Chapter 10 Construction NCERT Exemplar Solutions Exercise 10.2 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 10 Construction Exercise 10.2 
Book Name  NCERT Exemplar for Class 10 Maths 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 10.1 Solutions
Short Answer Questions with Reasoning
Write ‘True’ or ‘False’ and justify your answer in each of the following:
1. By geometrical construction, it is possible to divide a line segment in the ratio √3 : 1/√3.
Solution
True
Explanation:
As given in the question
Ratio = √3 : 1/√3.
On solving,
√3 : 1/√3. = 3 : 1
Therefore, geometrical construction is possible to divide a line segment in the ratio 3 : 1.
2. To construct a triangle similar to a given △ABC with its sides 7/3 of the corresponding sides of △ABC, draw a ray BX making acute angle with BC and X lies on the opposite side of A with respect to BC. The points B_{1} , B_{2} , ……, B_{7} are located at equal distance 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.
Solution
False
Steps of construction:
 Draw a line segment BC with suitable length.
 Taking B and C as centers draw two arcs of suitable radii intersecting each other at A.
 Join BA and CA. ∆ABC is the required triangle.
 From B draw any ray BX downwards making an acute angle CBX.
 Locate seven points B_{1} , B_{2} , B_{3} , .... B_{7} on BX such that BB_{1} = B_{1}B_{2} = B_{1}B_{3} = B_{3}B_{4} = B_{4}B_{1} = B_{5}B_{6} = B_{6}B_{7}
 Join B_{3}C and from B_{7} draw a line B_{7}C’ ॥ B_{3}C intersecting the extended line segment BC at C’.
 From point C’ draw C’A’ ॥ CA intersecting the extended line segment BA at A’.
∆A’BC’ is the required triangle whose sides are 7/3 of the corresponding sides of ∆ABC.
Given,
Segment B_{6}C’ is drawn parallel to B_{3}C.
But from our construction is never possible that segment B_{6}C’ is parallel to B_{3}C because the similar triangle A’BC’ has its sides 7/3 of the corresponding sides of triangle ABC.
Therefore, B_{7}C’ is parallel to B_{3}C.
Solution
False
As, the radius of the circle is 3.5 cm
r = 3.5 cm and a point P situated at a distance of 3 cm from the centre
So,
d = 3 cm.
We can see that r > d
Therefore, a point P lies inside the circle.
And, no tangent can be drawn to a circle from a point lying inside it.
4. A pair of tangents can be constructed to a circle inclined at an angle of 170°.
Solution
True
As, the angle between the pair of tangents is always greater than 0 but less than 180°.
Therefore, we can draw a pair of tangents to a circle inclined at an angle at 170°.