Construct a triangle with side 4cm, 5 cm, 6 cm. Then construct a triangle similar to it whose sides are 2/ 3 of the corresponding sides of the given triangle.
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Answered by
225
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Steps of Construction:
Step I: AB = 6 cm is drawn.
Step II: With A as a centre and radius equal to 4cm, an arc is draw.
Step III: Again, with B as a centre and radius equal to 5 cm an arc is drawn on same side of AB intersecting previous arc at C.
Step IV: AC and BC are joined to form ΔABC.
Step V: A ray AX is drawn making an acute angle with AB below it.
Step VI: 5 equal points (sum of the ratio = 2 + 3 =5) is marked on AX as A1 A2....A5
Step VII: A5B is joined. A2B' is drawn parallel to A5B and B'C' is drawn parallel to BC.
ΔAB'C' is the required triangle
Justification:
∠A(Common)
∠C = ∠C' and ∠B = ∠ B' (corresponding angles)
Thus ΔAB'C' ~ ΔABC by AAA similarity condition
From the figure,
AB'/AB = AA2/AA5 = 2/3
AB' =2/3 AB
AC' = 2/3 AC
_______________________________________
MY PLEASURE
Steps of Construction:
Step I: AB = 6 cm is drawn.
Step II: With A as a centre and radius equal to 4cm, an arc is draw.
Step III: Again, with B as a centre and radius equal to 5 cm an arc is drawn on same side of AB intersecting previous arc at C.
Step IV: AC and BC are joined to form ΔABC.
Step V: A ray AX is drawn making an acute angle with AB below it.
Step VI: 5 equal points (sum of the ratio = 2 + 3 =5) is marked on AX as A1 A2....A5
Step VII: A5B is joined. A2B' is drawn parallel to A5B and B'C' is drawn parallel to BC.
ΔAB'C' is the required triangle
Justification:
∠A(Common)
∠C = ∠C' and ∠B = ∠ B' (corresponding angles)
Thus ΔAB'C' ~ ΔABC by AAA similarity condition
From the figure,
AB'/AB = AA2/AA5 = 2/3
AB' =2/3 AB
AC' = 2/3 AC
_______________________________________
MY PLEASURE
Attachments:
Answered by
302
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.....
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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