Important Questions for Class 10 Maths Chapter 10 Circles

Important Questions for Class 10 Maths Chapter 10: Important questions for Class 10 Maths Chapter 10, "Circles," are provided here, based on the new exam pattern. These questions are important for students preparing for the 2023-2024 board exams to attain full marks in this chapter. Practicing these important questions will ensure students are well-prepared for the board exam, where they may face similar questions. Hence, thorough practice is important for exam preparation.

CBSE Class 10 Sample Paper

Important Questions for Class 10 Maths Chapter 10 Overview

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Important Questions for Class 10 Maths Chapter 10 Circles

Q.1: ABC is a right angled triangle, right angled at B such that BC = 6 cm, AB = 8 cm and AC = 10 cm. A circle with centre O is inscribed in ΔABC. The radius of the circle is:

Options

(a) 1 cm (b) 2 cm (c) 3 cm (d) 4 cm

Solution:

An incircle is drawn with centre O which touches the sides of the triangle ABC at P, Q and R. OP, OQ and OR are radii and AB, BC and CA are the tangents to the circle. OP ⊥ AB, OQ ⊥ BC and OR ⊥ CA. OPBQ is a square. ( ∵ ∠B - 90 ^o ) Let r be the radius of the circle PB = BQ = r AR = AP = 8 - r, CQ = CR = 6 - r AC = AR + CR ⇒ 10 = 8 - r + 6 - r ⇒ 10 = 14 - 2r ⇒ 2r = 14 - 10 = 4 ⇒ r = 2

Q.2: What should be the angle between the two tangents which are drawn at the end of two radii and are inclined at an angle of 45 degrees?

Solution:

Q.3: In the given figure, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC = 4 cm then the length of AP is

Solution:

Q.4: Shipra prepared a project for rain water harvesting. diagrammatic representation of the project is given in the figure. PQ and PR are the pipes touching the circular pit. Length of these pipes is 5 m each. What is the perimeter of ΔPMN?

Solution:

Q.5: In the given figure, PQ R is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°. Find ∠PCA.

Solution:

Q.6: I n the given figure, PA and PB are two tangents drawn from an external point P to a circle with centre C and radius 4 cm. If PA ⊥ PB, then find the length of each tangent.

Solution:

Construction: Join AC and BC. Proof: ∠1 = ∠2 = 90° ….[Tangent is I to the radius (through the point of contact ∴ APBC is a square. Length of each tangent = AP = PB = 4 cm = AC = radius = 4 cm

Q.7: In the given figure, PQ and PR are two tangents to a circle with Centre O. If ∠QPR = 46°, then calculate ∠QOR.

Solution:

∠OQP = 900 ∠ORP = 90° ∠OQP + ∠QPR + ∠ORP + ∠QOR = 360° …[Angle sum property of a quad. 90° + 46° + 90° + ∠QOR = 360° ∠QOR = 360° – 90° – 46° – 90° = 134°

Q.8: From an external point P, tangents PA and PB are drawn to a circle with centre 0. If ∠PAB = 50°, then find ∠AOB.

Solution:

PA = PB …[∵ Tangents drawn from external point are equal ∠PBA = ∠PAB = 50° …[Angles equal to opposite sides In ∆ABP, ∠PBA + ∠PAB + ∠APB = 180° …[Angle-sum-property of a ∆ 50° + 50° + ∠APB = 180° ∠APB = 180° – 50° – 50° = 80° In cyclic quadrilateral OAPB ∠AOB + ∠APB = 180° ……[Sum of opposite angles of a cyclic (quadrilateral is 180° ∠AOB + 80o = 180° ∠AOB = 180° – 80° = 100°

Q.9: If two tangents inclined at an angle ${60}^{\circ }$ are drawn to a circle of radius $3cm,$ then the length of each tangent is

(A) $\frac{3}{2}$ $\sqrt{3}cm$ (B) $6cm$ (C) $3cm$ (D) 3 $\sqrt{3}cm$

Solution:

Let $P$ be an external point and a pair of tangents is drawn from point P and angle between these two tangents is ${60}^{\circ }$ Radius of the circle $=3cm$ Join OA and OP Also, OP is a bisector line of $\angle$ APC $\therefore \angle APO=\angle CPO={30}^{\circ }OA\perp AP$ Also, tangents at any point of a circle is perpendicular to the radius through the point of contact. In right angled $\Delta OAP,$ we have $\tan {30}^{\circ }=\frac{OA}{AP}=\frac{3}{AP}$ $\Rightarrow \frac{1}{\sqrt{3}}=\frac{3}{AP}$ $\Rightarrow AP=3\sqrt{3}cm$ $AP=CP=3\sqrt{3}cm$ [Tangents drawn from an external point are equal] Hence, the length of each tangent is $3\sqrt{3}cm.$

Q.10: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Solution:

Given: $A$ circle with center $O$

with tangent $XY$ at point of contact $P.$

To Prove: $OP\perp XY$

Proof: Let $Q$ be a point on $XY$ connect $OQ$

Suppose it touches the circle at $R$

Hence,

$OQ>OR$

$OQ>OP$ $(\because OP=OR)$ (radius)

Same will be the case with all other points on the circle

Hence,

We get $OP$ is the smallest line that connects $XY$ .

Q.11. Two concentric circles are of radii 7 cm and r cm respectively, where r > 7. A chord of the larger circle, of length 48 cm, touches the smaller circle. Find the value of r.

Solution:

Given: OC = 7 cm, AB = 48 cm To find: r = ? ∠OCA = 90° ..[Tangent is ⊥ to the radius through the point of contact ∴ OC ⊥ AB AC = 1 2 (AB) … [⊥ from the centre bisects the chord ⇒ AC = 1 2 (48) = 24 cm In rt. ∆OCA, OA ² = OC ² + AC ² … [Pythagoras’ theorem r ² = (7) ² + (24) ² = 49 + 576 = 625 ∴ r= 625 −−−√ = 25 cm

Q.12: Prove that the parallelogram circumscribing a circle is a rhombus.

Solution:

ABCD is a ॥ ^gm . To prove. ABCD is a rhombus. Proof. In ॥ ^gm , opposite sides are equal

AB = CD and AD = BC ..(i) AP = AS …[Tangents drawn from an external point are equal in length PB = BQ CR = CO DR = DS By adding these tangents, (AP + PB) + (CR + DR) = AS + BQ + CQ + DS AB + CD = (AS + DS) + (BQ + CQ) AB + CD = AD + BC AB + AB = BC + BC … [From (i) 2AB = 2 BC AB = BC …(ii) From (i) and (ii), AB = BC = CD = DA ∴ ॥ ^gm ABCD is a rhombus.

Q. 13: In the given figure, an isosceles ∆ABC, with AB = AC, circumscribes a circle. Prove that the point of contact P bisects the base BC.

Solution:

The incircle of ∆ABC touches the sides BC, CA and AB at D, E and F F respectively. AB = AC To prove: BD = CD Proof: Since the lengths of tangents drawn from an external point to a circle are equal ∴ AF = AE … (i) BF = BD …(ii) CD = CE …(iii) Adding (i), (ii) and (iii), we get AF + BF + CD = AE + BD + CE ⇒ AB + CD = AC + BD But AB = AC … [Given ∴ CD = BD

Q.14: In the given figure, a circle inscribed in ∆ABC touches its sides AB, BC and AC at points D, E & F K respectively. If AB = 12 cm, BC = 8 cm and AC = 10 cm, then find the lengths of AD, BE and CF.

Solution:

Let AD = AF = x BD = BE = y …[Two tangents drawn from and an external point are equal CE = CF = z

AB = 12 cm …[Given ∴ x + y = 12 cm …(i) Similarly, y + z = 8 cm …(ii) and x + z = 10 cm …(iii) By adding (i), (ii) & (iii) 2(x + y + z) = 30 x + y + z = 15 …[∵ x + y = 12 z = 15 – 12 = 3 Putting the value of z in (ii) & (iii), y + 3 = 8 y = 8 – 3 = 5 x + 3 = 10 x = 10 – 3 = 7 ∴ AD = 7 cm, BE = 5 cm, CF = 3 cm

Q: 15. The incircle of an isosceles triangle ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. Prove that BD = DC.

Solution:

The incircle of ∆ABC touches the sides BC, CA and AB at D, E and F respectively. AB = AC To prove: BD = CD Proof: AF = AE ..(i) BF = BD …(ii) CD = CE …(iii) Adding (i), (ii) and (iii), we get AF + BF + CD = AE + BD + CE ⇒ AB + CD = AC + BD But AB = AC …[Given ∴ CD = BD

Benefits Of Solving Important Questions for Class 10 Maths Chapter 10

Conceptual Understanding: By solving these questions, students deepen their understanding of fundamental geometric concepts related to circles, such as tangents, chords, and theorems.

Exam Preparation: These questions are designed to align with the CBSE exam pattern, helping students familiarize themselves with the types of questions likely to appear in the board exams.

Increased Confidence: As students successfully solve these questions, their confidence in their mathematical abilities grows, boosting their overall performance in mathematics.

Time Management: Since the questions are structured similarly to those in the board exams, solving them helps students improve their time management skills, enabling them to tackle questions more efficiently during the exam.

Identification of Weak Areas: Working through these questions allows students to identify their strengths and weaknesses in the chapter, enabling them to focus their efforts on areas that need improvement.