Question

Find the area of a parallelogram ABCD in which AB = 14 cm, BC = 10 cm and AC = 16 cm. $(Given,\sqrt{3}=\mathrm{1.73.})$

Answer 1

In Δ ABC,
Using Heron's Formula,
S = (a + b + c)/2
⇒ S = (10 + 14 +16)/2
⇒ S = 40/2
⇒ S = 20

Area = $\sqrt{S(S-a)(S-b)(S-c)}$
∴ Area of the triangles = $\sqrt{20(20-10)(20-14)(20-16)}$
⇒ Area of Δ ABC = $\sqrt{20\left(10\right)\left(6\right)\left(4\right)}$
⇒ Area = √4800
⇒ Area = 10√48
⇒ Area = 10√(2 × 2 × 2 × 2 × 3)
⇒ Area = 10 × 2 × 2√3
⇒ Area of Δ ABC = 40√3 cm²
∴ Area of Δ ABC = 69.2 cm².

We know,
Area of the Δ ADC = Area of the ΔABC.
[∵ Both the triangles are congruent)

∴ Area of the Parallelogram = 2 × Area of Δ ABC.
⇒ Area of the Parallelogram = 2 × 69.2 cm²
∴ Area of the Parallelogram = 138.4 cm².


Hence, the Area of the Parallelogram is 138.4 cm².