Question

$\mathrm{In}\text{the}\mathrm{figure},\mathrm{if}\mathrm{LK}=6\sqrt{3,}\mathrm{find}\mathrm{MK},\mathrm{ML},\mathrm{KN}\text{and}\mathrm{MN}\mathrm{and}\mathrm{the}\mathrm{perimeter}\mathrm{of}\square \mathrm{MNKL}.$

Answer 1

In $△$ KLM, $\angle$MLK = 90°, $\angle$MKL = 30°. Then,
$\angle$KML = 180° – ($\angle$MLK +$\angle$MKL) = 180° – (90° + 30°) = 180° – 120° = 60°
Thus,$△$ KLM is a 30°-60°-90° triangle.

By the 30°-60°-90° triangle theorem, we get:
LK = $\frac{\sqrt{3}}{2}$ × MK
$\Rightarrow$ MK = $\frac{2}{\sqrt{3}}$ × LK = $\frac{2}{\sqrt{3}}$ × $6\sqrt{3}$ = 12
ML = $\frac{1}{2}$ × MK = $\frac{1}{2}$ × 12 = 6

Now, in$△$ MNK, $\angle$MKN = 90° and $\angle$MNK = 45°. Then,
$\angle$KMN = 180° – ($\angle$MNK +$\angle$MKN) = 180° – (45° + 90°) = 180° – 135° = 45°
Thus, $△$MNK is a 45°-45°-90° triangle.

By the 45°-45°-90° triangle theorem, we get:
MK = KN = $\frac{1}{\sqrt{2}}$× MN
$\Rightarrow$ MN = $\sqrt{2}$ × MK =12$\sqrt{2}$
Since MK = 12,
KN = MK = 12.

Therefore, the perimeter of quadrilateral MNKL is
MN + NK + KL + LM = 12$\sqrt{2}$ + 12 + $6\sqrt{3}$ + 6
= 18 +12$\sqrt{2}$ + $6\sqrt{3}$
= 6(3+2$\sqrt{2}$ +$\sqrt{3}$ ) units