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

Here you will get the CBSE Class 10 Mathematics chapter 6, Triangles: NCERT Exemplar Problems and Solutions (Part-IVA). This part contains solutions to Q. No. 1-6 from Exercise 6.4 that consists only of the Long Answer Type Questions. Every question is provided with an apt and simple solution.

Here you get the CBSE Class 10 Mathematics chapter 6, Triangles: NCERT Exemplar Problems and Solutions (Part-IVA) is available here. This part of the chapter includes solutions for question number 1-8from Exercise 6.4 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.

**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, Triangles:**

**Exercise 6.4**

**Long Answer Type Questions (Q. NO. 1-6):**

**Question. 1 **In given figure, if Ð*A* = ÐC*, AB *= 6cm, *BP *= 15cm, *AP = *12cm and *CP = *4cm, then find the lengths of *PD *and *CD.*

* *

**Question. 2** It is given that D*ABC* ~ D*EDF* such that *AB* = 5 cm, AC = 7cm, *DF* = 15cm and *DE = *12cm. Find the lengths of the remaining sides of the triangles.

**Solution.**

Hence, the lengths of remaining sides of given triangles are : *EF* = 16.8cm and *BC* = 6.25cm

**Question.**** 3** Prove that, if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.

**Solution. **Let a D*ABC* in which a line *DE *is drawn parallel to *BC *intersecting *AB *at *D *and *AC *at *E.*

**Hence proved.**

**Question.**** 4 **In the given figure, if *PQRS *is a parallelogram and *AB *|| *PS* then prove that *OC *|| *SR.*

**Solution. **

* *

**Question. ****5** A 5m long Ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4m high. If the foot of the ladder is moved 1.6m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.

**Solution. **

Let *AC *be the ladder and *BC *be the wall.

Let the two positions of ladder be represented by AC and DE.

* *

Hence, the top of the ladder would slide upwards on the wall by a distance equal to 0.8m.

**Question.**** 6** For going to a city *B *from city *A *there is a route via city *C *such that *AC* ⏊* **CB, AC = *2*x *km and *CB = *2(*x+ *7) km. It is proposed to construct a 26km highway which directly connects the two cities *A *and *B. *Find how much distance will be saved in reaching city *B *from city *A *after the construction of the highway.

**Solution. **

Using Pythagoras theorem in right angled D*ACB*, we have:

*AB*^{2} = *AC*^{2}* + BC ^{2}*

⟹ (26)^{2} *= *(2*x*)^{2}* + *{2(*x + *7)^{2}

⟹ 676 = 4*x*^{2}* + *4(*x ^{2} + *49 + 14

*x*)

⟹ 676 = 4*x*^{2}* + *4*x*^{2}* + *196 + 56

⟹ 676 = 8*x*^{2} + 56*x*+ 196

⟹ *8x*^{2}* + *56*x *- 480 = 0

⟹ *x*^{2}* + 7x *- 60 = 0

⟹ *x*^{2}* *+ 12*x* - 5*x *- 60= 0 [Solving by factorization method]

⟹ *x*(*x* + 12) - 5(*x* + 12) = 0

⟹ (*x + *12)(*x *- 5) = 0

⟹ *x* = -12 or 5

Ignoring *x* = -12, as distance cannot be negative, we have:

* x* = 5

∴* AC* = 2*x* = 10km

And *BC = *2(*x *+ 7) = 2(5 + 7) = 24km

So, distance covered from A to B via *C= AC + BC* = 10 + 24 = 34km

And, distance covered directly from A to B = *AB* = 26km

Hence, the distance saved = 34km - 26km =* *8km.

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