NCERT Exemplar Class 10 Maths Solutions Chapter 10 Constructions 2021

Q: Can we divide any line segment into equal parts with the help of construction?

Yes, we can divide any line segment into any number of parts with any length ratio.

Q: Can we draw a common tangent for two given circles?

Yes, we can draw a common tangent for two given circles with the help of construction.

Q: Is the chapter Constructions important for Board examinations?

The chapter Constructions is vital for Board examinations as it holds around 2-3% weightage of the whole paper.

Q: What is the pattern of question type in the chapter of Constructions that can be expected in the board examination?

Generally, you can expect to get either a Long answer or a Very Long answer question in the board examination. A thorough study from NCERT exemplar Class 10 Maths solutions chapter 10 can help you ace the questions on Constructions.

NCERT Exemplar Class 10 Maths Solutions Chapter 10 Constructions

Edited By Ravindra Pindel | Updated on Sep 05, 2022 06:08 PM IST | #CBSE Class 10th

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NCERT exemplar Class 10 Maths solutions chapter 10 enables students with techniques to draw angles and shapes without actual measurements. We need to use a compass, protractor, scale, and divider. The NCERT exemplar Class 10 Maths chapter 10 solutions are prepared by our subject-matter experts keeping in view the practical nature of this chapter so that students can learn the topics of NCERT Class 10 Maths related to constructions, seamlessly.

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These NCERT exemplar Class 10 Maths chapter 10 solutions provide step-by-step solutions to the construction problems to make it easier for students to understand the process flow. CBSE 10 Maths Syllabus has been followed while preparing these NCERT exemplar Class 10 Maths solutions chapter 10.

Question:1

To divide a line segment AB in the ratio $5:7$ , first a ray AX is drawn so that $\angle BAX$ 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) 12
Solution
Given: $\angle BAX$ is an acute angle.
The required ratio is $5:7$
Let m = 5, n = 7
57598

Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Locate 12 points on AX at equal distances. (Because here $m+n= 12$ )
3. Join $A_{12}B$
4. Through the point $A_{5}$ draw a line parallel to $A_{12}B$ intersecting AB at the point P.
Then $AP:PB= 5:7$
$\because A_{5}P\parallel A_{12}B$
$\therefore \frac{AA_{5}}{A_{5}A_{12}}= \frac{AP}{PB}$ (By Basic Proportionality theorem)
By construction $\frac{AA_{5}}{A_{5}A_{12}}= \frac{5}{7}$
$\therefore \frac{AP}{PB}= \frac{5}{7}$
Hence the number of points is 12.

Question:2

To divide a line segment AB in the ratio $4:7$ , a ray AX is drawn first such that $\angle BAX$ is an acute angle and then points $A_{1},A_{2},A_{3}\cdots$ are located at equal distances on the ray AX and the point B is joined to:
(A) A₁₂ (B) A₁₁ (C) A₁₀ (D) A₉

Answer(B) A₁₁
Solution
Given: $\angle BAX$ is an acute angle
The required ratio is $4:7$
Let m = 4, n = 7
$m+n= 4+7= 11$
57610

Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Locate 11 points on AX at equal distances (because m + n = 11)
3. Join $A_{11}B$
4. Through the point A4 draw a line parallel to $A_{11}B$ intersecting AB at the point P.
Then $AP:PB= 4:7$
Hence point B is joined to A₁₁.

Question:3

To divide a line segment AB in the ratio $5:6$ , draw a ray AX such that $\angle BAX$ is an acute angle, then draw a ray BY parallel to AX and the points $A_{1},A_{2},A_{3}\cdots$ and $B_{1},B_{2},B_{3}\cdots$ are located at equal distances on ray AX and BY, respectively. Then the points joined are
(A) A₅ and B₆ (B) A and B₅ (C) A₄ and B₅ (D) A₅ and B₄

Answer(A) A₅ and B₆
Solution
Given: $\angle BAX$ and $\angle ABY$ both are acute angles and AX parallel to BY
The required ratio is $5:6$
Let m = 5 , n = 6
57617

Steps of construction
1. Draw any ray AX making an acute angle with AB.
2. Draw a ray $BY\parallel AX$ .
3. Locate the points $A_{1},A_{2},A_{3},A_{4},A_{5}$ on AX at equal distances
4. Locate the points $B_{1},B_{2},B_{3},B_{4},B_{5}$ on BY at distance equal to the distance between points on AX line.
5. Join $A_{5}B_{6}$ .
Let it intersect AB at a point C in figure.
Then $AC:CB= 5:6$
Here $\bigtriangleup AA_{5}C$ is similar to $\bigtriangleup BB_{6}C$
Then $\frac{AA_{5}}{BB_{6}}= \frac{5}{6}= \frac{AC}{BC}$
$\therefore$ by construction $\frac{AA_{5}}{BB_{6}}= \frac{5}{6}$
$\therefore \frac{AC}{BC}= \frac{5}{6}$
$\therefore$ Points joined one A₅ and B₆.

Question:4

To construct a triangle similar to a given $\bigtriangleup ABC$ with its sides of $\frac{3}{7}$ the corresponding sides of $\bigtriangleup ABC$ , 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 B₁, B₂, B₃, ... 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

Answer(C) B₇ to C
Solution Given: $\angle CBX$ is an acute angle.
ccdd

Steps of construction
1. Draw any ray BX making angle with BC on the side opposite to vertex A.
2. Locate 7 points on BX in equidistant
3. Now join B₇to C
4. Draw a line through B₃ ${C}'$ parallel to B₇C .

Question:5

To construct a triangle similar to a given $\bigtriangleup ABC$ with its sides $\frac{8}{5}$ of the corresponding sides of $\bigtriangleup ABC$ draw a ray BX such that $\angle CBX$ 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) 8
Solution
To construct a triangle similar to a triangle, with its sides $\frac{8}{5}$ of the corresponding sides of given triangle, the minimum number of points to be located at an equal distance is equal to the greater of 8 and 5 in . $\frac{8}{5}$ Here $8> 5$
So, the minimum number of points to be located at equal distance 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 $60^{\circ}$ , it is required to draw tangents at end points of those two radii of The circle, the angle between them should be
(A) $135^{\circ}$ (B) $90^{\circ}$ (C) $60^{\circ}$ (D) $120^{\circ}$

Answer(D) 120°
Solution
According to question:-
57641

Given : $\angle QPR= 60^{\circ}$
Let $\angle QOR= x$
As we know that angle between tangent and radius of a circle is 90
$\angle PQO= \angle PRO= 90^{\circ}$
We know that $\angle PQO+ \angle PRO+ \angle QPR+\angle QOR= 360^{\circ}$
[ $\because$ sum of interior angles of quadrilateral is $360^{\circ}$ ]
$90^{\circ}+90^{\circ}+x+60^{\circ}= 360^{\circ}$ |
$240+x= 360^{\circ}$
$x= 120^{\circ}$