Question: The area of a right-angled triangle is 24 sq.cm. If the perpendicular of the triangle is 2cm shorter than its base, find two sides of a triangle.

Solution:
Given,

Area of a right-angled triangle (A) = 24 cm²

Let the length of the base (b) of the triangle be x cm.
According to the question, the length of the perpendicular (p) of the triangle is (x+2) cm.

We know,
$Area = \frac{1}{2} × p × b$
$or, 24 = \frac{1}{2} × (x+2)x$
$or, 24×2 = x² + 2x$
$or, 48 = x² +2x$
$or, x² +2x -48 = 0$
$or, x² +(8-6)x -48 = 0$
$or, x² +8x -6x -48 = 0$
$or, x(x +8) -6(x +8) = 0$
$or, (x -6)(x +8) = 0$

Either,
$x -6 = 0$
$\therefore x = 6$

Or,
$x +8 = 0$
$\therefore x = -8$

Since, distance is a scalar quantity, value of x is 6.

So, b = x cm = 6 cm
p = (x +2) cm = (6+2) cm = 8cm

Therefore, the required length of the perpendicular and base of the given triangle are 8 cm and 6 cm respectively.

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