AB is a chord of length 8 cm of a circle of radius 5 cm. The tangents to the circle at A and B intersect at P. Find the length of AP.
Answers
is had tried my best
Given,
The length of the chord, AB = 8 cm
The radius of the circle = 5 cm
The tangents to the circle at A and B intersect at P.
To find,
The length of AP.
Solution,
The length of AP will be 20/3 cm.
We can easily solve this problem by following the given steps.
Now, let's take the centre of the circle to be O.
Let's draw a perpendicular from the centre on the chord at a point C such that AC = CB = 4 cm.
Join OP.
OA = OB = 5 cm ( radius)
OA = OB = 5 cm ( radius)AC = CB = 4 cm
∆ OCA is a right-angled triangle.
Using the Pythagoras theorem in ∆ OCA,
OA² = AC² + OC²
(5)² = (4)² + OC²
25 = 16 + OC²
OC² = 25 - 16
OC² = 9
OC = √9
OC = 3 cm
Then, OP = (CP + 3) cm
Now, let's take AP to be b cm and CP to be a cm.
Then, OP = (a+3) cm
∆ ACP is a right-angled triangle.
Using the Pythagoras theorem in ∆ ACP,
AP² = AC² + CP² --- equation 1
b² = (4)² + a²
b² = 16 + a² --- equation 2
Now, we know that the tangents to the circle meet its radius at a right angle.
So, angle OAP is a right angle and ∆ OAP is a right-angled triangle.
Using the Pythagoras theorem in ∆ OAP,
OP² = AP² + OA²
OP² = (AC² + CP²) + OA² [ Putting the value of AP² from equation 1]
(a+3)² = (4)² + a² + (5)²
a² + (3)² + 2×a×3 = 16 + a² + 25 [ Using the identity, (a+b)² = a² + b² + 2ab]
a² + 9 + 6a = a² + 41
a² - a² + 9 + 6a = 41 ( Moving a² from the right-hand side to the left-hand side will result in the change of the sign from plus to minus.]
9 + 6a = 41
6a = 41-9
6a = 32
a = 32/6
a = 16/3 cm ( Converting 32/6 into the simplest form by dividing both the numerator and denominator by 2.)
Putting the value of a in equation 2,
b² = 16 + a² --- equation 2
b² = 16 + (16/3)²
b² = 16 + 256/9
b² = (16×9 + 256)/9 [Taking the LCM of 9 and 1 and adding the two fractions.]
b² = (144 + 256)/9
b² = 400/9
b =√400/√9
b = 20/3 cm
Hence, the value of AP is 20/3 cm.