Question

compute the are of trapezium pqrs​

Answer 1

Answer:

$from \: the \: figure \\ using \: the \: Pythagoras \: theorem \: in \: △ \: rtq \\ we \: get \\ {rt}^{2} + {TQ}^{2} = {qr}^{2} \\ by \: substituting \: the \: value \\ {rt}^{2} + {8}^{2} = {17}^{2} \\ on \: further \: calculation \\ {rt}^{2} = {17}^{2} + {8}^{2} \\ so \: we \: get \\ {rt}^{2} = 289 + 64 \\ by \: subtracting \: {rt}^{2} = 225 \\ by \: taking \: square \: root \\ rt \: = \sqrt{225} \\ rt \: = 15km \\ we \: can \: find \: the \: area \: of \: trapezium \\ area \: of \: trapezium \: PQRS = ½((8 + 16) \times 15) \\ on \: further \: calculation \: we \: get \\ area \: of \: trapezium \: PQRS \: = {180cm}^{2}$