# NCERT CBSE Class 10th Mathematics Chapter 11: Constructions

Safalta Expert Published by: Sylvester Updated Wed, 22 Jun 2022 06:51 PM IST

## Highlights

NCERT CBSE Class 10th Mathematics Chapter 11: Constructions

The eleventh chapter in Mathematics textbook is 'Constructions'.

### Determining a Point Dividing a given Line Segment, Internally in the given Ratio M : N

Let AB be the given line segment of length x cm. We are required to determine a point P dividing it internally in the ratio m : n.

Steps of Construction:

• Draw a line segment AB = x cm.
• Make an acute ∠BAX at the end A of AB.
• Use a compass of any radius and mark off arcs. Take (m + n) points A1, A2, … Am, Am+1, …, Am+n along AX such that AA1 = A1A2 = … = Am+n-1 , Am+n
• Join Am+nB.
• Passing through Am, draw a line AmP || Am+nB to intersect AB at P. The point P so obtained is the A required point which divides AB internally in the ratio m : n. ### Construction of a Tangent at a Point on a Circle to the Circle when its Centre is Known

Steps of Construction:

• Draw a circle with centre O of the given radius. • Take a given point P on the circle.
• Join OP. • Construct ∠OPT = 90°. • Produce TP to T’ to get TPT’ as the required tangent. ### The topics discussed in this chapter are as follows:

• Dividing a Line Segment
• Constructing Similar Triangles
• Drawing Tangents to a Circle
• Number of tangents to a circle from a given point
• Drawing tangents to a circle from a point outside the circle
• Drawing tangents to a circle from a point on the circle

## Draw a line segment of length 6 cm. Using compasses and ruler, find a point P on it which divides it in the ratio 3 : 4. Hence, PA : PB = 3 : 4

## Draw a triangle ABC in which AB = 5 cm, BC = 6 cm and ∠ABC = 60°. Then construct a triangle whose sides are 5/7 times the corresponding sides of ∆ABC

In ∆ABC
AB = 5 cm
BC = 6 cm
∠ABC = 60° Hence, ∆A’BC’ is the required ∆.

## Construct a triangle with sides 5 cm, 5.5 cm and 6.5 cm. Now construct another triangle, whose sides are 3/5 times the corresponding sides of the given triangle. ∴ ∆AB’C’ is the required ∆.

## Draw a triangle PQR such that PQ = 5 cm, ∠P = 120° and PR = 6 cm. Construct another triangle whose sides are 3/4 times the corresponding sides of ∆PQR.

In ∆PQR,
PQ = 5 cm, PR = 6 cm, ∠P = 120° ∴ ∆POʻR’ is the required ∆.

## Construct a triangle with sides 5 cm, 4 cm and 6 cm. Then construct another triangle whose sides are 2/3 times the corresponding sides of first triangle. Steps of Construction:

• Draw ∆ABC with AC = 6 cm, AB = 5 cm, BC = 4 cm.
• Draw ray AX making an acute angle with AÇ.
• Locate 3 equal points A1, A2, A3 on AX.
• Join CA3.
• Join A2C’ || CA3.
• From point C’ draw B’C’ || BC.

∴ ∆AB’C’ is the required triangle.