construction from class ix mathematic text book.

find the perimeter of rectangle abcd,given that ab=13cm,bc=8cm
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1. Steps,such as how to draw a perpendicular from a point on a line or perpendicular bisector of a line segment, and how to draw a line segment of a given length or an angle bisector, are skipped in here.
2. Mark a point B.  From B, draw a segment BC exactly 8cm long rightwards.  Mark C at the right end of the segment.  We are to construct vertex A above BC.
3. From B, draw a vertical approx. 1.5BC (=12cm) long upwards, that perpendicularly intersects a line BC at B. Label the upper end of the vert., K.  ∠KBC=90°
4. Bisect ∠KBC, and draw the angle bisector approx. 2BC (=16cm) long upwards.  Mark a point P on the bisector exactly 13cm apart from B.  ∠PBC=45°  Connect P to C.
5. Draw a bisector of segment CP, crossing CP at O and BP at A.  △OAC≡△OAP (SRS), so AC=AP. therfore AB+AC=AB+AP=BP=13cm.  △ABC is the required triangle with ∠ABC=45°, BC=8cm and AB+AC=13cm.  (AB=approx. 7.15cm, AC= approx. 5.85cm).

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solve the length of the sides of aabc are bc=12,ac=13, and ab=14. if M is the midpoint of ac and p is the point where ac is cut by the bisector of /b,find pm
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construct a right angle triangle abc with hypoternuse bc=8cm and ac =6cm
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triangle b=50 , a=40 , c=60
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