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A square and an equilateral triangle have the same perimeter, If the diagonal of the square is 6√2 cm, then what is the area of the triangle?
  • a)
    12√2 cm2
  • b)
    12√3 cm2
  • c)
    16√2 cm2
  • d)
    16√3 cm2
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A square and an equilateral triangle have the same perimeter, If the d...
Diagonal of the square = √2 × side of the square
∴ Side of the square = 6√2/√2 = 6 cm
4 × Side of the square = 3 × Side of the triangle
4 × 6 = 3 × Side of the triangle
∴ Side of the equilateral triangle = 8 cm
Area of the equilateral triangle = √3/4 × side2
⇒ √3/4 × 82 = 16√3 cm2
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Most Upvoted Answer
A square and an equilateral triangle have the same perimeter, If the d...
Let's denote the side length of the square as $s$ and the side length of the equilateral triangle as $t$.

The perimeter of the square is equal to four times its side length, so it is $4s$.

The perimeter of the equilateral triangle is equal to three times its side length, so it is $3t$.

Since the square and the equilateral triangle have the same perimeter, we have $4s = 3t$.

We are given that the diagonal of the square is 6. In a square, the diagonal is equal to the product of the side length and $\sqrt{2}$, so we have $s\sqrt{2} = 6$. Dividing both sides of this equation by $\sqrt{2}$ gives us $s = \frac{6}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}$.

Substituting this value of $s$ into the equation $4s = 3t$, we get $4(3\sqrt{2}) = 3t$. Simplifying this equation gives us $12\sqrt{2} = 3t$. Dividing both sides of this equation by 3 gives us $4\sqrt{2} = t$.

So, the side length of the equilateral triangle is $t = 4\sqrt{2}$.

In summary, the side length of the square is $s = 3\sqrt{2}$ and the side length of the equilateral triangle is $t = 4\sqrt{2}$.
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A square and an equilateral triangle have the same perimeter, If the diagonal of the square is 6√2 cm, then what is the area of the triangle?a)12√2 cm2b)12√3 cm2c)16√2 cm2d)16√3 cm2Correct answer is option 'D'. Can you explain this answer?
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