A Quadrilateral in a Square

First of all $6,8,10 \rightarrow 6^{2} + 8^{2} = 10^{2}$ Thus by the converse of Pytagoras theorem $\triangle ABC$ is rectangle. Since $AM$ is the median, and we know every triangle rectangle can be inscribed in the diameter of a circumference we can see the median of the right triangle equals half the diameter. Therefore $AM = \frac{1}{2}BC = 5.$ That means $AM = BM$ and $\triangle ABM$ is iscoseles. Let $\angle{ABM} = \alpha \rightarrow \angle{BAM} = \alpha \rightarrow \angle{FAM}= 90 - \alpha \rightarrow\angle{AFM} = \alpha$ . This means they are similar. $\\\frac{AM}{MF} = \frac{AC}{BA}$ $\frac{5}{MF} = \frac{8}{6}$ . $MF = \frac{15}{4}$ Area quadrilater {AEFD} = Area square {AEDM} - Area triangle {AMF} = $25 - \frac{75}{8} =\boxed{\frac{125}{8}}$

The cosine rule can also be used to find AM $\begin{array}{l} \cos \theta =\frac{a^{2}+b^{2}-c^{2}}{2\ast a\ast b} \\ \frac{6}{10} =\frac{6^{2}+5^{2}-c^{2}}{2\ast 6\ast 5} \\ \frac{6}{10} \ast 60=36+25-c^{2}\\ c^{2 }= 25\\ c = 5\\ ~ \end{array}$ The rest is quite simple

@Jordi Bosch great solution

(I.) The angle BAC is a right angle.

Proof:

10, 8 and 6 are a Pythagorean triple (multiples of the primitive (5,4,3)) therefore the triangle ABC is a right triangle. The right angle is the one opposite from the hypotenuse.

(II.) Draw a segment MK which is perpendicular to AC (not shown). Then the triangles KMC and ABC are similar (and KM=3)

Proof:

AB and MK are both perpendicular to AC; therefore they are parallel to each other. Hence, the angles ABC and KMC are equal. The triangles ABC and KMC share three angles, therefore they are similar. As AB=2MC by definition, the ratio is 2. Therefore KM=AB/2 =3

(III.) AM=5

Proof:

The triangles KMA and KMC share a side (KM); AK=KC from the similarity of ABC and KMC; both angles MKA and MKC are right angles by construction. Therefore the triangles are equal and AM-MC=5

(IV.) 3AF = 5MF

Proof:

From the sine rule we know sin(ACB)=AB/BC=3/5. The angles AMC and ACB are equal; therefore sin(AMC)=3/5 = MF/AF.

(V.) AF = 25/4

Proof:

As the triangle AMF is a right triangle, AF^2 = AM^2 + MF^2 AM^2 = 25 and MF = 3AF/5, which yields the desired result.

One can see from the diagram that

Area(AEDF) = Area(AEDM) - Area(AMF) = AM^2 - (AF MK/2) = 25 - (25 3)/8 = 125/8

yielding the desired result.

This question takes into account principles of similarity and trigonometry.

It is easy to see that the triangle is an Right angle triangle ABC (by the cosine identity, is you want it to be quite rigorous) is in fact a right angle triangle about A. Hence it can be inscribed inside a circle of diameter BC.

Then M would the centre of the circle, by the given information in the question and consequently AM = 5cm = MD =ED = AE = BM = MC = AM.

Now, for the trigonometric and similar part.

Tri(ABM) can now be see as an isosceles triangle. Hence, Angle(ABM) = Angle(BAM) = the larger acute angle of the 345 right angle triangle. therefore, Angle(MAF) = the smaller acute angle of the 345 right angle triangle. (because, Angle(BAM) + Angle(MAF) = 90 degrees. This consequently shows that Traingle(MAF) is similar to Triangle(BCA) (which already was similar to the 345 right angle triangle). therefore the in Triangle(MAF) MA = 5cm. and then one can work out its area (which would be (1/2) (15/4) (5) ).

All that remains is to calculate the area of the Square(AMDE) and subract from that area of the Triangle(AMF) which would give:

                                                                  125/8

AM=MC=MB = 5, so area of AMED= 5 5=25, tg ACB=6/8= tg MAF =MF/AM, thus MF= 15/4 Area of AFDE = area AMDE-area of AMF =5 5-1/2 5 15/4= 125/8

AB = 6, BC=10, AC=8...if we check by using Pythagorean Theorem, --> c2=a2+b2, this signify that triangle is a Right Triangle...

so therefore; Sin Ѳ = 4/5, Cos Ѳ = 3/5, tan Ѳ = 4/3... Area of ABM = AB x BM x sin Ѳ/2 = 6 5/2 4/5 = 12

Using Cosine Law; AM^2=AB^2+BM^2-2 AB AM cosѲ AM^2=6^2+5^2-2 6 5 3/5 AM^2=25 AM = 5 therefore Triangle ABM is Isosceles Tri., so, angle BAM = ABM, equal to Ѳ, Angle MAF =90-Ѳ, then

for Triangle AMF, since this is a Square, so angle AMF = 90, Angle MAF =90-Ѳ, then Angle AFM = Ѳ, for MF = ? for similar triangle, MF /5 = 3/4, so MF = 15/4

Area of AFM = AM MF/2 = 5 15/4/2= 75/8 Area of AEDF = Area of AEDM - Area of AFM Area of AEDF = 5*5-75/8

Area of AEDF = 25 - 75/8 ===>>>125/8..Ans..

1. Given: AB=6, AC=8 and BC=10. AM=MC=5 (Since, M is the midpoint of BC). Also, AMDE is a square.
2. It can be observed that: BC² =AB² +AC² . This says that ABC is right angled at A.
3. By using theorem "In a right angled triangle, the median from right angle to hypotenuse is half the length of hypotenuse", MC=MB=MA=5. Also, since AMDE is a square MA=AE=ED=DM=5. ( The angles of a square are all equal to 90°).
4. Now, ∠BAM+∠MAC=90°; ∠MAC+∠CAE = 90°. Thus, ∠BAM = ∠CAE---(1).
5. Since AE||MD with AC as a transversal, ∠MFA = ∠CAE (alternate interior angles)---(2).
6. From (1) and (2), ∠BAM=∠MFA=∠CAE---(3).
7. Triangle AMC being isosceles (AM=MC); ∠MAC=∠MCA. Also, triangle ABM is isosceles (MB=MA); ∠BAM=∠MBA ---(4).
8. From (3) and (4), ∠BAM=∠MFA=∠CAE=∠MBA ---(5)
9. Now, ∠MBA=∠MFA (from (5)); ∠BAC=∠AMF=90° and ∠MCA=∠MAF (from isosceles triangle AMC). By AAA theorem, triangles ABC and MAF are similar triangles. Thus, (AB/MF)=(AC/MA)=(BC/FA). Or, (6/MF)=(8/5)=(10/FA).
10. This gives: MF = 15/4 and FA =25/4.
11. Area of quadrilateral AEDF = Area of square AEDM - Area of triangle AMF = 5²-((1/2) 5 (15/4)) = 25-(75/8) = 125/8 or 15.625 square units.

first the triangle is right angled triangle and I draw m x orthogonal AC then XF =2.25 then the area of triangle AMF=75/8 then the area of figure AEDF =25-75/8=125/8

in triangle abc <bac=90 as 8^2+6^2=10^2,so cos <acb=8/10 and sin<acb=6/10,now consider ab along x and ac along y so considering a at(0,0) c=(0,8)and b=(6,0).now since m divides bc in 1:1 so m=(3,4)by mid point formula.so am =5.now in triangle acm <acm=<mac=<acb as equal side subtend equal angle.also side of square=5.so area of square =25.now for triangle amf tan<maf(=<mac)=mf/am=6/8.mf=6 5/8=15/4.so area of traingle afm=(1 5*15)/8.subtract this area from area of square 25-75/8=125/8 which is our answer.

The solution lies in the property of triangle and specifically the right angle tri....The main idea is to prove the tri ABC and AMF identical and than need to bring out the ration of sides.........AB, AM, MC, BM, AE, DE, MD are all equal i.e. = 5. Putting the property of ratio for the 2 right angle tri ABC and AMF we get ---> AC(height)/AB(base) = AM(height)/MF(base)...........Putting the value we get MF and than subtract the Area of Triangle AMF from the Are of Square AMDE..............We are with our solution..........

The triangle ABC is clearly right angled since it followed the ratio 3:4:5. AM is the median and the quadrilateral AMDE is a square, so the triangle AMF is right angled and the triangles AMF and CAB are similar. Therefore, CA/AM=AB/MF=BC/FA(=K) => AM=8K & MF=6K =>Area of triangle AMF = 24K K & Area of Square = 64K K =>After subtraction, area of the shaded region is 40K =>only option in the choices that can be obtained for the value of K is 125/8 :)

First. Let us deduce the relationship between different sides and angles of the triangles and quadrilaterals.

We have AM being the median of a right triangle, which by Thales theorem, means that, BC is the diamater of the Circumcircle and M is the mid-point of BC and hence AM = BM = CM = BC/2 = 5

Now since AEDM is a square, AE = ED = DM = AM = 5

Now let us see ∆ ABC. Since AngleBAC = 90˚, we can deduce that AngleABC = tan^-1 (3/5) ; AngleACB = tan^-1 (4/5)

Now see ∆ABM, since AM = BM, it is an isoceles triangle. So AngleABM = AngleBAM = tan^-1(3/5) Now since AngleBAC = 90˚ = AngleBAM + AngleCAM Therefore AngleCAM = tan^-1(4/5)

Similarly if you consider ∆AMC, you will find that AngleAFM = tan^-1(3/5)

Now compare ∆BAC and ∆AMF. By virtue of angles that we found out, we have ∆BAC ~ ∆AMF. Therefore: BC/AF = AC/AM = AB/MF 10/AF = 8/5 = 6/MF Therefore: AF= 25/4 ; MF = 15/4

NOW, Area(AEDF) = Area(AEDM) - Area(AMF) Area(AEDM) = 5 X 5 = 25 Area(AMF) = 1/2 X AM X MF = 1/2 X 15/4 X 5 = 75/8

Hence, AreaAEDF = 25 - 75/8 = 125/8

Using Apollonius Theorem, AM=5=MC. So, ar(AMDE)=25. Since AMC is isosceles, angles A and C are equal. tan A=3/4=MF/AM=MF/5. So, ar(AMF)=75/8 as M=90 deg and ar(triangle)=1/2 bh=1/2 AM*MF. So, ar(AFDE)=ar(AMDE)-ar(AMF)=25-75/8=125/8

6^2 + 8^2 = 10^2 (Pitagorasz) AM=BC /2 =10/2=5...................... 5/8 = MF/6 ……… MF = 30/8.............. 5/8 = AF/10 …….. AF = 50/8............... area of quadrilateral= area of square – area of small triangle.......... area of square=25 =200/8 ....... area of small triangle= 5 30/8 /2 =75/8 ......... area of quadrilateral=200/8 -75/8 = 125/8

Used coordinate geometry with AC and BC as axes.It was fun.

A(AFDE) = As(AMDE) - At(AMF);

|AB| = 6; |AC| = 8; |BC| = 10;

As(AMDE) = |AM|^2; At(AMF) = |AM||MF|/2;

Tr(ABC) is rectangle in A, angle(A) = 90°, because

|AB|^2 + |AC|^2 = |BC|^2, 6^2 + 8^2 = 10^2;

so M, midpoint of BC is the circumcenter of Tr(ABC),

|BC| = 2|AM| = 2r, r = |AM| = |BC|/2 = 5, is the radius of the circle;

Also with formula of medians:

|AM| = (√(2(|AC|^2 + |AB|^2) - |BC|^2))/2 =

√(2|BC|^2 - |BC|^2)/2;

|AM| = √(|BC|^2)/2 = |BC|/2 = 5;

As = 5^2 = 25;

angle(AMF) = 90°;

angle(MAB) = angle(MBA),

because |AM| = |BM|, Tr(AMB) is isosceles, and also Tr(AMC); so:

angle(CAM) = angle(BCA);

Tr(MAF) is rectangle and similar to Tr(ABC):

|MF|/|AM| = |AB|/|AC|;

|MF| = |AB||AM|/|AC| = 6 x 5/8 =15/4;

At(AMF) = |AM||MF|/2 = 5 x 15/(2 x 4) =75/8;

At(AFDE) = 25 - 75/8 = 125/8 cm^2;