If a secant and tangent share a point then the product of the secant and external part is equal to tangent squared.
Problem 1 :
Solution:
BC ⋅ BD = (BA)^{2}
10 ⋅ (10 + x) = (20)²
100 + 10x = 400
10x = 400 - 100
10x = 300
x = 30
Problem 2 :
Solution:
BC ⋅ BD = (BA)^{2}
0.8 ⋅ (0.8 + 1.0) = (x)²
0.8 (1.8) = x^{2}
x^{2} = 1.44
x = 1.2
Problem 3 :
Solution:
BC ⋅ BD = (BA)^{2}
4 ⋅ (4 + 5) = (x)²
4 (9) = x^{2}
x^{2} = 36
x = 6
Problem 4 :
Solution:
BC ⋅ BD = (BA)^{2}
5 ⋅ (5 + 3x) = (10)²
25 + 15x = 100
15x = 100 - 25
15x = 75
x = 75/15
x = 5
Problem 5 :
Solution:
BC ⋅ BD = (BA)^{2}
(x - 1) ⋅ (x - 1 + 5) = (x + 1)²
(x - 1) ⋅ (x + 4) = (x^{2} + 1 + 2x)
x^{2} + 4x - x - 4 = x^{2} + 2x + 1
x - 5 = 0
x = 5
Problem 6 :
Solution :
BC ⋅ BD = (BA)^{2}
x(x + 4) = (√12)^{2}
x^{2} + 4x = 12
x^{2} + 4x - 12 = 0
(x + 6)(x - 2) = 0
x = -6 and x = 2
So, the value of x is 2.
Problem 7 :
PAB is a secant and PT is a tangent to the circle from an external point. If PT = x cm, PA = 4 cm and AB = 5 cm, find x.
Solution:
PA ⋅ PB = (PT)^{2}
4 ⋅ (4 + 5) = x²
4 ⋅ 9 = x²
x² = 36
x = 6
Problem 8 :
Solution :
x^{2} = 3 (3 + 24)
x^{2} = 3 (27)
x = √3(27)
x = 9
Problem 9 :
Solution :
12^{2} = x (1 6)
x^{2} = 3 (27)
x = √3(27)
x = 9
Problem 10 :
Solution :
8^{2} = x (x + x)
8^{2} = x (2x)
2x^{2} = 64
x^{2} = 32
x = 4√2
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM