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By geometrical construction, it is possible to divide a line segment in the ratio √3:(1/√3)
True
Justification:
According to the question,
Ratio= √3 : ( 1/√3)
On simplifying we get,
√3/ (1/√3) = (√3 x √3)/1 = 3:1
Required ratio = 3:1
Hence,
Geometrical construction is possible to divide a line segment in the ratio 3:1.
To construct a triangle similar to a given △ABC with its sides 7/3 of the corresponding sides of △ABC, draw a ray BX making an acute angle with BC and X lying on the opposite side of A with respect to BC. The points B1, B2, …., B7 are located at equal distances on BX, B3 is joined to C, and then a line segment B6C‘ is drawn parallel to B3C where C‘ lies on BC produced. Finally, line segment A‘C‘ is drawn parallel to AC.
False
Justification:
Let us try to construct the figure as given in the question.
Steps of construction,
1. Draw a line segment BC.
2. With B and C as centres, draw two arcs of suitable radius intersecting each other at A.
3. Join BA and CA and we get the required triangle ∆ABC.
4. Draw a ray BX from B downwards to make an acute angle ∠CBX.
5. Now, mark seven points B1, B2, B3 …B7 on BX, such that BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7.
6. Join B3C and draw a line B7C’|| B3C from B7 such that it intersects the extended line segment BC at C’.
7. Draw C’A’ ||CA in such a way that it intersects the extended line segment BA at A’.
Then, ∆A’BC’ is the required triangle whose sides are 7/3 of the corresponding sides of
∆ABC.
According to the question,
We have,
Segment B6C’ || B3C. But it is clear in our construction that it is never possible that segment B6C’||B3C since the similar triangle A’BC’ has its sides 7/3 of the corresponding sides of triangle ABC.
So, B7C’ is parallel to B3C.