Question

Draw a ∆ABC with side BC = 7 cm, ∠B= 45° ,∠A105° .Then, construct a triangle whose sides are 4/3 times the corresponding sides of ABC.
(class 10 CBSE SAMPLE PAPER 2017-18 MATHS)

Answer 1

[FIGURE IS IN THE ATTACHMENT]

Given ,a ∆ABC , in which BC= 7 cm, ∠B = 45°, ∠A = 105°

∠A + ∠B + ∠C = 180°

[Angle sum property]

105°+ 45°+ ∠C = 180°
150° + ∠C = 180°
∠C = 180° - 150°

∠C= 30°

Steps of Construction:

1.Draw a line segment BC=  7 cm.

2. Make an angle of 45° at point B and 30° at point C, which intersect each other at point A.  Thus ABC is a given Triangle.

3. Now from B draw a ray BY by making an acute ∠CBY with base BC on the side opposite to the vertex A.

4.Now, locate 4 points B1 ,B2 , B3 ,B4 on BY such that BB1= B1B2= B2B3 =B3B4.

5.Join B3C, and draw a line through B4M || B3 C intersecting the extended line BC at M.

6. From M draw OM|| AC intersecting the extended line BA at O.

Thus, ∆OBM is the required triangle whose sides are 4/ 3 times of the corresponding sides of ∆ABC.

HOPE THIS WILL HELP YOU...

Answer 2

∠B = 45°, ∠A = 105°

Sum of all interior angles in a triangle is 180°.

∠A + ∠B + ∠C = 180°

105° + 45° + ∠C = 180°

∠C = 180° − 150°

∠C = 30°

The required triangle can be drawn as follows.

Step 1

Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.

Step 2

Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

Step 3

Locate 4 points (as 4 is greater in 4 and 3), B1, B2, B3, B4, on BX.

Step 4

Join B3C. Draw a line through B4 parallel to B3C intersecting extended BC at C'.

Step 5

Through C', draw a line parallel to AC intersecting extended line segment at C'. ΔA'BC' is the required triangle.



Justification

The construction can be justified by proving that



In ΔABC and ΔA'BC',

∠ABC = ∠A'BC' (Common)

∠ACB = ∠A'C'B (Corresponding angles)

∴ ΔABC ∼ ΔA'BC' (AA similarity criterion)

… (1)

In ΔBB3C and ΔBB4C',

∠B3BC = ∠B4BC' (Common)

∠BB3C = ∠BB4C' (Corresponding angles)

∴ ΔBB3C ∼ ΔBB4C' (AA similarity criterion)



On comparing equations (1) and (2), we obtain



⇒ 

This justifies the construction.