NCERT Exemplar Solutions for Class 10 Maths Chapter 10 Construction Exercise 10.2

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	Exercise 10.1 Exercise 10.3 Exercise 10.4
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 B₁ , B₂ , ……, B₇ are located at equal distance on BX, B₃ is joined to C and then a line segment B₆C'  is drawn parallel to B₃C where C' lies on BC produced.  Finally, line segment A'C' is drawn parallel to  AC.

Solution

False
Steps of construction:

1. Draw a line segment BC with suitable length.
2. Taking B and C as centers draw two arcs of suitable radii intersecting each other at A.
3. Join BA and CA. ∆ABC is the required triangle.
4. From B draw any ray BX downwards making an acute angle CBX.
5. Locate seven points B₁ , B₂ , B₃ , .... B₇  on BX such that BB₁ = B₁B₂ = B₁B₃ = B₃B₄ = B₄B₁ = B₅B₆  = B₆B₇
6. Join B₃C and from B₇ draw a line B₇C’ ॥ B₃C intersecting the extended line segment BC at C’.
7. From point C’ draw C’A’ ॥ CA intersecting the extended line segment BA at A’.
   

Then
∆A’BC’ is the required triangle whose sides are 7/3 of the corresponding sides of ∆ABC.
Given,
Segment B₆C’ is drawn parallel to B₃C.
But from our construction is never possible that segment B₆C’ is parallel to B₃C because the similar triangle A’BC’ has its sides 7/3 of the corresponding sides of triangle ABC.
Therefore, B₇C’ is parallel to B₃C.

3. A pair of tangents can be constructed from a point P to a circle of radius 3.5 cm situated at a distance of 3 cm from the centre.

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°.