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

In this article you will get the CBSE Class 10 Mathematics chapter 6, Triangles: NCERT Exemplar Problems and Solutions (Part-IIIB).  This part contains solutions to Q. No. 9-15 from Exercise 6.3 that consists of only the Short Answer Type Questions. These questions will prove to be very helpful while preparing for CBSE Class 10 Board Exam 2017-2018.

Created On: Jul 20, 2017 16:57 IST

Triangles NCERT Exemplar Problems, NCERT Exemplar Problems, Class 10 NCERT ExemplarCBSE Class 10 Mathematics chapter 6, Triangles: NCERT Exemplar Problems and Solutions (Part-IIIB) is available here. This part of the chapter includes solutions for question number 9-15 from 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 Solution for CBSE Class 10 Mathematics: Triangles (Part-IIIA)

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. 9-15):

Question. 9 ABCD is a trapezium in which AB ||DC and P,Q are points on AD and BC

respectively, such that PQ || DC, if PD = 18cm, BQ = 35cm and QC = 15cm, find AD.

Solution.

Question. 10 Corresponding sides of two similar triangles are in the ratio of 2:3. If the area of the smaller triangle is 48cm2, then find the area of the larger triangle.

Solution.

Let the two triangles be ABC and PQR, where ΔABC is smaller of the two.

Given, ΔABC~ ΔDEF

            AB : DE = 2:3

Area of smaller ΔABC = 48cm2

Since ΔABC~ ΔDEF

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

Question. 11 In a  ΔPQR, N is a point on PR, such that QN ^ PR. If PN.NR = QN2, then prove that ∠PQR = 90°.

Solution.

Question. 12 Areas of two similar triangles are 36cm2 and 100cm2. If the length of a side of the larger triangle is 20cm. Find the length of the corresponding side of the smaller triangle.

Solution.

Let the two triangles be, ΔABC and ΔDEF where ΔABC is the smaller of the two and

          Area of ΔABC = 36cm2

And,    Area ΔDEF = 100cm2

Also, DE = 20cm

Let length of the corresponding side of ΔABC, AB = x cm

As ΔABC ~ ΔDEF

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

Hence, the length of corresponding side of the smaller triangle is 12cm.

Question. 13 In given figure, if ÐACB = ÐCDA, AC = 8cm and AD = 3cm, then find BD.

Question. 14 A 15 metres high tower casts a shadow 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.

Solution.

Let BC = 15m be the tower and its shadow AB is 24m. Again, let EF = h be telephone pole and DE = 16m be its shadow.

Question. 15 Foot of a 10m long ladder leaning against a vertical wall is 6m away from the base of the wall Find the height of the point on the wall where the top of the Ladder reaches.

Solution.

Let BC be the vertical wall and AC = 10m is a ladder.

Distance of ladder from the base of the wall BC is 6m

Hence, the height of the point on the wall where the top of the ladder reaches is 8cm.

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