Question

One of the vertices of an equilateral triangle is on the vertex of a square and two other vertices are on the not adjacent sides of the same square. Find the side of the triangle if the side of the square is 10 ft.

Answer 1

Answer:

The side of the triangle is either 38.63ft or 10.35ft

Step-by-step explanation:

This problem can be translated as an image as shown in the Figure below. We know that:

• The side of the square is 10 ft.
• One of the vertices of an equilateral triangle is on the vertex of a square.
• Two other vertices are on the not adjacent sides of the same square.

Let's call:

Since the given triangle is equilateral, each side measures the same length. So:

x: The side of the equilateral triangle (Triangle 1)

y: A side of another triangle called Triangle 2.

That length is the hypotenuse of other triangle called Triangle 2. Therefore, by Pythagorean theorem:

$\mathbf{(1)} \ x^2=100+y^2$

We have another triangle, called Triangle 3, and given that the side of the square is 10ft, then it is true that:

$y+(10-y)=10$

Therefore, for Triangle 3, we have that by Pythagorean theorem:

$(10-y)^2+(10-y)^2=x^2 \\ \\ 2(10-y)^2=x^2 \\ \\ \\ \mathbf{(2)} \ x^2=2(10-y)^2$

Matching equations (1) and (2):

$2(10-y)^2=100+y^2 \\ \\ 2(100-20y+y^2)=100+y^2 \\ \\ 200-40y+2y^2=100+y^2 \\ \\ (2y^2-y^2)-40y+(200-100)=0 \\ \\ y^2-40y+100=0$

Using quadratic formula:

$y_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ y_{1,2}=\frac{-(-40) \pm \sqrt{(-40)^2-4(1)(100)}}{2(1)} \\ \\ \\ y_{1}=37.32 \\ \\ y_{2}=2.68$

Finding x from (1):

$x^2=100+y^2 \\ \\ x_{1}=\sqrt{100+37.32^2} \\ \\ x_{1}=38.63ft \\ \\ \\ x_{2}=\sqrt{100+2.68^2} \\ \\ x_{2}=10.35ft$

Finally, the side of the triangle is either 38.63ft or 10.35ft

Answer 2

Answer:

$10\sqrt{6} -10\sqrt{2} \approx 10.35 ft$

Step-by-step explanation:

1) Check the first picture, to visualize what the question has told. An equilateral triangle DLM inscribed in a square ABCD. A triangle like that has three 60º angles.

2) Looking to DCL and DBE we have two right triangles. The next natural step, is doing Pythagorean theorem.

$\triangle DCO \:a^{2}=10^{2}+x^{2}\Rightarrow a=\sqrt{100+x^{2}}\\\triangle DBE \:a^{2}=10^{2}+y^{2}\Rightarrow a=\sqrt{100+y^{2}}$

Since this is an equilateral triangle, all three sides have the same size, we can set them as equal:

$\sqrt{100+y^{2}}=\sqrt{100+x^{2}}\\\left ( \sqrt{100+y^{2}} \right )^{2}=\left ( \sqrt{100+x^{2}} \right)^{2}100+y^{2}=100+x^{2}\\y^{2}=x^{2}\\y=x$

3} This allows to replace x for y. We also have, since there is symmetry. We can check that's on the left, a right isosceles triangle (ALE) whose hypotenuse is $a\sqrt{2}$. Hence, the hypotenuse would be $\sqrt{2}(10-x)$

. Again starting from the fact this is an equilateral triangle

$\sqrt{2}(10-x)=\sqrt{100-x^2}\\\left ( \sqrt{2}(10-x) \right )^{2}=\left ( \sqrt{100+x^{2}} \right)^{2}\\2(100-20x+x^{2})=100+x^{2}\\200-40x+2x^{2}=100+x^{2}\Rightarrow x^{2}-40x+100$

$Completing \:the \: square\\ x^{2}-40x+100 -100 =0-100\\x^{2}-40x=-100\\\\ Adding \:the \:half \:of \:the \:absolute \:value \:of\: b, and \:square \:it\\x^{2}-40x +20^{2}= 20^{2}-100\\x^{2}-40x +400= 400-100\\ Rewrite\\(x-20)^2=300\\(x-20)=\sqrt{300}\:\rightarrow x-20=10\sqrt{3}\\\rightarrow x_{1}=20+10\sqrt{3} \approx37.32, x_{2}=20-10\sqrt{3}\approx 2.67$

4) One and only one value must fit. So let's try plugging the minor one.

$\sqrt{2}(10- (20-10\sqrt{3})=10\sqrt{6}-10\sqrt{2}\approx 10.35 \:ft$

$\sqrt{2}(10- (20+10\sqrt{3})=-10\sqrt{6}-10\sqrt{2}\approx -38.64 \:ft$

If one vertex of the triangle is on the square, it can't be an outlying value as the absolute value of 38.64.  Furthermore the greatest value for a line segment of this square would be the value of the diagonal, in this case$10\sqrt{2}\approx 14.14$

So, it is 10.35