NCERT Exemplar Solution for Class 10 Mathematics: Circles (Part-IVA)

Aug 9, 2017 13:41 IST

Circles Long Answer Type Questions, Circles NCERT Exemplar ProblemsHere 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 ΔABC 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 ΔABC, 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 ΔABC

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 ΔPCD

Proof:

Since, tangents drawn from an external point to a circle are equal,

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

Now, perimeter of DPCD = PC + CD+PD

                    = PC + CE + ED + PD

                    = PC + CA + DB + PD

                    = PA + PB

                    = 2PA = 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 DOPO’, 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 ΔABC 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

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