5 Mark Question

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 ⇒ 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 = acm, and DC = bcm. 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. ⇒ 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 = 16cm and DC = 24cm. 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. ⇒ 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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