The area of a right angled triangle is 600 cm $^{2}$ . If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle.

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Solution

Let the altitude BC of right-angled triangle ABC be x cm. Then,
Base = BC = (x+10) cm
Area = $\frac{1}{2} (x + 10) x$ $c m^{2}$

$\frac{1}{2} x (x + 10) = 600$
$x^{2} + 10 x = 1200$
$x^{2} + 10 x - 1200 = 0$
$x^{2} + 40 x - 30 x - 1200 = 0$
$x (x + 40) - 30 (x + 40) = 0$
$(x + 40) (x - 30) = 0$
$x = 30, - 40$
Hence, Base = 40 cm and Altitude = 30 cm.

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The area of a right angled triangle is 600 cm2. If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle.

Let the altitude BC of right-angled triangle ABC be x cm. Then,Base = BC = (x+10) cmArea = 12(x+10)x cm212x(x+10)=600x2+10x=1200x2+10x−1200=0x2+40x−30x−1200=0x(x+40)−30(x+40)=0(x+40)(x−30)=0x=30,−40Hence, Base = 40 cm and Altitude = 30 cm.

The area of a right angled triangle is 600 cm $^{2}$ . If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle.