Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
MediumIn the figure,$D$ and $E$ are two points on $BC$ such that $BD = DE = EC$. Show that $ar(ABD) = ar(ADE) = ar(AEC)$.
Can you now answer the question that you have left in the 'Introduction' of this chapter,whether the field of Budhia has been actually divided into three parts of equal area?
Can you now answer the question that you have left in the 'Introduction' of this chapter,whether the field of Budhia has been actually divided into three parts of equal area?
(N/A) Let us draw $AF$ perpendicular to $BC$ such that $AF$ is the height of $\Delta ABD$,$\Delta ADE$,and $\Delta AEC$.
We know that the area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Therefore,$ar(ABD) = \frac{1}{2} \times BD \times AF$.
Similarly,$ar(ADE) = \frac{1}{2} \times DE \times AF$.
And $ar(AEC) = \frac{1}{2} \times EC \times AF$.
Since it is given that $BD = DE = EC$,we can substitute these values.
Thus,$\frac{1}{2} \times BD \times AF = \frac{1}{2} \times DE \times AF = \frac{1}{2} \times EC \times AF$.
This implies $ar(ABD) = ar(ADE) = ar(AEC)$.
Yes,since the altitudes (heights) of all the triangles are the same and their bases are equal,the areas are equal. Therefore,Budhia can use this result to divide her land into three equal parts.
We know that the area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
Therefore,$ar(ABD) = \frac{1}{2} \times BD \times AF$.
Similarly,$ar(ADE) = \frac{1}{2} \times DE \times AF$.
And $ar(AEC) = \frac{1}{2} \times EC \times AF$.
Since it is given that $BD = DE = EC$,we can substitute these values.
Thus,$\frac{1}{2} \times BD \times AF = \frac{1}{2} \times DE \times AF = \frac{1}{2} \times EC \times AF$.
This implies $ar(ABD) = ar(ADE) = ar(AEC)$.
Yes,since the altitudes (heights) of all the triangles are the same and their bases are equal,the areas are equal. Therefore,Budhia can use this result to divide her land into three equal parts.
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Similar Questions
In the figure,$E$ is any point on the median $AD$ of a $\Delta ABC$. Show that $\text{ar} (ABE) = \text{ar} (ACE)$.
Easy
View SolutionIn the figure,$ABCD$ is a quadrilateral and $BE \parallel AC$. $BE$ meets $DC$ produced at $E$. Show that the area of $\Delta ADE$ is equal to the area of the quadrilateral $ABCD$.
Medium
View Solution$P$ and $Q$ are any two points lying on the sides $DC$ and $AD$ respectively of a parallelogram $ABCD$. Show that $\text{ar}(APB) = \text{ar}(BQC)$.
Medium
View SolutionDiagonals $AC$ and $BD$ of a quadrilateral $ABCD$ intersect each other at $P$. Show that $ar(APB) \times ar(CPD) = ar(APD) \times ar(BPC)$.
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
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