CBSE 10th Standard Maths Subject Circles Case Study Questions With Solution 2021 | QB365

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CBSE 10th Standard Maths Subject Circles Case Study Questions With Solution 2021

Published on: 22/05/2021

QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get more marks in Exams

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Std: 10th Standard CBSE - Maths

chapter: - Case study Questions

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Questions:

Case Study Questions

5 x 4 = 20

Following are questions of section-A in assessment test on circle that Eswar attend last month in school. He scored 5 out of 5 in this section. Answer the questions and check your score if 1 mark is allotted to each question.

(i) If two tangents AB and CDdrawn to a circle with centre 0 at P and Q respectively, are parallel to each other, then which of the following is correct?

(a) \(\angle\)POQ = 180°	(b) PQ is a diameter
(c) \(\angle\)APQ = \(\angle\)PQD = 90°	(d) All of these

(ii) If I is a tangent to the circle with centre 0 and line m is passing through 0 intersects the tangent I at point of contact, then

(a) I \|\| m	(b) l \(\perp\)m
(c) line I and line m intersects and makes an angle of 60°	(d) can't be determined

(iii) Number of tangents that can be drawn to a circle from a point inside it, is

(a) 1

(b) 2

(c) infinite

(d) 0

(iv) Which of the following is true?

(a) PQ is a tangent to both the circles	(b) Two circles are concentric
(c) PQ is a tangent to bigger circle only	(d) PQ is a tangent to smaller circle only

(v) A parallelogram circumscribing a circle is called a

(a) rhombus	(b) rectangle
(c) square	(d) none of these

In an online test, Ishita comes across the statement - If a tangent is drawn to a circle from an external point, then the square of length of tangent drawn is equal to difference of squares of distance of the tangent from the centre of circle and radius of the circle.

Help Ishita, in answering the following questions based on the above statement.
(i) If AB is a tangent to a circle with centre O at B such that AB = 10 cm and OB = 5 cm, then OA =

\((a) 3 \sqrt{5} \mathrm{~cm}\)

\((b) 5 \sqrt{5} \mathrm{~cm}\)

\((c) 4 \sqrt{5} \mathrm{~cm}\)

\((d) 6 \sqrt{5} \mathrm{~cm}\)

(ii) In the adjoining figure, radius of the circle is

(a) 8 cm

(b) 7 cm

(c) 9 cm

(d) 10 cm

(iii) In the adjoining figure, length of tangent AP is

(a) 12 cm

(b) 24 cm

(c) 30 cm

(d) None of these

(iv) PT is a tangent to a circle with centre 0 and diameter = 40 cm. If PT = 21 cm, then OP =

(a) 33 cm

(b) 29 cm

(c) 37 cm

(d) None of these

(v) In the adjoining figure, the length of the tangent is

(a) 15 cm

(b) 9 cm

(c) 8 cm

(d) 10 cm

In a maths class, the teacher draws two circles that touch each other externally at point K with centres A and B and radii 5 em and 4 em respectively as shown in the figure.

Based on the above information, answer the following questions.
(i) The value of PA =

(a) 12 cm

(b) 5 cm

(c) 13 cm

(d) Can't be determined

(ii) The value of BQ =

(a) 4 cm

(b) 5 cm

(c) 6 cm

(d) None of these

(iii) The value of PK =

(a) 13 cm

(b) 15 cm

(c) 16 cm

(d) 18 cm

(iv) The value of QY =

(a) 2 cm

(b) 5 cm

(c) 1 cm

(d) 3 cm

(v) Which of the following is true?

(a) PS²=PA.PK

(b) TQ²=QB.QK

(c) PS²=PX.PK

(d) TQ² = QA.QB

Smita always finds it confusing with the concepts of tangent and secant of a circle. But this time she has determined herself to get concepts easier. So, she started listing down the differences between tangent and secant of a circle along with their relation. Here, some points in question form are listed by Smita in her notes. Try answering them to clear your concepts also.

(i) A line that intersects a circle exactly at two points is called

(a) Secant

(b) Tangent

(c) Chord

(d) Both (a) and (b)

(ii) Number of tangents that can be drawn on a circle is

(a) 1

(b) 0

(c) 2

(d) Infinite

(iii) Number of tangents that can be drawn to a circle from a point not on it, is

(a) 1

(b) 2

(c) 0

(d) Infinite

(iv) Number of secants that can be drawn to a circle from a point on it is

(a) Infinite

(b) 1

(c) 2

(d) 0

(v) A line that touches a circle at only one point is called

(a) Secant

(b) Chord

(c) Tangent

(d) Diameter

In a park, four poles are standing at positions A, B, C and D around the fountain such that the cloth joining the poles AB, BC, CD and DA touches the fountain at P, Q, Rand S respectively as shown in the figure.

Based on the above information, answer the following questions.
(i) If 0 is the centre of the circular fountain, then \(\angle\)OSA =

(a) 60°	(b) 90°
(c) 45°	(d) None of these

(ii) Which of the following is correct?

(a) AS = AP

(b) BP= BQ

(c) CQ = CR

(d) All of these

(iii) If DR = 7 cm and AD = 11 ern, then AP =

(a) 4 cm

(b) 18 cm

(c) 7 cm

(d) 11 cm

(iv) If O is the centre of the fountain, with \(\angle\)QCR = 60°, then \(\angle\)QOR

(a) 60°

(b) 120°

(c) 90°

(d) 30°

(v) Which of the following is correct?

(a) AB + BC = CD + DA	(b) AB + AD = BC + CD
(c) AB + CD = AD + BC	(d) All of these

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Answers Key:

Case Study Questions

5 x 4 = 20

(i) (d):

Two tangents of a circle are parallel only when they are drawn at ends of a diameter.
So,PQ is the diameter of the circle.
(ii) (b)
(iii) (d)
(iv) (a): Here, the two circles have a common point of contact T and PQ is the tangent at T. So, PQ is the tangent to both the circles.
(v) (a)

(i) (b): OA²=AB²+OB²

\(\Rightarrow \quad O A=\sqrt{10^{2}+5^{2}}=5 \sqrt{5} \mathrm{~cm}\)
(ii) (a) : \(O A=\sqrt{O P^{2}-A P^{2}} \text { (Given) }\)
\(=\sqrt{17^{2}-15^{2}}=\sqrt{64}=8 \mathrm{~cm}\)
(iii) (b): Length of tangent \(A P=\sqrt{O P^{2}-O A^{2}} \text { (Given) }\)
\(=\sqrt{25^{2}-7^{2}}=\sqrt{576}=24 \mathrm{~cm}\)
(iv) (b):

\(\text { Since, } O P=\sqrt{(P T)^{2}+(O T)^{2}}=\sqrt{21^{2}+20^{2}}=29 \mathrm{~cm}\)
(v) (a): Since, OP² + PQ² = OQ²
\(\Rightarrow\) 8² + x² = (x + 2)²\(\Rightarrow\) 64 = 4x + 4\(\Rightarrow\) x = 15 cm
So, length of tangent, PQ = 15 cm.

Here, AS = 5 cm, BT = 4 cm [\(\therefore\)Radii of circles]
(i) (c): Since, radius at point of contact is perpendicular to tangent.
\(\therefore\) By Pythagoras theorem, we have
\(P A=\sqrt{P S^{2}+A S^{2}}=\sqrt{12^{2}+5^{2}}=\sqrt{169}=13 \mathrm{~cm}\)
(ii) (b): Again by Pythagoras theorem, we have
\(B Q=\sqrt{T Q^{2}+B T^{2}}=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5 \mathrm{~cm}\)
(iii) (d): PK = PA + AK = 13 + 5 = 18 cm
(iv) (c): QY = BQ - BY = 5 - 4 = 1 cm
(v) (c): PS² = PA² - AS² = PA² - AK²
= (PA + AK)(PA - AK) = PK.PX [\(\because\) AK = AX]

(i) (a)
(ii) (d)
(iii) (b)
(iv) (a)
(v) (c)

(i) (b):

Here, OS the is radius of circle.
Since radius at the point of contact is perpendicularto tangent.
So, \(\angle\)OSA = 90°
(ii) (d): Since, length of tangents drawn from an external point to a circle are equal.
\(\therefore\) AS=AP,BP=BQ,
CQ = CR and DR = DS
(iii) (a): AP = AS = AD _ DS = AD _ DR (Using (1)
= 11 - 7 = 4 cm
(iv) (b): In quadrilateral OQCR,

\(\angle\)QCR = 60° (Given)
And \(\angle\)OQC = \(\angle\)ORC = 90° [Since, radius at the point of contact is perpendicular to tangent.]
\(\therefore\) \(\angle\)QOR = 360° - 90° - 90° - 60° = 120°
(v) (c): From (1), we have AS = AP, DS = DR,
BQ = BP and CQ = CR
Adding all above equations, we get
AS + DS + BQ + CQ = AP + DR + BP + CR
\(\Rightarrow\) AD + BC = AB + CD