Areas in rectangles

We note that the top green triangle has an area $\frac 14$ of the rectangle or $\frac 14 \times 160 = \color{#20A900}40$ as labelled. Let the area of the four white triangles be $a$ , $b$ , $c$ , and $d$ as shown. Then we have:

$\begin{aligned} a+b+{\color{#D61F06}12} + \color{#20A900}40 & = \color{#3D99F6}\frac {160}2 \quad ...(1) & \small \color{#3D99F6} \text{Half the area of the rectangular} \\ c+d+{\color{#D61F06}12} & = \color{#3D99F6}\frac {160}4 \quad ...(2) & \small \color{#3D99F6} \text{Quarter the area of the rectangular} \end{aligned}$

$(1)+(2): \quad a+b+c+d = 56 = A_{\text{white}}$ , the area of white region. Therefore, the area of green region is given by:

$\begin{aligned} A_{\color{#20A900}\text{green}} & = A_{\text{rectangle}} - A_{\text{white}} - A_{\color{#D61F06}\text{red}} \\ & = 160 - 56 - 12 = \boxed{92} \end{aligned}$

Label each region $A$ through $H$ , as follows:

Then $C$ is a triangle that is $\frac{1}{4}$ the area of the whole rectangle, so $C = \frac{1}{4}160 = 40$ . Also, $A + B$ is a triangle that is $\frac{1}{4}$ the area of the whole rectangle, so $A + B = 40$ .

The triangle made from $B$ , $E$ , and $G$ and the triangle made from $G$ and $H$ have the same base and height, so their areas are the same, which means $B + E + G = G + H$ . We are given that $E = 12$ , so this simplifies to $B = H - 12$ .

Substituting $B = H - 12$ into $A + B = 40$ gives $A + (H - 12) = 40$ , or $A + H = 52$ .

The green areas are $A$ , $C$ , and $H$ , and $A + C + H = (A + H) + C = 52 + 40 = \boxed{92}$ .

Looking at the picture above:

Green area $G$ = Rectangle area - (White area $W$ + Red area $R$ ) ------- 1

Now to calculate the white area, we know that the sum area of the two triangles $afc$ and $bfe$ equal to the half of the rectangle area:

Triangles area $= 0.5 * ef * bc + 0.5 * fc * bc = 0.5 * (ef+fc) * bc = 0.5$ rectangle area, and also equal to sum of white plus two times the red.

$W + 2 R = 80$

$W = 80 - 2 * 12 = 56$

Back to formula 1

$G = 160 - (56 + 12) = \boxed{92}$

We have two overlapping white triangles with their bases side by side.

If we move the top corners of these triangles to the opposite side, they no longer overlap and the green area now has area $1/2*h*b$ = 80.

So in the original situation the area of the green part is 80 plus the overlap of the white triangles, which is the red part.

So the area of the green part is 92.

In the Diagram-1 & Diagram-2 the blue marked triangle has equal area. Both blue marked triangle has same base and height. If the common white area removed from the blue marked triangles then green area equal to red and white area. In the Diagram-3, the white and red area changed to the equivalent green area.

Therefore, the Area of green shaded region is half of the rectangle plus 12 which is equal to 92.

Note that $e+f=f+12+b$ , because they are triangles with the same height and base.

This simplifies to $b+12=e$ , and using the same logic, we can deduce that similarly $d+12=a$ .

$c+b+d+12=80$ , because it is half the rectangle. If we add 12 to both sides:

$c+b+12+d+12=92 \\ \rightarrow c+e+a=92$

The area in Green = Area of Rectangle - Area of two obtuse angle triangles + Area in Red (as it is counted twice while subtracting obtuse triangles)

Now Area of two obtuse triangles = $\frac{1}{2}$ a h + $\frac{1}{2}$ b h = $\frac{1}{2}$ (a+b)*h = $\frac{1}{2}$ * Area of Rectangle = $\frac{160}{2}$ = 80

so Area in Green = 160 - 80 + 12 = 92

Let $G_1 , G_2 , G_3 , W_1 , W_2 , \text{and} \space R \space \text{represents area of respective boundary}.$

We have $area(\Delta ABC) = 160/2 = 80 = area(\Delta ABD) = area(\Delta ABE)$

$area(\Delta ABD) + area(\Delta ABC) - area(\Delta ABE) = (G_1 + G_2 + W_1) + (G_2 + G_3 + W_2) - (G_2 + W_1 + W_2 + R)$

$\Rightarrow 80 + 80 - 80 = G_1 + G_2 + G_3 - R$

We are given $R = 12 \Rightarrow G_1 + G_2 + G_3 = 80 + 12 = 92$

So area in green is equal to $92$

AREA OF A TRIANGLE- 1/2 B * H SO SOME OF THE AREAS OF BOTH TRIANGLES (ECXEPT GREEN AREA) WILL BE- 1/2 ( BASE 1+ BASE 2) H = 1/2 L* B = 80.

AREA OF THE GREEN PART = AREA OF RECTANGLE - AREAS OF BOTH TRIANGLES = 160- (80-12) = 92

I just make 'x and y' axis and assume coordinates and find general area using coordinate geometry . It took me a page to solve this. #difficult method

There are two triangles whose bases together form the length of the rectangle and whose heights are both the width of the rectangle so their combined area if they didn’t overlap would be half of the rectangle. Therefor the remaining space in green is half the area of rectangle plus the red space as subtracting the two triangles from the total would subtract that space twice. This gives 80 + 12 = 92.

This problem is a special case of the more general situation where the vertices of the two top triangles still lie on the top edge of the rectangle, but are not confined to its corners. The answer remains the same.

General problem and solution: Consider a green rectangle. Consider two white triangles whose bases lie next to each other and together fill the bottom edge of the rectangle. The vertices of the triangles lie along the top edge of the rectangle. If the triangles do not overlap anywhere, then the sum of their areas is half the area of the rectangle. The area not covered by the triangles (the area remaining green) is also half the area of the rectangle. If the triangles do overlap then the area covered by the white triangles is reduced by the area of the overlap (the red area) and the green area is correspondingly enlarged.

Green area = half of rectangle area + red triangle overlap area = 160/2 + 12 = 80 + 12 = 92