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Question

In figure $$AB=35\ cm,\ AD=20\ cm$$ and area of the parallelogram is $$560\ cm^2$$, find $$LB$$.

A
23
Solution
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Correct option is A. 23
We have $$\text{ABCD}$$ is a parallelogram with base $$\text{AB}=35\space\mathrm{cm}$$ and corresponding attitude $$\text{DL}$$. The adjacent side of the parallelogram $$\text{AD}=20\space \mathrm{cm}$$
It is given that the area of the parallelogram $$\text{ABCD}=560\space \mathrm{cm}^{2}$$
Now,
Area of the parallelogram $$=$$ Base$$\times$$Height
$$\Rightarrow560 \space\mathrm{cm}^{2}=\text{AB} \times \text{DL}$$
$$\Rightarrow560\space\mathrm{cm^2}=35\space\mathrm{cm}\times\text{DL}$$
$$\therefore\text{DL}=\dfrac{560\space\mathrm{cm^2}}{35\space\mathrm{cm}}=16\space\mathrm{cm}$$
Again by Pythagoras theorem, we have
$$(A D)^{2}=(A L)^{2}+(D L)^{2}$$
$$\Rightarrow(20)^{2}=(A L)^{2}+(16)^{2}$$
$$\Rightarrow(A L)^{2}=(20)^{2}-(16)^{2}$$
$$\Rightarrow(A L)^{2}=400-256$$
$$\Rightarrow(AL)^{2}=144$$
$$\Rightarrow A L=12\space \mathrm{cm}$$
From the figure $$A B=A L+LB$$
$$\Rightarrow35\space\mathrm{cm}=12 \space\mathrm{cm}+LB$$
$$\Rightarrow LB=35 \space\mathrm{cm}-12 \space\mathrm{cm}$$
$$\Rightarrow LB=23 \mathrm{cm}$$
Hence, Length of $$LB$$ is $$23\space\mathrm{cm}$$.

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