NCERT Exemplar Solution for Class 10 Mathematics: Constructions (Part-II)
Here you will get the NCERT Exemplar Problems and Solutions for CBSE Class 10 Mathematics chapter 10, Constructions (Part-II). This exercise includes only the Very Short Answer Type Questions. Every question has been provided with a detailed solution. All the questions are very important to prepare for CBSE Class 10 Board Exam 2017-2018.
Here you get the CBSE Class 10 Mathematics chapter 10, Constructions: NCERT Exemplar Problems and Solutions (Part-II). This part of the chapter includes solutions for Exercise 10.2 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Constructions. This exercise comprises of only the Very Short Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.
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:
Very Short Answer Type Questions:
Write True or False and give reason for your answer in each of the following:
Hence, it is possible to divide a line segment in the ratio 3:1 by the geometrical construction.
Question. 2 To construct a triangle similar to a given DABC with its sides 7/3 of the corresponding sides of DABC,
draw a ray BX making an acute angle with BC and X lies on the opposite side of A with respect of BC. The
points B1, B2,…, B7 are located at equal distances on BX, B3is 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.
As the sides of new new triangle will be equal to 7/3 of the corresponding sides of ΔABC. Therefore, it will be larger than ΔABC. Hence, B7C’ (larger of 7/3) instead of B6C’, is drawn parallel to B3C.
Steps of construction are:
1. Draw the given ΔABC.
2. From B draw any ray BX downwards making an acute angle CBX.
3. Divide BX into 7 equal parts = B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7.
4. Join C and from B7 draw a line B7C’‖ B3C intersecting the extended line segment BC at C’.
5. From point C’ draw C’A’||CA intersecting the extended line segment BA at A’.
6. Then, DA’BC’ is the required triangle with sides equal to 7/3 of the corresponding sides of DABC.
Thus, B7C’ is parallel to B3C.
Question. 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.
Since, the point P is 3 cm away from the centre of a circle with radius 3.5 cm therefore, the point P lies inside the circle and we now that no tangent is possible from a point lying inside the circle. For a tangent to be drawn from a point, the distance of that point must be equal to or greater than the radius of circle. So, no tangent is possible here.
Question. 4 A pair of tangents can be constructed to a circle inclined at an angle of 170°.
It is possible to draw a pair of tangents to a circle, inclined at an angle between 0° to 180°. Hence, we can draw a pair of tangents to a circle inclined at an angle of 170°.