The total area

The yellow area and the green area equals to the entire area of the rectangle.

As the length and the width are both 15 and 20, and the area is the length and width multiplied, the total area is 300.

The red line

To find the length of the red line, we need to square the horizontal blue line’s length and add it to the square of the vertical blue line’s length. The length can be calculated by taking 8 away from 20, and 10 away from 15. If you square both the numbers, you get 144 (12 squared) and 25 (5 squared). Adding these together equals 169, which if you find the square root of, is 13. This shows that the red line is 13.

The green space

To find the area of the green space, we must take the yellow space away from the total. As I have said, the length and width of the triangle is 12 and 5, respectively. Finding the area, we multiply them together and divide by two. This gives us 30.

The total area has been mentioned above (300), so subtracting 30 from 300 equals 270. This is the green space’s area.

The final step

The question states that we need to add x and y. We have found the values of both (270 and 13), so the last thing to do is add them together.

270 + 13 = 283

$Our$ $answer$ $is$ $283$

Everyone, please don't delete solutions as it will confuse me and the Mathathon scoring system will be destroyed!!!!!!!!!!!!!!!!

Btw. there is no Latex used in this one ....

To figure out x, We split the shape into two parts. One part would be are rectangle. The other will be a trapezium. The dimensions of the rectangle is the height of it is 10 and the width is 20. The dimensions of the trapezium is that its height would be 15-10=5 and the two lengths of it would be 20 and 8.

This means the area of the shape, x, we add the area of the rectangle and the trapezium together. 10 * 20 + 5 * (20 + 8) / 2 = 200 + 70 = 270

So, the value of x is equal to 270.

To figure out y, We identify if we join up the sides of the rectangle into a corner it would be rectangle. Therefore, the triangle we have made by connecting the last bits of the sides is a right-angled one. We already have the height which is 15-10=5 and the base is 20-8=12. Therefore, we have the dimensions of the right-angled triangle other than the hypotenuse.

We calculate y, which is the hypotenuse of the triangle using the Pythagoras Theorem, a^2 + b^2 = y^2 This is where a and b are the sides of the triangle expect the hypotenuse. a = 12 and b = 5. We substitute. 12^2 + 5^2 = y^2 144 + 25 = y^2 169 = y^2 y = 13

x + y = 270 + 13 = 283

Therefore, the answer is 283.

$\large\text{Therefore, the final answer should be }270 + 13 = \boxed{283}$

$\text{\huge Completing the figure first.}$

We will draw the complete rectangle again, by bringing the cutout piece back and label the figure accordingly:

Image not to scale

$\text{\huge Calculating the length of y.}$

We will use the above image and solve for y, where y=XY. In the above image, $CX\bot CY$ as they are part of segments of a rectangle, so they are perpendicular.

Thus, $\triangle XCY$ is a right-angle triangle. We can now use Pythagoras theorem to find the length of XY.

We have, ${ XC }^{ 2 }+{ YC }^{ 2 }={ XY }^{ 2 }$

Substituting the length of XC and YC, we get, ${ 5 }^{ 2 }+{ 12 }^{ 2 }={ XY }^{ 2 }$

Solving further, 25 + 144 = 169 = ${ XY }^{ 2 }$

Taking the positive square root on both sides, $\sqrt { 169 }$ = 13 = XY.

• Thus, the length of XY i.e $\boxed{\text{y = 13}}$

$\text{\huge Solving for the area x.}$

We will again use the image above for reference, the area of the remaining part would be:

x = $Area(\Box ABCD)-Area(\triangle XCY)$

x =  $20\times 15-\frac { 12\times 5 }{ 2 }$ (As the length and breadth of the rectangle are 20 and 15, and the triangle's base is 12 and height is 5.)

x = 300-30

$\boxed{\text{x = 270}}$

$\text{\huge Adding the values.}$

x + y = 270 + 13

Thus, $\boxed{\text{x + y= 283}}$

As $ABCD$ is a rectangle $\therefore \overline{AD}=\overline{BC}=\overline{BE}+\overline{EC}=10cm+\overline{EC}$ $\Rightarrow 15cm=10cm+\overline{EC}\quad\blue{\because \overline{AD}=15cm}$ $\Rightarrow \overline{EC}=15cm-10cm=5(3)cm-5(2)cm=5(3-2)cm=5(1)cm=5cm\quad\blue{\because 5(3)=15 ;\space 5(2)=10 ;\space a(b)-a(c)=a(b-c)}$ Similarly, $\overline{AB}=\overline{DC}=\overline{DF}+\overline{FC}=8cm+\overline{FC}$ $\Rightarrow 20cm=8cm+\overline{FC}\quad\blue{\because \overline{AB}=20cm}$ $\Rightarrow \overline{FC}=20cm-8cm=4(5)cm-4(2)cm=4(5-2)cm=4(3)cm=12cm\quad\blue{\because 4(5)=20 ;\space 4(2)=8 ;\space a(b)-a(c)=a(b-c)}$ Also as $ABCD$ is a rectangle $\therefore \angle ECF=\dfrac{1}{2}\pi=90\degree$ $\Rightarrow$ $\triangle ECF$ is right angled at vertex $C$ $\therefore$ from Pythagorean theorem $\overline{FC}^2+\overline{EC}^2=\overline{EF}^2$ $\Rightarrow (12cm)^2+(5cm)^2=\overline{EF}^2$ $\Rightarrow 144cm^2+25cm^2=\overline{EF}^2$ $\Rightarrow 169cm^2=\overline{EF}^2$ $\Rightarrow \overline{EF}={^+_-}\sqrt{169cm^2}={^+_-}\sqrt{169}\sqrt{cm^2}={^+_-}13cm$ But as lengths are positive $\therefore \overline{EF}=13cm=y$ Area of $ABCD=BASE\times HEIGHT=20\red{cm}\times15\red{cm}=10\times2\times1.5\times10cm^2=100\times3=300cm^2$

Area of $\triangle ECF=\dfrac{1}{2}BASE\times HEIGHT=\dfrac{1}{2}\times12\red{cm}\times5\red{cm}=6\times5cm^2=3\times\red{2\times5}cm^2=3\times10cm^2=30cm^2$

$x=$ Area of $ADFEB=$ Area of $ABCD$ $-$ Area of $EFC$ = $300cm^2-30cm^2=30\times10^2cm^2-30cm^2=30(10-1)cm^2=30(9)cm^2=270cm^2$

$x+y=270\red{cm^2}+13\red{cm}=\red{283}$

A problem!, we can't add $cm^2$ to $cm$ !

First step: Use the Pythagorean Theorem to label the rectangle: Identities: $20-8=12,15-10=5,\color{#D61F06}\sqrt{5^2+12^2}=13$
So $y=13$ .
Now the rectangle’s area. This is given by $\displaystyle S_{\square}-S_{Rt\triangle}$ . So the area is $15\times 20-\frac{5\times 12}{2}=300-30=270.$ So the answer is $270+13=283.$

$y = \large{\sqrt{5^2 + 12^2}}$

$y = \large{\sqrt{25 + 144}}$

$y = \large{\sqrt{169}}$

$y = \large{\boxed{13}}$

---

$\large{x = (20 \times 15) - (\frac{1}{2} \times 5 \times 12)}$

$\large{x = 300 - 30}$

$\large{x = \boxed{270}}$

---

$\Huge{x + y = 270 + 13 = \boxed{283}}$

Notice that the original figure is a rectangle. This means that opposite sides are equal.

From this, we can deduce that the legs of the cut-out triangle are $5$ and $12$ . We know the triangle is right, so the hypotenuse is $13$ . $y = 13$ .

The original area of the rectangle is $300$ , and the area of the cut-out triangle is $\displaystyle \frac{5 \times 12}{2} = 30$ . So the remaining area is $300 - 30 = 270$ . $x = 270$ .

Then, $x + y = \boxed{283}$ .