Activity Based Questions.

Question 15 Marks

If ABCD is a parallelogram, then prove that$\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{BCD})\\ \ =\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ACD})=\frac{1}{2}\text{ar}$ $(||^{gm} ABCD)$

Answer

Given, ABCD is a parallelogram.
To prove: $\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{BCD})=\text{ar}(\triangle\text{ABC})\\ \ =\text{ar}(\triangle\text{ACD})=\frac{1}{2}\text{ar}$
$(||^{gm} ABCD)$
Proof: We know that diagonal of a parallelogram divides it into two equal triangles.
Since, AC is the diagonal
Then, $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ACD})=\frac{1}{2}\text{ar}$ $(||^{gm} ABCD)$ ...(1)
Since, BD is the diagonal
Then $\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{BCD})=\frac{1}{2}\text{ar}$ $(||^{gm} ABCD)$ ...(2)
Compare equation (1) and (2)
$\therefore\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ACD})=\text{ar}(\triangle\text{ABD})\\ \ =\text{ar}(\triangle\text{BCD})=\frac{1}{2}\text{ar}$ $(||^{gm} ABCD)$

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Question 25 Marks

ABCD is a parallelogram whose diagonals intersect at O .If P is any point on BO, prove that:

$\text{ar}(\triangle\text{ADO})=\text{ar}(\triangle\text{CDO})$
$\text{ar}(\triangle\text{ABP})=\text{ar}(\triangle\text{CBP})$

Answer

Given that ABCD is the parallelogram To Prove:

$\text{ar}(\triangle\text{ADO})=\text{ar}(\triangle\text{CDO}).$
$\text{ar}(\triangle\text{ABP})=\text{ar}(\triangle\text{CBP}).$

Proof: we know that diagonals of parallelogram bisect each other$\therefore$ AO = OC and BO = OD

In $\triangle\text{DAC},$ since DO is a median.

Then $\text{ar}(\triangle\text{ADO})=\text{ar}(\triangle\text{CDO}).$

In $\triangle\text{BAC},$ since BO is a median.

Then $\text{ar}(\triangle\text{BAO})=\text{ar}(\triangle\text{BCO})\ ....(1)$
In $\triangle\text{PAC},$ since PO is a median.
Then $\text{ar}(\triangle\text{PAO})=\text{ar}(\triangle\text{PCO})\ .....(2)$
Subtract equation 2 from 1.
$\Rightarrow\text{ar}(\triangle\text{BAO})−\text{ar}(\triangle\text{PAO})\\ \ =\text{ar}(\triangle\text{BCO})−\text{ar}(\triangle\text{PCO})$
$\Rightarrow\text{ar}(\triangle\text{ABP})=2\text{ar}(\triangle\text{CBP}).$

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Question 35 Marks

In figure, ABCD is a trapezium in which AB || DC and DC = 40cm and AB = 60cm. If X and Y are, respectively, the mid-points of AD and BC, prove that:

XY = 50cm
DCYX is a trapezium
$\text{ar}(\text{trap}.\ \text{DCYX})=\Big(\frac{9}{11}\Big)\text{ar}(\text{XYBA}).$

Answer

Join Dy and produce it to meet AB produced at P.

In triangles BYP and CYD we have,

$\angle\text{BYP}=\angle\text{CYD}$ [Vertically opposite angles]

$\angle\text{DCY}=\angle\text{PBY}$ [Since, DC || AP]

And BY = CY

So, by ASA congruence critrion, we have

$(\triangle\text{BYP})\cong(\triangle\text{CYD})$

⇒ DY = YP and DC = BP

⇒ Y is the midpoint of DP

Also, X is the midpoint of AD

Therefore, XY || AP and $\text{XY}||\Big(\frac{1}{2}\Big)\text{AP}$

$\Rightarrow\text{XY}=\Big(\frac{1}{2}\Big)(\text{AB}+\text{BP})$

$\Rightarrow\text{XY}=\Big(\frac{1}{2}\Big)(\text{AB}+\text{Dc})$

$\Rightarrow\text{XY}=\Big(\frac{1}{2}\Big)(60+40)$

$=50\text{cm}$

We have, XY || AP

⇒ XY || AB and AB || DC

⇒ XY || DC

⇒ DCYX is a trapezium

Since x and y are the midpoints of AD and BC respectively.

Therefore, trapezium DCYX and ABYX are of the same height say h cm

Now,

$\text{ar}(\text{trap}.\ \text{DCXY})=\Big(\frac{1}{2}\Big)(\text{DC}+\text{XY})\times\text{h}$

$\Rightarrow\text{ar}(\text{trap}.\ \text{DCXY})\\=\Big(\frac{1}{2}\Big)(50+40)\times\text{h}\text{ cm}^2=45\text{h}\text{ cm}^2$

$\Rightarrow\text{ar}(\text{trap}.\ \text{ABYX})=\Big(\frac{1}{2}\Big)(\text{AB}+\text{XY})\times\text{h}$

$\Rightarrow\text{ar}(\text{trap}.\ \text{ABYX})\\=\Big(\frac{1}{2}\Big)(60+50)\times\text{h}\text{ cm}^2=55\text{h}\text{ cm}^2$

$\text{ar}(\text{trap}.\ \text{DCYX})\ \text{ar}(\text{trap}.\ \text{ABYX})=\frac{45\text{h}}{55\text{h}}=\frac{9}{11}$

$\Rightarrow\text{ar}(\text{trap}.\ \text{DCYX})=\frac{9}{11}\text{ar}(\text{trap}.\ \text{ABYX})$

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