Chapter 10 Construction NCERT Exemplar Solutions Exercise 10.1 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 10 Construction Exercise 10.1 
Book Name  NCERT Exemplar for Class 10 Maths 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 10.1 Solutions
Multiple Choice Questions
Choose the correct answer from the given four options:
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
Solution
(D) 12
As given in the question,
A line segment AB in the ratio 5:7
So,
A : B = 5:7
We draw a ray AX making an acute angle ∠BAX,
And mark A+B points at equal distance.
A = 5 and B = 7
Therefore,
Minimum number of these points = A + B
= 5 + 7 = 12
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 A_{1} , A_{2} , A_{3} ... are located at equal distances on the ray AX and the point B is joined to
(A) A_{12}
(A) A_{12}
(B) A_{11}
(C) A_{10}
(D) A_{9}
Solution
(B) A_{11}
As given in the question,
A line segment AB in the ratio 4:7
So,
A:B = 4:7
Now,
Draw a ray AX making an acute angle BAX
Minimum number of points located at equal distances on the ray,
AX = A+B
= 4+7
= 11
A_{1} , A_{2} , A_{3} ... are located at equal distances on the ray AX.
Point B is joined to the last point is A_{11} .
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 A_{1} , A_{2} , A_{1} , ... and B_{1} , B_{2} , B_{3} , ... are located at 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}
(B) A_{6} and B_{5}
(C) A_{4} and B_{5}
(D) A_{5} and B_{4}
Solution
(A) A_{5} and B_{6}
(A) A_{5} and B_{6}
As given in the question,
A line segment AB in the ratio 5:7
So,
A:B = 5:7
Steps of construction:
 Draw a ray AX, an acute angle BAX.
 Draw a ray BY AX, angle ABY = angle BAX.
 Now, locate the points A_{1} , A_{2} , A_{3} , A_{4} ... and A5 on AX and B_{1} ,B_{2} ,B_{3} ,B_{4} ,B_{5} and B_{6}
(Because A: B = 5:7)  4. Join A_{5}B_{6} .
A_{5}B_{6} intersect AB at a point C.
AC: BC= 5:6
4. To construct a triangle similar to a given Î”ABC with its sides 3/7 of the corresponding sides of Î”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 B_{1} , B_{2} , B_{3} , ... on BX at equal distance 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
Solution
(C)
In this, we locate points B_{1} ,B_{2} ,B_{3} ,B_{4} ,B_{5} , B_{6} and B_{7} on BX at equal distance and in next step
join the last point B_{7} to C.
5. To construct a triangle similar to a given Î”ABC with its sides 8/5 of the corresponding sides of Î”ABC draw a ray BX such that ∠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
Solution
(B)
To construct a triangle similar to a given triangle, with its sides m/n of the n 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 . Here, m/n = 8/5 So, the minimum number of point to be located at equal distance on ray BX is 8.
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°
Solution
(D)
The angle between them should be 120° because in that case the figure formed by the
The angle between them should be 120° because in that case the figure formed by the
intersection point of pair of tangent, the two end points of those two radii (at which tangents
are drawn) and the centre of the circle is a quadrilateral.
From figure POQR is a quadrilateral,
∠POQ + ∠PRQ = 180° [as, sum of opposite angles are 180°]
⇒ 60° + Î¸ = 180°
⇒ Î¸ = 120°
Therefore, the required angle between them is 120.