 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

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) A12

(b) A11

(c) A10

(d) A9

Explanation:

Since 4 + 7 = 11points are to be located on AX  at equal distantes, so B is joined to last point, A11.

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) A5 and B6

(b) A6 and B5

(c) A4 and B5

(d) A5 and B4

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 AA1, A1A2, A2A3, A3A4 and A4A5

4. Divide BY into six (n = 6) equal parts and BB1, B1B2, B2B3, B3B4, B4B5 and B5B6.

4. Join B6 A5. Let it intersect AB at a point C.

Then,  AC : BC = 5 : 6

Question. 4 To construct a triangle similar to a given DABC with its sides 3/7  of the corresponding sides of DABC, 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) B10 to C

(b) B3 to C

(c) B7 to C

(d) B4 to C

Explanation:

As the sides of new triangle are 3/7  of the corresponding sides of DABC, so it will be smaller than DABC. So, after we locate points andon BX at equal distance, the next step is to join the last point B7 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 ΔABC with its sides 8/5of the corresponding sides of ΔABC

draw a ray BX such that ÐCBX 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

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°

Explanation:

Let PQ and PR be the two at points Q and R on circle with centre O. Given ∠RPQ = 60o

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

&there4; ∠P = ∠Q = 90°