Question

AB is a line segment of length 10cm.Locate a point C such that AC=1/3CB.Which is the ratio for construction

Answer 1

Solution; AB=10cm
Given: AC=1/3CB......1

or,. AC = 1/3(AB-AC)
or,. AC+AC = 1/3×10
or, 2AC=10/3
or,. AC=5/3
Put this value in 1.
AC=1/3CB
5/3=1/3CB
CB=5
So, the ratio in this construction,
AC/CB=5/3 ÷ 5
=5/15
=1/3=1:3
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Answer 2

Given:

Length of AB = 10cm

AC = 1/3CB

To Find:

The ratio of the construction

Solution:

Firstly draw a line segment AB of length 10cm. Now make an acute angle BAX on AB. Take any distance and mark three arcs such that $AA_{1}=A_{1}A_{2}=A_{2}A_{3}$. Connect $A_{3}$B.

Then, draw a line $A_{1$ parallel to $A_{3}B$, that connects AB at point C.  C is such on-point AB that AC = 1/3AB

Now,

In ΔAB$A_{3}$

To verify,

⇒ $\frac{BC}{AC} = \frac{A_{1}A_{3} }{AA_{1} }$

⇒ (BC/AC)+1 = $\frac{A_{1}A_{3} }{AA_{1} }+1$

⇒ $\frac{BC+AC}{AC}=\frac{A_{1}A_{3}+AA_{1} }{AA_{1} }$

⇒ AC/AB = 1/3

Putting the value,

⇒ AC = 1/3AB

⇒ 1:3 will be the ratio of construction

Therefore, the ratio of the construction is 1:3.