Andhra Pradesh Board Class 10th Maths Chapter 9 Tangents and Secants to a Circle Ex 9.1 Textbooks Solutions PDF
Andhra Pradesh State Board STD 10th Maths Chapter 9 Tangents and Secants to a Circle Ex 9.1 Books Solutions with Answers are prepared and published by the Andhra Pradesh Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of AP Board Class 10th Maths Chapter 9 Tangents and Secants to a Circle Ex 9.1 Books Answers Solutions, then you are in the right place. Here is a complete hub of Andhra Pradesh State Board Class 10th Maths Chapter 9 Tangents and Secants to a Circle Ex 9.1 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Andhra Pradesh Board STD 10th Maths Chapter 9 Tangents and Secants to a Circle Ex 9.1 Textbooks. These Andhra Pradesh State Board Class 10th Maths Chapter 9 Tangents and Secants to a Circle Ex 9.1 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Andhra Pradesh Board.Andhra Pradesh State Board Class 10th Maths Chapter 9 Tangents and Secants to a Circle Ex 9.1 Books Solutions
Board | AP Board |
Materials | Textbook Solutions/Guide |
Format | DOC/PDF |
Class | 10th |
Subject | Maths |
Chapters | Maths Chapter 9 Tangents and Secants to a Circle Ex 9.1 |
Provider | Hsslive |
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AP Board Class 10th Maths Chapter 9 Tangents and Secants to a Circle Ex 9.1 Textbooks Solutions with Answer PDF Download
Find below the list of all AP Board Class 10th Maths Chapter 9 Tangents and Secants to a Circle Ex 9.1 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.1 Textbook Questions and Answers
Question 1.
Fill in the blanks.
i) A tangent to a circle intersects it in ——— point(s). (one)
ii) A line intersecting a circle in two points is called a ———. (secant)
iii) The number of tangents drawn at the end of the diameter is ———. (two)
iv) The common point of a tangent to a circle and the circle is called ———. (point of contact)
v) We can draw ——— tangents to a given circle. (infinite)
Question 2.
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Find length of PQ.
Answer:
Given: A circle with centre O and radius OP = 5 cm
PQ⎯⎯⎯⎯⎯⎯⎯⎯ is a tangent and OQ = 12 cmWe know that ∠OPQ = 90°
Hence in △OPQ
OQ2 = OP2 + PQ2
[∵ hypotenuse2 = Adj. side2 + Opp. side2]
122 = 52 + PQ2
∴ PQ2 = 144 – 25 .
PQ2 = 119
PQ = √119
Question 3.
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
Answer:Steps:
- Draw a circle with some radius.
- Draw a chord of the circle.
- Draw a line parallel to the chord intersecting the circle at two distinct points.
- This is secant of the circle (l).
- Draw another line parallel to the chord, just touching the circle at one point (M). This is a tangent of the circle.
Question 4.
Calculate the length of tangent from a point 15 cm. away from the centre of a circle of radius 9 cm.
Answer:Given: A circle with radius OP = 9 cm
A tangent PQ from a point Q at a distance of 15 cm from the centre, i.e., OQ =15 cm
Now in △POQ, ∠P = 90°
OP2 + PQ2 – OQ2
92 + PQ2 = 152
PQ2 = 152 – 92
PQ2 = 144
∴ PQ = √144 = 12 cm.
Hence the length of the tangent =12 cm.
Question 5.
Prove that the tangents to a circle at the end points of a diameter are parallel.
Answer:
A circle with a diameter AB.
PQ is a tangent drawn at A and RS is a tangent drawn at B.
R.T.P: PQ || RS.
Proof: Let ‘O’ be the centre of the circle then OA is radius and PQ is a tangent.
∴ OA ⊥ PQ ……….(1)
[∵ a tangent drawn at the end point of the radius is perpendicular to the radius]
Similarly, OB ⊥ RS ……….(2)
[∵ a tangent drawn at the end point of the radius is perpendicular to the radius]
But, OA and OB are the parts of AB.
i.e., AB ⊥ PQ and AB ⊥ RS.
∴ PQ || RS.O is the centre, PQ is a tangent drawn at A.
∠OAQ = 90°
Similarly, ∠OBS = 90°
∠OAQ + ∠OBS = 90° + 90° = 180°
∴ PQ || RS.
[∵ Sum of the consecutive interior angles is 180°, hence lines are parallel]
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