Question

The lengths of the sides of a triangle are 33 cm, 44 cm and 55 respectively. Find the area of the triangle and hence find the height corresponding to the side measuring 44 cm.

Answer 1

Let a = 33 cm, b = 44 cm and c = 55 cm
Then, s = $\frac{a+b+c}{2}$ = $\left(\frac{33+44+55}{2}\right)$ cm = $\left(\frac{132}{2}\right)$ cm = 66 cm
∴ Area of the triangle = $\sqrt{\mathrm{s}\left(\mathrm{s}-\mathrm{a}\right)\left(\mathrm{s}-\mathrm{b}\right)\left(\mathrm{s}-\mathrm{c}\right)}$ sq. units
= $\sqrt{66\left(66-33\right)\left(66-44\right)\left(66-55\right)}$ cm²
= $\sqrt{66\times 33\times 22\times 11}$ cm²
= $\sqrt{6\times 11\times 3\times 11\times 2\times 11\times 11}$ cm²
= (6 $\times$ 11 $\times$ 11) cm² = 726 cm²

Let the height on the side measuring 44 cm be h cm.
Then, Area = $\frac{1}{2}\times b\times h$
⇒ 726 cm² = $\frac{1}{2}\times 44\times h$
⇒ h = $\left(\frac{2\times 726}{44}\right)$ cm = 33 cm.
∴ Area of the triangle = 726 cm²
Height corresponding to the side measuring 44 cm = 33 cm