Here you get the CBSE Class 10 Mathematics chapter 10, Constructions: NCERT Exemplar Problems and Solutions (Part-III). This part of the chapter includes solutions for Exercise 10.3 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Constructions. This exercise comprises of only the Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.
CBSE Class 10 Mathematics Syllabus 2017-2018
NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.
Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Constructions:
Exercise 10.3
Short Answer Type Questions:
Question. 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:
1. Draw a line segment AB = 7 cm.
2. Draw a ray AX, making an acute ∠BAX.
3. Divede AX, in 3 + 5 = 8 equal parts AA1, A1A2, A2A3, A3A4, A4A5, A5A6, A6A7 and A7A8.
4. Join A8B.
5. From A3, draw A3C || A8B intersecting AB at C.
6. Thus C is the required point on AB which divides it in the ratio 3:5.
i.e., AC : CB = 3 : 5
Question. 2 Draw a right ΔABC 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:
First we draw the right ΔABC by following the steps given below:
1. Draw a line segment BC = 12 cm.
2. From B draw a line AB = 5 cm such that AB ⏊ BC.
3. Join AC.
Thus given right DABC is obtained.
Now we construct another triangle similar to DABC with its sides equal to 2/3 of the corresponding sides of DABC, by following the steps given below:
4. From B draw an acute ∠CBY such that A and Y are in opposite direction with respect to BC.
5. Divide BY into three equal parts, BB1 = B1B2 = B2B3.
7. From point B2 draw B2C’ || B3C intersecting BC at C’.
8. From point C’ draw C’A’ || CA intersecting BA at A’.
DA’BC’ is the required triangle of scale factor 2/3.This is also a right angled triangle.
Question. 3 Draw a ΔABC in which BC = 6cm, CA = 5 cm and AB = 4 cm. Construct a triangle similar to it and of
scale factor.
Solution.
Steps of construction:
First we draw ΔABC by following steps given below:
1. Draw a line segment BC = 6 cm.
2. Taking B and C as centres, draw two arcs of radii 4 cm and 5 cm respectively, which intersect each other at A.
3. Join BA and CA. ΔABC is the required triangle.
Now we construct another triangle similar to DABC and of scale factor 5/3, by following the steps given below:
4. From B, draw an acute ∠CBY such that A and Y are in opposite direction with respect to BC.
5. Divide BY into five equal parts,
6. Join C and from B5 draw B5C’ || B3C intersecting the extended line segment BC at C’.
7. Again from point C’ draw C’A’ || CA intersecting the extended line segment BA at A’.
Thus, ΔA’BC’ is the required triangle similar to ΔABC and of scale factor 5/3.
Question. 4 Construct a tangent to a circle of radius 4 cm from a point which is at a distance of 6 cm from its centre.
Solution.
Steps of construction
1. Draw a line-segment OP= 6 cm.
2. Draw a circle of radius 4 cm taking O as centre.
2. Draw a perpendicular bisector of OP intersecting OP at M.
4. Taking M as centre and MO as radius draw a circle to intersect circle C (O, 4) at two points, Q and R.
5. Join PQ and PR.
Thus, PQ and PR are the required tangents.
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