# Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ΔABC

**Solution:**

- Draw the triangle with the given conditions.
- Then draw another line that makes an acute angle with the baseline. 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 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 construction:**

- Draw BC = 7cm and at B, make an angle ∠CBY = 45° and at C, make ∠BCZ = 30° [Since, 180° - (45° + 105° )]. Both BY and CZ intersect at A and thus ΔABC is constructed.
- Draw the ray BX so that ∠CBX is acute.
- Mark 4 (since 4 > 3 in 4/3) points B₁, B₂, B₃, B₄ on BX such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄
- Join B₃ third point on BX, (since 3 < 4 in 4/3) to C and draw B C' parallel to BC such that C’ lies on the extension of BC.
- Draw C’A’ parallel to CA to intersect the extension of BA at A’. Now, triangle A'BC' is the required triangle similar to ΔABC where, BA'/BA = BC'/BC = C'A'/CA = 4/3

Proof:

In ΔBB₄C', B₃C || B₄C'

Hence by Basic proportionality theorem,

B₃B₄/BB₃ = CC'/BC = 1/3

CC'/BC + 1 = 1/3 + 1 (Adding 1)

(BC + CC')/BC = 4/3

BC'/BC = 4/3

Consider ΔBA'C' and ΔBAC

∠A'BC' = ∠ABC = 45°

∠BC'A' = ∠BCA = 30° (Corresponding angles as CA|| C'A')

∠BA'C' = ∠BAC = 105° (Corresponding angles)

By AAA axiom, ΔBA'C' ~ ΔBAC

Hence corresponding sides are proportional

BA'/BA = BC'/BC = C'A'/CA = 4/3

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

**Video Solution:**

## Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ΔABC.

NCERT Solutions Class 10 Maths Chapter 11 Exercise 11.1 Question 6

**Summary:**

A triangle ABC with a side 7 cm,∠B = 45°,∠A = 105°, and another triangle A'BC' of sides that is 4/3 times the corresponding sides of triangle ABC have been constructed.

**☛ Related Questions:**

- 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.
- 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.