True False[1 Marks ]

Question 11 Mark

Write True or False and justify your answer: In the figure, $ABCD$ and $EFGD$ are two parallelograms and $G$ is the mid-point of $CD.$ Then, $\text{ar}(\triangle\text{DPC})=\frac{1}{2}\text{ar}(\text{EFGD}).$

Answer

In the given figure, join $PG.$
Since, $G$ is the mid-point of $CD.$
Thus, PG is a median of $ΔDPC$ and it divides the triangle into parts of equal areas.
Then, $\text{ar}(\triangle\text{DPG})=\text{ar}(\triangle\text{GPC})=\frac{1}{2}\text{ar}(\triangle\text{DPC})$
Also, we know that, if a parallelogram and a triangle lie on the same base and between the same parallel, then area of triangle is equal to triangle is equal to half of the area of parallelogram.
Here, parallelogram $FEGD$ and $\triangle\text{DPG}$ lie on the same base $DG$ and between the same parallel $DG$ and $EF.$
So, $\text{ar}(\triangle\text{DPG})=\frac{1}{2}\text{ar}(\text{EFGD})$ From eqs. $(i)$ and $(ii),$
$\frac{1}{2}\text{ar}(\triangle\text{DPC})=\frac{1}{2}\ \text{ar}(\text{EFGD})$
$\Rightarrow\text{ar}(\triangle\text{DPC})=\text{ar}(\text{EFGD})$

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Question 21 Mark

Write True or False and justify your answer: $PQRS$ is a rectangle inscribed in a quadrant of a circle of radius $13\ cm$ and $A$ is any point on $PQ.$ If $PS = 5\ cm,$ then ar $(\triangle\text{PAS})= 30\ cm^2$.

Answer

True Solution: Given, $P S=5 cm$
radius of circle $=S Q=13 cm$
In right angled $\triangle SPQ , SQ ^2=P Q^2+ PS ^2(13)^2= PQ ^2+(5)^2$
$ \Rightarrow P Q^2 169-25=144 $
$\Rightarrow P Q=12 cm$ [taking positive square root, because length is always positive]
Now, area of $\triangle APS =\frac{1}{2} \times$ Base $\times$ Height $\frac{1}{2} \times PS \times PQ$
$=\frac{1}{2} \times 5 \times 12=30 cm^2$ So, given statement is true, if $A$ coincides $Q .$

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Question 31 Mark

Write True or False and justify your answer: $ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC.$ Then $\text{ar}(\triangle\text{BDE})=\frac{1}{4}\text{ar}(\triangle\text{ABC}).$

Answer

Given, $\triangle\text{ABC}$ and $\triangle\text{BDE}$ are two equilateral triangle.
$\therefore$ Area of an equilateral
$\triangle\text{ABC}=\frac{\sqrt{3}}{4}\times(\text{sides})^2=\frac{\sqrt{3}}{4}(\text{BC})^2$
$[\because$ in equilateral $\triangle\text{ABC}, Ab = BC = AC]$
Also, given $D$ is the mid point of $BC.$
$\therefore\text{BD}=\text{DC}=\frac{1}{2}\text{BC}\ ...(\text{ii})$
Now, area of an equilateral $\triangle\text{BDE}=\frac{\sqrt{3}}{4}\times(\text{sides})^2$
$=\frac{\sqrt{3}}{4}\times(\text{BD})^2 [ \therefore$ in equilateral $\triangle\text{BDE} , BD = DE = BE]$
$=\frac{\sqrt{3}}{4}\times\Big(\frac{1}{2}\text{BC}\Big)^2 [$from eq. $(ii) =\frac{\sqrt{3}}{4}\times\frac{1}{4}\text{BC}^2=\frac{1}{4}\Big(\frac{\sqrt{3}}{4}\text{BC}^2\Big)$
$\text{Area of}\ \triangle\text{(BDE} )=\frac{1}{4}\text{Area of}\ \triangle\text{ABC}.$

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Question 41 Mark

Write True or False and justify your answer: $ABCD$ is a parallelogram and $X$ is the mid-point of $AB$. If ar $(AXCD) = 24\ cm^2,$ then ar $(ABC) = 24\ cm^2.$

Answer

Given, $ABCD$ is a parallelogram and ar $(AXCD) = 24\ cm^2$
Let area of parallelogram $ABCD$ is $24\ cm$ and join $AC. $
We know that, diagonals divides the area of parallelogram in two equal areas.
$\therefore\text{ar}(\triangle\text{ABC})=\text{ar}(\text{ACD})=\text{y}$
Also, $X$ is the mid $-$ point of $Ab.$
So, $\text{ar}(\triangle\text{ACX})=\text{ar}(\text{BCX})$
$[$since, $X$ is the median in $\triangle\text{ABC}]$
$=\frac{1}{2}\text{ar}(\text{ABC})=\frac{1}{2}\text{y}$
Now, $\text{ar}(\text{AXCD})=\text{ar}(\text{ADC})+\text{ar}(\text{ACX})$
$24=\text{y}+\frac{\text{y}}{2}$
$\Rightarrow24=\frac{3\text{y}}{2}$
$\Rightarrow\text{y}=\frac{24\times2}{3}=16\text{cm}^2$
Hence, $\text{ar}\triangle(\text{ABC})=16\text{cm}^2$

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