Question

The angle between a diagonal and the longer base of the isosceles trapezoid MNFD is 45°. If MK = 4 cm and the diagonal ND=4 sqrt6, find: NK​

Answer 1

The length of side NK in the given isosceles trapezoid is $4\sqrt{3} \ cm$

From the question, we are to determine the length of NK

In the given diagram, consider ΔNKD

ΔNKD is a right-angled triangle,

From the question

$ND = 4\sqrt{6} \ cm$

and

∠NDK = 45°

Let ∠NDK = θ

∴ θ = 45°

In the right-angled triangle, we can write that

$Sin\theta = \frac{NK}{ND}$

∴ $Sin45 ^{\circ} =\frac{NK}{4\sqrt{6} }$

Then,

$NK = 4\sqrt{6} \times sin45^{\circ}$

But,

$sin45^{\circ} = \frac{1}{\sqrt{2} }$

∴ $NK = 4\sqrt{6} \times \frac{1}{\sqrt{2} }$

$NK = \frac{4\sqrt{6} }{\sqrt{2} }$

Rationalizing

$NK = \frac{4\sqrt{6} }{\sqrt{2} } \times \frac{\sqrt{2} }{\sqrt{2} }$

$NK = \frac{4\sqrt{12} }{2}$

$NK = 2\sqrt{12}$

$NK = 2\sqrt{4\times3}$

$NK = 2\times\sqrt{4} \times \sqrt{3}$

$NK = 2\times2 \times \sqrt{3}$

∴ $NK = 4\sqrt{3} \ cm$

Hence, the length of side NK is $4\sqrt{3} \ cm$

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The diagram for the question is attached below