A square is divided, as shown. What fraction of the area of the square is shaded? Express your answer as a fraction.

Guest Jul 17, 2019

#1**+2 **

Call the side of the square , S

And its area = S^2

We can see that 3/4 of the area of the square is unshaded

So...we just need to find the area shaded in the remaining 1/4 th

Note that at the bottom right of the figure we have an isosceles right triangle with leg lengths of S / 2

So....the area of this triangle is (1/2) (product of the leg lengths) = (1/2) (S/2) (S/2) = S^2/8

But 1/2 of the area of this triangle plus the shaded area comprise the other 1/4th area of the square

So 1/2 of the area of the right triangle = (1/2)(S^2)/8 = S^2/16 = (1/16)S^2

So.....the shaded area = (1/4)S^2 - S^2/16 = (4/16)S^2 - (1/16)S^2 = (3/16)S^2

So....the shaded part is just 3/16 of the area of the square

CPhill Jul 17, 2019

#2**+3 **

**A square is divided, as shown. What fraction of the area of the square is shaded? Express your answer as a fraction.**

\(\text{Let $s$ is the side of the square }\\ \text{Let $s^2$ is the area of the square }\\ \text{Let $s\sqrt{2}$ is the diagonal of the square } \\ \text{Let $s\dfrac{\sqrt{2}}{2}$ is the parallel diagonal of the square } \)

\(\begin{array}{|rcll|} \hline \mathbf{A_{\text{trapezoid}}} &=& \left(\dfrac{s\sqrt{2}+s\dfrac{\sqrt{2}}{2}}{2}\right)\cdot s\dfrac{\sqrt{2}}{4} \quad &| \quad s\dfrac{\sqrt{2}}{4} \text{ is the height of the trapezoid } \\ &=& \left(\dfrac{1+ \dfrac{1}{2} }{2}\right)\cdot s\sqrt{2} s\dfrac{\sqrt{2}}{4} \\ &=& \left(\dfrac{\dfrac{3}{2} }{2}\right)\cdot s^2 \dfrac{2}{4} \\ &=& \left(\dfrac{3}{4}\right)\cdot s^2 \dfrac{1}{2} \\ &=& \mathbf{ \dfrac{3}{8} s^2} \\\\ \mathbf{A_{\text{shaded}}} &=& \dfrac{A_{\text{trapezoid}}}{2} \\ &=& \dfrac{\dfrac{3}{8} s^2}{2} \\ \mathbf{A_{\text{shaded}}} &=& \mathbf{ \dfrac{3}{16} s^2} \\ \hline \end{array}\)

The fraction of the area of the square shaded is \(\dfrac{\dfrac{3}{16} s^2}{s^2} = \dfrac{3}{16}\)

heureka Jul 17, 2019