Quadrilateral Area

By applying cosine rule in triangle $ADC$ , we get $AC^2=20^2 + 15^2 - 2*15*20*(\cos D) \qquad (1)$ . Also in triangle $ABC$ , we get $AC^2=7^2 + 24^2 - 2*7*24*(\cos B) \qquad (2)$ . Subtracting (2) from (1),we get $300*(\cos D)=168*(\cos B)$ . But since $ABCD$ is a cyclic quadrilateral, $\angle D + \angle B =180$ , so $\cos B = -\cos D$ . This implies that $\cos D =\cos B=0 \Rightarrow D=B=90$ .

Now, since $EF = (1/3) AC$ , by considering area of triangle as base*height/2, we see that $[BEF] = (1/3) [BAC] = 24*7/(2*3)$ . Also, $[DEF] = (1/3)[DAC] = 20*15/(3*2)$ . Hence $[BEDF] = [BEF] + [DEF] = 78$ .

[Edits for clarity - Calvin]

The length of AC can be determined using the cosine law on the triangles ABC and ADC, bearing in mind that the opposite angles of a cyclic quadrilateral are supplementary. This leads to AC = 25, while angles B and D = 90^\circ. The area of ABCD is therefore equal to (AB)(BC)/2 + (AD)(CD)/2 = 234. Draw altitudes from B and D to connect to AC at G and H, respectively. Then the total area of the figure would be equal to (AC)(BG)/2 + (AC)(DH)/2. When looking for the area of BEDF, EF = AC/3. The area of BEDF can then be considered as the total area of triangles BEF and DEF. Since the altitudes of these triangles remain the same, and the base length is one-third of AC (the total length), then it can be concluded that the area of BEDF is one-third the area of ABCD. 234/3 = 78.

Using Brahmagupta's formula for the area of a cyclic quadrilateral when the four side lengths a = AB, b = BC, c = CD, d = DA are given, where $s = \frac{a+b+c+d}{2}$ , we have area(ABCD) = $\sqrt{(s-a)(s-b)(s-c)(s-d)} = 234$ .

Now, we are asked to divide AC into three equal sections using the points E and F, and find the area of BEDF. ABCD consists of the two triangles ABC and ACD. The area of a triangle is equal to half its base times its height. Here, we choose AC as the base of ABC and the height is the line segment going through B and perpendicular to AC. Looking at EBF, the base is EF, and the height is the same as for ABC. Since EF is one third of the length of AC, the area of EBF is one third the area of ABC.

The same reasoning for ADC and EDF yields that the area of EDF is one third of the area of ADC. From this, the area of BEDF is one third of the sum of the areas of ADC and ABC, namely one third of the area of ABCD = 234/3 = 78.

From Ptolemy's Theorem, we get : AB.CD + AD.BC = AC.BD, so AC.BD = 585.

Let say point G is the intersection point between AC and BC.

From Similarity for triangle AGD and BGC, we get equation (1) and (2). Equation (1) : AD/BC = AG/BG --> 5/6 = AG/BG. Equation (2) : AD/BC = (BD - BG)/(AC - AG) --> 5/6 =.(BD - BG)/(AC - AG)

Again, from Similarity for triangle AGB and DGC, we get equation (3) and (4). Equation (3) : AB/CD = AG/(BD - BG) --> 7/15 = AG/(BD - BG) Equation (4) : AB/CD = BG/(AC - AG) --> 7/15 = BG/(AC - AG) --> 15/7 = (AC - AG)/BG.

Multiply equation (3) by (BD - BG)/BG, so that equation (3) is the same with equation (1), which is : (7/15) * ( (BD - BG)/BG ) = 5/6. From this process, we get equation (5) : 42BD = 117BG.

Again. multiply equation (4) by AG/(AC - AG), so that equation (4) is the same with equation (2), which is : 5/6 = (15/7) * ( AG/(AC - AG) ). From this process, we get equation (6) : 7AC = 25AG

Divide equation (5) with equation (6), and we get BD/AC = 117/125. We know that AC.BD = 585, so the length of AC and BD is 25 and 23.4 respectively. From valuae of AC = 25, we can say that triangle ADC and ABC are right angled triangle at B and D respectively because the length of the sides of those 2 triangle satisfy Phytagorean Triples.

Area of AED = 0.5 * AD * AE * sin(angle DAC) = 0.5 * 20 * 25/3 * 15/25 = 50.

Area of DCF = 0.5 * DC * CF * sin(angle DCA) = 0.5 * 15 * 25/3 * 20/25 = 50

Area of BAE = 0.5 * AB * AE * sin(angle BAC) = 0.5 * 7 * 25/3 * 24/25 = 28

Area of BCF = 0.5 * BC * CF * sin(angle BCA) = 0.5 * 24 * 25/3 * 7/25 = 28

Area of ABCD = 0.5 * AB * BC + 0.5 * AD * DC = 0.5 * 7 * 24 + 0.5 * 20 * 15 = 234

Finally, the Area of BEDF = Area of ABCD - (Area of AED + Area of DCF + Area of BAE + Area of BCF) = 234 - 50 - 50 - 28 - 28 = 78

Draw the cyclic quadrilateral as the problem state it, if needed, draw the circle too. First, let me find the $<ABC$

As we know, $<ABC$ is in the minor(mayor) arc of $AC$ , and $<ADC$ is in the mayor(minor) arc of $AC$

Then minor arc of $AC$ + mayor arc of $AC$ = $360^o$

And $<ABC + <ADC = 180^o$ or $<ADC = 180^o - <ABC$

Let me use the cosine rule of triangle ABC: $AC^2 = AB^2 + BC^2 - 2AB . BC cos(<ABC)$

$AC^2 = 7^2 + 24^2 - 2.7.24.cos(<ABC) = 625 - 336 cos(<ABC)$

Let me use the cosine rule of triangle ADC: $AC^2 = AD^2 + DC^2 - 2AD.DC cos(<ADC)$

$AC^2 = 15^2 + 20^2 - 2.15.20 cos(<ADC) = 625 - 600 cos(<ADC) = 625 - 600 cos(180^o - <ABC) = 625 + 600 cos(<ABC)$

Equalising $AC^2$ from the two equation above, we get: $625 -336 cos(<ABC) = 625 + 600 cos(<ABC)$

$cos(<ABC) = 0 ---> <ABC = 90^o and <ADC = 180^o - 90^o = 90^o$

So it is obvious that triangle ABC and ADC is a right triangle.

The area of ABC = $\frac{1}{2} AB.BC = \frac{1}{2} 7 . 24 = 84$ , The area of ADC = $\frac{1}{2} AD.DC = \frac{1}{2} 15 . 20 = 150$

And we know that $AE = EF = FC$ , so $EF = \frac{1}{3}AC$

Then by area ratio theorem: $[BEF] = \frac{1}{3}ABC = 28$ , $[DEF] = \frac{1}{3}[ACD] = 50$

So: $[BEDF]=[BEF] + [DEF] = 78$

First, we note that $[BEDF]=[BEF]+[DEF]$ , where $[PQR]$ stands for the area of $PQR$ . Since $E$ and $F$ trisect $AC$ , we can say that $[BEF]=\frac{1}{3}[ABC]$ and $[DEF]=\frac{1}{3}[ADC]$ , since the triangles have one-third the base of the larger triangles. Thus, we can say that $[BEDF]=\frac{1}{3}[ABC]+\frac{1}{3}[ADC]$ .

Next, we see that the quadrilateral is cyclic. Noting that $7^2+24^2=20^2+15^2=25^2$ , we can conclude that we can form such a quadrilateral by placing two right triangles ( $7-24-25$ and $15-20-25$ ) on a diameter of length $25$ (remembering that opposite angles of a cyclic quadrilaterals must sum to $180^\circ$ ). Now it is easy to compute the area:

$[BEDF]=\frac{1}{3} (\frac{7\cdot 24}{2})+\frac{1}{3}(\frac{15\cdot 20}{2})= 78$ .

In a cyclic quadrilateral with successive vertices A, B, C, D and sides a = AB, b = BC, c = CD, and d = DA, the length of the diagonals p = AC can be expressed in terms of the sides as: p^2= [(ac+bd)(ad+bc)] / (ab+cd)

Applying the lengths from the problem above we find that AC=25

Since AC=25 , CD=12 and AD=20 from the pythagoreum theorem it means we have a right triangle ADC.

The same happens with triangle ABC. It is a right one since AB=7, BC 24 and hypoteneuse AC = 25

Area of triangle ABC = (24*7) /2 = 84

Area of triangle ADC = (20*15) /2 = 150

Because we know also that AC is divided in 3 equal parts , the triangle DEF area is one third of 150 = 50

The same happens with triangle BEF where area is one third of 84 =38

So, area of BEDF = 50+38 = 78

Let $|X|$ denote the area of $X$ . By the Brahmagupta's formula, $|X|=234$ . Note that $3EF=AC$ , so $3|BEF|=|BAC|, 3|DEF|=|DAC$ by base and area relationships. Adding these two equations give us $3|BEDF|=234$ , so $|BEDF|=78$ .

the semiperimeter of ABCD is (7+24+15+20)/2=33 it is obvious that (ABCD)=(ACD)+(ACB)=3(EFD)+3(EFB)=3(BEFD)=((33-7)(33-15)(33-20)(33-24))^(1/2)=234 then (BEFD)=234/3=78

From Brahmagupta's formula, Area of a quadrilateral=sqrt{(s-a)(s-b)(s-c)(s-d)}, where s=1/2 (a+b+c+d). From the Geometry of the figure drawn according to the problem, area of BEDF=1/3 area of ABCD

Drawing a picture of the quadrilateral, it's easy to see how the area of $BEDF$ is one third of that of $ABCD$ .

Specifically, it's because you can imagine the diagonal $AC$ cutting them into triangles from which you can see how it forms two pairs of triangles with the same heights and bases in a ratio of $1:3$ .

Using Brahmangupta's formula for the area of a cyclic quadrilateral with sides $a$ , $b$ , $c$ , $d$ and semiperimeter $s$ : $A = \sqrt{(s-a)(s-b)(s-c)(s-d)}$

Plugging in our values, we find that the area of the figure $ABCD$ is 234 and the area of $BEDF$ must be 78.

since abcd is a cyclic quadrilateral and triangle ABC is a right angled triangle therefore ADC is also a right angled triangle therefor area of quadrilateral ABCD is =1/2( 7) (24) +1/2( 20)( 15)=234

since E and F trisect the triangle ABC and ADC

this implies area of traingle ADC=3(area of traingle EDF)

similarly area of traingle ABC=3(area of triangle FBE)

therefore area of BEDF=1/3 area of ABCD =1/3(234)=78

Note that $AB^2 + BC^2 = CD^2 + DA^2 = 25^2.$ Thus both ∆ABC and ∆ADC are right triangles.

E and F trisect the line segment AC. [BEDF] = [BEF] + [DEF]. Since EF is $\frac{1}{3} AC$ , [BEF] = [ABC]/3 and [DEF] = [ADC]/3.

Therefore [BEDF] = [ABCD]/3 = ( 84 + 150 ) / 3 = 78.

We notice that $AB^2 + BC^2 = 7^2 + 24^2 = 25^2 = 15^2 + 20^2 = CD^2 + DA^2$ . By cosine rule, we get that $AC^2 = AB^2 + BC^2 - 2 AB \cdot BC \cos \angle ABC$ and $AC^2 = CD^2 + DA^2 - 2 CD \cdot DA \cos \angle CDA$ . Since $\angle ABC + \angle CDA = 180^\circ$ , thus $\cos \angle ABC = - \cos \angle CDA$ $\Rightarrow AB \cdot BC \cos \angle ABC = -CD \cdot DA \cos \angle ABC$ . Since the coefficients are positive, this is only possible if $\cos \angle ABC = 0$ , thus $\angle ABC = \angle CDA = 90^\circ$ .

Let $[PQRS]$ denote the area of figure $PQRS$ . We have $[ABCD]= [ABC] + [BDC] = \frac{1}{2}\left(7\times 24 + 15\times 20\right) = 234$ . Since $3 [ BEF] = [ABC] \Rightarrow [BEF] = \frac{7\cdot 24}{6} = 28$ and $3[DEF] = [CDA]$ $\Rightarrow [DEF] = \frac{15\cdot 20}{6} = 50$ , thus $[BEDF]= [BEF] + [DEF] = 78$ .