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Try this beautiful Problem on Geometry based on Rectangular Piece of Paper from AMC 10 A, 2014. You may use sequential hints to solve the problem.

A rectangular piece of paper whose length is $\sqrt{3}$ times the width has area $A$. The paper is divided into three equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line to create a new shape with area $B$. What is the ratio $B: A ?$

,

- $2:3$
- $1:2$
- $1:3$
- $3:2$
- $2:5$

Geometry

Rectangle

Ratio

Pre College Mathematics

AMC-10A, 2014 Problem-23

$2:3$

we have to find out the The ratio of the area of the folded paper to that of the original paper.

At first we have to find out the midpoint of the dotted line. Now draw a line perpendicular to it. From the point this line intersects the top of the paper, draw lines to each endpoint of the dotted line. Now it will form a triangle . we have to find out the area of the triangle ....

Now can you finish the problem?

Let us assume the width of the paper is $1$ and the length is $\sqrt 3$.Now the side length of the triangle be $\frac{2 \sqrt{3}}{3}$, $\sqrt{\left(\frac{\sqrt{3}}{3}\right)^{2}+1}=\frac{2 \sqrt{3}}{3}$, and $\sqrt{\left(\frac{\sqrt{3}}{3}\right)^{2}+1}=\frac{2 \sqrt{3}}{3}$

Now for the area of the of the Paper,

It is an equilateral triangle with height $\frac{\sqrt{3}}{3} \cdot \sqrt{3}=1$ and the area will be $\frac{\frac{2 \sqrt{3}}{3} \cdot 1}{2}=\frac{\sqrt{3}}{3}$

Therefore the area of the paper will be $ 1 . \sqrt 3=\sqrt 3$

Now Can you finish the Problem?

Now the area of the folded paper is $\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2 \sqrt{3}}{3}$

Therefore the ratio of the area of the folded paper to that of the original paper is $2:3$

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=OvduZbqenWU

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