- About CBSE Exam
- CBSE Alerts | Results
- CBSE Class 9
- CBSE Class 10
- CBSE Class 11
- UP Board
- CBSE Class 12
- Jharkhand Board
- Gujarat Board
- CBSE Study Material
- Rajasthan Board
- CBSE Sample Papers
- Bihar Board
- CBSE Question Papers
- Uttarakhand Board
- West Bengal Board
- Himachal Pradesh Board
- Tamil Nadu Board
- Maharashtra Board
- Madhya Pradesh Board
- Orissa Board
- Andhra Pradesh Board
- Chhattisgarh Board
- Assam Board
- CBSE Mock Papers
- More In CBSE
- What? Why? How?
- CBSE Other Subjects

Here you get the CBSE Class 10 Mathematics chapter 9, Circles: NCERT Exemplar Problems and Solutions (Part-IVA). This part of the chapter includes solutions to Question Number 1 to 7 from Exercise 9.4 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Circles. This exercise comprises only the Long Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

**CBSE Class 10 Mathematics Syllabus 2017-2018**

NCERT exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

**Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Circles:**

**Exercise 9.4**

**Long Answer Type Questions**** (Q. No. 1-7)**

**Question. 1 **If a hexagon *ABCDEF *circumscribe a circle, prove that

*AB + CD* *+* *EF =* *BC + DE + FA*

**Solution. **

**Given: **A hexagon *ABCDEF *circumscribes a circle such that the sides, *AB*, *BC*, *CD*, *DE* and *DF* touches the circle at *P*, *Q*, *R*, *S*, *T* and *U* respectively.

**To prove: ***AB* *+* *CD* *+ EF = BC + DE+ FA*

**Proof: **

**As we know that the **tangents drawn from an external point to the circle are equal,

∴ *AP = AU*

* BP = BQ*

* CR = CQ*

* DS = DR*

* ES = ET*

* FT = FU *……(i)

Now,* AB* *+* *CD* *+* *EF = *(*AP *+ *BP*) + (*CR + DR*) + (*ET* +* FT*)

** ***= AU + BQ + CQ + DS + ES + FU *[Using equations in (i)]

* *= (*AU + FU*) + (*BQ + CQ*)* +* (*DS + ES*)

*⟹ AB + CD* *+ EF = AF *+ *BC + DE*

Hence proved.

**Question. 2 **Let *s* denotes the semi-perimeter of a Δ*ABC *in which *BC = a,* *CA = b *and *AB =* *c. *If a circle touches the sides *BC, CA, AB *at *D, E, F *respectively, prove that *BD = s* -* b.*

**Solution.**

Given:

A circle is inscribed in the Δ*ABC, *which touches *BC, CA *and *AB *at* D, E *and *F *respectively. Also *BC = a,* *CA = b *and *AB =* *c*.

And *s* is the semi-perimeter of a Δ*ABC*

**Question. 3 **From an external point *P, *two tangents, *PA *and *PB *are drawn to a circle with centre *O. *At one point *E *on the circle tangent is drawn which intersects *PA *and *PB *at *C *and *D*, respectively. If *PA *= 10 cm, find the perimeter of the triangle *PCD.*

**Solution.**

**Given:** *PA *and *PB* are two tangents drawn to a circle with centre O. A tangent *CD* at *E, *intersects *AP* and *PB* at *C* and *D* respectively.

**To find:** Perimeter of Δ*PCD*

**Proof:**

**Since, t**angents drawn from an external point to a circle are equal,

∴* CE *= *CA*, *DE *= *DB, PA = PB *....(i)* *

Now, perimeter of D*PCD = PC + CD+PD*

* = PC + CE* *+* *ED* *+* *PD*

* = PC + CA* + *DB* *+* *PD*

*= PA + PB*

= 2*PA* = 2 ×10

= 20 cm

**Question. 4 **If *AB *is a chord of a circle with centre *O, AOC *is a diameter and *AT *is the tangent at *A *as shown in figure. Prove that ∠*BAT = *∠*ACB.*

**Solution.**

**Given: **Chord *AB*, diameter *AOC* and tangents *AT* of a circle with centre *O*.

**To prove: **∠*BAT *= ∠*ACB*

Proof:

Here *AC *is a diameter and we know that diameter of a circle always subtends a right angle at any point on the circle.

∴ ∠*ABC *= 90°

**Question. 5 **Two circles with centres *O *and *O*’ of radii 3 cm and 4 cm, respectively intersect at two points *P *and *Q, *such that *OP *and *O’P *are tangents to the two circles. Find the length of the common chord *PQ.*

**Solution.**

Two circles with centres *O *and O’, and radii 3 cm and 4 cm respectively are drawn to intersect each other at points *P *and *Q*.* OP *and *PO’ *are two tangents drawn at point *P*.

Join *OO’ *such that it intersects *PQ* at *R*.

Since, tangent at any point of circle is perpendicular to radius through the point of contact,

∴ ∠*OPO’= *90°

Thus by using Pythagoras theorem in right angled D*OPO*’, we get:

Now, as the perpendicular drawn from the centre bisects the chord.

Therefore, length of common chord, *PQ *= 2 *PR = *2 × 2.4 = 4.8cm

**Question. 6 **In a right angle Δ*ABC* is which ∠*B* = 90°, a circle is drawn with *AB *as diameter intersecting the

hypotenuse *AC *at *P*. Prove that the tangent to the circle at *P *bisects *BC.*

**Solution.**

Given conditions can be drawn as the following diagram:

**Question. 7 **In figure, tangents *PQ *and *PR *are drawn to a circle such that ∠*RPQ *= 30°. A chord *RS *is drawn

parallel to the tangent *PQ. *Find the ∠*RQS.*

**Solution.**

Given, *PQ* and *PR* are tangents from an external point *P* to circle.

**CBSE Class 10 NCERT Textbooks & NCERT Solutions**

**NCERT Solutions for CBSE Class 10 Maths**

**NCERT Exemplar Problems and Solutions Class 10 Science: All Chapters**

Previous Article UP Board Class 12 Physics (Second) Solved Practice Paper: Set – 7

Copyright 2017 Jagran Prakashan Limited.

## Post Comment