23
The area of a right-angled triangle is $600$ cm$^2$. If the base of the triangle exceeds the altitude by $10$ cm, find all the three dimensions of the triangle.
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Let the altitude of triangle be $x$ cm
then Base of triangle = $(x + 10)$ cm
Area of triangle = $600$ cm$^2$
$\frac{1}{2} \times x \times (x + 10) = 600$ (1 1/2 Marks)
$\Rightarrow x^2+10x-1200 = 0$ (1 Mark)
$\Rightarrow (x + 40)(x – 30) = 0$ (1 Mark)
$\Rightarrow x = -40, x = 30$
$x = -40$ (rejected)
$x = 30$ (1/2 Mark)
$\therefore$ Altitude = $30$ cm
Base = $40$ cm (1/2 Mark)
Hypotenuse = $\sqrt{(30)^2 + (40)^2}= 50$ cm (1/2 Mark)
then Base of triangle = $(x + 10)$ cm
Area of triangle = $600$ cm$^2$
$\frac{1}{2} \times x \times (x + 10) = 600$ (1 1/2 Marks)
$\Rightarrow x^2+10x-1200 = 0$ (1 Mark)
$\Rightarrow (x + 40)(x – 30) = 0$ (1 Mark)
$\Rightarrow x = -40, x = 30$
$x = -40$ (rejected)
$x = 30$ (1/2 Mark)
$\therefore$ Altitude = $30$ cm
Base = $40$ cm (1/2 Mark)
Hypotenuse = $\sqrt{(30)^2 + (40)^2}= 50$ cm (1/2 Mark)