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  • The area of the parallelogram ABCD is 90 cm^2 (see Fig.9.13). Find (i) ar (ABEF) (ii) ar (ABD) (iii) ar (BEF)

The area of the parallelogram ABCD is 90 cm^{2} (see Fig.9.13). Find

(i) ar (ABEF)
(ii) ar (ABD)
(iii) ar (BEF)

Answers (1)

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Answer:   

(i) 90cm^{2}
(ii) 45cm^{2}
(iii) 45cm^{2}

Solution.
(i) Here, ABCD is a parallelogram with area 90cm^{2}
\therefore AB \parallel CD
So, AB \parallel CF [Extended part of CD is CF]
\therefore AB\parallel EF
Hence, ABEF is a parallelogram.
We know that parallelogram that lies between same parallel lines and having same base have equal areas.
Now, parallelogram ABCD and ABEF lies between same parallel lines AB and CF have same base AB. So these have equal area.
\therefore ar(ABCD) = ar(ABEF)
\therefore ar(ABEF) = 90 cm^{2}
(ii) \triangle ABD and parallelogram ABCD lies between same parallel lines and have same base AB. So we know that
\therefore ar(ABD)=\frac{1}{2}\times ar(ABCD)
Now, ar(ABCD)=90cm^{2}
\Rightarrow ar (ABD)=\frac{1}{2}\times 90cm^{2}
\therefore ar(ABD)=45cm^{2}
(iii) Here, \triangle BEF and parallelogram ABEF are on the same base EF and between the same parallels AB and EF. So we know that,
ar(\triangle BEF)=\frac{1}{2}ar(ABEF)=90cm^{2}
ar(\triangle BEF)=\frac{1}{2}\times 90=45cm^{2}

 

 

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