﻿ Class 10 NCERT Math Solution
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Introduction

Introduction of Constructions

Constructions using a straight edge (ruler) and a compass, e.g., bisecting an angle, drawing the perpendicular bisector of a line segment, some constructions of triangles etc. and also gave their justifications.

1. To divide a line segment in a given ratio.

Given a line segment AB, we want to divide it in the ratio m : n, where both m and n are positive integers. To help you to understand it, we shall take m = 3 and n = 2.

Steps of Construction : 1. Draw any ray AX, making an acute angle with AB.
2. Locate 5 (= m + n) points A1, A2, A3, A4 and A5 on AX so that AA1 = A1A2 = A2A3 = A3A4 = A4A5.
3. Join BA5.
4. Through the point A3 (m = 3), draw a line parallel to A5B (by making an angle equal to ∠ AA5B) at A3 intersecting AB at the point C (see Figure). Then, AC : CB = 3 : 2.
Let us see how this method gives us the required division. Since A3C is parallel to A5B, therefore,
AA3/A3A5 = AC/CB (By the Basic Proportionality Theorem)
By construction, AA3/A3A5 = 3/2
Therefore, AC/CB = 3/2
This shows that C divides AB in the ratio 3 : 2.

2. To construct the tangents to a circle from a point outside it.

We are given a circle with centre O and a point P outside it. We have to construct the two tangents from P to the circle.

Steps of Construction : 1. Join PO and bisect it. Let M be the midpoint of PO.
2. Taking M as centre and MO as radius, draw a circle. Let it intersect the given circle at the points Q and R.
3. Join PQ and PR.
Then PQ and PR are the required two tangents (see Figure).
Now let us see how this construction works. Join OQ. Then ∠ PQO is an angle in the semicircle and, therefore, ∠ PQO = 90°
Can we say that PQ ⊥ OQ?
Since, OQ is a radius of the given circle, PQ has to be a tangent to the circle.
Similarly, PR is also a tangent to the circle.

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