Explanation for Question 20 From the Math (No Calc) Section on the 2017 May Sat
Let's read question 20 and then determined how to solve questions. 2 When he says in the figure above the circle has center a and line 3 segment CB is tangent to the circle. 4 A point C if AB is 1.0. 5 So this hypotenuse is 1.0. 6 CB is 0.8. What is the length of the diameter 7 of the circle? So important here that they want under the diameter diameter 8 is any line that goes from one point to another point in 9 the circle and passes through the center. 10 So our best bet here would be to treat this as a triangle that's 11 in scrubbing the circle, um, 12 well, in, in the circle and we can find a C, 13 we can see that AC is the radius. 14 It goes from the center of the circle to a point. 15 So if we find a C, we can then multiply by two to find 16 the diameter because the diameter is always going to be twice the radius. 17 So we have a right triangle, 18 the T we know this right triangle because they tell us that it's tangent 19 to the circle. And we can just do Pythagoras theorem by 20 geography is a squared plus B squared equals C square 21 C squared, always being the hypotenuse. So we can plug in. 22 We can say eight is CB. So 0.8 squared plus 23 B squared equals 1.0 square 0.8 24 squared is going to be point 64, 25 just 0.8 times 0.8 plus B squared equals one. 26 Because again, we just have one times one, we have all the light comes 27 on one side. So we said, chocks points, explorative asides. 28 And we got B squared equals point 36 to 29 get rid of that squared, take the square root, 30 both sides. And we got B equals 0.6. 31 That's the square root of 0.36. Now that's not our final answer because 32 that was going to be our radius, but we weren't our diameter. 33 Again, the diameter is just twice the radius. 34 So we multiply this times two and we get 1.2 35 as our diameter.