Question

PA is a tangent with center O. If BC=3cm, AC=4cm and ΔACB~ΔPAO, then find OA and OP/AP.
Please solve this question with the figure.

Answer 1

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_____Here's your Answer _________

In ABC,

Angle C = 90°.

By Pythagoras Theorem,

AB^2 = BC^2 + AC^2

=> AB^2 = 3^2 + 4^2

=> AB^2 = 9 + 16


$= > \: ab = \sqrt{25}$

=> AB = 5 cm.

Now,

AB = 2OA

=> OA = 1/2 AB

=> OA = 5/2.


Since, ΔACB~ΔPAO

Therefore,

BA / CA = OP / AP

=> OP / AP = 5/4
✔✔✔✔

Answer 2

$\huge\textbf{Answer}$


» Given

PA is a tangent with centre O.
BC = 3cm, AC = 4cm and ∆ACB ~ ∆PAO

» To find
OA and $\dfrac{OP}{AP}$

» Solution

In ∆ABC

/_C = 90°

In ∆ABC

By Pythagoras theorem

AB² = BC² + AC²

AB² = (3)² + (4)²

AB² = 9 + 16

AB = √25

AB = 5 cm ....(1)

Now

OA = $\dfrac{1}{2}$ AB

OA = $\dfrac{1}{2}$ × 5 [From (1)]

OA = $\dfrac{5}{2}$

Now

∆ACB ~ ∆PAO [given]

Then

$\dfrac{BA}{CA}$ = $\dfrac{OP}{AP}$

$\dfrac{OP}{AP}$ = $\dfrac{5}{4}$