Here you get the CBSE Class 10 Mathematics chapter 10, Constructions: NCERT Exemplar Problems and Solutions (Part-II). This part of the chapter includes solutions for Exercise 10.2 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Constructions. This exercise comprises of only the Very 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 10.2 **

**Very Short Answer Type Questions:**

**Write True or False and give reason for your answer in each of the following:**

Hence, it is possible to divide a line segment in the ratio 3:1 by the geometrical construction.

**Question. 2 **To construct a triangle similar to a given D*ABC *with its sides 7/3 of the corresponding sides of D*ABC,*

draw a ray *BX *making an acute angle with *BC *and *X *lies on the opposite side of *A *with respect of *BC. *The

points *B*_{1}*, B*_{2},…,* B*_{7}* *are located at equal distances on *BX, B*_{3}is joined to *C *and then a line segment *B*_{6}*C’ *is drawn

parallel to *B*_{3}*C, *where *C’ *lies on *BC *produced. Finally line segment *A’C’ *is drawn parallel to *AC.*

**Answer: False **

**Explanation: **

As the sides of new new triangle will be equal to 7/3 of the corresponding sides of Δ*ABC*. Therefore, it will be larger than Δ*ABC*. Hence, *B*_{7}*C’* (larger of 7/3) instead of B_{6}C’, is drawn parallel to *B*_{3}*C. *

**Steps of construction are:**

**1. **Draw the given Δ*ABC*.

2. From *B *draw any ray *BX *downwards making an acute angle *CBX*.

3. Divide *BX *into 7 equal parts = *B _{2} = B*

_{2}

*B*

_{3}=

*B*

_{3}

*B*

_{4}=

*B*

_{4}

*B*

_{5}

*=*

*B*

_{5}

*B*

_{6}

*= B*

_{6}

*B*

_{7}

*.*

4. Join *C* and from *B*_{7} draw a line *B*_{7}*C*’‖ *B _{3}C *intersecting the extended line segment

*BC*at

*C*’.

5. From point *C*’ draw *C*’*A*’||*CA *intersecting the extended line segment *BA *at *A’.*

6. Then, D*A’BC’ *is the required triangle with sides equal to 7/3 of the corresponding sides of D*ABC*.

Thus, *B*_{7}*C’ *is parallel to *B*_{3}*C.*

**Question. 3 **A pair of tangents can be constructed from a point *P *to a circle of radius 3.5 cm situated at a distance of

3 cm from the centre.

**Answer: False**

**Explanation: **

Since, the point P is 3 cm away from the centre of a circle with radius 3.5 cm therefore, the point P lies inside the circle and we now that no tangent is possible from a point lying inside the circle. For a tangent to be drawn from a point, the distance of that point must be equal to or greater than the radius of circle. So, no tangent is possible here.** **

**Question. 4 **A pair of tangents can be constructed to a circle inclined at an angle of 170°.

**Answer: True**

**Explanation: **

It is possible to draw a pair of tangents to a circle, inclined at an angle between 0° to 180°. Hence, we can draw a pair of tangents to a circle inclined at an angle of 170°.

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