5 Marks 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 $(2)$
$\therefore\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ACD})=\text{ar}(\triangle\text{ABD})\\ \ =\text{ar}(\triangle\text{BCD})=\frac{1}{2}\text{ar}$ ($||^{gm} ABCD$)

View full question & answer→

Question 25 Marks

$\text{ABCD}$ is a parallelogram whose diagonals intersect at $O$ .If $P$ is any point on $BO,$ prove that:
$i. \text{ar}(\triangle\text{ADO})=\text{ar}(\triangle\text{CDO})$
$ii. \text{ar}(\triangle\text{ABP})=\text{ar}(\triangle\text{CBP})$

Answer

Given that $\text{ABCD}$ is the parallelogram To Prove:
$i. \text{ar}(\triangle\text{ADO})=\text{ar}(\triangle\text{CDO}).$
$ii. \text{ar}(\triangle\text{ABP})=\text{ar}(\triangle\text{CBP}).$
Proof: we know that diagonals of parallelogram bisect each other
$\therefore AO = OC$ and $BO = OD$
$i.$ In $\triangle\text{DAC},$ since $DO$ is a median.
Then $\text{ar}(\triangle\text{ADO})=\text{ar}(\triangle\text{CDO}).$
$ii.$ 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 $1.$
$\Rightarrow\text{ar}(\triangle\text{BAO})−\text{ar}(\triangle\text{PAO})\\ \ =\text{ar}(\triangle\text{BCO})−\text{ar}(\triangle\text{PCO})$
$\Rightarrow\text{ar}(\triangle\text{ABP})=2 \text{ar}(\triangle\text{CBP}).$

View full question & answer→

Question 35 Marks

In figure, $\text{ABCD}$ is a trapezium in which $AB \| DC$ and $DC = 40\ cm$ and $AB = 60\ cm$. If $X$ and $Y$ are, respectively, the mid$-$points of $AD$ and $BC,$ prove that:
$i. XY = 50\ cm$
$ii. \text{DCYX}$ is a trapezium
$iii. \text{ar}($trap.$\ \text{DCYX})=\Big(\frac{9}{11}\Big)\text{ar}(\text{XYBA}).$

Answer

$i.$ Join $Dy$ and produce it to meet $AB$ produced at $P.$
In triangles $\text{BYP}$ and $\text{CYD}$ 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 $\text{ASA}$ congruence critrion, we have
$(\triangle\text{BYP})\cong(\triangle\text{CYD})$
$\Rightarrow DY = YP$ and $DC = BP$
$\Rightarrow Y$ is the midpoint of $DP$
Also, $X$ is the midpoint of $AD$
Therefore, $XY \| AP$ and $\text{XY}\|\Big(\frac{1}{2}\Big)\text{AP}$
$\Rightarrow\text{XY}=\Big(\frac{1}{2}\Big)(\text{AB}+\text{BP})$
$\Rightarrow\text{XY}=\Big(\frac{1}{2}\Big)(\text{AB}+\text{Dc})$
$\Rightarrow\text{XY}=\Big(\frac{1}{2}\Big)(60+40)$
$=50\text{ cm}$
$ii.$ We have, $XY \| AP$
$\Rightarrow XY \| AB$ and $AB \| DC$
$\Rightarrow XY \| DC$
$\Rightarrow \text{DCYX}$ is a trapezium
$iii.$ Since $x$ and $y$ are the midpoints of $AD$ and $BC$ respectively.
Therefore, trapezium $\text{DCYX}$ and $\text{ABYX}$ are of the same height say $h\ cm$
Now,
$\text{ar}($trap.$\ \text{DCXY})=\Big(\frac{1}{2}\Big)(\text{DC}+\text{XY})\times\text{h}$
$\Rightarrow\text{ar}($trap.$\ \text{DCXY})$
$=\Big(\frac{1}{2}\Big)(50+40)\times\text{h}\text{ cm}^2=45\text{h}\text{ cm}^2$
$\Rightarrow\text{ar}($trap.$\ \text{ABYX})=\Big(\frac{1}{2}\Big)(\text{AB}+\text{XY})\times\text{h}$
$\Rightarrow\text{ar}($trap.$\ \text{ABYX})$
$=\Big(\frac{1}{2}\Big)(60+50)\times\text{h}\text{ cm}^2=55\text{h}\text{ cm}^2$
$\text{ar}($trap.$\ \text{DCYX})\ \text{ar}($trap.$\ \text{ABYX})=\frac{45\text{h}}{55\text{h}}=\frac{9}{11}$
$\Rightarrow\text{ar}($trap.$\ \text{DCYX})=\frac{9}{11}\text{ar}($trap.$\ \text{ABYX})$

View full question & answer→

Maths 5 Marks Questions

Test yourself on this topic