Question

In the given figure ∆ABC and ∆DBC have the same base BC such that AB = 120 m, AC = 122 m, BC = 22 m, BD = 24 m and CD = 26 m. Find the area of the shaded region.

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ a=22\mathrm{m},b=122\mathrm{m}\mathrm{and}c=120\mathrm{m} \\ s=\frac{a+b+c}{2}=\frac{22+122+120}{2}=132\mathrm{m} \\ \mathrm{By}\mathrm{Heron}\text{'}\mathrm{s}\mathrm{formula},\mathrm{we}\mathrm{have}: \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC=\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{132(132-22)(132-122)(132-120)} \\ =\sqrt{132\times 110\times 10\times 12} \\ =\sqrt{11\times 12\times 11\times 10\times 10\times 12} \\ =11\times 12\times 10 \\ =1320{\mathrm{m}}^2 \end{gathered}$$

Now,
$$\begin{gathered} \mathrm{Let}: \\ a=22\mathrm{m},b=24\mathrm{m}\mathrm{and}c=26\mathrm{m} \\ s=\frac{a+b+c}{2}=\frac{22+24+26}{2}=36\mathrm{m} \\ \mathrm{By}\mathrm{Heron}\text{'}\mathrm{s}\mathrm{formula},\mathrm{we}\mathrm{have}: \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}OBC=\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{36(36-22)(36-24)(36-26)} \\ =\sqrt{36\times 14\times 12\times 10} \\ =\sqrt{12\times 3\times 7\times 2\times 12\times 5\times 2} \\ =12\times 2\sqrt{105} \\ =24\sqrt{105} \\ =246{\mathrm{m}}^2 \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{shaded}\mathrm{portion}=\mathrm{Area}\mathrm{of}∆ABC-\mathrm{Area}\mathrm{of}∆OBC \\ =(1320-246){\mathrm{m}}^2 \\ =1074{\mathrm{m}}^2 \end{gathered}$$