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Given : Base BC = 7cm, angle B = 75° and sum of two sides AB + AC = 13cm
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 = 13cm ( because AB + AC = 13cm) 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 = 13cm,
Which is true as given.
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 = 13cm ( because AB + AC = 13cm) 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 = 13cm,
Which is true as given.
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