The diagonals of a square are 7√2 inches long. What is the length of each side of the square?
+118723 A square is a special type of rhombus. And the diagonals of all rhombuses bisect one another at right angles.
So the diagonals divide the rhombus into 4 congruent right angled triangles.
the short sides are each $${\frac{{\mathtt{7}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}}{{\mathtt{2}}}}$$ inches long. And the side of the rhombus is the hypotenuse
$$\\h^2=\left(\frac{7\sqrt2}{2}\right)^2+\left(\frac{7\sqrt2}{2}\right)^2\\\\
h^2=2\times \left(\frac{7\sqrt2}{2}\right)^2\\\\
h^2=2\times \frac{49\times 2}{4}\\\\
h^2=49\\\\
h=7\;inches$$
+130516 The diagonal of a square is √2 times as long as the length of a side.
So, dividing the diagonal length by √2, we have that the side is just 7 inches.
+118723 A square is a special type of rhombus. And the diagonals of all rhombuses bisect one another at right angles.
So the diagonals divide the rhombus into 4 congruent right angled triangles.
the short sides are each $${\frac{{\mathtt{7}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{2}}}}}{{\mathtt{2}}}}$$ inches long. And the side of the rhombus is the hypotenuse
$$\\h^2=\left(\frac{7\sqrt2}{2}\right)^2+\left(\frac{7\sqrt2}{2}\right)^2\\\\
h^2=2\times \left(\frac{7\sqrt2}{2}\right)^2\\\\
h^2=2\times \frac{49\times 2}{4}\\\\
h^2=49\\\\
h=7\;inches$$
