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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
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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,
$$s=\frac{a+b+c}{2}$$.

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

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

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

heureka  Jul 27, 2017

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