Length AB is 17 units
Endpoints: A (-2, -4) and B (-8, 4) Length of AB: 10 units
End points: (-2, -4) and (-8, 4) Length of line AB: 10
a = (-2,3)b = (5,-4)vector AB = b - a = (7,-7)Length of AB = sqrt( 72 + 72) = sqrt(98) = 7*sqrt(2)Midpoint of AB = a + (b-a/2) = (-2,3) + (7/2,-7/2)= (3/2,-1/2)
-1/2 or -0.50
Let the chord be AB and the centre of the circle be C. Draw a radius to points A and B. Draw a line perpendicular to and bisecting the chord at D - this will be another radius which bisects the angle formed by the other two radii. As the height of the segment is 2' then the length CD = 10 - 2 = 8' ∆CAD (& ∆CBD) are right angled triangles with CA, the hypotenuse = 10' and CD = 8' Therefore AD² + 8² = 10² : AD² = 100 - 64 = 36 : AD = √ 36 = 6 AD is half AB and therefore the length of the chord is 2 x 6 = 12'
Endpoints: A (-2, -4) and B (-8, 4) Length of AB: 10 units
End points: (-2, -4) and (-8, 4) Length of line AB: 10
Using Pythagoras Length AB = √((-8 - 2)² + (4 - -4)²) = √(6² + 8²) = √100 = 10 units.
The length of ab can be found by using the Pythagorean theorem. The length of ab is equal to the square root of (0-8)^2 + (0-2)^2 which is equal to the square root of 68. Therefore, the length of ab is equal to 8.24.
Using the distance formula the length of ab is 5 units
AB can be found by using the distance formula, which is the square root of (x2-x1)^2 + (y2-y1)^2. In this case, AB= the square root of (-2-(-8))^2 + (-4-(-4))^2 which AB= the square root of 64 + 0 which AB=8.
sqrt(ab^2 + bc^2)
To find the length of segment AB, we can use the distance formula. Given points A (-1, 3) and B (11, -8), the length of AB is calculated as follows: [ AB = \sqrt{(11 - (-1))^2 + (-8 - 3)^2} = \sqrt{(11 + 1)^2 + (-11)^2} = \sqrt{12^2 + (-11)^2} = \sqrt{144 + 121} = \sqrt{265} \approx 16.28. ] Therefore, the length of AB is approximately 16.28 units.
You take the Square root of A^2 + b^2 So Side A 10 Inches, b=10 Inches a. 10^2 = 100 b. 10^2 = 100 Total 200 Square root of 200 = 14.142......
To find the length of the line segment AB, you can use the distance formula: ( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ). For points A(0, 0) and B(6, 3), the calculation is ( AB = \sqrt{(6 - 0)^2 + (3 - 0)^2} = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} ). Therefore, the length of AB is ( 3\sqrt{5} ).
a = (-2,3)b = (5,-4)vector AB = b - a = (7,-7)Length of AB = sqrt( 72 + 72) = sqrt(98) = 7*sqrt(2)Midpoint of AB = a + (b-a/2) = (-2,3) + (7/2,-7/2)= (3/2,-1/2)
circumference of the circle = 2*pi*10 = 20pi units of measurement length of arc = (120/360)*20pi = 20.944 units (rounded to 3 decimal places)