# NCERT Exemplar Solution for Class 10 Mathematics: Constructions (Part-III)

CBSE Class 10 Mathematics chapter 10, Constructions: NCERT Exemplar Problems and Solutions (Part-III) is available here. This exercise comprises the Short Answer Type Questions that makes Exercise:10.3 of NCERT Exemplar Problems for calss 10 Maths, Constructions. Every question has been provided with a detailed solution.

Here you get the CBSE Class 10 Mathematics chapter 10, Constructions: NCERT Exemplar Problems and Solutions (Part-III). This part of the chapter includes solutions for Exercise 10.3 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Constructions. This exercise comprises of 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, Constructions:**

**Exercise 10.3 **

**Short Answer Type Questions:**

**Question. 1 **Draw a line segment of length 7 cm. Find a point *P* on it which divides it in the ratio 3:5.

**Solution.**

**Steps of construction:**

1. Draw a line segment *AB = *7 cm.

2. Draw a ray *AX*, making an acute ∠*BAX*.

3. Divede *AX*, in 3 + 5 = 8 equal parts *AA*_{1}, *A*_{1}*A*_{2}, *A*_{2}*A*_{3}*,* *A*_{3}*A*_{4}, *A*_{4}*A*_{5}, *A*_{5}*A*_{6}, *A*_{6}*A*_{7} and *A*_{7}*A*_{8}.

4. Join *A*_{8}*B*.

5. From *A*_{3}, draw *A*_{3}*C || A*_{8}*B* intersecting *AB *at *C*.

6. Thus *C* is the required point on *AB* which divides it in the ratio 3:5.

i.e., *AC *: CB = 3 : 5

**Question. 2 **Draw a right Δ*ABC* in which *BC* = 12 cm, *AB* = 5 cm and ∠*B = *90°*. *Construct a triangle similar to it and of scale factor 2/3. Is the new triangle also a right triangle?

**Solution.**

**Steps of construction:**

First we draw the right Δ*ABC* by following the steps given below:* *

1. Draw a line segment *BC = *12 cm.

2. From *B* draw a line *AB* *= *5 cm such that *AB* ⏊ *BC.*

3. Join *AC*.

Thus* *given right D*ABC *is obtained.

**Now we construct another ****triangle similar to ****D***ABC ***with its sides equal to 2/3 of the corresponding sides of ****D***ABC***, by following the steps given below:**

4. ** **From

*B*draw an acute ∠

*CBY*such that

*A*and

*Y*are in opposite direction with respect to

*BC*.

5. Divide *BY *into three equal parts, *BB*_{1} = *B*_{1}*B*_{2 }= *B*_{2}*B*_{3}.

7. From point *B*_{2} draw *B*_{2}*C’* || *B*_{3}*C *intersecting *BC *at *C’.*

8.** **From point *C’ *draw *C’A’* || *CA *intersecting *BA *at *A’. *

D*A’BC’ *is the required triangle of scale factor 2/3.This is also a right angled triangle.

**Question. 3 **Draw a Δ*ABC *in which *BC = *6cm, *CA* =* *5 cm and *AB* =* *4 cm. Construct a triangle similar to it and of

scale factor.

**Solution.**

**Steps of construction: **

First we draw Δ*ABC *by following steps given below:

1. Draw a line segment *BC = *6 cm.

2. Taking *B* and *C* as centres, draw two arcs of radii 4 cm and 5 cm respectively, which intersect each other at *A*.

3. Join *BA *and *CA. *Δ*ABC *is the required triangle.

**Now we construct another triangle similar to ****D***ABC ***and of scale factor 5/3, by following the steps given below:**

4.** **From *B*, draw an acute ∠*CBY *such that *A* and *Y* are in opposite direction with respect to *BC*.

5.** **Divide *BY *into five equal parts,

6. Join *C* and from *B*_{5} draw *B*_{5}*C’ *|| *B*_{3}*C *intersecting the extended line segment *BC *at *C’.*

7. Again from point *C’ *draw *C’A’ || CA *intersecting the extended line segment *BA *at *A’.*

Thus, Δ*A’BC’ *is the required triangle similar to Δ*ABC *and of scale factor 5/3*.*

**Question. 4 **Construct a tangent to a circle of radius 4 cm from a point which is at a distance of 6 cm from its centre.

**Solution. **

**Steps of construction**

1. Draw a line-segment *OP*= 6 cm.

2. Draw a circle of radius 4 cm taking *O* as centre.

2. Draw a perpendicular bisector of *OP *intersecting *OP *at *M*.

4. Taking *M *as centre and *MO *as radius draw a circle to intersect circle *C* (*O*, 4) at two points, *Q* and *R*.

5. Join *PQ *and *PR. *

* *

Thus, *PQ *and *PR* are the required tangents.

**CBSE Class 10 NCERT Textbooks & NCERT Solutions**

**NCERT Solutions for CBSE Class 10 Maths**

**NCERT Exemplar Problems and Solutions Class 10 Science: All Chapters**

## Comments