Question:
The area of a right angled triangle is $165 \mathrm{~m}^{2}$. Determine its base and altitude if the latter exceeds the former by $7 \mathrm{~m}$.
Solution:
Let the base of the right triangle be $=x$ metres and the altitude $=(x+7)$ metres Then
According to question,
Areas of the right triangle $=165 \mathrm{~m}^{2}$
And as we know that the area of the right triangle $=\frac{1}{2} \times$ base $\times$ height
$\frac{1}{2} \times x \times(x+7)=165$
$x^{2}+7 x=330$
$x^{2}+7 x-330=0$
$x^{2}-15 x+22 x-330=0$
$x(x-15)+22(x-15)=0$
$(x-15)(x+22)=0$
$(x-15)=0$
$x=15$
or
$(x+22)=0$
$x=-22$
Since negative value is not possible. So x =15 m
Therefore the altitude is
$=(x+7)$
$=15+7$
$=22$
Hence, base of the right triangle be $15 \mathrm{~m}$ and altitude be $22 \mathrm{~m}$