The sides of a triangle are 16 cm, 12 cm and 20 cm. Find:

(i) area of the triangle

(ii) height of the triangle corresponding to the largest side

(iii) height of the triangle corresponding to the smallest side.

(i) Let a = 16 cm, b = 12 cm and c = 20 cm.

$∵ s = \dfrac{a + b + c}{2}\\[1em] = \dfrac{16 + 12 + 20}{2}\\[1em] = \dfrac{48}{2}\\[1em] = 24$

∵ Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

= $\sqrt{24(24 - 16)(24 - 12)(24 - 20)}$ cm²

= $\sqrt{24 \times 8 \times 12 \times 4}$ cm²

= $\sqrt{9,216}$ cm²

= 96 cm²

Hence, area of triangle = 96 cm².

(ii) When, the largest side of the triangle = 20 cm

Let the height of the triangle = h cm

As we know, the area of the triangle = $\dfrac{1}{2}$ x side x height

= $\dfrac{1}{2}$ x 20 x h

[∵ Both areas remain same since they represent the area of the same triangle.]

⇒ $\dfrac{1}{2}$ x 20 x h = 96

⇒ 10 x h = 96

⇒ h = $\dfrac{96}{10}$

⇒ h = 9.6

Hence, height of the triangle corresponding to 20 cm side is 9.6 cm.

(iii) When, the smallest side of the triangle = 12 cm

Let the height of the triangle = h cm

As we know, the area of the triangle = $\dfrac{1}{2}$ x side x height

= $\dfrac{1}{2}$ x 12 x h

[∵ Both areas remain same since they represent the area of the same triangle.]

⇒ $\dfrac{1}{2}$ x 12 x h = 96

⇒ 6 x h = 96

⇒ h = $\dfrac{96}{6}$

⇒ h = 16 cm

Hence, height of the triangle corresponding to 12 cm side is 16 cm.