Maths M.C.Q

Figure obtained by joining the mid-points of adjacent sides of rectangle $ABCD$ is a rhombus $PQRS$.
$AB = 8\ cm, AD = 6\ cm$
$QS$ and $PR$ are diagonals of Rhombus $PQRS$.
$QS = AB = 8\ cm$
$PR = AD = 6\ cm$
Ar(Rhombus $PQRS$) $= 4 \times $ Area of $\triangle\text{POR}$
$=4\times\frac{1}{2}\times\text{OQ}\times\text{OP}$ $\Big(\triangle\text{POQ}$ is a Right $\triangle\text{OQ}=\frac{\text{QS}}{2},\ \text{OP}=\frac{\text{PR}}{2}\Big)$
$=\frac{{4}^2}{2}\times4\times3$
$\Rightarrow $ Ar(Rhombus $PQRS$) = $24 \mathrm{~cm}^2$
Hence, correct option is $(a)$.

$AQ, CP$ and $RB$ are medians of $\triangle\text{ABC}.$
Consider $\triangle\text{ACP}\ \&\ \triangle\text{ACQ}$
$\text{Ar}(\triangle\text{ACP})=\frac{27}{2}\text{cm}^2$ $\big[$Median divides a $\triangle$ into two equal Area$\big]$
$\text{Ar}(\triangle\text{ACQ})=\frac{27}{2}\text{cm}^2$ $\big[$Median divides a $\triangle$ into two equal Area$\big]$
$\Rightarrow\text{Ar}(\triangle\text{ACP})=\text{Ar}(\text{ACQ})$
$\text{Ar}(\triangle\text{AGC})$ is common in both the triangles.
$\Rightarrow\text{Ar}(\triangle\text{CGQ})=\text{Ar}(\triangle\text{AGP})\ ...(1)$
Similiarly $\text{Ar}\big(\triangle\text{ABR}\big)=\frac{27}{2}\text{cm}^2=\text{Ar}\big(\triangle\text{AQB}\big)$
$\text{Ar}\big(\triangle\text{ARG}\big)$ is commpn in both the triangle.
$\Rightarrow\text{Ar}\big(\triangle\text{ARG}\big)=\text{Ar}\big(\triangle\text{GQB}\big)\dots(2)$
From figure $GR, GP, GQ$ are also medians for $\triangle\text{AGC},\triangle\text{AGB }\&\triangle\text{CGB}$ respectively.
$\Rightarrow\text{Ar}\big(\triangle\text{AGC}\big)+\text{Ar}\big(\triangle\text{AGB}\big)+\text{Ar}\big(\triangle\text{CGB}\big)=27\text{cm}^2$
$\Rightarrow2\text{Ar}\big(\triangle\text{ARG}\big)+2\text{Ar}\big(\triangle\text{AGP}\big)+\text{Ar}\big(\triangle\text{BGC}\big)=27\text{cm}^2$
$\Rightarrow\big(2\text{Ar}\big(\triangle\text{ARG}\big)+\text{Ar}\big(\triangle\text{AGP}\big)+\text{Ar}\big(\triangle\text{BGC}\big)=27\text{cm}^2$
From equauations $(1)$ and $(2)$.
$2\big[\text{Ar}\big(\triangle\text{GQB}\big)+\text{Ar}\big(\triangle\text{AGB}\big)\big]+\text{Ar}\big(\triangle\text{CGB}\big)=27\text{cm}^2$
$\Rightarrow2\big[\text{Ar}\big(\triangle\text{BGC}\big)\big]+\text{Ar}\big(\triangle\text{BGC}\big)=27\text{cm}^2$
$\Rightarrow\text{Ar}\big(\triangle\text{BGC}\big)=9\text{cm}^2$
Hence, Correct option is $(b)$.

Diagonal $SQ$ divides $||^{\text{gm}}$ in two equal areas.
Hence $\text{Ar}(\triangle\text{QSR})=\frac{1}{2}\text{Ar}(\text{PQRS})$
Also $XY$ divides the $||^{\text{gm}}$ into two equal parts.
Hence, Ratio of $\text{Area}(||^{\text{gm}}\text{XQRY})=\frac{1}{2}\text{Ar}(\text{PQRS})$
Thus, Ratio of $\text{Area}(||^{\text{gm}}\text{XQRY}):\text{Ar}(\triangle\text{QSR})=1:1$
Hence, correct option is $(d)$.

$AC = 12\ cm$ and $BD = 16\ cm$ (given)
Now, consider $\triangle\text{ABC}.$
$P$ and $Q$ are mid-points of sides $AB$ and $BC$
So line joining them will be parallel to the third side $AC$ and equal to $\frac{1}{2}\text{AC}.$
$\Rightarrow\text{PQ}=\frac{1}{2}\text{AC}=6\text{cm}$
Similiary, in $\triangle\text{ABD},$ $PS\ ||\ BD$
$\Rightarrow\text{PS}=\frac{1}{2}\text{BD}=8\text{cm}$
Now we know that by joining mid-points of adjacent sidea of a Rhombus.
We get a Rectangle whose sides are $PQ$ and $PS$.
$\Rightarrow$ Area $=P Q \times P S=6 \times 8=48 \mathrm{~cm}^2$
Hence, correct option is $(b)$.

Area of $\triangle\text{ABC}$ = Area of $\triangle\text{AEF}$ + Area of trapazium $FBCE$
We know that any triangle formed by joining the mid-point of sides of triangle.
has area $=\frac{1}{4}\times(\text{parent}\triangle)$
$\Rightarrow\text{Area of }\triangle\text{AEF}=\frac{1}{4}\times\text{Ar}(\triangle\text{ABC})\\=\frac{1}{4}\times16\text{cm}^2=4\text{cm}^2$
$\Rightarrow$ Area of Trapezium $=(16-4) \mathrm{cm}^2=12 \mathrm{~cm}^2$
Hence, correct option is $(c)$.

Area of trapazium $ABCP$ = Area of $\triangle\text{APB}$ + ar of $\triangle\text{BPC}\ ...(1)$
$\triangle\text{APC}$ and $\triangle\text{BPC}$ have same base $PC$ and are batween same parallels.
$\Rightarrow $ Area of $\triangle\text{APC}$ = Area of $\triangle\text{BPC}$ = $20 \mathrm{~cm}^2$ $...(2)$
From figure, $\text{Ar}(\triangle\text{ADP})+\text{Ar}(\triangle\text{APC})=\frac{1}{2}\text{Ar}(||^{\text{gm}}\text{ABCD})$
$\Rightarrow\text{Ar}(||^{\text{gm}}\text{ABCD})=2(20+15)=70\text{cm}^2$
Area of trapazium ABCP $=\text{Ar}(||^{\text{gm}}\text{ABCD})-\text{Ar}(\triangle\text{ADP)}=70-15=55\text{cm}^2$
$\Rightarrow $ Area of $\triangle\text{APB}$ Area of trapezium $ABCP$ - Area of $\triangle\text{BPC}$
$=(55-20)\text{cm}^2$ [From $(1)$]
$=35\text{cm}^2$
Hence, correct option is $(c)$.