A) \[\frac{1}{2}\]
B) 4
C) 1
D) \[\frac{2}{3}\]
Correct Answer: C
Solution :
Given, in \[\Delta ABC,\]\[DE\,\,|\,\,\,|\,\,\,BC\] |
\[\therefore\] By Thales theorem, |
we get |
\[\frac{AD}{DB}=\frac{AE}{EC}\Rightarrow \,\,\frac{4x-3}{3x-1}=\frac{8x-7}{5x-3}\] |
\[\left[ \because \,\,\,AD=4x-3,\,DB=3x-1,\,AE=8x-7,\,EC=5x-3 \right]\] |
\[\Rightarrow \,\,\,\,\left( 4x-3 \right)\left( 5x-3 \right)=\left( 8x-7 \right)\left( 3x-1 \right)\] |
\[\Rightarrow \,\,\,20{{x}^{2}}-12x+9-15x=24{{x}^{2}}-21x\,-8x+7\] |
\[\Rightarrow \,\,\,\,\,4{{x}^{2}}-2x-2=0\] |
\[\Rightarrow \,\,\,\,2{{x}^{2}}-x-1=0\] |
[dividing both sides by 2] |
\[\Rightarrow \,\,\,\,2{{x}^{2}}-2x+x-1=0\] |
[by splitting the middle term] |
\[\Rightarrow \,\,\,\,2x\left( x-1 \right)+1\left( x-1 \right)=0\] |
\[\Rightarrow \,\,\,\,\,\,\left( 2x+1 \right)\left( x-1 \right)=0\] |
\[\Rightarrow \,\,\,\,\,\,x=-\frac{1}{2}\,\,or\,\,x=1\] |
If \[x=-\frac{1}{2}\], then \[AD=4\times \left( -\frac{1}{2} \right)-3=-5<0\] |
[not possible; since, length cannot be negative] |
Hence, x = 1 is the required value. |
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