Triangular Square

Geometry Level 2

Johnell cuts an equilateral triangular paper whose sides measure 2 cm into pieces. He then rearranges the pieces to form a square without overlapping. How long is the side of the square formed?

4 3 \large \sqrt[3]{4} 2 3 3 3 \large \sqrt[3]{3} 3 4 \large \sqrt[4]{3}

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Yasir Soltani
Feb 5, 2016

Let A t A_t be the area of the Triangle cut then we have A t = 1 2 × 2 3 A_t=\frac{1}{2}\times2\sqrt{3} A t = 3 A_t=\sqrt{3} As there was no overlapping of sides, this means total area is conserved , if A s A_s is the area of the square formed with side x x and x 0 x\geqslant 0 then A s = A t A_s=A_t x 2 = 3 x^2=\sqrt{3} x = 3 4 x=\boxed{\sqrt[4]{3}}

John Wyatt
Jan 20, 2016

Cut but no overlap means same amount of area

Area of Triangle = 0.5 a b*sin(C)

Area = 0.5 2 2*(60)

Area = 2*(SQRT(3)/2)

Area = 3^0.5

Area of Square = x^2

x^2 = 3^0.5

x = 3^0.25

Just for sake of curiosity, this link shows how to actually make the cuts to turn the equilateral triangle into a square.

Brian Charlesworth - 5 years, 4 months ago
Gean Llego
Jan 12, 2016

After cutting, the area of the figure is still the same. To find the area of an equilateral triangle, we use the following formula,

A=(a^2 √3)/4

Where a is the length of the side of an equilateral triangle. Finding the area we have,

A=(2^2 √3)/4

A=(4√3)/4

A= √3

Now, since the area of this triangle is equal to the area of the square, we can use the formula of the area of the square to find the length of the side.

A= s^2

Now, we equate the area of the triangle,

√(√3) = s^2

Since we are solving for s, we take the square root of both sides,

√(√3) =√(s^2 )

∜3=s

Or,

s=∜3

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...