Area of Triangle ABC

I noticed two things here which I am going to explain in $2$ solutions (as $1$ alternative solution).

• * 1st Solution : I noticed that $\bigtriangleup ABC$ is * Right-angled Triangle as $AB=5$ , $MB=6.5$ or $BC=6.5 \times 2 =13$ and $13^2-5^2=12^2$ (They're a pythagorean triplet).

Hence, $\implies[ABC]=\frac {1}{2} \times 12 \times 5=\boxed{30}.$

• * 2nd Solution(Aliter) : It's for those who didn't notice that $\bigtriangleup ABC$ is a * Right-angled Triangle .

First ,by using Cosine Formula in $\bigtriangleup ABM$ we get:

$\implies \frac {5^2+6.5^2-6.5^2}{65}=\frac{5}{13}=\cos \angle B$

$\implies \sin \angle B=\frac {12}{13}$

We then notice, $BC=13$ .

Hence, $[ABC]=\frac {1}{2} \times 5 \times 13 \times \frac{12}{13}=\boxed{30}.$

By applying Apollonius therorem we have AC^2 + AB^2 = 2(AM^2 + BM^2) AC^2 = 2(AM^2 + BM^2) - AB^2 = 4 * (6.5^2) - 5^2 AC^2 = 144 or, AC = 12 as the sides are 5,12 and 13 it becomes a right triangle Hence its area is 5*12/2 = 30

Because AM = MB = MC. So, point M is center point of radius in triangle ABC, If r = 6,5, then the d = 13 and because AB = 5, thus AC = 12. So, areas of triangle ABC is 5.12/2 = 5.6 = 30. Answer : 30

Let $\angle AMB = \alpha$ . Then $\angle BAM = \angle ABM = (\frac{180^\circ-\alpha}{2}) = (90^\circ - \frac{\alpha}{2})$

Using Sine Law,

$\frac{AB}{\sin AMB} = \frac {BM}{\sin BAM}$

$\Rightarrow \frac{AB}{\sin \alpha} = \frac {BM}{\sin(90^\circ - \frac{\alpha}{2})}$

$\Rightarrow \frac {5}{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}} = \frac{6.5}{\cos \frac{\alpha}{2}}$

$\Rightarrow \frac{5}{13} = \sin \frac{\alpha}{2}$

Then, $\cos \frac{\alpha}{2} = \sqrt{1 - \sin^2 \frac{\alpha}{2}} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$

$\sin \alpha = 2 \times \sin \frac{\alpha}{2} \times \cos \frac{\alpha}{2} = 2 \times \frac{5}{13} \times \frac{12}{13} = \frac{120}{169}$

Now, Draw a perpendicular $AD$ from $A$ to $BC$ .

$\sin AMB = \frac {AD}{AM}$

$\Rightarrow \frac{120}{169} = \frac {AD}{6.5}$

$\Rightarrow AD = \frac{60}{13}$

$[ABC] = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times 13 \times \frac{60}{13} = \boxed {30}$

$In \space a\space \Delta{ABC} \space and \space its \space lengths \space are \space a,b,c \space then\space we\space can\space write:$ $\cos{B}=\sqrt{\frac{a^{2}+c^{2}-b^{2}}{2*a*c}}$ $so\space in \space\Delta{ABM}\space we\space can\space write:$ $\cos{B}=\frac{BM^{2}+AB^{2}-AM^{2}}{2*AB*MB}$ $=>\cos{B}=\frac{6.5^{2}+5^{2}-6.5^{2}}{2*6.5*5}$ $=>\cos{B}=\frac{5}{13}$ $Now\space Draw\space a\space line\space AD\space on\space base\space BC\space such \space that \angle{ADB}=90^{o}$ $Now\space in\space \Delta{ADB}\space\space\space\space\space\space\sin{B}= \frac{AD}{AB}\space\space =>\space AD=\sin{B} \times{AB}$ $Again\space we\space know\space \sin\theta = \sqrt{1-\cos^{2}\theta}$ $so,\space \sin{B}=\sqrt{1-\cos^{2}{B}}$ $=>\sin{B}=\sqrt{1-\frac{5^{2}}{13^{2}}}$ $=>\sin{B}=\frac{12}{13}$ $AD \space = \space 5 \times{\frac{12}{13}}$ $so\space AD\space is\space Height \space of \Delta {ABC} \space and\space Base \space is\space BC = 13$ $\Delta \space \space = \frac{1}{2}\times {BC}\times{AD}$ $Area \space of \space \Delta {ABC}\space is \boxed{30}$

We have $BM=AM$ and we have $AM=MC$ . Hence, $M$ is the circumcentre of $\triangle ABC$ and since it lies on the midpoint of $BC$ , it is the diameter of the circumcircle and $\angle BAC=90^{\circ}$ . Using Pythagoras's theorem in $\triangle ABC$ we have $AC= 12$ and hence the area

= $\dfrac{1}{2} \times AB \times AC = \dfrac{1}{2} \times 5 \times 12 = \boxed{30}$

At first, we should know that a median divides the total area of the triangle into two equal parts. Here, M is the mid-pt of BC. So, we see that ar(ABC) $=2\times$ ar(AMB). Given that, AM=MB=6.5 and AB=5.

We use here the Heron's formula which states that a triangle having sides $a,b,c$ and semi-perimeter $(s)$ ,i.e, half of the total perimeter, has area $=\sqrt{s(s-a)(s-b)(s-c)}$ . For triangle AMB, we have,

Perimeter $=AM+MB+AB=6.5+6.5+5=18$ . So, semi-perimeter $(s)=\frac{18}{2}=9$ .

Let us take AM,MB and AB as $a,b,c$ , so $a=6.5,b=6.5,c=5$

ar(AMB) $=\sqrt{9\times (9-6.5)\times (9-6.5)\times (9-5)}=\sqrt{9\times 2.5\times 2.5\times 4} = 3\times 2.5\times 2=15$

Now, ar(ABC) $=2\times$ ar(AMB) $=2\times 15=\boxed{30}$

M is the circumcenter triangle ABC is right angle triangle. Remain side is 12

View this link for the diagram of the $\bigtriangleup ABC$ .

The perpendicular of $\bigtriangleup ABC AD$ is drawn. Let $AD = t$

Let, $BD = x and DM = (6.5 - x)$

$AM = MB = 6.5; M$ is the midpoint of $BC$ . So, $MB = MC = 6.5$

Now, in right angle triangle $\bigtriangleup ADB \implies t^2 + x^2 = 5^2 ....... i$

And, in right angle triangle $\bigtriangleup ADM \implies t^2 + (6.5 - x)^2 = 6.5^2 ........ii$

From $i$ and $ii$ we get the value of $x = \frac{25}{13}$ and $t = \frac{60}{13}$

Now, in triangle $\bigtriangleup ABC$ the base(BC) is $\implies BM + CM \implies 6.5 +6.5 \implies 13$ and the perpendicular(AD) is $\implies \frac{60}{13}$ .

So, the Area of the triangle $\bigtriangleup ABC$ is ( \implies \frac{1}{2} \times 13 \times \frac{60}{13} \implies \boxed{30}.

A median divides a triangle into two equal halves.The perimeter of one half is 5+6.5+6.5=18.now we can apply heron's formula.Area = (9 (9-6.5) (9-6.5)*(9-5))^0.5

SOlusi dengan Formula Heron

AB = 5 M adalah titik tengah garis BC. Karena M adalah mid point, maka CM = AM = BM = 6.5 Maka BC = BM + CM = 13

Cari AC dengan Phytagoras AC^2 = BC^2 - AB^2 AC = V(169 - 25) AC = V(144) AC = 12

Cari s dengan asumsi s = (a + b + c) /2 s = (12 + 13 + 5) /2 s = 15

[ABC] = V(s(s-a)(s-b)(s-c)) [ABC] = V(15(3)(2)(10)) [ABC] = V(45(20)) [ABC] = V(900) [ABC] = 30

in triangle ABC,let ANGLE MAB=ANGLE ABM=X.ANGLE CAM=ANGLE ACM=Y.SUMMING ALL THREE ANGLES OF TRIANGLE ABC WE GET (X+Y)=90.AB=5,BC=13,THEN AC=12.THEN AREA =0.5X12X5=30

M is the midpoint of BC.
Therefore, MB = MC
Given, AM = MB,
Hence, AM = MB = MC, which means that M is equidistant from A, B and C.
So if we make a circle passing through A, B and C, M will be its centre and BC will be its diameter.
So, measure of angle BAC is 90 degrees.
By pythagorus theorem, we get AC = 12 cm.
Therefore, area of triangle ABC = 0.5 X AB X AC = 0.5 X 5 X 12 = 30 sq.units

AM = BM = 6.5

Also, M is the midpoint of BC

Therefore, BM = MC = 6.5 = AM.

With the above result we can say that the point M is the circumcentre of the triangle ABC.

Also, the circumcentre lies inside or outside the circle when it is either acute or obtuse respectively. It will lie on a side (as in this case) only if it is a right angle triangle. Also, it will lie on the hypotenuse, BC.

BC = BM + BC = 13

AB = 5.

Applying Pythagoras Theorem we get AC = 12.

Area of the triangle = 1/2 x 12 x 5 = 30!!

I'm pretty sure there's an easier way out here, but here is my (lengthy) solution. Note: Draw your own illustration of the problem to understand it better. This. Is. Geometry.

Since $M$ is the midpoint of $\overline{BC}$ , it implies that $\overline{BM} \equiv \overline{MC}$ .

Also, since $AM = MB = 6.5$ , it then follows that $MC = 6.5$ also.

Using the cosine law where letting $\angle{AMB}$ as $\theta$ ,

$(AB)^2 = (AM)^2 + (MB)^2 - 2 (AM) (MB) \cos \theta$

$(5)^2 = (6.5)^2 + (6.5)^2 - 2 (6.5) (6.5) \cos \theta$

$(5)^2 = 2 (6.5)^2 - 2 (6.5)^2 \cos \theta$

$25 = 84.5 - 84.5 \cos \theta$

$-59.5 = -84.5 \cos \theta$

$0.704 = \cos \theta$

$\cos^{-1} 0.704 = \theta$

$\theta = 45.26$

We can obviously see in our diagram that points $C$ , $M$ and $B$ are collinear, that is, lying on the same line. Since $\angle AMB \approx 45.26 ^ \circ$ , $\angle AMC = 180 - \angle AMB \approx 134.73 ^ \circ$ .

By the cosine law (again), we can find for the length of the remaining side, that is, $AC$ .

$(AC)^2 = (AM)^2 + (MC)^2 - 2(AM)(MC) \cos 134.73$

$(AC)^2 = (6.5)^2 + (6.5)^2 - 2(6.5)(6.5) \cos 134.73$

$(AC)^2 = 2(6.5)^2 - 2(6.5)^2 (-0.703809)$

$(AC)^2 = 84.5 - 84.5 (-0.703809)$

$(AC)^2 = 84.5 + 54.47$

$(AC)^2 = 143.97$

$AC \approx 11.99 \approx 12$

The approximation above was done due to the fact that there was some inaccuracies regarding division (e.g.: not enough to show all digits of the decimal, "brilliant mathematics answers are integer values," easier to round off before the actual computation of the area). So let's use $12$ as the length of $AC$ .

Using Heron's (or Hero's) formula,

$s = \frac{AB+BC+AC}{2} = \frac{5+13+12}{2} = \frac{30}{2} = 15$

$A_\triangle = \sqrt{(15)(15-5)(15-13)(15-12)}$

$A_\triangle = \sqrt{(15)(10)(2)(3)}$

$A_\triangle = \sqrt{900}$

$A_\triangle = \boxed{30}$

Again, I'm pretty sure there's a better solution, but here's what I did to solve the problem.

Because the AM= MB and M is the midpoint of BC, so AM=MB=MC= 6.5 _ _(1)

Using the rule that stats that (If the length of a line which is drawn from an angle in the triangle to the midpoint of the opposite side is equal to the half length of that side, then the angle from which the line is drawn equals 90 degree. And Using Pythagorean : 13^{2} = 5^{2} + AC^{2} So AC=12. So [ABC]= 1/2 \times 12 \times 5 =30

Since, M is midpoint of BC, BM=MC=6.5, So, BC=12, Now since AM=MB=MC, Angle A=B+C, & since the sum of angles of atriangle is 180 degree, Angle A=90, Now, Using Pythagoras Theorem, we get, AC=12, So area of the Triangle is (1/2) 12 5=30.

Note that $|AM| = |MB|$ , hence $A$ lies on circle with center $M$ and radius $MB$ . Thus by Thales' theorem $\triangle ABC$ is right and since $BC = 2 \cdot 6.5 = 13$ we easily get from Pythagorean theorem $CA = \sqrt{13^2 - 5^2} = 12$ Hence the sought area is $[ABC] = \frac{5 \cdot 12}{2} = \boxed{30}$

Given AM=BM=6.5 So MC=6.5 (Since M is midpoint of BC) So angle(ABC)=angle(BAM) And, angle(BCA)=angle(MAC) Applying angle sum property in Triangle ABC, we have, angle(BAC) + angle(BCA) + angle(ABC) = 180 => angle(BAM) + angle(CAM) + angle(BCA) + angle(ABC) = 180 => angle(BAM) + angle(CAM) + angle(BAM) + angle(CAM) = 180 => 2( angle(BAM) + angle(CAM)) = 180 => angle(BAM) + angle(CAM) = 90 => angle(BAC) = 90 Applying pythagoras theorem in triangle ABC, AC = \sqrt{BC^2 - AB^2} = \sqrt{(2xBM)^2 - AB^2} = \sqrt{13^2 - 5^2} = 12 Ar(ABC) = 1/2 x 12 x 5 = \boxed{30}

Use Appolonia's theorem

Then 169/2+2(6.5)^2=25+(AC)^2 (AC)^2=144 (AC)=12

Then the triangle is a right angled triangle Ar(Triangle ABC)=(1/2)(12)(5)=30