• # question_answer In the given figure. JKLM is a square with sides of length 6 units. Points A and B are the mid-points of sides KL and LM respectively. If a point is selected at random from the interior of the square. What is the probability that the point will be chosen from the interior of $\Delta JAB$?   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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