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

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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

  • \[24\sqrt{2} c m^2\]

     

  • \[24\sqrt{3} c m^2\]

     

  • \[48\sqrt{3} c m^2\]

     

  • \[64\sqrt{3} c m^2\]

     

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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`

 

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Chapter 17: Heron’s Formula - Exercise 17.4 [Page 25]

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

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