construct triangle ABC in which BC=7cm, angle B=75°and AB +AC=12 cm
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Answer :
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
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
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