Question-90948

Question Number 90948 by tw000001 last updated on 27/Apr/20

Commented by tw000001 last updated on 27/Apr/20

Find ∣ PQ ∣ .

Commented by tw000001 last updated on 27/Apr/20

If △ APQ is right triangle,

∣ PQ ∣ should be \sqrt{38} .

However, what if △ APQ is not a right triangle ?

Answered by MJS last updated on 27/Apr/20

∡ A B C = \frac{π}{2} \Rightarrow A C = 13

the radius of the excircle is

\frac{\sqrt{(a + b + c) (- a + b + c) (a - b + c) (a + b - c)}}{2 (- a + b + c)} = 3

P C = Q C = 15

P Q = \frac{30}{\sqrt{26}}

Commented by tw000001 last updated on 27/Apr/20

{That}^{'} s correct, thank you .

Answered by mr W last updated on 27/Apr/20

A C = \sqrt{5^{2} + 12^{2}} = 13

C B + B Q = C A + A P

12 + B Q = 13 + A P \Rightarrow B Q = 1 + A P

B Q + A P = A B

1 + A P + A P = 5 \Rightarrow A P = 3

\Rightarrow B Q = 1 + 2 = 3

C P = C Q = 12 + 3 = 15

{P Q}^{2} = 15^{2} + 15^{2} - 2 \times 15 \times 15 \times \cos ∠ C

{P Q}^{2} = 2 (1 - \frac{12}{13}) \times 15^{2} = \frac{2 \times 15^{2}}{13}

\Rightarrow P Q = \frac{15 \sqrt{26}}{13} \approx 5.883

Commented by tw000001 last updated on 27/Apr/20

Your answer is correct, too .