CBSE 10th Standard Maths Subject Circles Case Study Questions 2021
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CBSE 10th Standard Maths Subject Circles Case Study Questions 2021
10th Standard CBSE
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Reg.No. :
Maths
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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 (a) -
A backyard is in the shape of a triangle with right angle at B, AB = 6 m and BC = 8 m. A pit was dig inside it such that it touches the walls AC, BC and AB at P, Q and R respectively such that AP = x m.
Based on the above information, answer the following questions.
(i) The value of AR =(a) 2x m (b) x/2 m (c) x m (d) 3x m (ii) The value of BQ =
(a) 2x m (b) (6-x) m (c) (2 - x) cm (d) 4x m (iii) The value of CQ =
(a) (4+x)m (b) (10 - x) m (c) (2+x)m (d) both (b) and (c) (iv) Which of the following is correct?
(a) Quadrilateral AROP is a square. (b) Quadrilateral BROQ is a square. (c) Quadrilateral CQOP is a square. (d) None of these (v) Radius of the pit is
(a) 2 cm (b) 3 cm (c) 4 cm (d) 5 cm (a) -
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 (a) -
Prem did an activity on tangents drawn to a circle from an external point using 2 straws and a nail for maths project as shown in figure.
Based on the above information, answer the following questions.
(i) Number of tangents that can be drawn to a circle from an external point is(a) 1 (b) 2 (c) infinite (d) any number depending on radius of circle (ii) On the basis of which of the following congruency criterion,\(\Delta \mathrm{OAP} \cong \Delta \mathrm{OBP} ?\)
(a) ASA (b) SAS (c) RHS (d) SSS (iii) If \(\angle\)AOB = 150°, then \(\angle\)APB =
(a) 75° (b) 30° (c) 60° (d) 100° (iv) If \(\angle\)APB = 40°, then \(\angle\)BAO =
(a) 40° (b) 30° (c) 50° (d) 20° (v) If \(\angle\)ABO = 45°, then which of the following is correct option?
(a) \(A P \perp B P\) (b) PAOB is square (c) \(\angle\)AOB = 90° (d) All of these (a) -
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 (a)
Case Study Questions
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CBSE 10th Standard Maths Subject Circles Case Study Questions 2021 Answer Keys
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(i) (a)
(ii) (d)
(iii) (b)
(iv) (a)
(v) (c) -
Here in right angled triangle ABC, AB = 6 m and BC= 8 cm.
\(\therefore\) By Pythagoras theorem \(A C=\sqrt{(A B)^{2}+(B C)^{2}}\)
\(=\sqrt{(8)^{2}+(6)^{2}}=\sqrt{100}=10 \mathrm{~m}\)
Also, AP = x m.
(i) (c):AR=AP = xm ..(1)
[Since, length of tangents drawn from an external point are equal]
(ii) (b): BQ = BR = AB - AR = (6 - x) m (Using (1))
(iii) (d): CQ = CP = AC - AP = (10 - x) m
Also, CQ = BC - BQ = BC - BR = 8 - (6 - x) = 2 + x
(iv) (b): Since, CQ = 10 - x = 2 + x
\(\Rightarrow\) 8 = 2x \(\Rightarrow\) x = 4
\(\therefore\) AR = AP = 4 m, BR = BQ = 2 m
and CP = CQ = 6 m
Also, OQ.\(\perp\)BQ and OR.l BR
\(\therefore\) BROQ is a square.
(v) (a): Radius of the circle, OR = BR = 2 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) (b)
(ii) (c): In \(\Delta\)OAP and \(\Delta\)OBP,
\(\angle\)OAP = \(\angle\)OBP = 90°
[Since, radius at the point of contact is perpendicular to tangent]
OP = OP (Common)
OA = OB (Radii of circle)
So, \(\angle\)OAP == \(\angle\)OBP (By RHS congruency criterion)
(iii) (b): In quadrilateral OAPB, \(\angle\)AOB = 150° [Given]
\(\angle\)OAP = \(\angle\)OBP = 90°
\(\therefore\) \(\angle\)APB = 360° - 90° - 90° - 150° = 30°
(iv) (d): We have, \(\angle\)APB = 40°
Now,PA =PB [Since, length of tangents drawn from an external point are equal]
In \(\Delta\)PAB, \(\angle\)PAB = \(\angle\)PBA = 70° [Angles opposite to equal sides are equal]
Also, \(\angle\)PAB + \(\angle\)BAO = 90°
[Since, radius at the point of contact is perpendicular to tangent]
\(\Rightarrow\) \(\angle\)BAO = 90° - 70° = 20°
(v) (d): We have, \(\angle\)ABO = 45°
\(\because\) AO = OB (Radii of circle)
\(\therefore\) \(\angle\)BAO = \(\angle\)ABO = 45° [Angles opposite to equal sides are equal]
Now, in \(\Delta\)OAB,
\(\angle\)AOB = 180° - 45° - 45° = 90°
Since, \(\angle\)APB = 360° - 90° - 90° - 90° = 90° i.e., AP\(\perp\) BP
So, OAPB is a square. -
(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.
Case Study Questions