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The area of a trapezium is (in cm2)

Option: 1

208


Option: 2

154


Option: 3

196


Option: 4

204


Answers (1)

Let ABCD be the fi eld in the form of a trapezium in which AB is parallel to CD

AB = 25cm, BC = 13cm, CD = 10cm and DA = 10cm

Draw CE parallel to DA and CF perpendicular to AB

Clearly, ADCE is a parallelogram.

\begin{aligned} &\therefore \quad C E=D A=14 \mathrm{cm} \text { and } A E=C D=10 \mathrm{cm}\\ &\therefore \quad E B=A B-A E=(25-10) \mathrm{cm}=15 \mathrm{cm} \end{aligned}

In triangle EBC, we have

EB = 15cm, BC = 13cm and CE = 14cm

\begin{array}{l} \therefore \quad a=15 \mathrm{cm}, b=13 \mathrm{cm} \text { and } c=14 \mathrm{cm} \\ \therefore \quad s=\frac{1}{2}(15+13+14) \mathrm{cm}=21 \mathrm{cm} \end{array}

\begin{array}{l} \therefore \quad(s-a)=(21-15) \mathrm{cm}=6 \mathrm{cm},(s-b)=(21-13) \mathrm{cm}=8 \mathrm{cm} \\ \text { and }(s-c)=(21-14) \mathrm{cm}=7 \mathrm{cm} \end{array}

\therefore \quad \text { area }(\triangle E B C)=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{21 \times 6 \times 8 \times 7} \mathrm{cm}^{2}

Area of triangle EBC = 84 cm2

\begin{array}{l} \text { Also, area }(\triangle E B C)=\left(\frac{1}{2} \times E B \times C F\right)=\left(\frac{1}{2} \times 15 \mathrm{cm} \times C F\right) \\ \therefore \quad \frac{1}{2} \times 15 \mathrm{cm} \times C F=84 \mathrm{cm}^{2} \Rightarrow C F=\frac{84 \times 2}{15} \mathrm{cm}=\frac{56}{5} \mathrm{cm}=11.2 \mathrm{cm} \\ \therefore \quad C F=11.2 \mathrm{cm} \end{array}\begin{aligned} \text { Area }(\operatorname{trap} . A B C D) &=\frac{1}{2} \times(A B+C D) \times C F \\ &=\left\{\frac{1}{2} \times(25+10) \times 11.2\right\} \mathrm{cm}^{2} \\ &=(35 \times 5.6) \mathrm{m}^{2}=196 \mathrm{cm}^{2} \end{aligned}

Posted by

Sumit Saini

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