Question
Find the area of a right triangle whose base is $1.2m$ and hypotenuse $3.7m.$
✓
Answer
In right angled $\triangle\text{ABC},$ Base $BC = 1.2m$
and hypotenuse $AC =3.7 m$
But $A C^2=A B^2+B C^2$ (Pythagoras Theorem)
$\Rightarrow(3.7)^2=A B^2+(1.2)^2$
$\Rightarrow 13.69=A B^2+1.44$
$\Rightarrow A B^2=13.69-1.44$
$\Rightarrow A B^2=12.25=(3.5)^2$
$\Rightarrow A B=3.5 m$
Now, area of $\triangle ABC =\frac{1}{2} \times$ Base $\times$ Altitude
$=\frac{1}{2} \times 1.2 \times 3.5 m^2=2.1 m^2$
and hypotenuse $AC =3.7 m$
But $A C^2=A B^2+B C^2$ (Pythagoras Theorem)
$\Rightarrow(3.7)^2=A B^2+(1.2)^2$
$\Rightarrow 13.69=A B^2+1.44$
$\Rightarrow A B^2=13.69-1.44$
$\Rightarrow A B^2=12.25=(3.5)^2$
$\Rightarrow A B=3.5 m$
Now, area of $\triangle ABC =\frac{1}{2} \times$ Base $\times$ Altitude
$=\frac{1}{2} \times 1.2 \times 3.5 m^2=2.1 m^2$
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