Triangolo noti due lati e mediana terzo lato. Calcola area e perimetro

Nel triangolo PQR il lato PQ=2√2, il lato QR=2 e la mediana RM=√3-1. Calcola l’area e il perimetro del triangolo.

$QM=MP=\frac{1}{2}QP=\frac{2\sqrt2}{2}=\sqrt2$

Considero il triangolo QRM e applico il teorema di Carnout
$RM^2=QR^2+QM^2-2\cdot QR\cdot QM\cdot cos\widehat{Q}$ calcolo COS(Q)

$cos\widehat{Q}=RM^2=QR^2+QM^2-2\cdot QR\cdot QM)$

$cos\widehat{Q}=\frac{QR^2+QM^2-RM^2}{2QR\cdot QM}=$

$\frac{4+2-(\sqrt3-1)^2}{2\cdot 2\cdot \sqrt2}=$

$\frac{6-(3+1-2\sqrt3)}{4\sqrt2}=$

$\frac{6-3-1+2\sqrt3)}{4\sqrt2}=$

$\frac{2+2\sqrt3}{4\sqrt2}=$

$\frac{2(1+\sqrt3)}{4\sqrt2}=$

$\frac{1+\sqrt3}{2\sqrt2}$ razionalizziamo

$\frac{1+\sqrt3}{2\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}=$

$\frac{\sqrt2+\sqrt6}{4}$

$\widehat{Q}=arccos\left(\frac{\sqrt6+\sqrt2}{4}\right)=15^\circ$

Applico Carnout al triangolo PQR
$PR=\sqrt{QR^2+QP^2-2\cdot QR\cdot QP\cdot cos\widehat{Q}}$

$PR=\sqrt{4+8-2\cdot 2\cdot 2\sqrt2\cdot \frac{\sqrt6+\sqrt2}{4}}$

$PR=\sqrt{12-8\sqrt2\cdot \frac{\sqrt6+\sqrt2}{4}}$

$PR=\sqrt{12-2\sqrt2\cdot (\sqrt6+\sqrt2)}$

$PR=\sqrt{12-2\sqrt12-4}$

$PR=\sqrt{8-2\sqrt12}=\sqrt{8-\sqrt48}$

sapendo che:
$\sqrt{a-\sqrt b}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$

avremo che:
$\sqrt{8-\sqrt48}=\sqrt{\frac{8+\sqrt{64-48}}{2}}-\sqrt{\frac{8-\sqrt{64-48}}{2}}$

$\sqrt{\frac{8+\sqrt{16}}{2}}-\sqrt{\frac{8-\sqrt{16}}{2}}$

$\sqrt{\frac{8+4}{2}}-\sqrt{\frac{8-4}{2}}$

$\sqrt{\frac{12}{2}}-\sqrt{\frac{4}{2}}$

$\sqrt{6}-\sqrt{2}$

Calcoliamo l’area:
$\mbox{Area}\;=\frac{1}{2}a\cdot b\cdot sen\gamma$

$\mbox{Area}\;=\frac{1}{2}QR\cdot QP\cdot sen\widehat{Q}$

$\mbox{Area}\;=\frac{1}{2}\cdot 2\cdot 2\sqrt2\cdot sen15^\circ$

$\mbox{Area}\;=\frac{1}{2}\cdot 2\cdot 2\sqrt2\cdot \frac{\sqrt6-\sqrt2}{4}$

$\mbox{Area}\;=\frac{4}{8}\cdot \sqrt2\cdot (\sqrt6-\sqrt2)$

$\mbox{Area}\;=\frac{1}{2}\cdot (\sqrt12-2)$

$\mbox{Area}\;=\frac{1}{2}\cdot (\sqrt{3\cdot2^2}-2)$

$\mbox{Area}\;=\frac{1}{2}\cdot (2\sqrt{3}-2)$

$\mbox{Area}\;=\frac{1}{2}\cdot 2(\sqrt{3}-1)$

$\mbox{Area}\;=\sqrt{3}-1$

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