle is 60 m. The area is
(a) $10\sqrt{3} {\mathrm{m}}^2$
(b) $15\sqrt{3} {\mathrm{m}}^2$
(c) $20\sqrt{3} {\mathrm{m}}^2$
(d) $100\sqrt{3} {\mathrm{m}}^2$

Answer 11:

Given: The perimeter of an equilateral triangle is 60 m.

Let the length of the side of an equilateral triangle be x m.

$$\begin{gathered} \mathrm{Perimeter}=60 \mathrm{m} \\ \Rightarrow 3x=60 \\ \Rightarrow x=\frac{60}{3} \\ \Rightarrow x=20 \mathrm{m} \\ \mathrm{Area} \mathrm{of} \mathrm{an} \mathrm{equilateral} \mathrm{triangl}e=\frac{\sqrt{3}}{4}\times x^2 \\ =\frac{\sqrt{3}}{4}{\left(20\right)}^2 \\ =\frac{\sqrt{3}}{4}\left(400\right) \\ =100\sqrt{3} {\mathrm{m}}^2 \end{gathered}$$

Hence, the correct option is (d).

Question 12:

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is

(a) $\sqrt{15} {\mathrm{cm}}^2$

(b) $\frac{\sqrt{15}}{2} {\mathrm{cm}}^2$

(c) $2\sqrt{15} {\mathrm{cm}}^2$

(d) $4\sqrt{15} {\mathrm{cm}}^2$

Answer 12:

Given:
The base of an isosceles triangle is 2 cm.
The length of one of the equal sides 4 cm.

Using Heron's formula:
If a, b and c are three sides of a triangle, then
area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.


Here, $s=\frac{2+4+4}{2}=\frac{10}{2}=5$

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\sqrt{5\left(5-2\right)\left(5-4\right)\left(5-4\right)} \\ =\sqrt{5\left(3\right)\left(1\right)\left(1\right)} \\ =\sqrt{15} {\mathrm{cm}}^2 \end{gathered}$$

Hence, the correct option is (a).

Question 13:

The length of each side of an equilateral triangle having an area of $9\sqrt{3} {\mathrm{cm}}^2$ is
(a) 8 cm
(b) 36 cm
(c) 4 cm
(d) 6 cm

Answer 13:

Given: The area of an equilateral triangle is $9\sqrt{3} {\mathrm{cm}}^2$.

Let the length of the side of an equilateral triangle be x cm.

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{an} \mathrm{equilateral} \mathrm{triangle}=\frac{\sqrt{3}}{4}\times x^2 \\ \Rightarrow 9\sqrt{3}=\frac{\sqrt{3}}{4}{\left(x\right)}^2 \\ \Rightarrow \frac{9\sqrt{3}\times 4}{\sqrt{3}}={\left(x\right)}^2 \\ \Rightarrow x^2=36 \\ \Rightarrow x=6 \mathrm{cm} \end{gathered}$$

Hence, the correct option is (d).

Page-17.25

Question 14:

If the area of an equilateral triangle is $16\sqrt{3} {\mathrm{cm}}^2$, then its perimeter is
(a) 48 cm
(b) 24 cm
(c) 12 cm
(d) 36 cm

Answer 14:

Given: The area of an equilateral triangle is $16\sqrt{3} {\mathrm{cm}}^2$.

Let the length of the side of an equilateral triangle be x cm.

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{an} \mathrm{equilateral} \mathrm{triangle}=\frac{\sqrt{3}}{4}\times x^2 \\ \Rightarrow 16\sqrt{3}=\frac{\sqrt{3}}{4}{\left(x\right)}^2 \\ \Rightarrow \frac{16\sqrt{3}\times 4}{\sqrt{3}}={\left(x\right)}^2 \\ \Rightarrow x^2=64 \\ \Rightarrow x=8 \mathrm{cm} \\ \mathrm{Thus}, \mathrm{Perimeter} \mathrm{of} \mathrm{triangle}=3x \\ =3\left(8\right) \\ =24 \mathrm{cm} \end{gathered}$$

Hence, the correct option is (b).

Question 15:

The sides of a triangle are 35 cm, 54 cm and 61 cm respectively. The length of its longest altitude is
(a) $16\sqrt{5} \mathrm{cm}$
(b) $10\sqrt{5} \mathrm{cm}$
(c) $24\sqrt{5} \mathrm{cm}$
(d) 28 cm

Answer 15:

Given:
The sides of a triangle are 35 cm, 54 cm and 61 cm respectively.



Let CD be the longest altitude of the triangle.

Using Heron's formula:
If a, b and c are three sides of a triangle, then
area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.


Here, $s=\frac{35+54+61}{2}=\frac{150}{2}=75$

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\sqrt{75\left(75-35\right)\left(75-54\right)\left(75-61\right)} \\ =\sqrt{75\left(40\right)\left(21\right)\left(14\right)} \\ =\sqrt{\left(5\times 5\times 3\right)\left(5\times 2\times 2\times 2\right)\left(3\times 7\right)\left(2\times 7\right)} \\ =\left(5\times 3\times 7\times 2\times 2\right)\sqrt{5} \\ =420\sqrt{5} {\mathrm{cm}}^2 \end{gathered}$$


We know,
$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ \Rightarrow 420\sqrt{5}=\frac{1}{2}\times 35\times CD \\ \Rightarrow 12\sqrt{5}=\frac{1}{2}CD \\ \Rightarrow CD=24\sqrt{5} \end{gathered}$$

Thus, the length of its longest altitude is $24\sqrt{5} \mathrm{cm}$.

Hence, the correct option is (c).

Question 16:

The sides of a triangle are 56 cm, 60 cm and 52 cm. Area of the triangle is
(a) 1322 cm²
(b) 1311 cm²
(c) 1344 cm²
(d) 1392 cm²

Answer 16:

Given:
The sides of a triangle are 56 cm, 60 cm and 52 cm.

Using Heron's formula:
If a, b and c are three sides of a triangle, then
area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.


Here, $s=\frac{56+60+52}{2}=\frac{168}{2}=84$

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\sqrt{84\left(84-56\right)\left(84-60\right)\left(84-52\right)} \\ =\sqrt{84\left(28\right)\left(24\right)\left(32\right)} \\ =\sqrt{3\times 7\times 2\times 2\left(7\times 2\times 2\right)\left(3\times 2\times 2\times 2\right)\left(2\times 2\times 2\times 2\times 2\right)} \\ =\left(3\times 7\times 2\times 2\times 2\times 2\times 2\times 2\right) \\ =1344 {\mathrm{cm}}^2 \end{gathered}$$

Hence, the correct option is (c).

Question 17:

The edges of a triangular board are 6 cm, 8 cm and 10 cm long. The cost of painting it at the rate of 9 paise per  cm² is
(a) ₹ 2
(b) ₹ 2.16
(c) ₹ 2.48
(d) ₹ 3

Answer 17:

Given:
The edges of a triangular board are 6 cm, 8 cm and 10 cm long.
The cost of painting per cm² is 9 paise.


Using Heron's formula:
If a, b and c are three sides of a triangle, then
area of the triangle = $\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ , where $s=\frac{a+b+c}{2}$.


Here, $s=\frac{6+8+10}{2}=\frac{24}{2}=12$

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{triangle}=\sqrt{12\left(12-6\right)\left(12-8\right)\left(12-10\right)} \\ =\sqrt{12\left(6\right)\left(4\right)\left(2\right)} \\ =\sqrt{\left(3\times 2\times 2\right)\left(3\times 2\right)\left(2\times 2\right)\left(2\right)} \\ =\left(3\times 2\times 2\times 2\right) \\ =24 {\mathrm{cm}}^2 \end{gathered}$$

The cost of painting per cm² = 9 paise.
The cost of painting 24 cm² = 24 × 9 paise
                                             = 216 paise
                                             = ₹ 2.16


Hence, the correct option is (b).

Question 18:

The area of an equilateral triangle with side $2\sqrt{3}$ cm is
(a) 5.196 cm²
(b) 0.866 cm²
(c) 3.496 cm²
(d) 1.732 cm²

Answer 18:

Given: The side of an equilateral triangle is $2\sqrt{3}$ cm.

$$\begin{gathered} \mathrm{Area} \mathrm{of} \mathrm{an} \mathrm{equilateral} \mathrm{triangl}e=\frac{\sqrt{3}}{4}\times x^2 \\ =\frac{\sqrt{3}}{4}{\left(2\sqrt{3}\right)}^2 \\ =\frac{\sqrt{3}}{4}\left(12\right) \\ =3\sqrt{3} \\ =5.196 {\mathrm{cm}}^2 \end{gathered}$$

Hence, the correct option is (a).

Question 19:

If the area of a regular hexagon is $54\sqrt{3} {\mathrm{cm}}^2$, then the length of its each side is
(a) 3 cm
(b) $2\sqrt{3} \mathrm{cm}$
(c) 6 cm
(d) $6\sqrt{3} \mathrm{cm}$

Answer 19:

Given: The area of a regular hexagon is $54\sqrt{3} {\mathrm{cm}}^2$.

We know, a hexagon is formed by joining 6 equilateral triangles together.
Therefore, Area of 6 equilateral triangles = $54\sqrt{3} {\mathrm{cm}}^2$

$$\begin{gathered} \mathrm{Area} \mathrm{of} 6 \mathrm{equilateral} \mathrm{triangles}=6\times \frac{\sqrt{3}}{4}\times x^2 \\ \Rightarrow 54\sqrt{3}=6\times \frac{\sqrt{3}}{4}\times x^2 \\ \Rightarrow \frac{54\sqrt{3}\times 4}{6\times \sqrt{3}}=x^2 \\ \Rightarrow 9\times 4=x^2 \\ \Rightarrow 36=x^2 \\ \Rightarrow x=6 \mathrm{cm} \end{gathered}$$

Hence, the correct option is (c).

Question 20:

If the length of each edge of a regular tetrahedron is 'a', then its surface area is
(a) $\sqrt{3} a^2 \mathrm{sq}. \mathrm{units}$
(b) $3\sqrt{2} a^2 \mathrm{sq}. \mathrm{units}$
(c) $2\sqrt{3} a^{\mathit{2}} \mathrm{sq}. \mathrm{units}$
(d) $\sqrt{6} a^2 \mathrm{sq}. \mathrm{units}$

Answer 20:

Given: The length of each edge of a regular tetrahedron is 'a' units.

We know, the surface area of tetrahedron = area of 4 equilateral triangles

$$\begin{gathered} \mathrm{Thus}, \\ \mathrm{Surface} \mathrm{Area}=4\times \frac{\sqrt{3}}{4}\times a^2 \\ =\sqrt{3} a^2 \mathrm{sq}. \mathrm{units} \end{gathered}$$

Hence, the correct option is (a).

Question 21:

If the area of an isosceles right triangle is 8 cm², what is the perimeter of the triangle?

(a) 8 + $\sqrt{2}$ cm²

(b) 8 + 4$\sqrt{2}$ cm²

(c) 4 + 8$\sqrt{2}$ cm²

(d) 12$\sqrt{2}$ cm²

Answer 21:

We are given the area of an isosceles right triangle and we have to find its perimeter. 
Two sides of isosceles right triangle are equal and we assume the equal sides to be the base and height of the triangle. We are asked to find the perimeter of the triangle
Let us take the base and height of the triangle be x cm. 
Area of a isosceles right triangle, say A having base x cm and height x cm is given by

A = 8 cm²; Base = Height = x cm

Using Pythagorean Theorem we have;


Let ABC be the given triangle
Perimeter of triangle ABC, say P is given by

AB = 4 cm; BC = 4 cm; AC =

Therefore, the answer is (b). 

Question 22:

The lengths of the sides of Δ ABC are consecutive integers. It Δ ABC has the same perimeter as an equilateral triangle with a side of length 9 cm, what is the length of the shortest side of ΔABC?

(a) 4

(b) 6

(c) 8

(d) 10

Answer 22:

We are given that triangle ABC has equal perimeter as to the perimeter of an equilateral triangle having side 9 cm. The sides of triangle ABC are consecutive integers. We are asked to find the smallest side of the triangle ABC 
Perimeter of an equilateral triangle, say P having side 9 cm is given by



Let us assume the three sides of triangle ABC be x, x+1, x−1
Perimeter of triangle ABC, say P₁ is given by
P₁ = AB + BC + AC
AB = x; BC = x +1; AC = x−1. Since P₁ = P. So

By using the value of x, we get the sides of triangle as 8 cm, 9 cm and 10 cm
Therefore, the answer is (c).

Question 23:

In the given figure, the ratio AD to DC is 3 to 2. If the area of Δ ABC is 40 cm², what is the area of Δ BDC?

Answer 23:

Area of triangle ABC is given 40 cm².

Also
We are asked to find the area of the triangle BDC
Let us take BE perpendicular to base AC in triangle ABC.
We assume AC equal to y and BE equal to x in triangle ABC
Area of triangle ABC, say A is given by

We are given the ratio between AD to DC equal to 3:2
So,

In triangle BDC, we take BE as the height of the triangle
Area of triangle BDC, say A₁ is given by

Therefore, the answer is (a).

Page-17.26

Question 24:

If the length of a median of an equilateral triangle is x cm, then its area is

(a) x²

(b) $\frac{\sqrt{3}}{2}{x}^2$

(c) $\frac{x^2}{\sqrt{3}}$

(d) $\frac{x^2}{2}$

Answer 24:

We are given the length of median of an equilateral triangle by which we can calculate its side. We are asked to find area of triangle in terms of x
Altitude of an equilateral triangle say L, having equal sides of a cm is given by, where, L = x cm

Area of an equilateral triangle, say A₁ having each side a cm is given by 

Since .So

Therefore, the answer is (c).

Question 25:

If every side of a triangle is doubled, then increase in the area of the triangle is

(a) $100\sqrt{2}\%$

(b) 200%

(c) 300%

(d) 400%

Answer 25:

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,
, where

We take the sides of a new triangle as 2a, 2b, 2c that is twice the sides of previous one
Now, the area of a triangle having sides 2a, 2b, and 2c and as semi-perimeter is given by,

Where,

Now,

Therefore, increase in the area of the triangle

Percentage increase in area 

Therefore, the answer is (c).

Question 26:

A square and an equilateral triangle have equal perimeters. If the diagonal of the square is $12\sqrt{2}$ cm, then area of the triangle is

(a) $24\sqrt{2} c{m}^2$

(b) $24\sqrt{3} c{m}^2$

(c) $48\sqrt{3} c{m}^2$

(d) $64\sqrt{3} c{m}^2$

Answer 26:

It is given the perimeter of a square ABCD is equal to the perimeter of triangle PQR.
The measure of the diagonal of the square is given .We are asked to find the area of the triangle
In square ABCD, we assume that the adjacent sides of square be a.
Since, it is a square then
By using Pythagorean Theorem

Therefore, side of the square is 12 cm.
Perimeter of the square ABCD say P is given by

Side = 12 cm

Perimeter of the equilateral triangle PQR say P₁ is given by



The side of equilateral triangle PQR is equal to 16 cm.
Area of an equilateral triangle say A, having each side a cm is given by 

Area of the given equilateral triangle having each equal side equal to 4 cm is given by
a = 16 cm

Therefore, the answer is (d).