Question
Find the area of a right angled triangle whose hypotenuse is $15\ cm$ and the base is $9\ cm.$
✓
Answer
The perpendicular of a right triangle whose hypotenuse is $h$ and base is $b$, is given by
$\sqrt{h^2-b^2}$
The perpendicular of a right triangle whose hypotenuse is $15$ and base is $9 $, is given by
$\sqrt{15^2-9^2}$
$=\sqrt{225-81}$
$=\sqrt{144}$
$=12 \ cm$
We also know that, Area of a Triangle
$=\frac{1}{2} \text { b.h}.$ i.e. $\frac{1}{2} ($Base $\times$ Height$)$
Area of a Triangle with bas e
$=9 \ cm$ and height
$=$ perpendicular
$=12 \ cm$
$\Rightarrow \frac{1}{2} b \cdot h$
$=\frac{1}{2} \times 9 \times 12$
$=54 \ cm ^2$.
$\sqrt{h^2-b^2}$
The perpendicular of a right triangle whose hypotenuse is $15$ and base is $9 $, is given by
$\sqrt{15^2-9^2}$
$=\sqrt{225-81}$
$=\sqrt{144}$
$=12 \ cm$
We also know that, Area of a Triangle
$=\frac{1}{2} \text { b.h}.$ i.e. $\frac{1}{2} ($Base $\times$ Height$)$
Area of a Triangle with bas e
$=9 \ cm$ and height
$=$ perpendicular
$=12 \ cm$
$\Rightarrow \frac{1}{2} b \cdot h$
$=\frac{1}{2} \times 9 \times 12$
$=54 \ cm ^2$.
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