Subject: equilateral triangle on a square
Who is asking: Teacher
My Grade 8 class and I were discussing the solution to the following problem:
What is the area of the largest equilateral triangle that can be drawn on a 5 cm square.
We used 5 cm as the base of our triangle and then drew the other two legs of 5 cm each to make the equilateral triangle. We then drew an altitude from the upper vertex to the base of the triangle. Using the law of Pythagoras with side a of 2.5 and side c of 5 we calculated side b to be 4.3 cm (the altitude). Therefore the area of the triangle would be 5 x 4.3 divided by 2 or 10.75 square cm.
The answer key to this resource says I am wrong.
What do you think? Have we interpreted the question incorrectly?
Go to Math Central
Try to draw the largest equilateral triangle you can on the square with no side parallel to a side of the square. If the picture you draw looks like the diagram on the right then it seems you can make a larger triangle by sliding it up to put the vertex at the vertex A of the square, and then rotating the triangle clockwise to allow it to grow larger. The resulting picture looks like the second diagram. If the triangle is equilateral then |AC| = |EA| and since triangles ABC and EFA are right angled, they must be congruent. Thus angle CAB = angle FAE = 15o. The question then is can you put an equilateral triangle in this position?
Starting with the aquare ABCD Draw the lines AE and AC so that angle CAB = angle FAE = 15o. Hence angle EAC = 60o. Since |FA| = |AB|,
angle CAB = angle FAE and triangles ABC and EFA are right angled these triangles are congruent and thus |EA| = |AC|. Hence angle CEA = angle ACE. Since angle EAC = 60o it follows that angle CEA = angle ACE = 60o. Thus triangle ACE is an equilateral triangle.
Chris and Harley
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