Question

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the lengths of the other two sides (in cm).

• A. 2, 5
• B. 5, 3
• C. 7, 2
• D. 12, 5

Answer 1

The correct option is D

12, 5

Let the length of the base be x cm​.

Given that the altitude of the right triangle is 7 cm less than its base

$\Rightarrow$ Altitude = x - 7 cm

Hypotenuse = 13 cm ...[given]

Applying Pythagoras theorem,

$(base{)}^2$+ $(altitude{)}^2$ = $(hypotenuse{)}^2$

On substituting the values in the above equation, we get

$x^2$ + $(x-7{)}^2$ = ${13}^2$

$\Rightarrow$ $x^2$ + $x^2$ + 49 – 14x = 169

$\Rightarrow$ 2$x^2$ – 14x + 49 – 169 = 0

$\Rightarrow$ 2$x^2$ – 14x – 120 = 0

Dividing by 2 on both sides, we get

$x^2$ – 7x – 60 = 0

$\Rightarrow$ $x^2$ – 12x + 5x – 60 = 0

$\Rightarrow$ x(x – 12) + 5(x – 12) = 0

$\Rightarrow$ (x – 12)(x + 5) = 0

$\Rightarrow$ x – 12 = 0 or x + 5 = 0

$\Rightarrow$ x = 12 or x = –5

Length cannot be negative.

Therefore, x cannot be – 5.

Hence, base is 12 cm long and altitude is 12 – 7 = 5 cm long.