Question

Find the area of the following figure.​

Answer 1

Given that,

• $\sf\ \: Parallel \: sides \: of \: a \: trapezium \: are \: 18cm \: and \: 30cm.$
• $\sf\ \: Non \: parallel \: side \: is \: 15cm.$

Firstly, we've to find the distance between the parallel sides.

• $\sf \: Construct \: a \: line \: from \: Q \: to \: T \: and \: join \: TQ.$
• $\sf \: TQ \: is \: the \: perpendicular.$
• $\sf \: Now, \: PQ \: = \: ST$
• $\sf \: So, \: TR \: = \: SR \: - \: PQ \: = \: 30 \: - \: 18 \: = \: 12cm.$

$\sf As \: we \: know \: that$ :

$\boxed{\boxed{\bold\purple{ {\bigstar \: \:hypotenuse }^{2} = {perpendicular}^{2} + {base}^{2} }}}$

$\underline{\bold{According \: to \: question \: - \: \: \: }}$

$\sf\implies\: \sf\purple{(QR)^{2} = {(TQ)}^{2} + {(TR)}^{2} } \\ \\ \sf\implies(15) {}^{2} = (TQ) {}^{2} + (12) {}^{2} \\ \\ \sf\implies \: 225 = (TQ) ^{2} \: + 144 \\ \\ \sf\implies \: 225 - 144 = {(tQ)}^{2} \\ \\ \sf\implies \: 81 = {(TQ)}^{2} \\ \\ \sf\implies \: TQ \: = \sqrt{81} \\ \\ \sf\implies \: TQ \: = \boxed{\bold\purple{9cm}} \: \bigstar$

$\sf\therefore\underline{Distance \: between \: the \: parallel \: sides \: is \: \bold{9cm.}}$

$\sf \: As \: we \: know \: that$ :

$\boxed{\boxed{\bold\purple{\bigstar \: area \: of \: a \: trapezium \: = \frac{1}{2} \times (sum \: of \: parallel \: sides \:) \times distance }}}$

$\sf \: Now$

$\sf\implies \: area \: = \frac{1}{2} \times \: (18 + 30) \times 9 \\ \\ \sf\implies \: area \: = \frac{1}{2} \times 48 \times 9 \\ \\ \sf\implies \: area \: = \: \frac{48}{2} \times 9 \\ \\ \sf\implies \: \: area \: = \cancel \dfrac{48}{2} \times 9 \: \\ \\ \sf\implies \: area \: = 24 \times 9 \: \\ \\ \sf\implies \: area \: = \: \boxed{\bold\purple{216 {cm}^{2} }} \bigstar$

$\sf\therefore\underline{Hence, \: the \: area \: of \: the \: trapezium \: is \: \bold{216cm².}}$