construct a triangle of side 5 centimetre 4 cm and 6 CM then construct another triangle whose side are 2/3 times the corresponding sides of first angle
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Given , a ∆ ABC , in which BC = 5 cm , AC= 4 cm & AC = 6 cm
Steps of construction:
1. Draw a line segment BC =5 cm.
2.Taking B & C as draw two arcs of radii 4 cm and 6 cm to intersect each other at A
3. Join BA and CA. Thus ∆ABC is the given Triangle.
4. Now from B draw any ray BX making an acute ∠CBX with base BC on the side opposite to the vertex A.
5. Along BX Mark 3 points B1,B2,B3, on BX such that BB1 = B1B2 = B2B3 .
6.Join B3C and from B2
draw a line B2N || B3C intersecting BC at N.
7. From point N, draw MN || CA intersecting AB at M. Thus , ∆MBN is the required triangle whose sides are 2/3 of the corresponding sides of ∆ABC.
HOPE THIS WILL HELP YOU.....
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