R.d Sharma 2022 _mcqs for Class 9 Maths Chapter 17 - Heron S Formula

Answer:

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).

Answer:

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


Hence, the correct option is (d).

Answer:

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).

Answer:

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).

Answer:

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).

Answer:

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).

Answer:

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).

Answer:

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).

Answer:

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). 

Answer:

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).
 

Answer:

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).

Answer:

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).

Answer:

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).

Answer:

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).

Answer:

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 =  , where .


Here, 




We know,


Thus, the length of its longest altitude is .

Hence, the correct option is (c).

Answer:

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).

Answer:

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).

Answer:

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).

Answer:

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).

Answer:

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).

Answer:

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). 

Answer:

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).

Answer:

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).

Answer:

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).

Answer:

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).

Answer:

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).

Answer:

Statement-2 (Reason): The area ∆ of a triangle with semi-perimeter s and sides a, b and c is given by $∆=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$.

According to Heron's formula, the area of a triangle with sides a, b and c is given by $∆=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ where the semi-perimeter is $s=\frac{a+b+c}{2}$.

Thus, Statement-2 is true.

Statement-1 (Assertion): The area ∆ of an isosceles triangle with base a and each equal side b is given by $∆=\frac{a}{4}\sqrt{4b^2-a^2}$.
Let an isosceles triangle with base a and each equal side b.

According to Statement-2,
$$\begin{gathered} s=\frac{a+b+b}{2} \\ \Rightarrow s=\frac{a+2b}{2} \\ \Rightarrow s=\frac{a}{2}+b .....\left(1\right) \end{gathered}$$


And,  
$\begin{array}{rcl}\mathrm{area} \mathrm{of}∆& =& \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-b\right)}\\ & =& \sqrt{\left(\frac{a}{2}+b\right)\left(\frac{a}{2}+b-a\right)\left(\frac{a}{2}+b-b\right)\left(\frac{a}{2}+b-b\right)} \left[\mathrm{From} \left(1\right)\right]\\ & =& \sqrt{\left(\frac{a}{2}+b\right)\left(b-\frac{a}{2}\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)}\\ & =& \left(\frac{a}{2}\right)\sqrt{b^2-{\left(\frac{a}{2}\right)}^2}\\ & =& \left(\frac{a}{2}\right)\sqrt{\frac{4b^2-a^2}{4}}\\ & =& \left(\frac{a}{4}\right)\sqrt{4b^2-a^2}\\ & & \end{array}$
Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

 

Answer:

Statement-2 (Reason): If p is the altitude of an equilateral triangle, then its area A is given by $∆=\frac{p^2}{\sqrt{3}}$.
Let each side of the equilateral triangle be a and p be the altitude of an equilateral triangle.
According to Heron's formula, the area of a triangle with sides a, b and c is given by $∆=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ where the semi-perimeter is $s=\frac{a+b+c}{2}$.
Here,
$$\begin{gathered} s=\frac{a+a+a}{2} \\ \Rightarrow s=\frac{3a}{2} .....\left(1\right) \end{gathered}$$


And,  
$\begin{array}{rcl}\mathrm{area} \mathrm{of}∆& =& \sqrt{s\left(s-a\right)\left(s-a\right)\left(s-a\right)}\\ & =& \sqrt{\left(\frac{3a}{2}\right)\left(\frac{3a}{2}-a\right)\left(\frac{3a}{2}-a\right)\left(\frac{3a}{2}-a\right)} \left[\mathrm{From} \left(1\right)\right]\\ & =& \sqrt{\left(\frac{3a}{2}\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)}\\ & =& \left(\frac{a^2}{4}\right)\sqrt{3} .....\left(2\right)\end{array}$
$$\begin{gathered} \mathrm{area} \mathrm{of}∆=\frac{1}{2}\times \mathrm{base}\times \mathrm{height} \\ \Rightarrow \frac{\sqrt{3}{a}^2}{4}=\frac{1}{2}\times a\times p \left[\mathrm{From} \left(2\right)\right] \\ \Rightarrow \frac{\sqrt{3}\mathrm{a}}{2}=p \end{gathered}$$
$\Rightarrow \frac{2p}{\sqrt{3}}=a .....\left(3\right)$

Substituting (3) in (2), we get
$$\begin{gathered} \mathrm{area} \mathrm{of}∆=\frac{\sqrt{3} }{4}\times {\left(\frac{2p}{\sqrt{3}}\right)}^2 \\ \Rightarrow \mathrm{area} \mathrm{of}∆=\frac{\sqrt{3} }{4}\times \frac{4p^2}{3} \\ \Rightarrow \mathrm{area} \mathrm{of}∆=\frac{p^2}{\sqrt{3}} \end{gathered}$$

Thus, Statement-2 is true.

Statement-1 (Assertion): The altitude p of an equilateral triangle having each side a is given by $p=\frac{\sqrt{3}}{2}a$.

Let each side of the equilateral triangle be a and p be the altitude of an equilateral triangle.

According to Statement-2, if p is the altitude of an equilateral triangle, then its area A is given by
 $∆=\frac{p^2}{\sqrt{3}}$       .....(1)
According to Heron's formula, the area of a triangle with sides a, b and c is given by $∆=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ where the semi-perimeter is $s=\frac{a+b+c}{2}$.
Here,
$$\begin{gathered} s=\frac{a+a+a}{2} \\ \Rightarrow s=\frac{3a}{2} .....\left(2\right) \end{gathered}$$


And,  
$\begin{array}{rcl}\mathrm{area} \mathrm{of}∆& =& \sqrt{s\left(s-a\right)\left(s-a\right)\left(s-a\right)}\\ & =& \sqrt{\left(\frac{3a}{2}\right)\left(\frac{3a}{2}-a\right)\left(\frac{3a}{2}-a\right)\left(\frac{3a}{2}-a\right)} \left[\mathrm{From} \left(2\right)\right]\\ & =& \sqrt{\left(\frac{3a}{2}\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)}\\ & =& \left(\frac{a^2}{4}\right)\sqrt{3} .....\left(3\right)\end{array}$

From (1) and (4), we get
$$\begin{gathered} \frac{\sqrt{3}}{4}{a}^2=\frac{p^2}{\sqrt{3}} \\ \Rightarrow \frac{3}{4}{a}^2=p^2 \\ \Rightarrow \frac{\sqrt{3}}{2}a=p \end{gathered}$$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).













 

Answer:

Statement-2 (Reason): The area of an equilateral triangle with altitude p is $∆=\frac{p^2}{\sqrt{3}}$.
Let each side of the equilateral triangle be a and p be the altitude of an equilateral triangle.
According to Heron's formula, the area of a triangle with sides a, b and c is given by $∆=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ where the semi-perimeter is $s=\frac{a+b+c}{2}$.
Here,
$$\begin{gathered} s=\frac{a+a+a}{2} \\ \Rightarrow s=\frac{3a}{2} .....\left(1\right) \end{gathered}$$


And,  
$\begin{array}{rcl}\mathrm{area} \mathrm{of}∆& =& \sqrt{s\left(s-a\right)\left(s-a\right)\left(s-a\right)}\\ & =& \sqrt{\left(\frac{3a}{2}\right)\left(\frac{3a}{2}-a\right)\left(\frac{3a}{2}-a\right)\left(\frac{3a}{2}-a\right)} \left[\mathrm{From} \left(1\right)\right]\\ & =& \sqrt{\left(\frac{3a}{2}\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)}\\ & =& \left(\frac{a^2}{4}\right)\sqrt{3} .....\left(2\right)\end{array}$
$$\begin{gathered} \mathrm{area} \mathrm{of}∆=\frac{1}{2}\times \mathrm{base}\times \mathrm{height} \\ \Rightarrow \frac{\sqrt{3}{a}^2}{4}=\frac{1}{2}\times a\times p \left[\mathrm{From} \left(2\right)\right] \\ \Rightarrow \frac{\sqrt{3}\mathrm{a}}{2}=p \end{gathered}$$
$\Rightarrow \frac{2p}{\sqrt{3}}=a .....\left(3\right)$

Substituting (3) in (2), we get
$$\begin{gathered} \mathrm{area} \mathrm{of}∆=\frac{\sqrt{3} }{4}\times {\left(\frac{2p}{\sqrt{3}}\right)}^2 \\ \Rightarrow \mathrm{area} \mathrm{of}∆=\frac{\sqrt{3} }{4}\times \frac{4p^2}{3} \\ \Rightarrow \mathrm{area} \mathrm{of}∆=\frac{p^2}{\sqrt{3}} \end{gathered}$$

Thus, Statement-2 is true.

Statement-1 (Assertion): The area of an equilateral triangle the length of whose altitude is 6 cm, is $12\sqrt{3} {\mathrm{cm}}^2$.

Here, altitude of equilateral triangle (p) = 6 cm
According to Statement-2, if p is the altitude of an equilateral triangle, then its area A is given by
 $∆=\frac{p^2}{\sqrt{3}}$            .....(1)
Now, 
$\begin{array}{rcl}∆& =& \frac{{\left(6\right)}^2}{\sqrt{3}}\\ & =& \frac{36}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ & =& \frac{36\sqrt{3}}{3}\\ & =& 12\sqrt{3} {\mathrm{cm}}^2\end{array}$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

Statement-2 (Reason): If the perimeter of an equilateral triangle is 72 cm, then its altitude is $8\sqrt{3}$ cm.

Let each side of the equilateral triangle be a and p be the altitude of an equilateral triangle.
Given that, the perimeter of an equilateral triangle is 72 cm.
⇒ a + a + a = 72
⇒ 3a = 72
​⇒ a = 24

Since, the altitude p of an equilateral triangle having each side a is given by $p=\frac{\sqrt{3}}{2}a$.
$$\begin{gathered} \therefore p=\frac{\sqrt{3}}{2}\times 24 \\ \Rightarrow p=12\sqrt{3} \mathrm{cm} \end{gathered}$$

Thus, Statement-2 is false.

Statement-1 (Assertion): If the area of an equilateral triangle is $36\sqrt{3} {\mathrm{cm}}^2$, then its perimeter is 36 cm.
Let each side of the equilateral triangle be a.
Given that, the area of an equilateral triangle is $36\sqrt{3} {\mathrm{cm}}^2$.

Since, a is the length each side of the equilateral triangle then its area is given by $∆=\frac{\sqrt{3}}{4}{a}^2$.
$$\begin{gathered} \therefore 36\sqrt{3}=\frac{\sqrt{3}}{4}{a}^2 \\ \Rightarrow 36=\frac{1}{4}{a}^2 \\ \Rightarrow 36\times 4=a^2 \\ \Rightarrow \sqrt{36\times 4}=a \\ \Rightarrow 6\times 2=a \\ \Rightarrow a=12 \mathrm{cm} \end{gathered}$$
Now, perimeter of the triangle = a + a + a
                                                 = 3a
                                                 = 3 × 12
                                                 = 36 cm

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is false.

Hence, the correct answer is option (c).





 

Answer:

Statement-2 (Reason): The area of an isosceles triangle with base a and each equal side b is $\frac{b}{4}\sqrt{4a^2-b^2}$.
According to Heron's formula, the area of a triangle with sides a, b and c is given by $∆=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ where the semi-perimeter is $s=\frac{a+b+c}{2}$.
Let an isosceles triangle with base a and each equal side b.
Here, 
$$\begin{gathered} s=\frac{a+b+b}{2} \\ \Rightarrow s=\frac{a+2b}{2} \\ s=\frac{a}{2}+b .....\left(1\right) \end{gathered}$$


And,  
$\begin{array}{rcl}\mathrm{area} \mathrm{of}∆& =& \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-b\right)}\\ & =& \sqrt{\left(\frac{a}{2}+b\right)\left(\frac{a}{2}+b-a\right)\left(\frac{a}{2}+b-b\right)\left(\frac{a}{2}+b-b\right)} \left[\mathrm{From} \left(1\right)\right]\\ & =& \sqrt{\left(\frac{a}{2}+b\right)\left(b-\frac{a}{2}\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)}\\ & =& \left(\frac{a}{2}\right)\sqrt{b^2-{\left(\frac{a}{2}\right)}^2}\\ & =& \left(\frac{a}{2}\right)\sqrt{\frac{4b^2-a^2}{4}}\\ & =& \left(\frac{a}{4}\right)\sqrt{4b^2-a^2}\\ & & \end{array}$
Thus, Statement-2 is false.

Statement-1 (Assertion): The area of an isosceles triangle having base 24 cm and each of the equal sides equal to 13 cm is 60 cm².

Given that, an isosceles triangle with base a  = 24 cm and each equal side b = 13 cm.
Since, the area ∆ of an isosceles triangle with base a and each equal side b is given by $∆=\frac{a}{4}\sqrt{4b^2-a^2}$.

$$\begin{gathered} \therefore ∆=\frac{24}{4}\sqrt{4{\left(13\right)}^2-{\left(24\right)}^2} \\ ∆=6\sqrt{4\left(169\right)-{\left(24\right)}^2} \\ ∆=6\sqrt{676-576} \\ ∆=6\sqrt{100} \\ ∆=60 {\mathrm{cm}}^2 \end{gathered}$$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is false.

Hence, the correct answer is option (c).





 

Answer:

Statement-2 (Reason): The base and the corresponding altitude of a parallelogram are 10 cm and 3.5 cm respectively. The area of the parallelogram is 30 cm².

Given that, the base of a parallelogram (b) = 10 cm and the corresponding altitude of the parallelogram (h) = 3.5 cm.

∵ Area of a parallelogram = b × h
                                          = 10 × 3.5 
                                          = 35 cm²

Thus, Statement-2 is false.

Statement-1 (Assertion): If the side of a rhombus is 10 cm and one diagonal is 16 cm, the area of the rhombus is 96 cm².

Consider the ABCD be the rhombus such that AB = BC = CD = DA = 10 cm and AC = 16 cm.


According to Heron's formula, the area of a triangle with sides a, b and c is given by $∆=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ where the semi-perimeter is $s=\frac{a+b+c}{2}$.
Here, a = AB = 10 cm, b = BC = 10 cm and  c = AC = 16 cm.

$$\begin{gathered} s=\frac{10+10+16}{2} \\ \Rightarrow s=\frac{36}{2} \\ \Rightarrow s=18 \mathrm{cm} .....\left(1\right) \end{gathered}$$


And,  
$\begin{array}{rcl}\mathrm{Area} \mathrm{of}∆\mathrm{ABC}& =& \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\\ & =& \sqrt{18\left(18-10\right)\left(18-10\right)\left(18-16\right)} \left[\mathrm{From} \left(1\right)\right]\\ & =& \sqrt{2\times 9\times 8\times 8\times 2} \\ & =& 2\times 8\times 3\\ & =& 48 {\mathrm{cm}}^2 .....\left(2\right)\end{array}$
$$\begin{gathered} \mathrm{Now}, \mathrm{area} \mathrm{of}∆\mathrm{ADC}=\mathrm{area} \mathrm{of}∆\mathrm{ABC} .....\left(3\right) \\ \mathrm{Area} \mathrm{of} \mathrm{parallelogram} \mathrm{ABCD}=\mathrm{area} \mathrm{of}∆\mathrm{ABC}+\mathrm{area} \mathrm{of}∆\mathrm{ADC} \\ \Rightarrow \mathrm{Area} \mathrm{of} \mathrm{parallelogram} \mathrm{ABCD}=48+48 \left[\mathrm{From} \left(2\right) \mathrm{and} \left(3\right)\right] \\ \Rightarrow \mathrm{Area} \mathrm{of} \mathrm{parallelogram} \mathrm{ABCD}=96 {\mathrm{cm}}^2 \end{gathered}$$

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is false.

Hence, the correct answer is option (c).