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# plz help and explain thoroughly helpp!

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1.  Find the area of the triangle below.

2.  The longer leg of a right triangle is three times as long as the shorter leg. The hypotenuse is sqrt(5). What is the area of this triangle?

3. We know BE=CE and AE=2DE. What is AD/BC?

4. Given that AF=4sqrt(3) and FC=5sqrt(3), what is BC?

Dec 3, 2018

#1
+1035
-1

We see the triangle is a right triangle, so $$r^2+(2r)^2=10, r^2+4r^2=10, 5r^2=10, r^2=2, r=\sqrt2, \sqrt2\times 2\sqrt2=4, \frac{4}{2}=2$$

The area of the triangle is 2.

Use the Pythagorean theorem.

$$x^2+9x^2=5, 10x^2=5, x^2=\frac{1}{2}, x=\frac{\sqrt2}{2}$$

Then $$\frac{\sqrt2}{2}\times \frac{3\sqrt2}{2}=\frac{3}{2}, \frac{3}{2}\times \frac{1}{2}=\frac{3}{4}$$

The area of the triangle is 0.75.

$$\Delta ABE, \Delta ECD$$ are similar.

$$\frac{BE}{CD}=2$$, so $$CD=\frac{BE}{2}=\frac{EC}{2}$$.

Using Pythagorean theorem, $$AD=\sqrt5DE$$$$DE=BE\sqrt5/2$$

$$\frac{AD}{BC}=5DE/4DE$$

5/4

#4: I think there is insufficient explanation here.

You are very welcome!

:P

Dec 3, 2018
#2
+107656
+1

3.

Triangle ABE is simlar to Triangle  ECD

AB /AE = EC / ED

AB / (2DE) = EC / DE

AB = 2EC

AB = 2EB

And

AE = 2DE

Therefore.....AD = sqrt ( AE^2 + DE^2) = sqrt [ (2DE)^2 + DE^2) = sqrt(5)DE

So....AD = sqrt(5) (AE/2) = [ sqrt(5)/2] AE

And

AE = sqrt (AB^2 + BE^2) = sqrt [ (2BE)^2 + BE^2] = sqrt(5)BE

BE = AE/sqrt(5)

2BE = 2AE/sqrt(5) = BC

Therfore

AD /BC  =   [ sqrt(5)/2] AE ] / [2AE/sqrt(5)] =  [ sqrt (5) / 2  ] / [ 2 /sqrt(5) ] =

sqrt(5) * sqrt(5) / [ 2 * 2 ] =

5 /4

Dec 3, 2018
#3
+24133
+12

4.

Given that $$AF=4\sqrt3$$ and $$FC=5\sqrt{3}$$, what is $$BC$$?

$$\text{Let BC=x}\\ \text{Let AB=y}\\ \text{Let ED=z}\\ \text{Let BD=H}\\ \text{Let EF=h}\\ \text{Let AF=4\sqrt3} \\ \text{Let FC=5\sqrt{3}}\\ \text{Let AC=AF+FC=9\sqrt{3}}$$

$$\begin{array}{|lrcll|} \hline & \cos{(\alpha)} = \dfrac{H}{x} &=& \dfrac{z}{H} \\ (1) & \mathbf{H^2} & \mathbf{=} & \mathbf{xz} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline & 2A = H\cdot AC &=& xy \\ (2) & \mathbf{H} & \mathbf{=} & \mathbf{\dfrac{xy}{AC}} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline & x^2+y^2 &=& (AC)^2 \\ (3) & \mathbf{y^2} & \mathbf{=} & \mathbf{(AC)^2-x^2} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline & \cos{(\alpha)} &=& \dfrac{h}{z} \\ & \tan{(\alpha)} &=& \dfrac{h}{AF} \\ & h = z\cdot \cos{(\alpha)} &=& AF\cdot \tan{(\alpha)} \\ & z &=& AF \cdot \dfrac{\sin{(\alpha)}} {\cos^2{(\alpha)}} \\\\ &&& \sin{(\alpha)} = \dfrac{x}{AC} \\ &&& \cos^2{(\alpha)} = 1-\sin^2{(\alpha)} =\dfrac{(AC)^2-x^2}{(AC)^2} \\\\ (4) & \mathbf{ z } & \mathbf{=} & \mathbf{ AF\cdot \dfrac{AC\cdot x}{(AC)^2-x^2} } \\ \hline \end{array}$$

We substitute in (1):

$$\begin{array}{|rcll|} \hline \mathbf{H^2} & \mathbf{=} & \mathbf{xz} \quad & | \quad \mathbf{H} \mathbf{=} \mathbf{\dfrac{xy}{AC}} \\ \dfrac{x^2y^2}{(AC)^2} &=& xz \quad & | \quad \mathbf{ z } \mathbf{=} \mathbf{ AF\cdot \dfrac{AC\cdot x}{(AC)^2-x^2} } \\ \dfrac{x^2y^2}{(AC)^2} &=& \dfrac{AF\cdot AC\cdot x^2}{(AC)^2-x^2} \\ \dfrac{y^2}{(AC)^2} &=& \dfrac{AF\cdot AC}{(AC)^2-x^2} \quad & | \quad \mathbf{y^2} \mathbf{=} \mathbf{(AC)^2-x^2} \\ \dfrac{(AC)^2-x^2}{(AC)^2} &=& \dfrac{AF\cdot AC}{(AC)^2-x^2} \\ \left((AC)^2-x^2\right)^2 &=& AF\cdot (AC)^3 \quad & | \quad \text{sqrt both sides} \\ (AC)^2-x^2 &=& AC\cdot \sqrt{AF\cdot AC} \\ x^2 &=& (AC)^2 - AC\cdot \sqrt{AF\cdot AC} \\ x &=& \sqrt{ (AC)^2 - AC\cdot \sqrt{AF\cdot AC} } \\ \mathbf{x} &\mathbf{=}& \mathbf{\sqrt{ AC\cdot \left( AC - \sqrt{AF\cdot AC} \right)} } \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{x} &\mathbf{=}& \mathbf{\sqrt{ AC\cdot \left( AC - \sqrt{AF\cdot AC} \right)} } \quad & | \quad AC = 9\sqrt{3} \qquad AF = 4\sqrt{3} \\ x &=& \sqrt{ 9\sqrt{3}\cdot \left(9\sqrt{3} - \sqrt{108} \right)} \\ x &=& \sqrt{ 243- 9\sqrt{3}\cdot \sqrt{108} } \\ x &=& \sqrt{ 243- 9\sqrt{324} } \\ x &=& \sqrt{ 243- 9\cdot 18 } \\ x &=& \sqrt{ 243- 162 } \\ x &=& \sqrt{ 81 } \\ \mathbf{x} &\mathbf{=}& \mathbf{9} \\ \hline \end{array}$$

BC  is 9

Dec 3, 2018
#4
+107656
+2

Whoa!!!......very well done......that was a difficult one!!!

CPhill  Dec 3, 2018
#5
+24133
+14

Thank you, CPhill !

heureka  Dec 4, 2018