Question

Match the figure is Column-I with their shaded areas given the Column-II.

• A. $\left(P\right)\to \left(3\right),\left(Q\right)\to \left(1\right),\left(R\right)\to \left(2\right),\left(S\right)\to \left(4\right)$
• B. $\left(P\right)\to \left(1\right),\left(Q\right)\to \left(3\right),\left(R\right)\to \left(3\right),\left(S\right)\to \left(4\right)$
• C. $\left(P\right)\to \left(2\right),\left(Q\right)\to \left(3\right),\left(R\right)\to \left(1\right),\left(S\right)\to \left(4\right)$
• D. $\left(P\right)\to \left(3\right),\left(Q\right)\to \left(1\right),\left(R\right)\to \left(4\right),\left(S\right)\to \left(2\right)$

Answer 1

Answer: (D)

$\left(P\right)\to \left(3\right),\left(Q\right)\to \left(1\right),\left(R\right)\to \left(4\right),\left(S\right)\to \left(2\right)$

(P) Area of $△ABC$
$=\frac{1}{2}\times 25\times 18=225c{m}^2$
$=$ Area of $△BDC=\frac{1}{2}\times 18\times 6=54c{m}^2$
$\therefore$ Area of shaded region $=225-54$
$=171c{m}^2$

(Q) Area of shaded region
$=\frac{1}{2}\times AB\times AE+\frac{1}{2}\times AB\times DE$ {Because ED parallel to BC therefore height is AB}
$=\frac{1}{2}\times AB\times (AE+DE)=\frac{1}{2}\times 6\times 18=54c{m}^2$

(R) Area of shaded region$=$ Area of $△AED+$ Area of square $ABCD-$ Area of circle
$=\frac{1}{2}\times AD\times ED+DC\times DC-\pi \times (7{)}^2$
$=\frac{1}{2}\times 24\times 24+(24{)}^2-\frac{22}{7}\times (7{)}^2$
$288+576-154=710c{m}^2$

(S) Shaded area $=$ Area of rectangle $ABCD-$ Area of rectangle $EFGH$
$=AB\times AD-EF\times FG$
$=(18+3)\times (9+3)-18\times 9$
$=252-162=90c{m}^2$.