NCERT solutions for class 10 Maths Chapter 11 Exercise 11.1, 11.2 of Constructions in PDF format to free download. Some examples of constructions are also given on this page to understand the steps of constructions. In these * solutions *the steps of constructions of each question of

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**CLICK HERE**## NCERT solutions for class 10 Maths Chapter 11

### Solutions of Class 10 Maths – Constructions

##### English Medium

**Exercise 11.1****Exercise 11.2****NCERT Book in Hindi & English Medium**- Hindi Medium solutions will be uploaded in June, 2017.

Just like in **Class 9**, there are also the construction using a ruler and compass in class 10. In class 10 constructions we will learn to divide a line segment into given ratio, constructing a similar triangle to a given triangle and tangent to a circle from a given external point. These all constructions in class 10 requires the basic knowledge of class 9 constructions.

#### Construction 1: To divide a line segment in a given ratio.

In this case, a line segment is to be divided in to two part of given ratio with the help of ruler and compass. The concept of **Basic Proportionality Theorem** (**Thales** Theorem) is applied here for justification.

#### Construction 2: To construct a triangle similar to a given triangle as per given scale factor.

There are two cases for these type of constructions. One, the triangle to be constructed larger than the given triangle (dividing the sides into larger ratio) and two, the triangle to be constructed smaller than the given triangle (dividing the sides into smaller ratio).

Example: Construct a triangle similar to a given triangle ABC with its sides equal to 3/5 of the corresponding sides of the triangle ABC.

**Steps of Constructions:**

- Let ABC be the given Δ. Draw any ray BX making an acute angle with BC on the side opposite to vertex A.
- Locate 5 points B1, B2, B3, B4 and B5 on BX so that BB1 = B1B2 = B2B3 = B3B4 = B4B5
- Join B5C and draw a line through B3 parallel to B5C to meet BC at C′.
- Draw a line though C′ parallel to CA to meet AB in A′.
- Then ΔA′BC′ is the required Triangle.

#### Construction 3: To construct the tangents to a circle from a point outside it.

There are also two cases depending whether the point lies on the circle or lying outside of the circle. If the point lies on the circle, we draw radius through this point and draw a line perpendicular to this radius through this point. If the point lies outside of the circle, there would be two tangents through this point.

Example: To draw tangents to a circle from a given point outside it.

Suppose C be the given circle with centre O and a point P outside it. We have to draw tangents to the circle from the point P. For that, we go through the following steps:

- Join PO and bisect it. Let M be the midpoint of PO.
- Taking M as centre and MO as radius, draw a circle. Let it intersect the given circle at the points Q and R.
- Join PQ and PR. Then PQ and PR are the required two tangents.

**Historical Facts!**

The ancient Greek mathematician **Euclid** is the acknowledged inventor of geometry. He did this over 2300 years ago and his book ‘Element’ is still regarded as the ultimate geometry reference. In that work, he uses these construction techniques extensively and so they have become a part of the field of study in **geometry**. They also provide a greater insight into geometric concepts and give us tools to draw figures when direct measurement is not appropriate.