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In the 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 $\triangle JAB$?"

Given:

In the 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. 

A point is selected at random from the interior of the square.

To do:

We have to find the probability that the point will be chosen from the interior of $\triangle JAB$.

Solution:

Length of the side of square $JKLM =6$ units

Area of the square $=(6)^{2}=36$ sq. units.
A and B are the midpoints of sides $\mathrm{KL}$ and $\mathrm{LM}$ respectively.

$\mathrm{AL}=\mathrm{AK}=\mathrm{BM}=\mathrm{BL}=\frac{6}{2}=3 \mathrm{units}$

Area of $\Delta \mathrm{AJK}=\frac{\mathrm{JK} \times \mathrm{AK}}{2}$

$=\frac{6 \times 3}{2}$

$=9$ sq. units.

Area of $\Delta \mathrm{JMB}=\frac{\mathrm{JM} \times \mathrm{MB}}{2}$

$=\frac{6 \times 3}{2}$

$=9$ sq. units.

Area of $\Delta \mathrm{LAB}=\frac{\mathrm{LA} \times \mathrm{LB}}{2}$

$=\frac{3 \times 3}{2}$

$=\frac{9}{2}$ sq. units.
Area of 3 triangles $=9+9+\frac{9}{2}$

$=\frac{18+18+9}{2}$ sq. units

$=\frac{45}{2}$ sq. units

Therefore,

Area of $\Delta \mathrm{JAB}=$ Area of square  $-$ Area of 3 triangles

$=36-\frac{45}{2}$

$=\frac{72-45}{2}$

$=\frac{27}{2}$ sq. units

Therefore,

Probability that the point will be chosen from the interior of $\triangle JAB=\frac{\text { Area of } \Delta \mathrm{JAB}}{\text { Area of square JMLK }}$

$=\frac{27}{2 \times 36}$

$=\frac{3}{8}$

The probability that the point will be chosen from the interior of $\triangle JAB$ is $\frac{3}{8}$. 

Updated on: 10-Oct-2022

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