Question

5. Find the area of a trapezium PQRS such that PS IPQ,
QR = 5 cm, PQ = 7 cm, SR= 10 cm.
6. One of the parallel sides of a trapezium is thrice the
11.(plz solve it)​

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

\begin{gathered}x = \sqrt{ {13}^{2} - {5}^{2} } \\ x = \sqrt{169 - 25 } \\ x = \sqrt{144} \\ x = 12m \\ \end{gathered}

x=

13

2

−5

2

x=

169−25

x=

144

x=12m

TR = PQ = 12m

\begin{gathered}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} \end{gathered}

aroftrapezium=

2

1

×sumofparallelsides×height

=

2

1

×(12+7)×12

=19×6=114m

2