Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
MediumIn $\Delta ABC$,point $D$ lies on side $BC$. $E$ is the midpoint of $AD$. Prove that,$ar(\Delta EBC) = \frac{1}{2} ar(\Delta ABC)$.
(N/A) Given: In $\Delta ABC$,$D$ is a point on $BC$ and $E$ is the midpoint of $AD$.
To prove: $ar(\Delta EBC) = \frac{1}{2} ar(\Delta ABC)$.
Proof:
$1$. In $\Delta ABD$,$E$ is the midpoint of $AD$. Since $BE$ is a median of $\Delta ABD$,it divides the triangle into two triangles of equal area. Therefore,$ar(\Delta EBD) = \frac{1}{2} ar(\Delta ABD)$.
$2$. Similarly,in $\Delta ADC$,$E$ is the midpoint of $AD$. Since $CE$ is a median of $\Delta ADC$,it divides the triangle into two triangles of equal area. Therefore,$ar(\Delta ECD) = \frac{1}{2} ar(\Delta ADC)$.
$3$. Adding the two equations: $ar(\Delta EBD) + ar(\Delta ECD) = \frac{1}{2} ar(\Delta ABD) + \frac{1}{2} ar(\Delta ADC)$.
$4$. $ar(\Delta EBC) = \frac{1}{2} [ar(\Delta ABD) + ar(\Delta ADC)]$.
$5$. Since $ar(\Delta ABD) + ar(\Delta ADC) = ar(\Delta ABC)$,we get $ar(\Delta EBC) = \frac{1}{2} ar(\Delta ABC)$.
Hence proved.
To prove: $ar(\Delta EBC) = \frac{1}{2} ar(\Delta ABC)$.
Proof:
$1$. In $\Delta ABD$,$E$ is the midpoint of $AD$. Since $BE$ is a median of $\Delta ABD$,it divides the triangle into two triangles of equal area. Therefore,$ar(\Delta EBD) = \frac{1}{2} ar(\Delta ABD)$.
$2$. Similarly,in $\Delta ADC$,$E$ is the midpoint of $AD$. Since $CE$ is a median of $\Delta ADC$,it divides the triangle into two triangles of equal area. Therefore,$ar(\Delta ECD) = \frac{1}{2} ar(\Delta ADC)$.
$3$. Adding the two equations: $ar(\Delta EBD) + ar(\Delta ECD) = \frac{1}{2} ar(\Delta ABD) + \frac{1}{2} ar(\Delta ADC)$.
$4$. $ar(\Delta EBC) = \frac{1}{2} [ar(\Delta ABD) + ar(\Delta ADC)]$.
$5$. Since $ar(\Delta ABD) + ar(\Delta ADC) = ar(\Delta ABC)$,we get $ar(\Delta EBC) = \frac{1}{2} ar(\Delta ABC)$.
Hence proved.
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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 $\operatorname{ar}(\triangle DPC) = \frac{1}{2} \operatorname{ar}(EFGD).$
In the figure,$ABCD$ and $EFGD$ are two parallelograms and $G$ is the mid-point of $CD.$ Then $\operatorname{ar}(\triangle DPC) = \frac{1}{2} \operatorname{ar}(EFGD).$
Medium
View SolutionIn the figure,$l, m,$ and $n$ are straight lines such that $l \parallel m$ and $n$ intersects $l$ at $P$ and $m$ at $Q$. $ABCD$ is a quadrilateral such that its vertex $A$ is on $l$. The vertices $C$ and $D$ are on $m$ and $AD \parallel n$. Show that $\operatorname{ar}(ABCQ) = \operatorname{ar}(ABCDP).$
Medium
View SolutionIn the figure,$CD \parallel AE$ and $CY \parallel BA$. Prove that $\operatorname{ar}(\triangle CBX) = \operatorname{ar}(\triangle AXY)$.
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View SolutionThe area of the parallelogram $ABCD$ is $90 \, cm^{2}$ (see figure). Find:
$(i) \; ar(ABEF)$
$(ii) \; ar(ABD)$
$(iii) \; ar(BEF)$
$(i) \; ar(ABEF)$
$(ii) \; ar(ABD)$
$(iii) \; ar(BEF)$
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View Solution$(1)$ In $\Delta ABC$,$AD$ is an altitude. If $BC = 8 \text{ cm}$ and $AD = 5 \text{ cm}$,then $\text{ar}(\Delta ABC) = \dots \text{ cm}^2$.
$(2)$ $A$ $\dots$ of a triangle divides the triangle into two triangles of equal area.
$(2)$ $A$ $\dots$ of a triangle divides the triangle into two triangles of equal area.
Easy
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