Question

The area of a parallelogram with base $24$cm is equal to the area of a triangle of sides $10$cm,$24$cm and $26$cm. Find the altitude of the parallelogram.

Answer 1

Area of a parallelogram$=bh=24\times h=24h$

Perimeter of triangle with sides $10$cm,$24$cm and $26$cm is

$s=\frac{a+b+c}{2}=\frac{10+24+26}{2}=\frac{60}{2}=30$cm

Area of triangle using Heron's formula$=\sqrt{s(s-a)(s-b)(s-c)}$

where $s=30$cm , $a=10$cm,$b=24$cm and $c=26$cm

Area of triangle$=\sqrt{30(30-10)(30-24)(30-26)}$

$=\sqrt{30\times 20\times 6\times 4}$

$=\sqrt{6\times 5\times 5\times 4\times 4\times 6}=6\times 5\times 4=120$sq.m

Given Area of parallelogram$=$ Area of triangle

$\Rightarrow 24h=120$

$\Rightarrow h=\frac{120}{24}=5$m