# Construct a Triangle Abc with Sides Bc = 7 Cm, ∠B = 45° and ∠A = 105°. Then Construct a Triangle Whose Sides Are 3/4 Times the Corresponding Sides of ∆Abc. - Mathematics

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

#### Solution

Steps of construction:

1. Draw line BC = 7 cm

2. At B, construct ∠CBY = 45° and at C, construct 180° −(105° + 45°) = 30°.

3. Mark the point of intersection of ∠B = 45° and ∠C = 30° as A. Thus, ∆ABC is obtained.

4. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

5. Locate 4 (3 < 4) points B1, B2, B3 and B4 on BX such that  BB1 = B1B2 = B2B3 = B3B4.

6. Join B4C and draw a line through B3 parallel to B4C to intersect BC at C'.

7. Draw a line through C′ parallel to the line CA to intersect BA at A′

Then, ΔA′BC′ is the required triangle similar to the ΔABC

Is there an error in this question or solution?
2016-2017 (March) Delhi Set 3

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