Area Without Numbers

Some of the other solutions are somewhat tedious in finding the intersection point, so I'm presenting an approach using Menelaus.

By considering a base along the line AD and a height of FE, we see that
$\frac{ \text{Area of } AFGE} { \text{Area of } AFE} = \frac { \text{Area of } AFE + \text{ Area of }GFE} { \text{Area of } AFE } = \frac{ AH + HG } { AH } = \frac{ AG } { AH }$

We know that the area of $AFE$ is $\frac{b^2}{2}$ , and need to find $\frac{AG}{AH}$ .

Applying menelaus' theorem on triangle $ADB$ with transversal $FGE$ , we obtain

$\frac{AG}{GD} \times \frac{DE}{EB} \times \frac{BF}{FA} = 1.$

Thus $\frac{AG}{GD} = \frac{2}{1} \times \frac{a-b}{b} = \frac{2b}{a-b}$ , so $\frac{AG}{AD} = \frac{2b}{a+b}$ , so $\frac{AG}{AH} = \frac{AG}{AD} \times \frac{AD}{AH} = \frac{2b}{a+b} \times \frac{a}{b} = \frac{2a}{a+b}$ .

Hence, Area of $AFGE = \frac{b^2}{2} \times \frac{2a}{a+b} = \frac{ab^2}{a+b}$ .

Given - ABC & BDE are two triangles which are overlapped over each other as shown in the given figure . AB = BD = $a$ and BE = BC = $b$ . Also, $a$ > $b$ . To Find - Area of quadrilateral BCOE .

Construction :- Draw OM || AB to meet BD at M. Join EC .

In the figure , let AO = DO = $x$ and EO = CO = $y$ .

It is clear that △ ABC ~ △ OMC

=> $\frac{AB}{OM}$ = $\frac{AC}{OC}$

=> $\frac{a}{OM}$ = $\frac{x + y}{y}$

=> OM = $\frac{ay}{x + y}$ ...................... (1)

and △ EBD ~ △ OMD

=> EB/OM = ED/OD

=> b / OM = (x+y) / x

=> OM = $\frac{bx}{x + y}$ ........................ (2)

From (1) and (2) $\frac{ay}{x + y}$ = $\frac{bx}{x + y}$ => ay = bx

=> $x$ = $\frac{ay}{b}$

Now , in right angled triangle ABC , AB ^2 + BC^{2} = AC^2 (By Pythagoras ' theorem ) => $a$ ^2 + $b$ ^2 = { $x$ + $y$ } ^2 => $a$ ^2 + $b$ ^2 = [ $\frac{ay}{b}$ + $y$ ] ^2

Solving this equation we get : -
$y$ ^2 = $\frac{(a^2 + b^2)b^2}{(a+b)^2}$ ................... (3)

Also, in right angled triangle BEC , BE^2 + BC^2 = CE^2 (By Pythagoras ' theorem ) => $b$ ^2 + $b$ ^2 = CE ^2 => CE = √(2b^2) ..................... ( 4 )

Now, Area of triangle COE = 1/4 * base * √{4 * ( equal sides)^2 - (base)^2} { '.' It is an isosceles triangle}

= 1/4 * EC * √{4 * y^2 - CE^2}

= 1/4 * b√2 * √{4 * $\frac{(a^2 + b^2)b^2}{(a+b)^2}$ - (b√2)^2 }

= $\frac{b^2(a-b)}{2(a+b)}$

Also , Area of triangle BEC = $\frac{1}{2}$ * base * height = $\frac{1}{2}$ BC *BE = $\frac{1}{2}$ * $b$ * $b$ = $\frac{1}{2}$ $b$ ^2

.'. Area quadrilateral BCOE = Area of triangle COE + Area of triangle BEC

= $\frac{b^2*(a-b)}{2*(a+b)} + \frac{b^2}{2}$

= $\frac{b^2}{2} * \frac{a+b+a-b}{a-b}$

= $\frac{(ab^2)}{a-b}$

................................................................................................................................... By Vishwash Kumar

$\tan c=\frac{a}{b}$ and from areas of regions, we have:

$2m+2n=ab$ and $2n+m+k=\frac{a^{2}}{2}$ , from where we need the expression for $m$ . Eliminate $2n$ between the equations to get $\boxed{m=ab-\frac{a^{2}}{2} -k}$ .

Using cosine law in region $k$ , we have:

$2x^{2}[1-cos(270-2c)]=2a^{2}$

$\implies x^{2}[1+sin(2c)]=a^{2}$

$\implies x^{2}[1+\frac{2\tan c}{1+\tan^{2} c}]=a^{2}$

$\implies x^{2}=\frac{a^{2}(a^{2}+b^{2})}{(a+b)^{2}}$

Now, area of $k$ :

$= \frac{1}{2} \times x^{2} \times \sin(270-2c)$

$= \frac{1}{2} \times \frac{a^{2}(a^{2}+b^{2})}{(a+b)^{2}} \times (- \cos 2c)$

$= \frac{1}{2} \times \frac{a^{2}(a^{2}+b^{2})}{(a+b)^{2}} \times \frac{\frac{a^{2}}{b^{2}}-1}{\frac{a^{2}}{b^{2}}+1}$

$= \frac{1}{2} \times \frac{a^{2}(a^{2}-b^{2})}{(a+b)^{2}}$

Substituting this value of $k$ in the above boxed equation gives us $m=\boxed{\frac{ab^{2}}{a+b}}$

We proceed by cheating with coordinate geometry. Let the bottom-left corner be the origin.

The gray area can be found by taking the area of one of the right triangles (say, the bottom one), and subtracting the area of the obtuse green triangle shown.

The area of one of the right triangles would very obviously be $\frac{ab}{2}$ .

The green triangle's length would be $a-b$ . It's height will be the y - coordinate of the intersection of the hypotenuses of the right triangles.

Instead of finding the coordinates of the intersection by finding the equations for both of the hypotenuses, we note that the intersection lies on the line $y=x$ , which can be shown by symmetry.

We find the equation of the hypotenuse of the bottom-most right triangle. Using slope and y - intercept, we get:

$y = \frac{-b}{a}x+b$

The point in question lies on $y=x$ , so substitute to get:

$y = \frac{-b}{a}y+b$

Solve for y:

$y=\frac{b}{\frac{b}{a}+1}$

This is the height of the green triangle. The area would then be:

$\frac{1}{2}\frac{(a-b)b}{ \frac{b}{a}+1 } = \frac{ ab(a-b) }{2( a+b )}$

Recall the area of the right triangle: $\frac{ab}{2}$

Subtract:

$\frac{ab}{2} - \frac{ ab(a-b) }{ 2(a+b) }$

$=\frac{ab(a+b)}{2(a+b)} - \frac{ ab(a-b) }{ 2(a+b) }$

$=\frac{ab(a+b)-ab(a-b)}{2(a+b)}$

$=\frac{ab(a+b-a+b)}{2(a+b)} =\frac{ab(2b)}{2(a+b)} =\boxed{\frac{ab^2}{a+b}}$