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Question 73
In the given figure, EADF is a rectangle and ABC is a triangle whose vertices lie on the sides of EADF and AE = 22, BE = 6, CF = 16 and BF = 2. Find the length of the line joining the mid-points of the sides AB and BC.
Solution
From the above figure, $\triangle$ EBF and $\triangle$ABC are similar.
$\frac{AC}{EF}$=$\frac{AB}{EB}$
From the above figure, $\triangle$ EBF and $\triangle$ABC are similar.
$\frac{AC}{EF}$=$\frac{AB}{EB}$
$$AC^2 = 8^2 + 6^2 = 100$$
So AC = 10
$$\frac{10}{EF}$$ = 2
EF = 5
Length of line joining mid-points will be = 5
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