Pythagoras is Always Right

When the original right triangle is cut into two smaller right triangles, both of the small triangles are similar to the original because they each share two angles with the original triangle. Therefore, they are similar with each other and we can set up the ratio $\frac{BD}{AB} = \frac{AD}{AC}$ .

We also know that because $\triangle ABD$ is right, $AD = 6$ . (This can be computed using the Pythagorean Theorem, or recognized because $\triangle ABD$ is similar to a 3-4-5 right triangle.)

Therefore, $\frac{BD}{AB} = \frac{AD}{AC} \rightarrow \frac{8}{10} = \frac{6}{x} \rightarrow x=7.5$

Therefore, the area of the triangle is $\dfrac{\text{base} \times \text{height}}{2} = \dfrac{1}{2} (10)(7.5) = \dfrac{75}{2}$ .

From the properties of a $3-4-5$ right triangle, $AD=6$ .

Apply pythagorean theorem at $\triangle ADC$ .

$DC=\sqrt{(AC)^2-36}$

Since $\triangle ABC \sim \triangle ADC$ , then $\dfrac{10}{AD}=\dfrac{BC}{AC}$ $\large \implies$ $\dfrac{10}{6}=\dfrac{8+DC}{AC}$ $\large \implies$ $\dfrac{5}{3}=\dfrac{8+DC}{AC}$

$\large \implies$ $5AC=24+3DC$ $\large \implies$ $5AC=24+3\sqrt{(AC)^2-36}$ $\large \implies$ $5AC-24=3\sqrt{(AC)^2-36}$

Squaring both sides, we obtain

$(5AC-24)^2=(3\sqrt{AC^2-36})^2$ $\large \implies$ $16AC^2-240AC+900=0$

Using the quadratic formula to solve for $AC$ , we get

$AC=7.5$

Finally, the area of $\triangle ABC$ is

$A=\dfrac{1}{2}(10)(7.5)=\dfrac{75}{2}$

I've done it in the following way,

• In right triangle ABD by Pythagorean theorum, AD=6 units.

• Let AC be x and DC be y,

Then in right triangle ABC, by Pythagorean theorum

(8+y)^2=10^2 + x^2

simplifying you'll get x^2= -36 + 16y + y^2----------(1)

• Next in right triangle ADC,by Pythagorean theorum

  x^2= 6^2 + y^2= 36 + y^2-------------------------------(2)

• Equating (1) and (2) 72=16y

Therefore y= $\frac{9}{2}$

• Now substitute this in equation (2),

x^2=36+[ $\frac{9}{2}$ ]^2

x^2= $\frac{225}{4}$

So x= $\frac{15}{2}$

• Now find the area by $\frac{1}{2}$ bh

Area(ABC)= $\frac{1}{2}$ * 10 * $\frac{15}{2}$ = $\frac{150}{4}$ = $\frac{75}{2}$

Therefore answer is $\frac{75}{2}$ units

Noticed that $\triangle ABD$ and $\triangle ABC$ are similar triangles.

Therefore, we have

$\dfrac{AC}{AD}=\dfrac{AB}{BD}=\dfrac{BC}{AB}$

Note that $Area=\dfrac{AC\times\ AB}{2}$

So we need to find the value of $AC$

We can use $\dfrac{AC}{AD}=\dfrac{AB}{BD}$ , substitute $AB=10$ , $AD=6$ , and $BD=8$ , then solve for $AC$ you'll get, $AC=\dfrac{15}{2}$

$Area=\dfrac{AC\times\ AB}{2}=\dfrac{15\times\ 10}{4}=\dfrac{75}{2}$

AD=√(100-64)=√36=6 so the ratio of triangle ABD to triangle ACD is 6/8 or 3/4. That means triangle ACD's base =3/4 of triangle ABD's base and the same goes for the height. So triangle ACD's area is 3/4* 3/4=9/16 of that of triangle ABD. The total area of both triangles then is (1+9/16)* 24=(25/16)*24=75/2.

∆ABD and ∆ABC are similar,
So, BC/AB = AB/BD,
BC = 10×10/8 = 12.5,
And AD = 6 by Pythagoras theorem.
So area = (1/2)×AD×BC = (1/2)×6×12.5,
= 37.5

AD^2=10^2-8^2=36 so AD=6. Cos(ABC))=sin(ACB) ∆ABC is right angle. Sin(ACB)=0.8 so AC=6÷0.8=7.5 so area of ∆ABC=0.5×10×7.5×sin90=75÷2#######

By pythagoras theorem, AD=6:Triangle BDA Similar Triangle ADC. So BD / AD = AD / DC; 8/6=6/DC. ; DC=(6x6)/8 = 36/8 = 9/2; Ar(ABC) = Ar(ABD) + Ar(ADC) = (1/2 x 8 x 6) + (1/2 x 9/2 x 6) = 24 + 27/2 = 75/2

Using Euclid's theroem in right triangles you can calculate the hypotenuse with the formula AB^2=BC BD to get the result 12.5. The next step is to use Pythagoras theorem to calculate the side AC, which is 7.5. To calculate the area you then need to use the formula AB AC/2 and get the result 75/2