Question 136

In ΔABC, ∠BCA = 90°, AC = 24 cm and BC = 10 cm. What is the radius (in cm) of the circum-circle of ΔABC?

Solution

Given : In ΔABC, ∠BCA = 90°, AC = 24 cm and BC = 10 cm

To find : OB = ?

Solution : In $$\triangle$$ ABC,

=> $$(AB)^2=(AC)^2+(BC)^2$$

=> $$(AB)^2=(24)^2+(10)^2$$

=> $$(AB)^2=576+100=676$$

=> $$AB=\sqrt{676}=26$$ cm

Also, in a right angled triangle, circumradius is half the hypotenuse of the triangle.

$$\therefore$$ OB = $$r=\frac{26}{2}=13$$ cm

=> Ans - (B)

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