Question

Construct a triangle ABC in which BC = 7 cm, angle B = 75°, and AB + AC = 13 cm.

Answer 1

Given : Base BC = 7cm, angle B = 75° and sum of two sides AB + AC = 12 cm.

Required : To Construct ∆ABC

STEPS OF CONSTRUCTION :

1. Draw a ray BX and cut off a line segment BC = 7cm; from it.

2. At B; construct angle YBX = 75°

3. With B as centre and radius = 12 cm (because AB + AC = 12cm) draw an arc to meet BY at D.

4. Join CD

5. Draw Perpendicular bisector PQ of CD intersecting BD at A.

6. Join AC

Then ABC is the required triangle.

A lies on perpendicular bisector of CD.

Therefore, AC = AD

=> AB = BD - AD

=> AB = BD - AC

=> AB + AC = BD = 12cm,

Which is true as given.

HOPE IT WOULD HELP YOU

Answer 2

Given :–

• BC = 7 cm.
• ∠B = 75°
• AB + AC = 13 cm.

Required :–

• Construct a ∆ABC.

★ Construction is refer to the attachment ★

From construction we get,

• AC = 7.6 cm
• AB = 5.4 cm

Steps of Construction :–

1. Draw BC = 7 cm.
2. Draw ∠B = 75°.
3. Cut an arc of the length 13 cm from point B on BX, we get point D.
4. Join CD.
5. Draw perpendicular bisector of CD.
6. Now, join AC.

we get required ∆ABC

→ AB + AC

→ 5.4 + 7.6

→ 13.0

→ 13 cm.

Justification :–

AC = AD [points are equidistant from perpendicular bisector]

→ BD = BA + AD

→ AB + AC

→ 13 cm.