Find the perimeter of a triangle whose vertices are the points A(-2,1) B(1,3) and C(4,-3)?

Darian Hubbard
2022-07-28
Answered

Find the perimeter of a triangle whose vertices are the points A(-2,1) B(1,3) and C(4,-3)?

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sweetwisdomgw

Answered 2022-07-29
Author has **20** answers

First you must find the length of the 3 lines and then add them

$\text{from A to B}\sqrt{(1-(-2){)}^{2}+(3-1{)}^{2}}=\sqrt{{3}^{2}+{2}^{2}}=\sqrt{9+4}=\sqrt{13}\sim 3.6\phantom{\rule{0ex}{0ex}}\text{from B to C}\sqrt{(4-1{)}^{2}+(-3-3{)}^{2}}=\sqrt{32+(-6{)}^{2}}=\sqrt{9+36}=\sqrt{45}\sim 6.7\phantom{\rule{0ex}{0ex}}\text{from C to A}\sqrt{(4-(-2){)}^{2}+(-3-1{)}^{2}}=\sqrt{{6}^{2}+(-4{)}^{2}}=\sqrt{36+16}=\sqrt{52}\sim 7.21\phantom{\rule{0ex}{0ex}}\text{the perimeter}\sim 3.6+6.7+7.21\sim 17.51$

$\text{from A to B}\sqrt{(1-(-2){)}^{2}+(3-1{)}^{2}}=\sqrt{{3}^{2}+{2}^{2}}=\sqrt{9+4}=\sqrt{13}\sim 3.6\phantom{\rule{0ex}{0ex}}\text{from B to C}\sqrt{(4-1{)}^{2}+(-3-3{)}^{2}}=\sqrt{32+(-6{)}^{2}}=\sqrt{9+36}=\sqrt{45}\sim 6.7\phantom{\rule{0ex}{0ex}}\text{from C to A}\sqrt{(4-(-2){)}^{2}+(-3-1{)}^{2}}=\sqrt{{6}^{2}+(-4{)}^{2}}=\sqrt{36+16}=\sqrt{52}\sim 7.21\phantom{\rule{0ex}{0ex}}\text{the perimeter}\sim 3.6+6.7+7.21\sim 17.51$

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