In a parallelogram ABCD,AB = 10 , cm and AD = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) $ABCD$ is a parallelogram,in which $AB = 10 \, cm$ and $AD = 6 \, cm$. The bisector of $\angle A$ meets $DC$ in $E$. $AE$ and $BC$ produced meet a

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) $ABCD$ is a parallelogram,in which $AB = 10 \, cm$ and $AD = 6 \, cm$. The bisector of $\angle A$ meets $DC$ in $E$. $AE$ and $BC$ produced meet at $F$.Since $AE$ is the bisector of $\angle A$,we have $\angle BAE = \angle EAD \quad \dots(1)$.Since $AD \parallel BC$ and $AF$ is a transversal,$\angle EAD = \angle EFB$ (Alternate interior angles) $\dots(2)$.From $(1)$ and $(2)$,$\angle BAE = \angle EFB$.In $	riangle ABF$,since $\angle BAE = \angle EFB$,the sides opposite to these angles are equal,so $BF = AB$.Given $AB = 10 \, cm$,therefore $BF = 10 \, cm$.We know $BF = BC + CF$. Since $BC = AD = 6 \, cm$ (opposite sides of a parallelogram are equal),$10 \, cm = 6 \, cm + CF$.$CF = 10 \, cm - 6 \, cm = 4 \, cm$.

In a parallelogram $ABCD$,$AB = 10 \, cm$ and $AD = 6 \, cm$. The bisector of $\angle A$ meets $DC$ in $E$. $AE$ and $BC$ produced meet at $F$. Find the length of $CF$ (in $cm$).

Similar Questions

The figure obtained by joining the mid-points of the sides of a rhombus,taken in order,is

In $\Delta XYZ$,$A$ and $B$ are the midpoints of $XY$ and $XZ$ respectively. If $AB = 7.5 \, cm$,then find the length of $YZ$ in $cm$.

$ABCD$ is a parallelogram. If its diagonals are equal,then find the value of $\angle ABC$. (in $^{\circ}$)

Show that the quadrilateral formed by joining the mid-points of the sides of a rhombus,taken in order,is a rectangle.

One angle of a quadrilateral is $108^{\circ}$ and the remaining three angles are equal. Find each of the three equal angles. (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module