Question

Find the area of the trapezium $PQRS$ with height $PQ$ using Heron's formula.

• A. $75$ sq m
• B. $63$ sq m
• C. $114$ sq m
• D. None of these

Answer 1

Answer: (C)

$114$ sq m

In $△$STR, using Pythajoras theorem,

TR = $\sqrt{{SR}^2-{ST}^2}=\sqrt{{\left(13\right)}^2-{\left(5\right)}^2}=\sqrt{144}=12cm$

$\therefore$ $△$STR = $\sqrt{s(s-a)(s-b)(s-c)},s=\frac{a+b+c}{2}=\frac{13+12+15}{2}=15m$

ar $△$STR = $\sqrt{15(15-13)(15-12)(15-5)}=\sqrt{15\times 2\times 3\times 10}=30m^2$

Area rectangle PQRT = PQ $\times$ QR = 12 $\times$ 7 = 84$m^2$

$\therefore$ Area of trapezium $= 30 + 84 = 114{ m }^{ 2 }$