# Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle

**Solution:**

- Draw the line segment of the largest length 6 cm. Measure 5 cm and 4 cm separately and cut arcs from 2 ends of the line segment such that they cross each other at one point. Connect this point from both ends.
- Then draw another line that makes an acute angle with the given line segment (6 cm).
- Divide the line into (m + n) parts where m and n are the ratios given.
- Two triangles are said to be similar if their corresponding angles are They are said to satisfy Angle-Angle-Angle (AAA) Axiom.
- The basic proportionality theorem states that “If a straight line is drawn parallel to a side of a triangle, then it divides the other two sides proportionally ".

Steps of constructions:

- Draw BC = 6 cm. With B and C as centres and radii, 5 cm and 4 cm respectively draw arcs to intersect at A. ΔABC is obtained.
- Draw ray BX making an acute angle with BC.
- Mark 3 (3 > 2 in the ratio 2/3) points B
_{1}, B_{2}, B_{3}on BX such that BB_{1}= B_{1}B_{2}= B_{2}B_{3}. - Join B
_{3}C and draw the line through B_{2}(2nd point where 2 < 3 in the ratio) parallel to B_{3}C meeting BC at C'. - Draw a line thorough C' parallel to CA to meet BA at A’. Now ΔA'BC' is therequired triangle similar to ΔABC where BC'/BC = BA'/BA = C'A'/CA = 2/3

Proof:

In ΔBB_{3}C, B_{2}C' is parallel to B_{3}C.

Hence by Basic proportionality theorem,

B_{2}B_{2}/BB_{2} = C'C/BC' = 1/2

Adding 1,

C'C/BC' + 1 = 1/2 + 1

(C'C + BC')/BC' = 3/2

BC/BC' = 3/2

or BC'/BC = 2/3 ....(1)

Consider ΔBA'C' and ΔBAC

∠A'BC' = ∠ABC (Common)

∠BA'C' = ∠BAC (Corresponding angles ∵ C' A' ||CA)

∠BA'C' = ∠BCA (Corresponding angles ∵ C' A' ||CA)

Hence by AAA axiom, ΔBA'C' ~ ΔBAC

Corresponding sides are proportional

BA'/BA = C'A'/CA = BC'/BC = 2 (from(1))

**Video Solution:**

## Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle

### NCERT Solutions Class 10 Maths - Chapter 11 Exercise 11.1 Question 2:

Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle

A triangle BAC of sides 4 cm, 5 cm, and 6 cm and a triangle B'A'C having sides measuring 2/3 of the ones of triangle BAC have been constructed