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case /data -based (4 Marks)

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1. (b): Perimeter of farmland $X =6 a^2+2 a^2 b-2 a b-4 b^2$.
Length of farmland $X =3 a^2+a^2 b-2 a b$.
Now, $2(l+b)=$ perimeter [for a rectangle]
$
\begin{aligned}
\Rightarrow \quad b & =\left(\frac{1}{2} \times \text { perimeter }\right)-l=\frac{1}{2}\left(6 a^2+2 a^2 b-2 a b-4 b^2\right)-\left(3 a^2+a^2 b-2 a b\right) \\
& =\left(3 a^2+a^2 b-a b-2 b^2\right)-\left(3 a^2+a^2 b-2 a b\right) \\
& =\left(3 a^2+a^2 b-a b-2 b^2-3 a^2-a^2 b+2 a b\right)=a b-2 b^2
\end{aligned}
$
2. (a): Length of the combined farmland ( $X + Y$ )
$
\begin{array}{l}
=\left(3 a^2+a^2 b-2 a b\right)+\left(a^2+a b\right) \\
=3 a^2+a^2 b-2 a b+a^2+a b=4 a^2+a^2 b-a b=4 a^2-a b+a^2 b .
\end{array}
$
3. $\begin{array}{l}\text { (d): Perimeter of combined farmland (X }+ Y ) \\ \qquad \begin{aligned} & =2 \text { (length }+ \text { breadth })=2\left\{\left(4 a^2-a b+a^2 b\right)+\left(a b-2 b^2\right)\right\} \\ & =2\left\{4 a^2-a b+a^2 b+a b-2 b^2\right\}=2\left(4 a^2+a^2 b-2 b^2\right)=8 a^2+2 a^2 b-4 b^2\end{aligned}\end{array}$
4. (d): Area of farmland $Y =($ length of farmland Y $) \times($ breadth of farmland Y $)$
$\begin{array}{l}=\left(a^2+a b\right)\left(a b-2 b^2\right)=a^3 b-2 a^2 b^2+a^2 b^2-2 a b^3 \\ =a^3 b-a^2 b^2-2 a b^3 .\end{array}$
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case /data -based (4 Marks) - MATHS STD 7 Questions - Vidyadip