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 = 90^o

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.

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: