Question

find the area of the trapezium PQRS with height PQ. where SR is 13m, RQ is 7m and PS is 12m and the value of PQ is unknown.

Answer 1

In the figure ,,
SR = 13m
QR = PT = 7m
PS = 12m
so,TS = PS-PT = 12-7 = 5m
Now In ∆STR
let TR = x m
$x = \sqrt{ {13}^{2} - {5}^{2} } \\ x = \sqrt{169 - 25 } \\ x = \sqrt{144} \\ x = 12m$
TR = PQ = 12m
$ar \: of \: trapezium = \frac{1}{2} \times sum \: of \: parallel \: sides \times height \\ = \frac{1}{2} \times (12 + 7) \times 12 \\ = 19 \times 6 = 114 {m}^{2}$

Answer 2

Answer:

Step-by-step explanation: