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# Can you help me with this please?

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421
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A square is divided, as shown. What fraction of the area of the square is shaded? Express your answer as a fraction.

Jul 17, 2019

### 2+0 Answers

#1
+111435
+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

Jul 17, 2019
#2
+25474
+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 $$\dfrac{\dfrac{3}{16} s^2}{s^2} = \dfrac{3}{16}$$

Jul 17, 2019