Question

A square KLMN is inscribed in an equilateral triangle
PQR. If diagonal of the square is 1472 cm, then by
how much square cm, the area of the APQR is more
than that of the square?​

Answer 1

Answer:

Area of Triangle PQR more than Area of square by 1094339.1 square centimeter .

Step-by-step explanation:

Given as :

A square KLMN is inscribed in an equilateral triangle PQR

The measure of diagonal of square = NL = KM = 1472 cm

Now, Let the each side of square = a unit

So, For square

NK² + KL² = NL²

Or, a² + a² = (1472)²

Or,  a² = $\dfrac{1472^{2} }{(\sqrt{2} )^{2} }$

Or, a²  = ( $\dfrac{1472}{\sqrt{2} }$ )²

And a = 1040.8 cm

Since Area of square = side²

So, The area of square KLMN = ( $\dfrac{1472}{\sqrt{2} }$ )²

Or, The area of square KLMN = 1083392 cm²

Again

In Δ NQK

Since The triangle is equilateral

So, The angle ∠NQK = 60°

Or, Tan 60° = $\dfrac{NK}{QK}$

Or, √3 = $\dfrac{a}{QK}$

Or, QK = $\dfrac{1040.8}{1.732}$

Or, QK = 600.9 cm

Similarly

The measure of LR = 600.9 cm

So, The measure of side QR = QK + KL + LR

Or, QR = 600.9 cm + 1040.8 cm + 600.9 cm

Or, QR = 2242.6 cm

Again

Area of equilateral triangle = $\dfrac{\sqrt{3} }{4}$ ×  side²

Or, Area of equilateral triangle = $\dfrac{\sqrt{3} }{4}$ × QR²

Or, Area of equilateral triangle = $\dfrac{\sqrt{3} }{4}$ × (2242.6)²

Or, Area of equilateral triangle PQR = 2177731.1 cm²

Again

Area of Triangle PQR more than Area of square = 2177731.1 cm² - 1083392 =  1094339.1 cm²

Hence, Area of Triangle PQR more than Area of square by 1094339.1 square centimeter . Answer