the perimeter will then just be the sum of the distances of A, B and C, namely AB + BC + CA.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\qquadB(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\AB=\sqrt{[0-(-2)]^2+[5-(-2)]^2}\implies AB=\sqrt{(0+2)^2+(5+2)^2}\\\\\\AB=\sqrt{4+49}\implies \boxed{AB=\sqrt{53}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\B(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad C(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\\\\\\BC=\sqrt{(3-0)^2+(1-5)^2}\implies BC=\sqrt{3^2+(-4)^2}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%5C%5CA%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5CqquadB%28%5Cstackrel%7Bx_2%7D%7B0%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%5Cqquad%20%5Cqquadd%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%5C%5C%5C%5C%5C%5CAB%3D%5Csqrt%7B%5B0-%28-2%29%5D%5E2%2B%5B5-%28-2%29%5D%5E2%7D%5Cimplies%20AB%3D%5Csqrt%7B%280%2B2%29%5E2%2B%285%2B2%29%5E2%7D%5C%5C%5C%5C%5C%5CAB%3D%5Csqrt%7B4%2B49%7D%5Cimplies%20%5Cboxed%7BAB%3D%5Csqrt%7B53%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5CB%28%5Cstackrel%7Bx_2%7D%7B0%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%5Cqquad%20C%28%5Cstackrel%7Bx_1%7D%7B3%7D~%2C~%5Cstackrel%7By_1%7D%7B1%7D%29%5C%5C%5C%5C%5C%5CBC%3D%5Csqrt%7B%283-0%29%5E2%2B%281-5%29%5E2%7D%5Cimplies%20BC%3D%5Csqrt%7B3%5E2%2B%28-4%29%5E2%7D)
![\bf BC=\sqrt{9+16}\implies \boxed{BC=5}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\C(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\\\\\\CA=\sqrt{(-2-3)^2+(-2-1)^2}\implies CA=\sqrt{(-5)^2+(-3)^2}\\\\\\CA=\sqrt{25+9}\implies \boxed{CA=\sqrt{34}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\~\hfill \stackrel{AB+BC+CA}{\approx 18.11}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20BC%3D%5Csqrt%7B9%2B16%7D%5Cimplies%20%5Cboxed%7BBC%3D5%7D%5C%5C%5C%5C%5B-0.35em%5D%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5CC%28%5Cstackrel%7Bx_2%7D%7B3%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%5Cqquad%20A%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5C%5C%5C%5C%5C%5CCA%3D%5Csqrt%7B%28-2-3%29%5E2%2B%28-2-1%29%5E2%7D%5Cimplies%20CA%3D%5Csqrt%7B%28-5%29%5E2%2B%28-3%29%5E2%7D%5C%5C%5C%5C%5C%5CCA%3D%5Csqrt%7B25%2B9%7D%5Cimplies%20%5Cboxed%7BCA%3D%5Csqrt%7B34%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C~%5Chfill%20%5Cstackrel%7BAB%2BBC%2BCA%7D%7B%5Capprox%2018.11%7D~%5Chfill)
Answer:
The answer is $19.24
Step-by-step explanation:
$18.50/100
0.185 = 1%
0.185 x 4 = 0.74
18.50 + 0.74 = $19.24
Answer:
Answer:
Length of leg of triangle =
Step-by-step explanation:
We are given a 45°-45°-90° triangle with hypotenuse measuring 18 cm.
Let us assume the following labeling of the diagram, as attached in the answer area.
The two angles here are equal to 45° so two sides of the triangle opposite to the equal angles will also be equal to each other i.e. it is an isosceles triangle.
i.e. The sides XY and YZ will be equal.
Let XY = YZ = x cm
It is known that in a right angled triangle, pythagorean theorem holds well.
As per pythagoream theorem:
Putting the values:
So, the length of leg of triangle =
Step-by-step explanation:
It becomes a liquid
hope this helped
