Chapter 10 Construction NCERT Exemplar Solutions Exercise 10.3 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 10 Construction Exercise 10.3 
Book Name  NCERT Exemplar for Class 10 Maths 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 10.3 Solutions
Short Answer Questions
1. Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio 3:5.
Solution
Steps of construction:
 Draw a line segment AB = 7 cm.
 Draw a ray AX, making an acute ∠BAX.
 Along AX, mark 3 + 5 = 8 points A_{1} , A_{2} , A_{3} , A_{4} , A_{5} , A_{6} , A_{7} , A_{8}
Such that AA_{1} = A_{1}A_{2} = A_{2} A_{3} = A_{3}A_{4} = A_{4}A_{5} = A_{6}A_{7} = A_{7}A_{8}  Join A_{8}B.
 From A3, draw A_{3}P ॥ A_{8}B meeting AB at P.
[by making an angle equal to ∠BA_{8}A at A_{3}]
Explanation :
Let
AA_{1} = A_{1}A_{2} = A_{2} A_{3} = A_{3}A_{4} ........... = A_{7}A_{8} = x
In ∆ABA_{8} , we have
A_{3}P  A_{8}B
AP/PB = AA_{3}/A_{3}A_{8} = 3x/5x
Therefore, AP : PB = 3 : 5
2. Draw a right triangle ABC in which BC = 12 cm, AB = 5 cm and ∠B = 90°. Construct a triangle similar to it and of scale factor 2/3 . Is the new triangle also a right triangle?
Solution
Steps of construction:
 Draw a line segment BC = 12 cm.
 From B draw a line AB = 5 cm which makes right angle at B.
 Join AC, ∆ABC is the given right triangle.
 From B draw an acute ∠CBX downwards.
 On ray BX, mark three points B_{1} , B_{2} and B_{3} , such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} .
 Join B_{3}C.
 From point B_{2} draw B_{2}N ॥ B_{3}C intersect BC at N.
 From point N draw NM ॥ CA intersect BA at M. ∆MBN is the required triangle. ∆MBN is also a right angled triangle at B.
3. Draw a triangle ABC in which BC = 6 cm, CA = 5 cm and AB = 4 cm.
Construct a triangle similar to it and of scale factor 5/3 .
Solution
Steps of construction:
 Draw a line segment BC = 6 cm.
 Taking B and C as centers, draw two arcs of radii 4 cm and 5 cm intersecting each other at A.
 Join BA and CA. ∆ABC is the required triangle.
 From B, draw any ray BX downwards making at acute angle ∠CBX
 Mark five points B_{1} , B_{2} , B_{3} , B_{4} and B_{5} on BX, such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5}.
 Join B3C and from B_{5} draw B_{5}M ॥ B_{3}C intersecting the extended line segment BC at M.
 From point M draw MN ॥ CA intersecting the extended line segment BA at N.
Therefore, ∆NBM is the required triangle whose sides is equal to 5/3 of the corresponding sides of the ∆ABC.
4. Construct a tangent to a circle of radius 4 cm from a point which is at a distance of 6 cm from its center.
Solution
We have, a point M’ is at a distance of 6 cm from the centre of a circle of radius 4 cm.
Steps of construction:
 Draw a circle of radius 4 cm. Let the centre of this circle be O.
 Join OM’ and bisect it. Let M be midpoint of OM’.
 Taking M as centre and MO as radius draw a circle to intersect circle (0, 4) at two points, P and Q.
 Join PM’ and QM’. PM’ and QM’ are the required tangents from M’ to circle C(0, 4).