[2 Mark Question Answer]

Question 12 Marks

In the following figure$, \text{CE}$ is drawn parallel to diagonals $\text{DB}$ of the quadrilateral $\text{ABCD}$ which meets $AB$ produced at point $E.$Prove that $\triangle ADE$ and quadrilateral $\text{ABCD}$ are equal in area.

Answer

Since $\triangle DCB$ and $\triangle DEB$ are on the same base $DB$ and between the same parallels
i.e. $DB \| CE,$ therefore we get
Ar.$ ( \triangle DCB) =$ Ar.$ ( \triangle DEB )$
Ar. $( \triangle DCB + \triangle ADB )$ = Ar. $(\triangle DEB + \triangle ADB )$
Ar.$ \text{( ABCD )} =$ Ar. $( \triangle ADE )$
Hence proved.

View full question & answer→

Question 22 Marks

In the given figure $, D$ is mid $-$ point of side $\text{AB}$ of $\triangle ABC$ and $\text{BDEC}$ is a parallelogram.

Prove that$:$ Area of $ABC =$ Area of $\| gm \ \text{BDEC}.$

Answer

Here $\text{AD} = \text{DB}$ and $\text{EC} =\text {DB},$ therefore $\text{EC} =\text {AD}$
Again$,$
$\angle EFC = \angle AFD .....($ Opposite angles $)$
Since $\text{ED}$ and $\text{CB}$ are parallel lines and $\text{AC}$ cut this line$,$ therefore
$\angle ECF = \angle FAD$
From the above conditions$,$ we have
$\triangle EFC = \triangle AFD$
Adding quadrilateral $\text{CBDF}$ in both sides$,$ we have
Area of $\| gm\ \text{BDEC} =$ Area of $\triangle {ABC}.$

View full question & answer→

MATHEMATICS [2 Mark Question Answer]

Test yourself on this topic