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

False
Solution:
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:
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

True Solution: 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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MATHS True False[1 Marks ]

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