The area of a right-angled triangle is 165sq.meters. If the altitude of the triangle exceeds the base by 7m then the base and altitude are

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Base =22m and Altitude =15m

Base =15m and Altitude =22m

Base =17m and Altitude =20m

Base =20m and Altitude =17m

answer is A.

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Detailed Solution

Given that area of a right-angled triangle = 165sq.m and altitude exceeds the base by 7m.
Let’s consider that the base and altitude of the triangle are

p

and

q

respectively.
According to the given condition,

q = p + 7

- eq. 1)
Then,
Area of the triangle =

\frac{1}{2} \times base \times altitude

⟹ \frac{1}{2} \times p \times q = 165

⟹ \frac{1}{2} \times p \times (p + 7) = 165

{Using (1)}

⟹ \frac{1}{2} \times (p^{2} + 7 p) = 165

⟹ p^{2} + 7 p = 330

⟹ p^{2} + 7 p = 330 = 0

⟹ p^{2} + 7 p - 330 = 0

⟹ p^{2} + 22 p - 15 p - 330 = 0

⟹ p (p + 22) - 15 (p + 22) = 0

⟹ (p - 15) (p + 22) = 0

⟹ p - 15 = 0 or p + 22 = 0

∴ p = 15

p = - 22

∴

The base of the right-angled triangle is 15m. (the base of the triangle cannot be negative so will ignore another value we got).
Then from equation (1),