+0  
 
0
581
1
avatar+62 

Let triangle ABC be a triangle such that AB=13, BC=14, and AC=15. Meanwhile, D is a point on BC such that AD bisects angle A. Find the area of triangle ADC

bbelt711  Jul 26, 2017
 #1
avatar+20153 
+2

Let triangle ABC be a triangle such that AB=13, BC=14, and AC=15. 

Meanwhile, D is a point on BC such that AD bisects angle A.

Find the area of triangle ADC

 

Let area of triangle \(\text{ADC} = A_\text{ADC} \)

Let area of triangle \( \text{ADB} = A_\text{ADB}\)
Let area of triangle \(\text{ABC} = A\)

 

Let \(A_\text{ADB} = A - A_\text{ADC}\)

 

\(\begin{array}{|rcll|} \hline \frac { A_\text{ADB} } { A_\text{ADC} } &=& \frac{ \overline{AD}\cdot 13 \cdot \sin(\frac{A}{2}) \cdot \frac12 } { \overline{AD}\cdot 15 \cdot \sin(\frac{A}{2}) \cdot \frac12 } \\ \frac { A_\text{ADB} } { A_\text{ADC} } &=& \frac{ 13 } { 15 } \quad & | \quad A_\text{ADB} = A - A_\text{ADC} \\ \frac { A - A_\text{ADC} } { A_\text{ADC} } &=& \frac{ 13 } { 15 } \\ \frac { A } { A_\text{ADC} } - 1 &=& \frac{ 13 } { 15 } \\ \frac { A } { A_\text{ADC} } &=& 1+ \frac{ 13 } { 15 } \\ \frac { A } { A_\text{ADC} } &=& \frac{ 28 } { 15 } \\ \frac { A_\text{ADC} } { A } &=& \frac{ 15 } { 28 } \\\\ \mathbf{ A_\text{ADC } } & \mathbf{=} & \mathbf{ \frac{ 15 } { 28 } A } \\ \hline \end{array} \)

 

Heron's formula states that the area of a triangle whose sides have lengths a, b, and c is

\( A = \sqrt{s(s-a)(s-b)(s-c)}\)

 

where s is the semiperimeter of the triangle; that is,
\(s=\frac{a+b+c}{2}\).

 

\(\begin{array}{|rcll|} \hline s &=& \frac{a+b+c}{2} \quad & | \quad a= 14, \ b= 15, \ c= 13 \\ s &=& \frac{15+15+13}{2} \\ s &=& \frac{42}{2} \\ \mathbf{ s }& \mathbf{=} & \mathbf{21} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline A &=& \sqrt{s(s-a)(s-b)(s-c)} \quad & | \quad s= 21, \ a= 14, \ b= 15, \ c= 13 \\ A &=& \sqrt{21(21-14)(21-15)(21-13)} \\ A &=& \sqrt{21\cdot 7 \cdot 6 \cdot 8 } \\ A &=& \sqrt{3\cdot 7 \cdot 7 \cdot 2\cdot 3 \cdot 2\cdot 4 } \\ A &=& \sqrt{4^2\cdot 7^2 \cdot 3^2 } \\ A &=& 4\cdot 7 \cdot 3 \\ \mathbf{ A }& \mathbf{=} & \mathbf{84} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{ A_\text{ADC } } & \mathbf{=} & \mathbf{ \frac{ 15 } { 28 } A } \quad & | \quad \mathbf{ A } \mathbf{=} \mathbf{84} \\ A_\text{ADC } & = & \frac{ 15 } { 28 } \cdot 84 \\ A_\text{ADC } & = & 15 \cdot 3 \\\\ \mathbf{ A_\text{ADC } }& \mathbf{=} & \mathbf{45} \\ \hline \end{array}\)

 

laugh

heureka  Jul 27, 2017

16 Online Users

avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.