# A Square and an Equilateral Triangle Have Equal Perimeters. If the Diagonal of the Square is 12 √ 2 Cm, Then Area of the Triangle is - Mathematics

MCQ

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

#### Options

• $24\sqrt{2} c m^2$

• $24\sqrt{3} c m^2$

• $48\sqrt{3} c m^2$

• $64\sqrt{3} c m^2$

#### Solution

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 12 sqrt(2) cm.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  a= b

By using Pythagorean Theorem

a^2 +b^2 = (12sqrt(2))^2

a^2 +a^2 = 288

2a^2 = 288

a^2 = 288/2

 a = sqrt(144)

a = 12 cm

Therefore, side of the square is 12 cm.

Perimeter of the square ABCD say P is given by

p = 4 × side

Side = 12 cm

p = 4 ×  12

p = 48 cm

Perimeter of the equilateral triangle PQR say P1 is given by

p1= 3 ×  side

p = p1

p = 3 ×  side

48 = 3 ×  side

side = 48/3

side = 16 cm

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

A = sqrt(3)/4 a^2

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

a = 16 cm

A = sqrt(3)/4 (16)^2

A = sqrt(3)/4 xx 256

A=64 sqrt(3)  cm^2

Is there an error in this question or solution?
Chapter 17: Heron’s Formula - Exercise 17.4 [Page 25]

#### APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 17 Heron’s Formula
Exercise 17.4 | Q 15 | Page 25

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