Bunuel wrote:

An equilateral triangle and three squares are combined as shown above, forming a shape of area 48+4√3. What is the perimeter of the shape formed by the triangle and squares?

A. 18

B. 27

C. 36

D. 48

E. 64

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The attachment **T6027.png** is no longer available

Quick answer Assume the second term,

\(4\sqrt{3}\), of the given area is a clue: it's likely to be the area of the equilateral triangle.

Divide 48 by 3 (squares) = 16. Square side = 4.

Check: Length = 4 must equal equilateral triangle side. Area of equilateral triangle if side = 4 is

\(\frac{4^2\sqrt{3}}{4} = 4\sqrt{3}\) - Correct

Square side = 4, there are nine sides on the perimeter, 9 * 4 = 36

Answer C

Longer versionArea generally: formulas and given valueKnowing the formula for the area of an equilateral triangle makes this problem a lot easier.*

Total area = (area of three squares) + (area of equilateral triangle)

Area of three square

\((3)s^2\)Area of equilateral triangle =

\(\frac{s^2\sqrt{3}}{4}\)Total area:

\(3s^2 + \frac{s^2\sqrt{3}}{4}\)Total area =

\(48 + 4√3\)Find side length of square from area of square set equal to 48Gamble a little. The second given term,

\(4\sqrt{3}\) is likely to be the area of the equilateral triangle.

So assume that the integer portion of the given area will yield a square's side length

Set just the integer portion of the area, 48, equal to the area of the three squares

\(3s^2 = 48\)

\(s^2 = 16\)

\(s = 4\)Check: square side = triangle side, find triangle areaEach square: has equal side lengths that are equal to the side of the equilateral triangle (each square shares a side with the triangle)

If square side length = 4, triangle side length must = 4

Set the other given term equal to the area of the equilateral triangle

\(\frac{s^2\sqrt{3}}{4} =4√3\)

\({s^2\sqrt{3}}=16√3\)

\(s^2 = 16\)

\(s = 4\) Correct: the side of square = side of triangle = 4

PerimeterPerimeter = 9 square side lengths:

\(9 * 4 = 36\)Answer C *

If you don't remember the formula for the area of an equilateral triangle, draw one. Drop an altitude, which is a perpendicular bisector of the opposite side and of the vertex.

That altitude creates two congruent right 30-60-90 triangles with side lengths that correspond to 30-60-90, in ratio \(x : x\sqrt{3} : 2x\)

Side lengths? Side opposite the 90° angle = \(2x = 4\). Side opposite 30° angle is half of that, i.e., \(x, x = 2\). Side opposite 60° angle = height of triangle = \(x\sqrt{3}\) or \(2\sqrt{3}\)

Area\(=\frac{b*h}{2} = (4*2\sqrt{3})*\frac{1}{2} = 4\sqrt{3}\). Correct: it is equal to the second term in the area given.Attachment:

equitriarea.png [ 40.47 KiB | Viewed 1311 times ]