Solution: Pick's theorem

Pick's Theorem states $\text{Area} = {\color{#D61F06}{\text{Interior Points}}} + \dfrac{{\color{#EC7300}{\text{Boundary Points}}}}{2} - 1$

• For Triangle A: ${\color{#D61F06}{\text{I = 2}}}, {\color{#EC7300}{\text{B = 4}}}$ $\implies \text{Area} = {\color{#D61F06}{\text{2}}} + \dfrac{{\color{#EC7300}{\text{4}}}}{2} - 1 = \boxed{3}$

• For Triangle B: ${\color{#D61F06}{\text{I = 1}}}, {\color{#EC7300}{\text{B = 6}}}$ $\implies \text{Area} = {\color{#D61F06}{\text{1}}} + \dfrac{{\color{#EC7300}{\text{6}}}}{2} - 1 = \boxed{3}$

$\boxed{A = B}$

${\text{Area of }}\vartriangle A = {\text{Area of }}\vartriangle CDF + {\text{Area of }}\vartriangle CFE = \frac{1}{2}CF \cdot DG + \frac{1}{2}CF \cdot EL = \frac{1}{2} \times 3 \times 1 + \frac{1}{2} \times 3 \times 1 = 3$

${\text{Area of }}\vartriangle B = \frac{1}{2}HI \cdot JK = \frac{1}{2} \times 3 \times 2 = 3$

Hence, $\boxed{{\text{Area of }}\vartriangle A = {\text{Area of }}\vartriangle B}$ .

We can cut the top half of the orange triangle $A$ into two smaller triangles $a$ and $b$ . Flip $a$ over and place it on top of the bottom half of the orange triangle. Place $b$ by the side of the bottom half of the orange triangle. We get a reflection of the blue triangle $B$ . Therefore $\boxed {A = B}$ in area.

The blue (B) triangle has a base length 3 (left side) and height 2 (from right vertex to base). Triangle area = $\frac{b*h}{2}$ = $\frac{2*3}{2}$ = 3. The gold (A) triangle area can be found by subtraction. Look at the rectangle starting at the left vertex of the triangle and 4 units tall and 2 units wide ending at the lower right vertex. that rectangle consists of 4 areas - the gold triangle and 3 right triangular spaces surrounding it. The rectangle is area = 8. The upper left space is a triangle with base = 1 height = 2, area = 1. The upper right space is a triangle with base = 1 height = 4, area = 2. The lower left space is a triangle with base = 2 height = 2, area = 2. So by subtraction, the gold (A) triangle's area = 8 - 1 -2 - 2 = 3. The triangles have the same area, 3. The gold (A) triangle area can also be found using Pick's Theorem. Arrea = 2 + $\frac{4}{2}$ - 1 = 3.

$\cfrac{1}{2}\left \| \begin{matrix}x_1&x_2&x_3\\y_1&y_2&y_3\\1&1&1\end{matrix}\right \|=Area$ $\begin{array}{lrl}\therefore&\cfrac{1}{2}\left \| \begin{matrix}0&1&2\\2&4&0\\1&1&1\end{matrix}\right \|&=\cfrac{1}{2}\left \| \begin{matrix}0&0&2\\0&3&2\\1&1&1\end{matrix}\right \|\\[20px] \because&\left \| \begin{matrix}2&4&0\\0&1&2\\1&1&1\end{matrix}\right \|&=\left \| \begin{matrix}1&1&1\\0&3&2\\0&0&2\end{matrix}\right \|\\[20px] \because&\left \| \begin{matrix}2&4&0\\0&1&2\\0&1&-1\end{matrix}\right \|&=1\times3\times2\\[20px] \because&\left \| \begin{matrix}2&4&0\\0&1&2\\0&0&-3\end{matrix}\right \|&=6\\[20px] \because& 2\times1\times3&=6\end{array}$

Or a simple version:

$Area=\cfrac{1}{2}(x_2y_3-x_3y_2)$

If the third( $x_1;y_1$ )point's coordinates are $(0;0)$ .

Pick's Theorem states that -- Area = Interior Points(I) + Boundary Points(B)/2 - 1

Using Pick's Theorem we get area as

For Triangle A -- I = 2, B = 4 Area = 2 + 4/2 - 1 = 2+2-1 = 3

For Triangle B -- I = 1, B = 6 Area = 1 + 6/2 - 1 = 1+3-1 = 3

Area of Triangle A = Area of Triangle B

Dimensions of $\color{#EC7300}A$ : Base $=2\sqrt{2}$ , Height $=1.5\sqrt{2}\Rightarrow$ Area $=3$

Dimensions of $\color{#3D99F6}B$ : Base $=2$ , Height $=3\Rightarrow$ Area $=3$

Therefore $\boxed{A_{\color{#EC7300}A}\color{#333333}=A_\color{#3D99F6}B}$

Dunno aby theorems, dunno how to put the pure logic and simple maths into these fancy formulas. Got to the correct answer by just ADDING UP/SUBSTRACTING THE HALVES AND QUARTERS of grid point squares and rectangles. I wish I could do it the ‘academic way’ though

The both have same base that is 3 and altitude of first triangle is 2 and altitude of the orange triangle is sum of two altitudes both equal to 1 Hence both blue and orange triangles have same area.