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?
+1253 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
+128475 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
+26367 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
