Question

The base of a triangle is $\frac{3}{4}th$ the length of the corresponding altitude. If the altitude is increased by $3cm$ and the base is decreased by $2cm$, the area of the triangle remains the same find the base and altitude of the triangle.

Answer 1

Let the area of triangle be $A$ base be $b$ and height be $h$.

$\therefore A=\frac{1}{2}\times b\times h$

also $b=\frac{3}{4}h$

$\therefore A=\frac{1}{2}\times \frac{3}{4}\times h=\frac{3h^2}{8}c{m}^2$...(1)

Now,

New $h^{\prime }=h+3$ (new height)

New $b^{\prime }=b-2$ (new base)

$\therefore A=\frac{1}{2}{h}^{\prime }{b}^{\prime }$ (Area remain some)

$\therefore A=\frac{1}{2}\times (h+3)/(b-2)$.....(2)

From (1) and (2)

$\frac{(h+3)(b-2)}{2}=\frac{3h^2}{8}$

$\frac{(h+3)(\frac{3}{4}h-2)}{2}=\frac{3h^2}{8}$

$\therefore \frac{3h^2}{4}-2h+\frac{9h}{4}-6=\frac{3h^2}{4}$

$\therefore \frac{h}{4}=6;h=6\times 4=24cm$

$\therefore b=\frac{3}{4}\times 24=18cm$

The base is $18cm$ and altitude is $24cm$