# 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 both the ends of the line segment such that they cross each other at one point. Connect this point from both ends.
- Then draw another ray 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 equal they are similar by AA criteria.
- 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 (since, 3 > 2 in the ratio 2/3) points B₁, B₂, B₃ on BX such that BB₁ = B₁B₂ = B₂B₃.
- Join B₃C and draw a line through B₂ (second point where 2 < 3 in the ratio) parallel to B₃C meeting BC at C'.
- Draw a line thorough C' parallel to CA to meet BA at A’. Now ΔA'BC' is the required triangle similar to ΔABC where BC'/BC = BA'/BA = C'A'/CA = 2/3

Proof:

In ΔBB₃C, B₂C' is parallel to B₃C.

Hence by Basic proportionality theorem,

B₂B₃/BB₂ = C'C/BC' = 1/2

Adding 1 to both the sides of C'C/BC' = 1/2,

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)

Hence by AA similarity, ΔBA'C' ~ ΔBAC

Corresponding sides are proportional

BA'/BA = C'A'/CA = BC'/BC = 2/3 [From equation(1)]

**☛ Check: **Class 10 Maths NCERT Solutions Chapter 11

**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

**Summary:**

A triangle BAC of sides 4 cm, 5 cm, and 6 cm and a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle is constructed.

**☛ Related Questions:**

- Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.
- Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1(1/2) times the corresponding sides of the isosceles triangle.
- Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/2 of the corresponding sides of the triangle ABC.
- Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/2 of the corresponding sides of the triangle ABC.

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