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 ΔJAB?
Given: JKLM is a square with sides of length 6units. Points A and B are the midpoints of sides KL and ML respectively. If a point is selected at random from the interior of the square
To find: Probability that the point will be chosen from the interior of ΔJAB.
We the following figure
Area of square JLKM is equal to
$=6^{2}$
$=36 \mathrm{sq}$ units
Now we have
$\operatorname{ar}(\Delta \mathrm{KAJ})=\frac{1}{2} \times \mathrm{AK} \times \mathrm{KJ}$
$=\frac{1}{2} \times 3 \times 6$
$=9$ units $^{2}$
$\operatorname{ar}(\Delta \mathrm{JMB})=\frac{1}{2} \times \mathrm{JM} \times \mathrm{BM}$
$=\frac{1}{2} \times 6 \times 3$
$=9$ units $^{2}$
$\operatorname{ar}(\Delta \mathrm{ALB})=\frac{1}{2} \times \mathrm{AL} \times \mathrm{BL}$
$=\frac{1}{2} \times 3 \times 3$
$=\frac{9}{2}$ units $^{2}$
Now area of the triangle AJB
$\operatorname{ar}(\Delta \mathrm{AJB})=36-9-9-\frac{9}{2}$
$=\frac{27}{2}$ units $^{2}$
We know that Probability
$=\frac{\text { Number of favourable event }}{\text { Total number of event }}$
$=\frac{\frac{27}{2}}{36}$
$=\frac{27}{2 \times 36}$
$=\frac{3}{8}$
Hence the Probability that the point will be chosen from the interior of $\triangle \mathrm{AJB}$ is $\frac{3}{8}$
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