In-a-triangle-whose-sides-measure-5-cm-6-cm-and-7-cm-a-point-P-inside-the-triangle-is-2-cm-from-de-long-side-5-cm-and-3-cm-from-the-long-side-6-cm-What-is-the-distance-from-P-beside-7-cm-side-

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Question Number 76455 by Maclaurin Stickker last updated on 27/Dec/19

$$Inatrianglewhosesidesmeasure$$$$5cm,6cm,and7cmapointP,inside$$$$thetriangleis2cmfromdelongside$$$$5cmand3cmfromthelongside$$$$6cm.WhatisthedistancefromP$$$$beside7cmside?$$

Commented by MJS last updated on 27/Dec/19

$$\mathrm{we}\mathrm{can}\mathrm{construct}\mathrm{it}\Rightarrow \mathrm{we}\mathrm{can}\mathrm{calculate}\mathrm{it}$$$$$$$$\left(1\right)\mathrm{draw}\mathrm{the}\mathrm{triangle}$$$$\left(2\right)\mathrm{draw}\mathrm{parallels}\mathrm{to}\mathrm{the}\mathrm{sides}\Rightarrow P$$$$\left(3\right)\mathrm{measure}\mathrm{the}\mathrm{distance}\mathrm{between}P\mathrm{and}$$$$\mathrm{the}{3}^{\mathrm{rd}}\mathrm{side}$$

Commented by Maclaurin Stickker last updated on 27/Dec/19

$$thanks$$

Answered by mr W last updated on 27/Dec/19

$$area=\frac{1}{2}(a\times d_a+b\times d_b+c\times d_c)$$$$area=\sqrt{s(s-a)(s-b)(s-c)}$$$$withs=\frac{a+b+c}{2}=\frac{5+6+7}{2}=9$$$$\frac{1}{2}(5\times 2+6\times 3+7\times d_c)=\sqrt{9(9-5)(9-6)(9-7)}$$$$\Rightarrow d_c=\frac{12\sqrt{6}}{7}-4\approx 0.2$$

Commented by mr W last updated on 27/Dec/19

$$youevenmustsay!$$

Commented by Maclaurin Stickker last updated on 27/Dec/19

$$howdidyoufindthefirstexpression?$$

Commented by mr W last updated on 27/Dec/19

Commented by mr W last updated on 27/Dec/19

$$\mathrm{\Delta }ABC=\mathrm{\Delta }BCP+\mathrm{\Delta }ACP+\mathrm{\Delta }BAP$$$$=\frac{1}{2}{ad}_a+\frac{1}{2}{bd}_b+\frac{1}{2}{cd}_c$$$$=\frac{1}{2}({ad}_a+{bd}_b+{cd}_c)$$

Commented by Maclaurin Stickker last updated on 27/Dec/19

$$CanIjustsaythatthedistancesare$$$$perpendicular?$$

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