Maths M.C.Q

$ABCD$ is a $||\ gm$ and $AC$ and $BD$ are the diagonals.
So, $\text{ar}(\triangle\text{ABD})=\frac{1}{2}\text{ar}(||\text{gm ABCD})$
We know that, the diagonals of a parallelogram bisect each other.
So, $O$ is the mid-point of $AC$ and $BD.$
In $\triangle\text{ABD},$
$AO$ is the median on $BD.$
So, $\text{ar}(\triangle\text{OAB})=\frac{1}{2}\text{ar}(\triangle{\text{ABD}})$
$=\frac{1}{2}\times\frac{1}{2}\text{ar}(|| \text{gm ABCD})$
$=\frac{1}{2}\times\frac{1}{2}\times52$
$=13\text{cm}^2$

Clearly, $PQRS$ is a rhombus whose area $=\frac{1}{2}\times\text{PR}\times\text{SQ}$
$=\frac{1}{2}\times6\times8$
$=24\text{cm}^2$

In right $\triangle \mathrm{BMC}$,
By pythagoras theorem,
$\mathrm{BC}^2=\mathrm{BM}^2+C M^2$
$\Rightarrow \mathrm{CM}^2=\mathrm{BC}^2-\mathrm{BM}^2$
$\Rightarrow \mathrm{CM}^2=5^2-4^2$
$\Rightarrow \mathrm{CM}^2=25-16$
$\Rightarrow C M^2=9$
$\Rightarrow \mathrm{CM}=3 \mathrm{~cm}$
In right $\triangle \mathrm{ALD}$
By pythagoras theorem,
$A D^2=A L^2+D L^2$
$\Rightarrow D L^2=A D^2-A L^2$
$\Rightarrow D L^2=5^2-4^2$
$\Rightarrow D L^2=25-16$
$\Rightarrow D L^2=9$
$\Rightarrow D L=3 \mathrm{~cm}$
Since $AMML$ forms a rectangle, the opposite sides are equal.
So, $\mathrm{AB}=\mathrm{LM}=7 \mathrm{~cm}$
Area of a trapezium $=\frac{1}{2} \times$ (Sum of the parallel sides) $\times$ Distance between the parallel sides
$=\frac{1}{2} \times(\mathrm{AB} \times \mathrm{CD}) \times \mathrm{AL}$
$=\frac{1}{2} \times(3+7+3) \times 4$
$=40 \mathrm{~cm}^2$

$ABCD$ is a $\| \mathrm{gm}$ and $\mathrm{DL} \perp \mathrm{AB}$.
So, $D L$ is the height of the $\| \mathrm{gm} A B C D$.
$\operatorname{ar}(\| \mathrm{gm} \mathrm{ABCD})=\text { base } \times \text { height }$
$=\mathrm{AB} \times \mathrm{DL}$
$=10 \times 4$
$=40 \mathrm{~cm}^2$

From the given figure, we have $||\ gm\ ABCD$ and $||\ gm\ ABFE$ are lying on the same base and between the same parallels.
$\Rightarrow\ \text{ar}(||\text{gm ABCD})=\text{ar}(||\text{gm ABFE})=25\text{cm}^2$
$\Rightarrow\ \text{ar}(\triangle\text{BCF})=\text{ar}(||\text{gm ABFE})-\text{ar(quadrilateral EABC)}$
$=25-17$
$=8\text{cm}^2$

Area of a trapezium $=\frac{1}{2}\times ($Sum of the parallel sides$)\ ×$ Distance between the parallel sides
$=\frac{1}{2}\times(12+8)\times6.5$
$=65\text{cm}^2$

Since $A B C D$ is a trapezium, $A B \| C D$, that is $A E \| C D$.
Also, $A E=C D=8 \mathrm{~cm}$
So, $A E C D$ is a square.
In right $\triangle \mathrm{CEB}$,
By Pythagoras theoram,
$\mathrm{BC}^2=\mathrm{EC}^2+E \mathrm{~EB}^2$
$\Rightarrow \mathrm{EC}^2=\mathrm{BC}^2-\mathrm{EB}^2$
$\Rightarrow \mathrm{EC}^2=17^2-15^2$
$\Rightarrow \mathrm{EC}^2=289-225$
$\Rightarrow E C^2=64$
$\Rightarrow E C^2=8 \mathrm{~cm}$
Which is the required height of trapezium $A B C D$.
$\operatorname{ar}($ trapezium $ABCD )$
$=\frac{1}{2} \times$ (Sum of the parallel sides) $\times$ height
$=\frac{1}{2} \times(\mathrm{AB}+\mathrm{CD}) \times \mathrm{EC}$
$=\frac{1}{2} \times(8+15+8) \times 8$
$=124 \mathrm{~cm}^2$

Area of quadrilateral $ABCD =\text{ar}(\triangle\text{ABD})+\text{ar}(\triangle\text{BCD})$
$=\frac{1}{2}\times\text{BD}\times\text{AL}+\frac{1}{2}\text{BD}\times\text{CM}$
$=\frac{1}{2}\times16\times9+\frac{1}{2}\times16\times7$
$=72+56$
$=128\text{cm}^2$

$a.$ The area of the rhombus $=\frac{1}{2} \times$ product of the diagonals
$=\frac{1}{2} \times 18 \times 14$
$=126 \mathrm{~cm}^2$
So, the statement is true.
$b.$ area of a $\| g m=$ base $\times$ height
So, the statement is false.
$c.$ Since parallelograms between the same parallel lines and on the same base are equal in area, and a rectangle is a $ll\ gm,$ the statement is true.
$d.$ area of a $\| g m=$ base $\times$ height
$\Rightarrow 192=24 \times h$
$\Rightarrow h=8 \mathrm{~cm}$
So, the statement is true.

Given that $ABCD$ is a rhombus.
$BC = CD [$Since all the sides of a rhombus are equal$]$
$\Rightarrow\ \angle\text{BDC}=\angle\text{BCD} [$Angles opposite equal sides are equal$]$
In $\triangle\text{BCD},$
$\angle\text{BDC}+\angle\text{BCD}+\angle\text{CBD}=180^\circ[ $Angle sum property$]$
$\Rightarrow\ \angle\text{BDC}+60^\circ+\angle\text{BDC}=180^\circ$
$\Rightarrow\ 2\angle\text{BDC}=120^\circ$
$\Rightarrow\ \angle\text{BDC}=60^\circ$
$\Rightarrow\ \angle\text{BDC}=\angle\text{BCD}=60^\circ$
So, $\triangle\text{BCD},$ is an equilateral triangle.
$\therefore B D=B C=C D=x \text { (say) }$
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
In right $\triangle \mathrm{AOB}$,
$A B^2=A O^2+O B^2[B y$ Pythagoras theoram $]$
$\Rightarrow A O^2=A B^2-O B^2$
$\Rightarrow\ \text{AO}^2=\text{x}^2-\Big(\frac{\text{x}}{2}\Big)^2$
$\Rightarrow\ \text{AO}^2=\text{x}^2-\Big(\frac{\text{x}^2}{4}\Big)$
$\Rightarrow\ \text{AO}^2=\frac{3\text{x}^2}{4}$
$\Rightarrow\ \text{AO}=\frac{\sqrt{3}\text{x}}{2}$
So, $\text{AC}=\sqrt{3}\text{x}$
Thus,
$\frac{\text{AC}}{\text{BD}}=\frac{\sqrt{3}\text{x}}{\text{x}}=\frac{\sqrt{3}}{1}$
Hence,
$\text{AC}:\text{BD}=\sqrt{3}:1.$

Since $ABCD$ is a rectangle, $\text{AD}=\text{BC}=2\sqrt{5}\text{cm}$
In right $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\ \text{AB}^2=\text{AC}^2-\text{BC}^2$
$\Rightarrow\ \text{AB}^2=10^2-(2\sqrt{5})^2$
$\Rightarrow\ \text{AB}^2=100-20$
$\Rightarrow\ \text{AB}=4\sqrt{5}\text{cm}$
$\therefore\ \text{ar(rectangle ABCD)}=\text{lenght}\times\text{breadth}$
$=\text{AB}\times\text{AD}$
$=4\sqrt{5}\times2\sqrt{5}$
$=40\text{cm}^2$