Question

ABCD is a rectangle, AD = 22 cm. AB = 8 cm. $△DEF$ is a triangle whose vertices lie on the sides of ABCD such that FB = 2 cm and BE = 16 cm. Then the length of the line joining the mid points of the sides DF and FE is:

• A. 3 cm
• B. 4 cm
• C. 5 cm
• D. 6 cm

Answer 1

The correct option is C 5 cm

Given that,
AD = 22 cm and AB = 8 cm
So,
BC = 22 cm and DC = 8 cm [Since, AD || BC and AB || DC]
Now,
BC = BE + EC
$\Rightarrow$22 = 16 + EC
$\Rightarrow$ EC = 6 cm
$\text{In}△DCE,\text{we have}$

$D{E}^2=E{C}^2+D{C}^2\text{[Pythagoras theorem]}$
$\Rightarrow D{E}^2=6^2+8^2$
$DE=\sqrt{36+64}=\sqrt{100}=10cm$

From the given rectangle,
FH = HE ; FG = GD
$\therefore$ GH || DE
Therefore,
$GH=\frac{1}{2}DE=\frac{10}{2}=5cm$
So, the length of the line joining the mid points of the sides DF and FE is 5 cm.