Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
MediumIn 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:
$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$
$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$
$(v)$ $\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$
$(vi)$ $\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$
$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$
$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$
$(v)$ $\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$
$(vi)$ $\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$
(A) Let the side of equilateral triangle $ABC$ be $2x$. Then $BC = 2x$. Since $D$ is the mid-point of $BC$,$BD = DC = x$. Since $BDE$ is an equilateral triangle,$BD = DE = BE = x$.
$(i)$ $\operatorname{ar}(ABC) = \frac{\sqrt{3}}{4} (2x)^2 = \sqrt{3}x^2$. $\operatorname{ar}(BDE) = \frac{\sqrt{3}}{4} x^2$. Thus,$\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$.
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \times \text{base} \times \text{height}$. Since $BE$ and $BA$ are sides of equilateral triangles,$\triangle BAE$ has base $BE=x$ and height equal to the altitude of $\triangle ABC$,which is $\sqrt{3}x$. $\operatorname{ar}(BAE) = \frac{1}{2} \times x \times \sqrt{3}x = \frac{\sqrt{3}}{2}x^2 = 2 \operatorname{ar}(BDE)$. Hence,$\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$.
$(iii)$ Since $D$ is the mid-point of $BC$,$\triangle BEC$ and $\triangle ABC$ share the same altitude from $A$ to $BC$. Thus,$\operatorname{ar}(BEC) = \frac{1}{2} \operatorname{ar}(ABC)$,which implies $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$.
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(ABE) - \operatorname{ar}(ABF)$. Using properties of triangles on the same base and between same parallels,it can be shown that $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$.
$(v)$ Since $F$ is the centroid of $\triangle ABC$ (or by geometric ratios),$\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$.
$(vi)$ By calculating the areas relative to the total area,$\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$.
$(i)$ $\operatorname{ar}(ABC) = \frac{\sqrt{3}}{4} (2x)^2 = \sqrt{3}x^2$. $\operatorname{ar}(BDE) = \frac{\sqrt{3}}{4} x^2$. Thus,$\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$.
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \times \text{base} \times \text{height}$. Since $BE$ and $BA$ are sides of equilateral triangles,$\triangle BAE$ has base $BE=x$ and height equal to the altitude of $\triangle ABC$,which is $\sqrt{3}x$. $\operatorname{ar}(BAE) = \frac{1}{2} \times x \times \sqrt{3}x = \frac{\sqrt{3}}{2}x^2 = 2 \operatorname{ar}(BDE)$. Hence,$\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$.
$(iii)$ Since $D$ is the mid-point of $BC$,$\triangle BEC$ and $\triangle ABC$ share the same altitude from $A$ to $BC$. Thus,$\operatorname{ar}(BEC) = \frac{1}{2} \operatorname{ar}(ABC)$,which implies $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$.
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(ABE) - \operatorname{ar}(ABF)$. Using properties of triangles on the same base and between same parallels,it can be shown that $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$.
$(v)$ Since $F$ is the centroid of $\triangle ABC$ (or by geometric ratios),$\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$.
$(vi)$ By calculating the areas relative to the total area,$\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$.
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