What's the area of this rectangle?

Relevant wiki: Menelaus' Theorem

Solve this quadratic equation and we get x=63, y=12, S=336.

Wikipedia for Menelaus's Theorem

Since $\Delta BEG = \Delta HEG = 21$ and both traingles have the same height, then $BG = GH$

We can deduce that $\Delta BAG = \Delta HAG$ since they have the same height.

Let $\Delta BAG = \Delta HAG = a$ .

Also $\Delta HFI = 6; \Delta DFI = 8$ and both traingles have the same height, then $\dfrac{HI}{ID} = \dfrac{6}{8} = \dfrac{3}{4}$

We can deduce that $\Delta AHI = 3b; \Delta ADI = 4b$ for some positive number $b$ .

Note that $\Delta AED = \dfrac{1}{2}ABCD = \Delta BFA$

$\implies 2a + 3b + 6 = a + 7b +21$

$\implies a = 4b + 15$

Also notice that $\Delta CBF + \Delta DAF = \Delta AED$

$\implies \Delta BEH + ECFH + \Delta DAF = \Delta AED$

$\implies 42 + ECFH + 4b + 8 = 4b + 15 + 7b + 21$

$\implies ECFH = 7b - 14$

$\implies ABCD = 22b + 72$

Since $E$ is the midpoint, Let $AB=x, BC=y, BE = \dfrac{y}{2}$

$\implies xy = ABCD = 16b + 144$

$\implies b = 12$

$\therefore 22(12) + 72 = \boxed{336}$

By applying a horizontal scale of factor $\lambda$ , coupled with a vertical scale of factor $\lambda^{-1}$ , we can retain the areas of the various regions, but have the whole problem inside a square.

Assume then that the points $A,B,C$ have coordinates $(X,0)$ , $(X,X)$ and $(0,X)$ respectively, and that the points $D,E$ have coordinates $(uX,X)$ and $(X,vX)$ for some $0 < u,v < 1$ . Then the points $P,Q,R$ have coordinates $\left(\tfrac{uX}{1+u-uv},\tfrac{X}{1+u-uv}\right) \hspace{0.5cm} \left(\tfrac{uX}{u+v-uv},\tfrac{vX}{u+v-uv}\right) \hspace{0.5cm} \left(\tfrac{X}{1+v-uv},\tfrac{vX}{1+v-uv}\right)$ respectively (these are found by solving suitable pairs of simultaneous equations).

Thus we have triangle areas: $21 \; = \; \tfrac12 \times uX \times \tfrac{u(1-v)X}{1+u-uv} \; = \; \tfrac{u^2(1-v)X^2}{2(1+u-uv)} \hspace{1cm} 21 + 21 \; = \; \tfrac12 \times uX \times \tfrac{u(1-v)X}{u+v-uv} \; = \; \tfrac{u^2(1-v)X^2}{2(u+v-uv)}$ and these equations tell us that $1 + u - uv \; = \; 2(u + v - uv) \; \Leftrightarrow \; v \; = \; \frac{1-u}{2-u}$ We also have the triangle areas: $6+8 \; = \; \tfrac12 \times vX \times \tfrac{(1-u)vX}{u+v-uv} \; = \; \tfrac{(1-u)v^2X^2}{2(u+v-uv)} \hspace{1cm} 8 \; = \; \tfrac12 \times vX \times \tfrac{(1-u)vX}{1+v-uv} \; =\; \tfrac{(1-u)v^2X^2}{2(1+v-uv)}$ and so $4(1+v-uv) \; = \; 7(u+v - uv) \; \Leftrightarrow \; v \; = \; \frac{4-7u}{3(1-u)}$ Thus we deduce that $\frac{1-u}{2-u} = \frac{4-7u}{3(1-u)}$ , which implies that $u = \tfrac12$ or $\tfrac52$ . Since $0 < u < 1$ we deduce that $u = \tfrac12$ , and hence $v = \tfrac13$ . We then obtain the area equations $21 \; = \; \tfrac{1}{16}X^2 \hspace{1cm} 6 \; = \; \tfrac{1}{56}X^2 \hspace{1cm} 8 \; = \; \tfrac{1}{42}X^2$ and hence we see that all is well, and that the area of the rectangle is $X^2 = \boxed{336}$ .

Apex of upright triangle is at midpoint of top. Apex of other triangle is $\dfrac{1}{3}$ of the height.

For sake of ease of computations, assume height $= 21$ , width $= 16$ , but any rectangle of area $336$ will work.

Easy to see:

Area = 21 x 4 x 4 = 336

This is all you need if you assume G is the midpoint of the top.

B=(0,b), A=(a,0), the area of the rectangle sought is ab.

Line FG is $y=\frac{2b}{a}x$ , line GA is $y=\frac{-2b}{a}x+2b$

Since BG = a/2 to make the area of triangle BHG be 21, its height must be 84/a and so the y-coordinate of H is b-84/a

$H=(\frac{a}{2}-\frac{42}{b}, b-\frac{84}{a})$

Since triangle GHJ also has area 21, BJ=2*BH

$J=(a-\frac{84}{b},b-\frac{168}{a})$

This point must lie on GA. Substituting into the equation for this line gives

$b-\frac{168}{a}=\frac{-2b}{a}(a-\frac{84}{b})+2b$

$b=\frac{336}{a}$

$ab=\boxed{336}$

Let a and b be the length and width of the rectangle, such that b = x * a ==> S0 = a * b = x * a ^ 2. In an orthonormal reference of origin O (0,0), the different points have coordinates: A (0, x * a); B (y * a, x * a); C (a, x * a); D (a, x * t * a); E (a, 0) with 0 <x, y, t <1.It is necessary to calculate the equations of the different lines passing through these points in accordance with the statement.Thus, we will determine the coordinates of the 3 internal points of intersection.After calculation, F (y * a / (1-y * t + y), xa / (1-y * t + y)); G (y * a / (t-y * t + y), x * t * a / (t-y * t + y)); H (a / (1-y * t + y), x * t * a / (1-y * t + y)) The triangles (A, B, F) and (A, B, G) have the following meanings: base [AB] = y * a and respective heights [x * a-xa / (1-y * t + y)] and [x * a- x * t * a / (ty * t + y)]. Which is translated in other words by: 21 = S (A, B, F) and 21 + 21 = S (A, B, G). And the other 2 known triangles S (D, E, G) = 14 and S (D, E, H) = 8 are based on [DE] and for respective heights [ay * a / (ty * t + y)] and [a- a / (1-y * t + y) After solving the system of equations, we get: y = 1/2; t = 1/3 and (x * a ^ 2) / 24 = S0 / 24 = 14 or (x * a ^ 2) / 42 = S0 / 42 = 8 ===> S0 = 336