TANGENT AND SECANT OF A CIRCLE TO FIND AREA OF CIRCLE

In each circle, C is the center and Ab is tangent to the circle at point B. Find the area of each circle.

Problem 1 :

Solution:

Since, the tangent is perpendicular to the circle.

So, AB ⊥ BC

From Δ ABC we have

AC² = AB² + BC²

30² = 25² + BC²

900 = 625 + BC²

Area of circle = πr²

= π(5√11)²

= 275 π sq.units

Hence, area of circle is 275 π sq.units.

Problem 2 :

Solution:

From Δ ABC we have

AC² = AB² + BC²

45² = 12² + BC²

2025 = 144 + BC²

Area of circle = πr²

= π(√1881)²

= 1881 π sq.units

Hence, area of circle is 1881 π sq.units.

Problem 3 :

Solution:

∠ABC = 90°, ∠BCA = 60°

In special right triangle, shorter side = BC

Side opposite to 60 = BA = 6√3

6√3 = √3(Shorter side)

shorter side (BC) = 6

Area of circle = πr²

= π(6)²

= 36 π sq.units

Hence, area of circle is 36 π sq.units.

Problem 4 :

Solution:

∠ABC = 90°

In 45-45-90 right triangle, shorter side = AB = 18

BC = 18

Area of circle = πr²

= π(18)²

= 324 π sq.units

Hence, area of circle is 324 π sq.units.

Problem 5 :

Solution:

From Δ ABC we have

AC² = AB² + BC²

16² = 12² + BC²

256 = 144 + BC²

Area of circle = πr²

= π(4√7)²

= 112 π sq.units

Hence, area of circle is 112 π sq.units.

Problem 6 :

Solution:

From Δ ABC we have

AC² = AB² + BC²

86² = 56² + BC²

7396 = 3136 + BC²

Area of circle = πr²

= π(√4260)²

= 4260 π sq.units

Hence, area of circle is 4260 π sq.units.

Problem 7 :

Solution:

From Δ ABC we have

AC² = AB² + BC²

90² = 28² + BC²

8100 = 784 + BC²

Area of circle = πr²

= π(√7316)²

= 7316 π sq.units

Hence, area of circle is 7316 π sq.units.

Problem 8 :

Solution:

From Δ ABC we have

AC² = AB² + BC²

(AD + DC)² = 24² + BC²

Since CD = BC = radius

(18 + r)² = 24² + r²

324 + r² + 36r = 576 + r²

252 - 36r = 0

36r = 252

r = 252/36

r = 7

Area of circle = πr²

= π(7)²

= 49 π sq.units

Hence, area of circle is 49 π sq.units.