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Triangles — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTriangles3 Marks
Question
In a $\triangle\text{ABC,D}\ \text{and E}$ are points on the sides AB and AC respectively such that DE || BC.If AD = 4x - 3, AE = 8x - 7, BD = 3x - 1 and CE = 5x - 3, find the value of x.
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Answer

Given: AD = 4x - 3, AE = 8x - 7, BD = 3x - 1 and CE = 5x - 3
by thales theorem
$\frac{\text{AD}}{\text{BD}}=\frac{\text{AE}}{\text{CE}}$
$\Rightarrow\frac{4\text{x}-3}{3\text{x}-1}=\frac{8\text{x}-7}{5\text{x}-3}$
$\Rightarrow(4\text{x}-3)(5\text{x}-3)=(8\text{x}-7)(3\text{x}-1)$
$\Rightarrow20\text{x}^2+9-15\text{x}-12\text{x}=24\text{x}^2-21\text{x}-8\text{x}+7$
$\Rightarrow9-7-27\text{x}+29\text{x}=24\text{x}^2-20\text{x}^2$
$\Rightarrow2+2\text{x}=4\text{x }^2$
$\Rightarrow4\text{x}^2-2\text{x }-2=0$
$\Rightarrow2\text{x}^2-2\text{x}+\text{x}-1=0$
$\Rightarrow2\text{x}(\text{x}-1)+(\text{x}-1)=0$
$\Rightarrow(2\text{x}+1)(\text{x}-1)=0$
$\Rightarrow2\text{x}+1=0\ \text{or}\ \text{x}-1=0$
$\Rightarrow2\text{x}=-1\ \text{or x}=1$
$\Rightarrow\text{x}=\frac{-1}{2}$
$\text{Thus, x}=\frac{-1}{2}\ \text{and x}=1$

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