Question 68

PR is a tangent to a circle with centre O and radius 4 cm at point Q. If ∠POR = 90°, OR = 5 cm and OP = 20/3 cm, then the length of PR is:

Solution

Given : ∠POR = 90°, OR = 5 cm and OP = 20/3 cm

To find : PR = ?

In $$\triangle$$ POR, PR is the tangent,

=> $$(PR)^2=(OR)^2+(OP)^2$$

=> $$(PR)^2=(5)^2+(\frac{20}{3})^2$$

=> $$(PR)^2=25+\frac{400}{9}$$

=> $$(PR)^2=\frac{225+400}{9}=\frac{625}{9}$$

=> $$PR=\sqrt{\frac{625}{9}}=\frac{25}{3}$$ cm

=> Ans - (D)

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