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Question Number 36328 by Rio Mike last updated on 31/May/18

find the perimeter of a right angle   triangle 4xcm high and 5xcm base

$$\mathrm{find}\:\mathrm{the}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{a}\:\mathrm{right}\:\mathrm{angle}\: \\ $$$$\mathrm{triangle}\:\mathrm{4}{xcm}\:{high}\:{and}\:\mathrm{5}{xcm}\:{base} \\ $$

Commented by Rasheed.Sindhi last updated on 02/Jun/18

Hypotenuse=(√((5x)^2 +(4x)^2 ))=(√(25x^2 +16x^2 ))=(√(41)) x  Perimeter=5x+4x+(√(41)) x= 15.40x

$$\mathrm{Hypotenuse}=\sqrt{\left(\mathrm{5x}\right)^{\mathrm{2}} +\left(\mathrm{4x}\right)^{\mathrm{2}} }=\sqrt{\mathrm{25x}^{\mathrm{2}} +\mathrm{16x}^{\mathrm{2}} }=\sqrt{\mathrm{41}}\:\mathrm{x} \\ $$$$\mathrm{Perimeter}=\mathrm{5x}+\mathrm{4x}+\sqrt{\mathrm{41}}\:\mathrm{x}=\:\mathrm{15}.\mathrm{40x} \\ $$

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