CBSE 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:
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.
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 ΔNSQ≅ΔMTR, then prove that ΔPTS ~ ΔPRQ.
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.
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.
Question. 6 Find the altitude of an equilateral triangle of side 8cm.
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.
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).
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: