Problem 2 In the above diagram, circles O and O' are tangent at X, and PQ is tangent to both circles. Given that
OX= 3 and O'X = 8. find PQ.

√96

Step-by-step explanation:

PQ is tangent to both lines, so PQ is perpendicular to PO and QO'.

The radius of the smaller circle is 3, and the radius of the larger circle is 8.

If we draw a line from O to O', and another line from point O to line QO' that is parallel to PQ, we get a right triangle where OO' is the hypotenuse, the short leg is 8−3=5, and the long leg is the same length as PQ.

Using Pythagorean theorem:

x² + 5² = 11²

x = √96

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

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