5 Marks Questions

Question 15 Marks

$P, Q, R, S$ are respectively the midpoints of the sides $AB, BC, CD$ and $DA$ of $||gm$ $ABCD$.
Show that $PQRS$ is a parallelogram and also show that $\text{ar}(||\text{gm PQRS})=\frac{1}{2}\times\text{ar}(||\text{gm ABCD}).$

Answer

Given: $ABCD$ is a parallelogram and $P, Q, R$ and $S$ are the moidpoints of sides $AB, BC, CD$ and $DA$, respectively.
To prove: $\text{ar}(||\text{gm PQRS})=\frac{1}{2}\times\text{ar}(||\text{gm ABCD})$
Proof: In $\triangle\text{ABC},$ $PQ\ ||\ AC$ and $\text{PQ}=\frac{1}{2}\times\text{AC}$ [By mid-point theorem] Again, in $\triangle\text{DAC},$ the points $S$ and $R$ are the mid-points of $AD$ and $DC$, respectively.
$\therefore$ $SR\ ||\ AC$ and $\text{SR}=\frac{1}{2}\times\text{AC}$ [By mid-point theorem]
Now, $PQ\ ||\ AC$ and $SR\ ||\ AC \Rightarrow PQ\ ||\ SR$ Also,
$\text{PQ}=\text{SR}=\frac{1}{2}\times\text{AC}$
$\therefore$ $PQ || SR$ and $PQ = SR$
Hence, $PQRS$ is a parallelogram Now, ar(parallelogram $PQRS$) $=\text{ar}(\triangle\text{PSQ})+\text{ar}(\triangle\text{SRQ})\dots(\text{i})$
Also, ar(parallelogram $ABCD$) = ar(parallelogram $ABQS$) + ar(parallelogram $SQCD$) $...(ii)$
$\triangle\text{PSQ}$ and parallelogram $ABQS$ are on the same base and between the same parallel lines.
So, $\text{ar}(\triangle\text{PSQ})=\frac{1}{2}\times\text{ar}(\text{parallelogram ABQS})\dots(\text{iii})$
Similary, $\triangle\text{SRQ}$ parallelogram $SQCD$ are on the same base and between the same parallel lines.
So, $\text{ar}(\triangle\text{SQR})=\frac{1}{2}\times\text{ar}(\text{parallelogram SQCD})\dots(\text{iv})$
Putting the value from $(iii)$ and $(iv)$ in $(i)$,
we get: ar(parallelogram $PQRS$) $=\frac{1}{2}\times\text{ar}(\text{parallelogram ABQS})\\+\frac{1}{2}\times\text{ar(parallelogram SQCD)}$ From $(ii)$,
we get: ar(parallelogram $PQRS$) $=\frac{1}{2}\times\text{ar(parallelogram ABCD)}$

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Question 25 Marks

In a trapezium $ABCD, AB\ ||\ DC, AB = a\ cm$, and $DC = b\ cm$. If $M$ and $N$ are the midpoints ofthe nonparallel sides, $AD$ and $BC$ respectively then find the ratio of $ar(DCNM)$ and $ar(MNBA)$.

Answer

Construction: Join $DB$. Let $DB$ cut $MN$ at point $Y$.

$M$ and $N$ are the mid-points of $AD$ and $BC$ respectively.
$\Rightarrow$ $MN\ ||\ AB ||\ CD$ In $\triangle\text{ADB},$ $M$ is the mid-point of $AD$ and $MY\ ||\ AB$.
$\therefore$ $Y$ is the mid-point of $DB$.
$\Rightarrow\ \text{MY}=\frac{1}{2}\text{AB}$ Similarly, in $\triangle\text{BDC},$
$\Rightarrow\ \text{NY}=\frac{1}{2}\text{CD}$ Now, $MN = MY + YN$
$\Rightarrow\ \text{MN}=\frac{1}{2}\text{AB}+\frac{1}{2}\text{CD}$
$\Rightarrow\ \text{MN}=\frac{1}{2}(\text{AB}+\text{CD})$
$\Rightarrow\ \text{MN}=\frac{\text{a}+\text{b}}{2}$
Construction: Draw $\text{DQ}\perp\text{AB}.$ Let $DQ$ cut $MN$ at point $P$.
Then, $P$ is the mid-point of $DQ$. i.e. $DP = PQ = h$ (say) Now,
$\text{ar(trapezium DCNM)}=\frac{1}{2}\times(\text{MN}+\text{CD})\times\text{DP}$
$=\frac{1}{2}\times\Big(\frac{\text{a+b}}{2}+\text{b}\Big)\text{h}=\frac{\text{h}}{4}(\text{a}+3\text{b})$
$\text{ar(trapezium MNBA)}=\frac{1}{2}\times(\text{MN}+\text{AB})\times\text{PQ}$
$=\frac{1}{2}\times\Big(\frac{\text{a+b}}{2}+\text{a}\Big)\text{h}=\frac{\text{h}}{4}(3\text{a}+\text{b})$
$\Rightarrow\ \frac{\text{ar(trapezium DCNM)}}{\text{ar(trapezium MNBA)}}=\frac{\frac{\text{h}}{4}(\text{a}+3\text{b})}{\frac{\text{h}}{4}(3\text{a}+\text{b})}=\frac{\text{a}+3\text{b}}{3\text{a}+\text{b}}$
$\Rightarrow\ \text{ar(trapezium DCNM)}:{\text{ar(trapezium MNBA)}}$
$={\text{a}+3\text{b}}:{3\text{a}+\text{b}}$

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Question 35 Marks

$ABCD$ is a trapezium in which $AB\ ||\ CD, AB = 16\ cm$ and $DC = 24\ cm$. If $E$ and $F$ are respectively the midpoints of $AD$ and $BC$, prove that $\text{ar(ABFE)}=\frac{9}{11}\text{ar(EFCD)}.$

Answer

Construction: Join $AC$. Let $AC$ cut $EF$ at point $Y$.

$E$ and $F$ are the mid-points of $AD$ and $BC$ respectively.
$\Rightarrow EF\ ||\ AB ||\ CD$ In $\triangle\text{ADC},$ $E$ is the mid-point of $AD$ and $EY\ ||\ CD$.
$\therefore$ $Y$ is the mid-point of $AC$.
$\Rightarrow\ \text{EY}=\frac{1}{2}\text{CD}$ Similarly, in $\triangle\text{ABC},$
$\Rightarrow\ \text{FY}=\frac{1}{2}\text{AB}$
Now, $EF = EY + YF$ $\Rightarrow\ \text{EF}=\frac{1}{2}\text{CD}+\frac{1}{2}\text{AB}$
$\Rightarrow\ \text{EF}=\frac{1}{2}(\text{CD}+\text{AB})$
$\Rightarrow\ \text{MN}=\frac{24+16}{2}=20\text{cm}$
Construction: Draw $\text{AQ}\perp\text{DC}.$
Let $AQ$ cut $EF$ at point $P$. Then, $P$ is the mid-point of $AQ$. i.e. $AP = PQ = h$ (say) Now, $\text{ar(trapezium ABFE)}=\frac{1}{2}\times(\text{EF}+\text{AB})\times\text{AP}$ $=\frac{1}{2}\times(20+16)\text{h}=18\text{h}\text{cm}^2$ $\text{ar(trapezium EFCD)}=\frac{1}{2}\times(\text{EF}+\text{CD})\times\text{PQ}$ $=\frac{1}{2}\times(20+24)\text{h}=22\text{h}\text{cm}^2$ $\Rightarrow\ \frac{\text{ar(trapezium ABFE)}}{\text{ar(trapezium EFCD)}}=\frac{18\text{h}}{22\text{h}}=\frac{9}{11}$ $\Rightarrow\ \text{ar(trapezium ABFE)}=\frac{9}{11}\times\text{ar(trapezium EFCD)}$

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