At noon three students, Abby, Ben, and Cassie, are standing so that Abby is m west of Ben and Cassie is m east of Ben. While Ben stays in his initial position, Abby begins walking south at a constant rate of m/min and Cassie begins walking north at a constant rate of m/min. In how many minutes will the distance between Cassie and Ben be twice the distance between Abby and Ben?

Let represent the number of minutes until Cassie’s distance to Ben is twice that of Abby’s distance to Ben. In minutes Abby will walk m and Cassie will walk m. The following diagram contains the information showing Abby’s position, , Ben’s position, , and Cassie’s position, , at time .

A right-angled triangle is formed by joining Abby's initial position to points A and B. Its hypotenuse is A B and its other sides are of length 100 and 20 times t. Another right-angled triangle is formed by joining Cassie's initial position to points C and B. Its hypotenuse is B C and its other sides are of length 160 and 41 time. Represent Abby, Ben and Cassie’s respective positions at noon as points on the -axis so that Ben is positioned at the origin , Abby is positioned units left of Ben at and Cassie is positioned units right of Ben at .

solution 2

Let represent the number of minutes until Cassie’s distance to Ben is twice that of Abby’s distance to Ben.

In minutes Abby will walk south m to the point . In minutes Cassie will walk north m to the pointTherefore, in

minutes ( minutes seconds), Cassie’s distance to Ben will be twice that of Abby’s distance to Ben. .

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

50-38=12

step-by-step explanation: