Areas in a Quadrilateral

Join $AK$ . Let $[ADK] = x$ and $[AEK] = y$ .

Now consider triangles $BKD$ and $AKD$ which have common base $DK$ , and triangles $BKC$ and $AKC$ which have common base $CK$ respectively.

So, $\frac{[AKD]}{[AKC]} = \frac{56}{84}$ .

Therefore, $\frac{x}{y+90} = \frac{56}{84} = \frac{2}{3}$ .

Similarly, consider triangles $ABK$ and $CBK$ on the common base $BK$ , and triangles $AKE$ and $CKE$ on the common base $KE$ .

So, $\frac{[ABK]}{[AKE]}= \frac{84}{90}$ .

Therefore, $\frac{56+x}{y}= \frac{84}{90} = \frac{14}{15}$ .

Thus, we get two linear equations in $x$ and $y$ , after solving them we can get $x+y$ , which is the required area.

We will first discuss a powerful tool for finding area/side ratios in triangles, then we will use these tools to solve the problem in question.

The theorem:

The area of any triangle can be expressed as $\frac{1}{2}bh$ , where $b$ is the length of the side that the altitude, $h$ , is dropped to. It follows that if two triangles have the same base length then the ratio of their areas is equal to the ratio of their altitudes. Similarly if two triangles have the same altitude then the ratio of their areas is equal to the ratio of their bases. (prove these!)

Solving the problem:

First we draw in $\overline{AK}$ to create triangles which we can use the theorem on, namely $\triangle{AKD}$ and $\triangle{AEK}$ . Furthermore we let these triangles be $x,y$ , respectively.

(Note: $[PQRS]$ is used to denote the area of quadrilateral $PQRS$ , etc.)

By the same base same altitude theorem we have

$\frac{[AKB]}{[AEK]}=\frac{[KCB]}{[ECK]}\implies{\frac{x+56}{y}}=\frac{14}{15}$

and

$\frac{[AKD]}{[ACK]}=\frac{[DKB]}{[KCB]}\implies\frac{x}{y+90}=\frac{14}{21}$

Solving these simultaneous equations we obtain the solution $(x,y)=(350,435)$ , thus $[AEKD]=x+y=\boxed{785}$ .

DK / CK = [BDK] / [BCK] = 56/84 = 2/3. [DKE] / [CKE] = DK / CK = 2/3, therefore [DKE] = 2/3 * [CKE] = 2/3 * 90 = 60.

Let x = [ADE], we have: AD / BD = [ADE] / [BDE] = x / (56 + 60) = x / 116. AD / BD = [ADC] / [BDC] = (x + 60 + 90) / (56 + 84) = (x + 150) / 140.

Then we've got x / 116 = (x + 150) / 140, or x = 725. Finally, [ADKE] = [ADE] + [DEK] = x + 60 = 725 + 60 = 785.

Now, $\frac {AE}{CE} = \frac{[AEK]}{[CEK]} = \frac{[AEB]}{[CEB]}$ . Denoting this ratio by r:

$\begin{aligned} &[AKB] = [AEB] - [AEK] = r[CEB] - r[CEK] = r[CKB]\\ &\implies \frac {[AKB]}{[CKB]} = \frac{[AEK]}{[CEK]}.\end{aligned}$

Likewise, we have $\frac{[AKC]}{[BKC]} = \frac{[ADK]}{[BDK]}$ . If we denote $[AEK] = a$ and $[ADK] = b$ , then this gives the equations:

$\frac{84}{56+b} = \frac{90}{a}, \ \frac{84}{90+a} = \frac{56}b.$

Solving gives $a=435, b=350$ , so the area of $PQRS$ is $a+b = 785$ .

Let $[\text{AKD}]=a$ , $[\text{AKE}]=b$ , $BD=e$ , $DA=c$ , $AE=d$ , and $EC=f$ .

Since triangles $BKD$ and $AKD$ have the same altitude with respect to AB, their bases are in the same ratio as their areas, giving us $\frac{e}{c}=\frac{56}{a}$

Repeating this with triangles $ABE$ and $CEB$ , $AKE$ and $CKE$ , and $BCD$ and $ACD$ gives us

$\frac{f}{90}=\frac{d}{b}$

$\frac{d}{a+b+56}=\frac{f}{174}$

$\frac{e}{140}=\frac{c}{a+b+90}$

Rearranging some of these equations will allow us to substitute out all the variables besides $a$ and $b$ .

$\frac{e}{c}=\frac{56}{a}$

$\frac{b}{90}=\frac{d}{f}$

$\frac{d}{f}=\frac{a+b+56}{174}$

$\frac{e}{c}=\frac{140}{a+b+90}$

Substituting out $\frac{e}{c}$ and $\frac{d}{f}$ gives us

$\frac{56}{a}=\frac{140}{a+b+90}$

$\frac{b}{90}=\frac{a+b+56}{174}$

Cross-multiplying,

$56a+56b+5040=140a$

$174b=90a+90b+5040$

Solving this system, we find that $a=350$ and $b=435$ . Therefore, the answer to our original problem is $a+b=\boxed{785}$ .

Let the area of ADKE be x. AD/DB = (x+90)/140 AE/EC = (x+56)/174 x = Area of ADK + Area of AKE = ((x+90)/140) 56 + ((x+56)/174) 90 = (2x+180)/5 + (15x+840)/29 After manipulation, you arrive at 12x = 9420, and thus x = 785.

We will construct a line segment passing through K from A, and meeting BC.Now AEKD, is divided into two parts, let [AKE]=x, [ADK]=y.We have to find x+y.Now in triangle DBC, line segment BK,divide line segment DC,in the ratio 56/84= 2/3.So DK/KC=2/3.Now looking on other side of DC,in triangle DAC, point K will also divide base DC of DAC, into ratio 2/3.So y/(90+x)=2/3. Now looking in the triangle AKC, AC is divided by point E in the ratio x/90, as triangles AKE and KEC have same height so base is divided in the ratio of there areas.Now looking in the big triangle ABC, triangles BAE and BEC, have same height so base AC is divided in the ratio of their areas, which gives [BAE]/[BEC]=AE/EC= (x+y+56)/174=x/90........so we have two equations and two variables, on solving we get x=435, and y=350.........hence the answer......

Draw AK and let the area of triangle ADK be x and the area of triangle AKE be y. Denote by [ABC] the area of triangle ABC. Then 14/15=84/90=[BKC]/[CKE]=BK/KE=[ABK]/[AKE]=(56+x)/y. Similarly, 2/3=56/84=[BKD]/[BKC]=KD/KC=[AKD]/[AKC]=x/(90+y). Clearing denominators gives 14y=15x+840 and 2y=3x-180. Solving gives x=435 and y=350. So x+y=785 which is our answer.

Let the area of ADKE is x, the area of triangle AKE is a, and triangle ADK is b.So a + b = x ..........(1).Consider triangle CEB and AEB . EC/AE = (174)/(56+x) ..... (2). Consider triangles ECK and AEK. EC/AE = 90/b ........(3). From eqs. (2) and (3) we get 45 x - 87 b = 2520. .............(4). Consider triangles ACD and BCD. AD/DB = (90+x)/140 .........(5). Consider triangles ADK and BDK. AD/DB = a/56 ............(6). From eqs. (5) and (6), we get : 28 x - 70 a = 2520...........(7). Further by elimination of eqs (4) and (7), we get : 70 a - 87 b + 17 x = 0 ...... (8).Further we can use eq.(1), to find that b = (87/70) a., and x = (157/70) a. So we can find a = 350, and b = 435. So x = 485

Construction to be done:

1. Join $D$ to $E$ .

2.Through $D$ draw a line parallel to $BE$ such that it intersects $AC$ at, say, $M$ .

Since triangles $KEC$ and $BKC$ have the same altitude over the sides $KE$ and $BK$ respectively, we can say that:

$\frac{EK}{BK} = \frac {[KEC]}{[BKC]} = \frac{90}{84} = \frac{15}{14}$

Similarly,

$\frac{DK}{KC} = \frac{[DKB]}{[BKC]} = \frac{56}{84} = \frac{2}{3}$

Let $[DKE] = x$ . Using similar method as above, we can say that,

$\frac{90}{x} = \frac{[CKE]}{[EKD]} = \frac{CK}{KD} = \frac{3}{2} \implies x = 60$ i.e. $[DKE] = 60$

Now, since triangles KEC and DMC are similar, we can say that

$\frac{[KEC]}{[DMC]} = \frac{90}{90+60+[DME]} = (\frac{KC}{DC})^2\ = \frac{9}{25} \implies [DME]=100$ Now $\frac{KE}{DM}=\frac{15\alpha}{DM}$ and $\frac{KE}{DM}=\frac{CK}{CD}=\frac{3}{5} \implies \frac{15\alpha}{DM}=\frac{3}{5}\implies DM=25\alpha$

Since triangles ADM and ABE are similar, we can say that,

$\frac{[ADM]}{[ABE]}=\frac{[ADM]}{[ADM]+56+60+100}=(\frac{DM}{BE})^2=\frac{625}{841}\implies [ADM]=625$ .

Therefore $[ADKE]=[ADM]+[DME]+[DKE]$

$\implies [ADKE]=625+100+60=785$

$[ADKE] = \delta$

$\angle ACD= \phi$

Drop perpendiculars DM & KN on BC.

Since $\triangle CKN$ and $\triangle CDM$ are similar, and because $\triangle BKC$ and $\triangle BDC$ are on same base BC, we get,

$\frac{CK}{CD}=\frac{KN}{DM}=\frac{84}{85+56}$

$\frac{CK}{CD}=\frac{84}{140}=\frac{3}{5}$ ..................(1)

$\frac{1}{2}.EC.CB.sinC=174$ and

$\frac{1}{2}.AC.CB.sinC=230+\delta$

$\frac{EC}{AC}=\frac{174}{230+\delta}$ ...................(2)

Similarly,

$\frac{1}{2}.EC.CK.sin\phi=90$ and

$\frac{1}{2}.AC.CD.sin\phi=90+\delta$

Dividing & using (1) and (2),

$\frac{174}{230+\delta}.\frac{3}{5}=\frac{90}{90+\delta}$

This on solving gives $\delta$ = 785