# Find the area of the shaded region in the figure below:

Find the area of the shaded region in the figure below:

I am completely stuck on how to start off this question. Please help on some guidance on how to start it off.

• Unless I am very mistaken, Mind Your Decisions made a video on this exact problem. – Mohammad Zuhair Khan Oct 27 '18 at 8:08
• @Raptor, I've checked his YouTube channel, you're right thanks. An alternative approach will also be appreciated. – Meghan C Oct 27 '18 at 8:16
• Hint: the sum of areas of two quadrilaterals on upper-left and lower-right equal to the sum of areas of the two quadrilaterals on upper-right and lower-left. – achille hui Oct 27 '18 at 8:16
• @achillehui, thanks, I will attempt that approach. May I ask why this is true? – Meghan C Oct 27 '18 at 8:20

## 2 Answers

Split the square into $$8$$ triangles, convince yourself you can group them into 4 pairs and each pair has same area. Let the area of the triangles be $$a, b, c, d$$ as illustrated above.

You are given $$c + d = 20$$, $$b + c = 32$$ and $$a + d = 16$$. The area of the quadrilateral (in cyan) is $$a + b = (a + d) + ( b + c) - ( c + d) = 16 + 32 -20 = 28$$

• +1 Nice Answer!! – the_candyman Oct 27 '18 at 8:57
• @achillehui, I really appreciate your help. Thank you. – Meghan C Oct 27 '18 at 9:18

Alternatively, refer to the figure:

$$\hspace{4cm}$$

Let $$x$$ be the half of the side of the large square. Then the side of the smaller oblique square is $$x\sqrt{2}$$, how:

Pythagorean theorem.

The total green area is $$x^2$$, how:

$$\frac12 \cdot x\sqrt{2}\cdot h_1+\frac12 \cdot x\sqrt{2} \cdot h_2=\frac12\cdot x\sqrt{2}\cdot (h_1+h_2)=\frac12\cdot x\sqrt{2}\cdot x\sqrt{2}=x^2.$$

The total area of grey and green regions is $$2x^2=16+32=48$$, how:

Green area is $$x^2$$ and grey area is $$2\cdot \frac{x^2}{2}=x^2$$.

Hence, the required area is $$96-(16+20+32)=28$$, how:

the area of the large square $$(4x^2)$$ minus the total area of grey, green and white regions $$16+20+32$$.