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

Jagranjosh brings you the Class 10 Mathematics chapter 10, Constructions: NCERT Exemplar Problems and Solutions (Part-I). Every question has been provided with a detailed solution. All the questions given in this article are very important to prepare for CBSE Class 10 Board Exam 2017-2018.

Here you get the CBSE Class 10 Mathematics chapter 10, Constructions: NCERT Exemplar Problems and Solutions (Part-I). This part of the chapter includes solutions for Exercise 10.1 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Constructions. This exercise comprises of only the Multiple Choice Questions (MCQs) 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 10.1 **

**Multiple Choice Questions (MCQs)**

**Question. 1** To divide a line segment *AB *in the ratio 5:7, first a ray *AX *is drawn, so that *∠**BAX *is an acute angle and

then at equal distances points are marked on the ray *AX *such that the minimum number of these points is

(a) 8

(b) 10

(c) 11

(d) 12

**Answer. (d)**

**Explanation:*** *

To divide a line segment *AB* in the ratio *m *: *n, *a ray *AX* making an acute ∠*BAX, *is drawn and then *m + n *points are marked at equal distances on the ray *AX*.

Here, *m = *5, *n = *7

Therefore, minimum number of points to be marked on *AX* = *m* + *n* = 5 + 7 = l2.

**Question. 2** To divide a line segment *AB *in the ratio 4:7, a ray *AX *is drawn first such that* **∠BAX *is an acute angle and then points * *are located at equal distances on the ray *AX* and the point *B *is joined to

(a) *A*_{12}

(b) *A*_{11 }

(c) *A*_{10}

(d) *A*_{9}

**Answer. (b)*** *

**Explanation:**

Since 4 + 7 = 11points are to be located on *AX* at equal distantes, so *B* is joined to last point, *A*_{11}.

**Question. 3 **To divide a line segment *AB *in the ratio 5:6, draw a ray *AX *such that* *∠*BAX *is an acute angle, then draw a ray *BY *parallel to *AX *and the points * *and are located to equal distances on ray *AX *and *BY, *respectively. Then, the points joined are

(a) *A*_{5} and *B*_{6}

(b)* A*_{6} and* B*_{5}

(c) *A*_{4 }and *B*_{5}

(d) *A*_{5} and *B*_{4}

**Answer. (a)**

**Explanation:**

To divide line segment *AB* in the ratio 5:6.

** **

**Steps of construction**

1. Draw a ray *AX* making an acute ∠*BAX.*

2. Draw a ray *BY *parallel to *AX* by taking ∠*ABY *equal to ∠*BAX.*

3.* *Divide *AX *into five (*m* = 5) equal parts *AA*_{1}, *A*_{1}*A*_{2}, *A*_{2}*A*_{3}, *A*_{3}*A*_{4 }and *A*_{4}*A*_{5}

4. Divide *BY *into six (*n* = 6) equal parts and *BB*_{1}, *B*_{1}*B*_{2}, *B*_{2}*B*_{3}, *B*_{3}*B*_{4}, *B*_{4}*B*_{5} and *B*_{5}*B*_{6}.

4. Join *B*_{6 }*A _{5}. *Let it intersect

*AB*at a point

*C*.

Then, *AC *: *BC *= 5 : 6

**Question. 4 **To construct a triangle similar to a given D*ABC *with its sides 3/7 of the corresponding sides of D*ABC, *first draw a ray *BX *such that ∠*CBX *is an acute angle and *X *lies on the opposite side of *A *with respect to *BC. *Then, locate points on *BX* at equal distances and next step is to join

(a) *B*_{10} to *C *

(b) *B*_{3} to *C*

(c) *B*_{7 }to *C*

(d) *B*_{4} to *C*

**Answer. (c)**

**Explanation:**

As the sides of new triangle are 3/7 of the corresponding sides of D*ABC*, so it will be smaller than D*ABC*. So, after we locate points andon *BX* at equal distance, the next step is to join the last point *B*_{7} to *C* so, that parallel line from third part of *BX* meet on *BC* without producing.

**Question. 5 **To construct a triangle similar to a given Δ*ABC *with its sides 8/5of the corresponding sides of Δ*ABC*

draw a ray *BX *such that Ð*CBX *is an acute angle and *X *is on the opposite side of *A *with respect to *BC*. The minimum number of points to be located at equal distances on ray *BX *is

(a) 5

(b) 8

(c) 13

(d) 3

**Answer. (b)**

**Explanation:**

To* *construct a triangle similar to a given triangle, with its sides *m*/*n* of the corresponding sides of given triangle the minimum number of points to be located at equal distance is equal to the greater of *m *and *n*

in *m*/*n*. As, here* m*/*n* = 8/5.

So, the minimum number of points to be located at equal distances on ray *BX *is 8.

**Question. 6 **To draw a pair of tangents to a circle which are inclined to each other at an angle of 60°, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be

(a) 135°

(b) 90°

(c) 60°

(d) 120°

**Answer. (d)**

**Explanation:**

Let *PQ* and *PR* be the two at points *Q *and *R* on circle with centre *O*.

Given ∠*RPQ* = 60^{o}

We know that the tangent at a point on a circle is perpendicular to the radius through that point.

∴ ∠*P* = ∠*Q* = 90°

In quadrilateral PQRS,

∠*P* +∠*Q* + ∠*R* + ∠*O* = 360°

⟹ 90° + 90° + 60°+ ∠*O* = 360°

⟹ ∠*O* = 360° − 240 = 120°

Hence, the required angle between two radii is 120°.

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