Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
MediumIn the figure,$ABCD$ and $AEFD$ are two parallelograms. Prove that $\operatorname{ar}(\triangle PEA) = \operatorname{ar}(\triangle QFD)$.
(N/A) Given: $ABCD$ and $AEFD$ are two parallelograms.
To prove: $\operatorname{ar}(\triangle PEA) = \operatorname{ar}(\triangle QFD)$.
Proof:
In $\triangle PEA$ and $\triangle QFD$:
$1$. $\angle APE = \angle DQF$ (Corresponding angles are equal as $AB \parallel CD$ and $PQ$ is a transversal).
$2$. $\angle AEP = \angle DFQ$ (Corresponding angles are equal as $AE \parallel DF$ and $PQ$ is a transversal).
$3$. $AE = DF$ (Opposite sides of parallelogram $AEFD$ are equal).
Therefore,by $AAS$ congruence rule,$\triangle PEA \cong \triangle QFD$.
Since congruent triangles have equal areas,$\operatorname{ar}(\triangle PEA) = \operatorname{ar}(\triangle QFD)$.
To prove: $\operatorname{ar}(\triangle PEA) = \operatorname{ar}(\triangle QFD)$.
Proof:
In $\triangle PEA$ and $\triangle QFD$:
$1$. $\angle APE = \angle DQF$ (Corresponding angles are equal as $AB \parallel CD$ and $PQ$ is a transversal).
$2$. $\angle AEP = \angle DFQ$ (Corresponding angles are equal as $AE \parallel DF$ and $PQ$ is a transversal).
$3$. $AE = DF$ (Opposite sides of parallelogram $AEFD$ are equal).
Therefore,by $AAS$ congruence rule,$\triangle PEA \cong \triangle QFD$.
Since congruent triangles have equal areas,$\operatorname{ar}(\triangle PEA) = \operatorname{ar}(\triangle QFD)$.
Explore More
Similar Questions
The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides $8 \, cm$ and $6 \, cm$ is:
Easy
View SolutionIn trapezium $PQRS, PQ || RS$ and diagonals $PR$ and $QS$ intersect at point $M$. Prove that,$ar(PQS) = ar(QPR)$.
Easy
View Solution$(1)$ The part of the plane enclosed by a simple closed figure is called a $\ldots \ldots \ldots$
$(2)$ $\ldots \ldots \ldots$ of the planar region corresponding to a closed figure is called its area.
$(2)$ $\ldots \ldots \ldots$ of the planar region corresponding to a closed figure is called its area.
Easy
View Solution$D, E$ and $F$ are the midpoints of the sides $BC, CA$ and $AB$ respectively of $\Delta ABC$. Show that:
$(i)$ $BDEF$ is a parallelogram.
$(ii)$ $ar(DEF) = \frac{1}{4} ar(ABC)$
$(iii)$ $ar(BDEF) = \frac{1}{2} ar(ABC)$
$(i)$ $BDEF$ is a parallelogram.
$(ii)$ $ar(DEF) = \frac{1}{4} ar(ABC)$
$(iii)$ $ar(BDEF) = \frac{1}{2} ar(ABC)$
Difficult
View SolutionState whether each of the following statements is true or false:
$(1)$ If $ar(ABC) = 96 \, cm^2$ for the parallelogram $ABCD$,then $ar(ABCD) = 192 \, cm^2$.
$(2)$ Area of a right triangle = Product of the sides forming the right angle.
$(1)$ If $ar(ABC) = 96 \, cm^2$ for the parallelogram $ABCD$,then $ar(ABCD) = 192 \, cm^2$.
$(2)$ Area of a right triangle = Product of the sides forming the right angle.
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo