+586 In the diagram below, points A, B, C, and D are situated so that PA=2, PB=3, PC=4, and BC=5. What is the maximum possible area of \(\triangle ABC\)?
+26388 In the diagram below, points A, B, C, and P are situated so that PA=2, PB=3, PC=4, and BC=5.
What is the maximum possible area of \(\triangle ABC\)?
\(\text{Let $\triangle BPC = 90^\circ$ } \\ \text{Let $\triangle BPA = \alpha$ } \\ \text{Let $\triangle CPA = 270^\circ-\alpha$ } \\ \text{Let $ A_1 = \text{area }\triangle BPA$ } \\ \text{Let $ A_2 = \text{area }\triangle CPA$ } \\ \text{Let $ A_3 = \text{area }\triangle BPC=\dfrac{3\cdot 4}{2}= 6 $ } \\ \)
\(\begin{array}{|lrcll|} \hline & A_1 &=& \dfrac{2\cdot 3 \cdot \sin(\alpha)}{2} \\ (1) & A_1 &=& 3 \cdot \sin(\alpha) \\\\ & A_2 &=& \dfrac{2\cdot 4 \cdot \sin(270^\circ-\alpha)}{2} \\ & A_2 &=& 4 \cdot \sin(270^\circ-\alpha) \quad | \quad \sin(270^\circ-\alpha) = -\cos(\alpha) \\ (2) & A_2 &=& -4 \cdot \cos(\alpha) \\ \hline & f(\alpha) &=& A_1+A_2 \\ & &=& 3 \sin(\alpha)-4 \cos(\alpha) \\ & f'(\alpha) &=& 3 \cos(\alpha)+4 \sin(\alpha) = 0 \\ & 3 \cos(\alpha)+4 \sin(\alpha)& =& 0 \\ & 4 \sin(\alpha)& =& -3 \cos(\alpha) \\ & \tan(\alpha)& =& -\dfrac{3}{4} \\ & \alpha &=& \arctan \left(-\dfrac{3}{4} \right) +180^\circ\\ & \alpha &=& -\arctan (\dfrac{3}{4}) +180^\circ\\ & \alpha &=& -36.8698976458 +180^\circ\\ &\mathbf{ \alpha } &=& \mathbf{143.130102354^\circ \quad | \quad \text{area }(A_1+A_2)_{\text{max.}} }\\ \hline \end{array}\)
Area of \(\triangle ABC_{\text{max.}}\)
\(\begin{array}{|rcll|} \hline A_{\text{max.}} &=& A_1+A_2+A_3 \\ &=& 3 \sin(\alpha)-4 \cos(\alpha) + \dfrac{3\cdot 4}{2} \\ &=& 3 \sin(143.130102354^\circ)-4 \cos(143.130102354^\circ) + \dfrac{3\cdot 4}{2} \\ &=& 3\cdot (0.6) -4\cdot (-0.8) + 6 \\ &=& 1.8 + 3.2 + 6 \\ &=& 5 + 6 \\ \mathbf{A_{\text{max.}}} &=& \mathbf{ 11 } \\ \hline \end{array}\)
The maximum possible area of \(\triangle ABC =11\)
The maximum area of ABC is when it is equilateral, which leads to an answer of 16 sqrt(3).
Using Law of Cosines:
Side AB =4.635
Side AC =5.596
Side BC =5.00
Solving for SSS triangle and using Heron's formula:
Area of ABC = 10.95 sq. units.
+26388 In the diagram below, points A, B, C, and P are situated so that PA=2, PB=3, PC=4, and BC=5.
What is the maximum possible area of \(\triangle ABC\)?
\(\text{Let $\triangle BPC = 90^\circ$ } \\ \text{Let $\triangle BPA = \alpha$ } \\ \text{Let $\triangle CPA = 270^\circ-\alpha$ } \\ \text{Let $ A_1 = \text{area }\triangle BPA$ } \\ \text{Let $ A_2 = \text{area }\triangle CPA$ } \\ \text{Let $ A_3 = \text{area }\triangle BPC=\dfrac{3\cdot 4}{2}= 6 $ } \\ \)
\(\begin{array}{|lrcll|} \hline & A_1 &=& \dfrac{2\cdot 3 \cdot \sin(\alpha)}{2} \\ (1) & A_1 &=& 3 \cdot \sin(\alpha) \\\\ & A_2 &=& \dfrac{2\cdot 4 \cdot \sin(270^\circ-\alpha)}{2} \\ & A_2 &=& 4 \cdot \sin(270^\circ-\alpha) \quad | \quad \sin(270^\circ-\alpha) = -\cos(\alpha) \\ (2) & A_2 &=& -4 \cdot \cos(\alpha) \\ \hline & f(\alpha) &=& A_1+A_2 \\ & &=& 3 \sin(\alpha)-4 \cos(\alpha) \\ & f'(\alpha) &=& 3 \cos(\alpha)+4 \sin(\alpha) = 0 \\ & 3 \cos(\alpha)+4 \sin(\alpha)& =& 0 \\ & 4 \sin(\alpha)& =& -3 \cos(\alpha) \\ & \tan(\alpha)& =& -\dfrac{3}{4} \\ & \alpha &=& \arctan \left(-\dfrac{3}{4} \right) +180^\circ\\ & \alpha &=& -\arctan (\dfrac{3}{4}) +180^\circ\\ & \alpha &=& -36.8698976458 +180^\circ\\ &\mathbf{ \alpha } &=& \mathbf{143.130102354^\circ \quad | \quad \text{area }(A_1+A_2)_{\text{max.}} }\\ \hline \end{array}\)
Area of \(\triangle ABC_{\text{max.}}\)
\(\begin{array}{|rcll|} \hline A_{\text{max.}} &=& A_1+A_2+A_3 \\ &=& 3 \sin(\alpha)-4 \cos(\alpha) + \dfrac{3\cdot 4}{2} \\ &=& 3 \sin(143.130102354^\circ)-4 \cos(143.130102354^\circ) + \dfrac{3\cdot 4}{2} \\ &=& 3\cdot (0.6) -4\cdot (-0.8) + 6 \\ &=& 1.8 + 3.2 + 6 \\ &=& 5 + 6 \\ \mathbf{A_{\text{max.}}} &=& \mathbf{ 11 } \\ \hline \end{array}\)
The maximum possible area of \(\triangle ABC =11\)
