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NCERT Exemplar Solution for CBSE Class 10 Mathematics: Triangles (Part-IIIA)

Here you will get the CBSE Class 10 Mathematics chapter 6, Triangles: NCERT Exemplar Problems and Solutions (Part-IIIA).  This part contains solutions to Q. No. 1-7 from Exercise 6.3 that consists only of the Short Answer Type Questions. These questions will prove to be very helpful while preparing for CBSE Class 10 Board Exam 2017-2018.

Jul 20, 2017 13:10 IST
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Triangles Class 10 Important Questions, Class 10 Maths Chapter 6 NCERTCBSE Class 10 Mathematics chapter 6, Triangles: NCERT Exemplar Problems and Solutions (Part-IIIA) is available here. This part of the chapter includes solutions for question number 1-8from Exercise 6.3 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Triangles. This exercise comprises only the Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

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, Triangles:

Exercise 6.3

Short Answer Type Questions (Q. NO. 1-8):

Question. 1 In a ΔPQR, PR2 - PQ2 = QR2 and M is a point on side PR such that QM PR. Prove that QM2 = PM × MR.

Solution.

Given: ln ΔPQR,

            PR2 - PQ2 = QR2

And     QM PR

To prove: QM2 =PM × MR

Proof: Since, PR2 - PQ2 = QR2

⟹       PR2 = PQ2 + QR2

⟹       ΔPQA is right angled triangle at Q with PR being its hypotenuse.

Also,    QM PR

⟹      ∠QMR = ∠QMP = 90o

Question. 2 Find the value of x for which ΔEAB in given figure.

Hence, the required value of x is 2

Question. 3 In figure, if ∠l = ∠2 and ΔNSQΔMTR, then prove that ΔPTS ~ ΔPRQ.

             

Hence proved.

Question. 4 Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3RS. Find the ratio of the areas of ΔPOQ and ΔROS.

Solution.

Hence, the required ratio is 9:1

Question. 5 In figure, if AB || DC and AC, PQ intersect each other at the point O. Prove that OA.CQ = OC.AP.   

   

Hence proved.

Question. 6 Find the altitude of an equilateral triangle of side 8cm.

Solution.

Let ABC be an equilateral triangle.

Question. 7 If ΔABC ~ ΔDEF, AB = 4cm, DE = 6, EF = 9cm and FD = 12cm, then find the perimeter of ΔABC.

Solution.

Given,  ΔABC~ ΔDEF

           AB = 4cm, DE = 6cm

And     EF = 9cm, FD = 12cm.

Thus, perimeter of D ABC = AB + BC + AC = 4 + 6 + 8 = 18cm

Question. 8 In figure, if DE || BC, then find the ratio of ar (ΔADE) and ar (DECB).

                                  

Solution.

Given, in ΔABC

   DE || BC

   DE = 6cm and BC =12cm

Now, in ΔABC and ΔADE,

            ∠ABC = ADE                      [Corresponding angles]

            ∠ACB = ∠AED                      [Corresponding angles]

ΔABC ~ ΔAED                               [By AA similarity criterion]

Using the property, “The ratio of areas of similar triangles is equal to the ratio of the squares of their corresponding sides,” we have:

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