In the adjoining figure, show that ABCD is a parallelogram.
Calculate the area of || gm ABCD.
Given: A quadrilateral ABCD and BD is a diagonal.
To prove: ABCD is a parallelogram.
Construction: Draw AM ⊥ DC and CL ⊥ AB (extend DC and AB). Join AC, the other diagonal of ABCD.
Proof: ar(quad. ABCD) = ar(∆ABD) + ar(∆DCB)
= 2 ar(∆ABD) [∵ ar(∆ABD) = ar(∆DCB)]
$\therefore \operatorname{ar}(\Delta A B D)=\frac{1}{2} \operatorname{ar}($ quad. $A B C D)$ ...(i)
Again, ar(quad. ABCD) = ar(∆ABC) + ar(∆CDA)
= 2 ar(∆ ABC) [∵ ar(∆ABC) = ar(∆CDA)]
$\therefore \operatorname{ar}(\Delta A B C)=\frac{1}{2} \operatorname{ar}($ quad. $A B C D)$ ...(ii)
From (i) and (ii), we have:
$\operatorname{ar}(\Delta A B D)=\operatorname{ar}(\Delta A B C)=\frac{1}{2} \mathrm{AB} \times \mathrm{BD}=\frac{1}{2} \mathrm{AB} \times \mathrm{CL}$
⇒ CL = BD
⇒ DC || AB
Similarly, AD || BC.
Hence, ABCD is a paralleogram.
∴ ar(|| gm ABCD) = base ⨯ height = 5 ⨯ 7 = 35 cm2
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