A) 5/8
B) 7/8
C) 3/4
D) 3/8
Correct Answer: D
Solution :
Area of square JMLK \[={{6}^{2}}=36\text{ }sq.\]units A and B are the mid-points of sides KL and LM. i \[\therefore \] AL = KA = LB = BM = 3 units Now, Area of \[\Delta ALB=\frac{1}{2}\times AL\times LB\] \[=\frac{1}{2}\times 3\times 3=\frac{9}{2}\,sq.\]units Area of \[\Delta JMB=\frac{1}{2}\times BM\times JM\] \[=\frac{1}{2}\times 3\times 3=\frac{9}{2}sq.\] units. Area of \[\Delta KAJ=\frac{1}{2}\times KJ\times KA\] \[=\frac{1}{2}\times 6\times 3=9\,sq.\]units Total area of all the three triangles \[=\left( \frac{9}{2}+9+9 \right)=\frac{45}{2}\,\,sq.\] units \[\therefore \] Area of \[\Delta JAB=\left( 36-\frac{45}{2} \right)=\frac{27}{2}\] sq. units \[\therefore \] Required probability \[=\frac{\frac{27}{2}}{36}=\frac{27}{2\times 36}=\frac{3}{8}\]
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