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
- 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
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
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
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.
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