Get here NCERT Solutions Class 10 Maths Chapter 10. These NCERT Solutions for Class 10 of Maths subject includes detailed answers of all the questions in Chapter 10 – Circles Tangents Perpendicular provided in NCERT Book which is prescribed for class 10 in schools.
Resource: National Council of Educational Research and Training (NCERT) Solutions
Class: 10th Class
Subject: Maths
Chapter: Chapter 10 – Circles Tangents Perpendicular
NCERT Solutions Class 10 Maths Chapter 10 – Free Download PDF
NCERT Solutions Class 10 Maths Chapter 10 – Circles Tangents Perpendicular
Students of Class 10 can either use these solutions, which consists of solved questions, online or download the PDF files and prepare for their board exams. The solutions of 10th Standard Maths Chapter 10 are formulated by our experts, which are based on CBSE syllabus. Circles class 10 chapter consists of topics such as;
- Introduction to Circles
- Tangent to a circle
- Number of Tangents from a point on a circle
- Summary of the Whole Chapter
Apart from referring to solutions, students can also use previous year question papers as a tool to practice and prepare for their Class 10 board exams, which gives an idea of the paper pattern as well.
The circle is a part of Geometry, which is considered to be one of the major topics usually asked in the examination. Students, who aim to secure excellent results need to have a good understanding of this chapter 10 – circles. Circles basically consist of various terminologies, such as radius, diameter, chords, segment, angles etc. and its properties that need to be understood for solving various questions based on circles. These theorems and formulas are easy to remember and also can also be proven. The students preparing for the board examination must have a good practice of this circles chapter, as many questions can be framed from this chapter. Student’s can also refer to NCERT Notes For Class 10 to prepare for board exams.
This Circles chapter of class 10 maths deals with the various types of questions such as finding the angles, length of a chord, length of a tangent, tangent drawn from a point to a circle, etc., which should be practised thoroughly to have a better learning experience and also to excel in the examination.
NCERT Solutions for class 10 Maths Chapter 10 -Circles Exercises
- NCERT Solutions Class 10 Maths Chapter 10 Circles Tangents Perpendicular Exercise 10.1
1. In the figure, PQ is a chord of a circle and PT is the tangent at P such that ∠QPT = 60°. Then find the measure of ∠PRQ.
Solution.
Since OP is perpendicular to PT.
∠OPT = 90°
∠OPQ = 90°- ∠QPT
∠OPQ = 90 – 60 = 30°.
In ΔOPQ, OP= OQ = r ( Radius of the circle )
∠OPQ= ∠OQP = 30.
And,
∠POQ = 180 – ∠OPQ- ∠OQP
= 180° – 30° – 30° = 120°
Also, reflex ∠POQ = 360° – 120° = 240°
Now, ∠PRQ = reflex ∠POQ
= 12x 240° = 120°
2. If the angle between two radii of a circle is 130°, then find the degree measure of the angle between the tangents at the ends of the radii.
Solution.
It is already known that angle between two radii and the angle between the tangents at the ends of the radii are supplementary.
Hence, Angle between the tangents at the ends of the radii is 180° – 130° , i.e., 50°.
3. In the figure, if ∠AOB = 125°, then find the degree measure of ∠
Solution.
It is already known that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
∠AOB + ∠COD = 180°.
125° + ∠COD = 180°
∠COD = 180° – 125° = 55°.
4. In the figure, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then find the measure of ∠BAT.
Solution.
∠BAT= 50°.
5. At one end A of a diameter AB of a circle of radius 5cm, tangent XAY is drawn to the circle. Find the length of the chord CD parallel to XY and at a distance 8cm from A.
Solution.
Since XAY is a tangent through one end A of a diameter of a circle,
AB is perpendicular to XY
And also, CD is parallel to XAY ⇒ AB is perpendicular to CD
Since OM is perpendicular from centre O to the chord CD.
⤑ OM is perpendicular bisector of chord CD.
That is., CM = MD = 12CD.
Now, AM =8cm (given)
⇒ OM = AM – AO = 8-5 = 3cm.
In OMC, we obtain
CM = CM=sqrtOC2−OM2
= sqrt52−32
= sqrt16
= 4 cm.
Hence, CD = 2cm = 2 x 4 = 8 cm.
6. In the given figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Find the length of the segment AT.
Solution.
In tight triangle ∆OTA, ∠OTA = 30°.
fracOAOT=sin30circ
OA=frac12OT=frac12imes4=2cm
Again, AT=sqrtOT2−OA2
AT = sqrt42−22
=sqrt16−4
sqrt12
= 2sqrt3cm.
7. In the figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, find the measure of ∠POQ.
Solution.
Since OP is perpendicular to PR
∠OPR = 90°
∠OPQ + ∠QPR = 90°
∠OPQ + 50° = 90°
→ ∠OPQ = ∠OQP = 40°
Again,
∠POQ + ∠OPQ + ∠OQP = 180°
→ ∠POQ + 40° + 40° = 180°
→ ∠POQ = 180° – 40° – 40°
= 100°
8. In the figure, if PA and PB are tangents to the circle with centre O such that ∠APB = 50°, find the degree measure of ∠OAB.
Solution.
Since OA is perpendicular to PA and also, OB is perpendicular to PB
∠APB + ∠AOB = 180°
50°+ ∠AOB = 180°
∠AOB = 180° – 50° = 130°
In △AOB,
OA = OB = radii of same circle
∠OAB = ∠OBA = x ( say )
Again, ∠OAB + ∠OBA + ∠AOB = 180°
x +x + 130° = 180°
2x = 180° – 130° = 50°
X = 25°
Hence, ∠OAB =25°
9. In the figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, find the measure of ∠AQB.
Solution.
Since AB is parallel to PQR
∠B =∠BQR
→ ∠B =70°
Also, ∠A = ∠BQR [∠s in the corresponding alternate segments are equal ]
→ ∠A = 70°
Now, in △ABQ, we have
∠A + ∠B + ∠AQB = 180°
70° + 70° + ∠AQB = 180°
∠AQB=180° – 70° – 70°
=40°
Hence, ∠AQB = 40°
10. In the figure, DE and DF are tangents from an external point D to a circle with centre A. If DE=5cm and DE is perpendicular DF, find the radius of the circle.
Solution.
Join AE and AF → AE is perpendicular to DE and AF is perpendicular to DF.
AEDF is a square of side 5cm [DE = 5cm(given)]
Hence, radius of the circle is 5 cm.
11. If two tangents inclined at an angle of 60° are drawn to a circle of radius 3cm, find the length of each tangent.
Solution.
.
Here, let PA and PB be two tangents to a circle with centre O.
OA is perpendicular to AP.
Since OP bisects ∠APB
∠OPA = 30°
Now, in △OAP, we have
fracAPOA=cot30circ
→ fracAP3=sqrt3
→ AP=3sqrt3cm
12. From a point P which is at a distance of 13cm from the centre O of a circle of radius 5cm, a pair of tangents PQ and PR to the circle are drawn. Find the area of the quadrilateral PQOR.
Solution.
In △OPQ, we have
OQ perpendicular to PQ
OP2=PQ2+OQ2
132=PQ2+52
PQ2=169–25 = 144
PQ = 12cm
Now, (△OPQ) = x PQxOQ
= x12x5 = 30cm2
Thus, (quad, PQOR) = 2 x ar(△OPQ)
= 2×30 cm2
= 60cm2
NCERT solutions class 10 maths chapter 10 pdf are prepared by our subject experts under the guidelines of NCERT to assist students in their board exam preparations.
Apart from notes and books, we are also providing here NCERT Exemplar books for student’s reference. Prepare with these materials provided by us and score well in your board exams. Also, these materials are helpful for students appearing for various competitive exams which are going to be conducted this year, in 2019.
Students can find a good number of questions from this chapter in pdf, thus one should have a good practice of circles. Circles hold various important theorems in mathematics and it should be thoroughly practiced.
With good practice, students can get to know all the method of solving the questions of a different kind as well as the various important part from which students can get the question. Thus practicing questions and reviewing the various important questions back from the textbook can help students to have a better way of solving questions.