Construct a ∆ABC in which BC =5 cm, CA = 6 cm and AB=7 cm.
Construct a ∆A' BC' similar to ∆ ABC, each of whose sides is times
7/5 the corresponding side of ∆ABC.
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Step-by-step explanation:
Given , a ∆ ABC , in which BC = 5 cm , AC= 6 cm & BC = 7 cm
STEPS OF CONSTRUCTION:
1. Draw a line segment BC =5 cm.
2.Taking B & C as draw two arcs of radii 7 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. Mark 7 points B1,B2,B3,B4,B5,B6,B7 on BX such that BB1 = B1B2 = B2B3 = B3B4= B4B5 = B5B6 = B6B7.
6.Join B5C and from B7
draw a line B7C’ || B5C intersecting BC at C’
7. From point C, draw A’C’ || CA intersecting AB at A’. Thus , ∆A’BC’ is the required triangle whose sides are 7/5 of the corresponding sides of ∆ABC.
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