A square and an equilateral triangle have the same perimeter. If the diagonal of the square is $12 \sqrt{2}$ $12sqrt2$ cm, what is the area of the triangle?

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A square and an equilateral triangle have the same perimeter. If the diagonal of the square is $12 \sqrt{2}$ $12sqrt2$ cm, what is the area of the triangle?

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Answer 1 · 2015-12-09T15:14:26

$64 \sqrt{3} {cm}^{2}$ $64 sqrt(3) " cm"^2$

Explanation:

Let $A_{s}$ $A_s$ be the area of the square, $P_{s}$ $P_s$ be the perimeter of the square and $a_{s}$ $a_s$ be the length of a side of the square. (All sides have equal lengths.)

Let $A_{t}$ $A_t$ be the area of the triangle, $P_{t}$ $P_t$ be the perimeter of the triangle and $a_{t}$ $a_t$ be the length of a side of the triangle. (All sides have equal lengths.)

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1) As we know the length of the diagonal of the square, we can compute the length of a side of the square using the Pythagoras formula:

$a_{s}^{2} + a_{s}^{2} = d^{2}$ $a_s^2 + a_s^2 = d^2$

$\Rightarrow 2 a_{s}^{2} = {(12 \sqrt{2})}^{2}$ $=> 2 a_s^2 = (12 sqrt(2) )^2$

$\Rightarrow a_{s}^{2} = 12^{2}$ $=> a_s^2 = 12^2$

$\Rightarrow a_{s} = 12 cm$ $=> a_s = 12 "cm"$

2) Knowing the length of one side of the square (and thus knowing all lengths of a square), we can easily compute the square's perimeter:

$P_{s} = 12 \cdot 4 = 48 cm$ $P_s = 12 * 4 = 48 "cm"$

3) We know that the square and the equilateral triangle have the same perimeter, thus

$P_{t} = 48 cm$ $P_t = 48 "cm"$

4) As all sides have the same length in an equilateral triangle, the length of one side is

$a_{t} = \frac{P_{t}}{3} = 16 cm$ $a_t = P_t / 3 = "16 cm"$

5) Now, to compute the area of the equilateral triangle, we need the height $h$ $h$ which can be computed with the Pythagoras formula again:

$h^{2} + {(\frac{a_{t}}{2})}^{2} = a_{t}^{2}$ $h^2 + (a_t/2)^2 = a_t^2$

$\Rightarrow h^{2} + 8^{2} = 16^{2}$ $=> h^2 + 8^2 = 16^2$

$\Rightarrow h^{2} = 192 = 64 \cdot 3$ $=> h^2 = 192 = 64 * 3$

$\Rightarrow h = 8 \sqrt{3} cm$ $=> h = 8 sqrt(3) "cm"$

6) At last, we can compute the area of the triangle:

$A_{t} = \frac{1}{2} \cdot h \cdot a_{t} = \frac{1}{2} \cdot 8 \sqrt{3} \cdot 16 = 64 \sqrt{3} {cm}^{2}$ $A_t = 1/2 * h * a_t = 1/2 * 8 sqrt(3) * 16 = 64 sqrt(3) " cm"^2$

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A square and an equilateral triangle have the same perimeter. If the diagonal of the square is $12 \sqrt{2}$ $12sqrt2$ cm, what is the area of the triangle?

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Explanation:

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