Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
MediumIn the figure,$AP || BQ || CR$. Prove that $\operatorname{ar}(AQC) = \operatorname{ar}(PBR)$.
(N/A) Given: $AP || BQ || CR$.
Step $1$: Since $BQ || CR$,$\Delta BCQ$ and $\Delta BQR$ are on the same base $BQ$ and between the same parallels $BQ$ and $CR$.
Therefore,$\operatorname{ar}(\Delta BCQ) = \operatorname{ar}(\Delta BQR)$ --- $(1)$
Step $2$: Since $AP || BQ$,$\Delta ABQ$ and $\Delta PBQ$ are on the same base $BQ$ and between the same parallels $AP$ and $BQ$.
Therefore,$\operatorname{ar}(\Delta ABQ) = \operatorname{ar}(\Delta PBQ)$ --- $(2)$
Step $3$: Adding equations $(1)$ and $(2)$,we get:
$\operatorname{ar}(\Delta BCQ) + \operatorname{ar}(\Delta ABQ) = \operatorname{ar}(\Delta BQR) + \operatorname{ar}(\Delta PBQ)$
From the figure,$\operatorname{ar}(\Delta BCQ) + \operatorname{ar}(\Delta ABQ) = \operatorname{ar}(\Delta AQC)$ and $\operatorname{ar}(\Delta BQR) + \operatorname{ar}(\Delta PBQ) = \operatorname{ar}(\Delta PBR)$.
Thus,$\operatorname{ar}(\Delta AQC) = \operatorname{ar}(\Delta PBR)$.
Hence proved.
Step $1$: Since $BQ || CR$,$\Delta BCQ$ and $\Delta BQR$ are on the same base $BQ$ and between the same parallels $BQ$ and $CR$.
Therefore,$\operatorname{ar}(\Delta BCQ) = \operatorname{ar}(\Delta BQR)$ --- $(1)$
Step $2$: Since $AP || BQ$,$\Delta ABQ$ and $\Delta PBQ$ are on the same base $BQ$ and between the same parallels $AP$ and $BQ$.
Therefore,$\operatorname{ar}(\Delta ABQ) = \operatorname{ar}(\Delta PBQ)$ --- $(2)$
Step $3$: Adding equations $(1)$ and $(2)$,we get:
$\operatorname{ar}(\Delta BCQ) + \operatorname{ar}(\Delta ABQ) = \operatorname{ar}(\Delta BQR) + \operatorname{ar}(\Delta PBQ)$
From the figure,$\operatorname{ar}(\Delta BCQ) + \operatorname{ar}(\Delta ABQ) = \operatorname{ar}(\Delta AQC)$ and $\operatorname{ar}(\Delta BQR) + \operatorname{ar}(\Delta PBQ) = \operatorname{ar}(\Delta PBR)$.
Thus,$\operatorname{ar}(\Delta AQC) = \operatorname{ar}(\Delta PBR)$.
Hence proved.
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Similar Questions
The side $AB$ of a parallelogram $ABCD$ is produced to any point $P$. $A$ line through $A$ and parallel to $CP$ meets $CB$ produced at $Q$ and then parallelogram $PBQR$ is completed. Show that $\text{ar}(ABCD) = \text{ar}(PBQR)$.
[Hint: Join $AC$ and $PQ$. Now compare $\text{ar}(ACQ)$ and $\text{ar}(APQ)$.]
[Hint: Join $AC$ and $PQ$. Now compare $\text{ar}(ACQ)$ and $\text{ar}(APQ)$.]
Difficult
View SolutionIn the figure,$ABC$ is a right triangle right-angled at $A$. $BCED$,$ACFG$,and $ABMN$ are squares on the sides $BC$,$CA$,and $AB$ respectively. Line segment $AX \perp DE$ meets $BC$ at $Y$. Show that: $\operatorname{ar}(CYXE) = 2 \operatorname{ar}(FCB)$.
Medium
View Solution$D$ and $E$ are points on sides $AB$ and $AC$ respectively of $\Delta ABC$ such that $\operatorname{ar}(DBC) = \operatorname{ar}(EBC)$. Prove that $DE \parallel BC$.
Medium
View SolutionIn the figure,$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC$. If $AE$ intersects $BC$ at $F$,show that:
$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$
$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$
$(v)$ $\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$
$(vi)$ $\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$
$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$
$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$
$(v)$ $\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$
$(vi)$ $\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$
Medium
View Solution$A$ farmer has a field in the form of a parallelogram $PQRS$. She took any point $A$ on $RS$ and joined it to points $P$ and $Q$. In how many parts is the field divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
Medium
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