Question
Calculate the area of a right$-$angled triangle whose hypotenuse is $65\ cm$ and one side is $16\ cm$.
✓
Answer
Hypotenuse $= 65\ cm$
One side $= 16\ cm$
Let the other side be of length $x \ cm$
By Pythagoras theorem,
$(65\ cm)^2= (16\ cm)^2 + (x\ cm)^2$
$(x \ cm)^2 = 4225\ cm^2 - 256\ cm^2$
$= 3969\ cm^2$
$= (63\ cm)^2$
$\Rightarrow x = 63\ cm$
Area of the triangle
$=\frac{1}{2} \times($Base $\times$ Height$)$
$=\frac{1}{2} \times 16 \ cm \times 63 \ cm$
$=504 \ cm ^2 .$
One side $= 16\ cm$
Let the other side be of length $x \ cm$
By Pythagoras theorem,
$(65\ cm)^2= (16\ cm)^2 + (x\ cm)^2$
$(x \ cm)^2 = 4225\ cm^2 - 256\ cm^2$
$= 3969\ cm^2$
$= (63\ cm)^2$
$\Rightarrow x = 63\ cm$
Area of the triangle
$=\frac{1}{2} \times($Base $\times$ Height$)$
$=\frac{1}{2} \times 16 \ cm \times 63 \ cm$
$=504 \ cm ^2 .$
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