Get the solution to "The perpendicular bisector of hypotenuse AB in the..."

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atodion23

Answered question

2021-11-29

The perpendicular bisector of hypotenuse AB in the right triangle $△ A B C$ forms the triangle with the area 3 times smaller than the area of $△ A B C$ . Find the measures of acute angles in $△ A B C$ .(Solve this tasked without using Trigonometry)

Answer & Explanation

Leory2000

Beginner2021-11-30Added 10 answers

Let the length be CB =l and AC=h in $△ A B C$ .

So, by Pythagoras theorem,
$A B = \sqrt{h^{2} + l^{2}}$
Also, P is a perpendicular bisector of AB.
So, AP=PB
Hence, $A P = P B = \frac{\sqrt{h^{2} + l^{2}}}{2}$
Step2
Here, we see that $△ A B C$ is similar to $△ A P Q$ .
Since, both are right angle triangles.
So, ratio of their sides is
$\frac{△ A C B}{△ A P Q} = \frac{3}{1}$
$\frac{3}{1} = {(\begin{matrix} \frac{A C}{A P} \end{matrix})}^{2}$
$= {(\begin{matrix} \frac{l}{\frac{\sqrt{l^{2} + k^{2}}}{2}} \end{matrix})}^{2}$
$l = \frac{l}{\sqrt{3}} h$
From the ratio of side one can clearly say that $∠ A = 30^{\circ}, ∠ B = 60^{\circ}$ .

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