Q. 135.0( 2 Votes )

# Hypotenuse of a right angled triangle is 2 less than 3 times its shortest side. If the remaining side is 2 more than twice the shortest side, find the area of the triangle.

Answer :

Let length of shortest side(base) be x.

⇒ hypotenuse = 3x – 2.

⇒ remaining side(perpendicular) = 2x + 2

Now, According to Pythagoras theorem,

Hypotenuse^{2} = Base^{2} + Perpendicular^{2}

⇒(3x – 2)^{2} = x^{2} + (2x + 2)^{2}

⇒ 9x^{2} + 4 – 12x = x^{2} + 4x^{2} + 4 + 8x

⇒ 4x^{2} – 20x = 0

Dividing the whole equation by 4, we get –

⇒ x^{2} – 5x = 0

⇒ x(x – 5) = 0

⇒ x = 0 or x – 5 = 0

But side cannot be zero

⇒ x(Base) = 5

⇒Perpendicular = 2x + 2

= 2(5) + 2

= 12

⇒ hypotenuse = 3x – 2

= 3(5) – 2

= 13

Now, area of right triangle –

= 30 sq. unit.

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