Here we will solve different types of Problems on relation between tangent and secant.
1. XP is a secant and PT is a tangent to a circle. If PT = 15 cm and XY = 8YP, find XP.
Solution:
XP = XY + YP = 8YP + YP = 9YP.
Let YP = x. Then XP = 9x.
Now, XP × YP = PT^{2}, as the product of the segments of a secant is equal to the square of the tangent.
Therefore, 9x ∙ x = 15^{2} cm^{2}
⟹ 9x^{2} = 15^{2} cm^{2}
⟹ 9x^{2} = 225 cm^{2}
⟹ x^{2} = \(\frac{225}{9}\) cm^{2}
⟹ x^{2} = 25 cm^{2}
⟹ x = 5 cm.
Therefore, XP = 9x = 9 ∙ 5 cm = 45 cm.
2. XYZ is an isosceles triangle in which XY = XZ. If N is the mid point of XZ, prove that XY = 4 XM.
Solution:
Let XY = XZ = 2x.
Then XN = \(\frac{1}{2}\)XZ = x.
XY is a secant and XN is a tangent.
Therefore, XM × XY = XN^{2} (Product of segments of secant = square of tangent).
Therefore, XM × 2x = x^{2}
⟹ XM = \(\frac{x}{2}\).
Therefore, XY = 2x = 4 ∙ \(\frac{x}{2}\) = 4XM
From Problems on Relation Between Tangent and Secant to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
May 24, 24 06:42 PM
May 24, 24 06:23 PM
May 24, 24 06:22 PM
May 24, 24 05:37 PM
May 24, 24 05:09 PM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.