Math 3118 Practice Exam 2 Solutions

Solutions to practice exam 2-1

1. The area of the parallelogram is A=b*h=6*18=108. The area of the circle is A=\pi*r^2=\pi*3^2=9*pi. So the answer is 108-9*\pi.

2. We need b=2*r*s=48, where r and s are relatively prime, one odd one even, r 3. See the textbook how to change the area of the parallelogram into the area of a rectangle.

4. TRUE. If the hypotenuse has lenght z, then x^2+y^2=z^2. So (2x)^2+(2y)^2=(2z)^2, then new hypotenuse has length 2z.

5. We must find the height h. Drop a perpendicular to a base, it cuts off a 45-45-90 right triangle whose hypotenuse has length s. So each leg has length s/\sqrt(2)=h, thus A=b*h=s*s/\sqrt(2).

6. The diagonal of the square has length 2, so the Pythagorean theorem says each side of the square has length \sqrt(2). So the square has area 2. Since the square lies inside the circle, its area must be less than \pi, so 2<\pi.

Solutions to Practice Exam 2-2

1. If b=24=2*r*s, then r*s=12, but 12=1*12=2*6=3*4, so the only choices for relatively prime r and s, one odd and one even are
r=1 s=12, so a=s^2-r^2=143, b=24, c=s^2+r^2=145
r=3 s=4 so a=4^2-3^2=7, c=4^2+3^3=25, b=24.

2. FALSE. Try (a,b,c)=(3,4,5) and (a',b',c')=(5,12,13). Then (8,16,18) does not work because 8^2+16^2=64+256=320 while 18^2=324 is not equal to 320.

3. (a) Let's start at a vertex of the triangle and move a distance b to another vertex. Next we move distance c from this vertex to find the third vertex. Then form our original vertex we must be at a distance (which is a) of less than b+c.

(b) FALSE. Just think what happens as a gets closer and closer to b+c. The triangle gets thinner and thinner, and its area is getting close to 0. (It would be zero for c=a+b from the Heron formula.) So the area could not keep on increasing.

4. Take a regular hexagon inscribed in a circle of radius 1. The circle has a circumference 2*pi. The hexagon has a perimeter of 6. Since the shortest path between two points is a straight line between them, we must have 6 is less than 2*\pi, or \pi is greater than 3.

Solutions to Practice Exam 2-3

1. The trapezoid has area 2*4=8. We have equilateral triangles of sides 2\sqrt(2) and 6. The area of an equilateral triangle of side s is s^2*\sqrt(3)/4. (We did this in class.) So the answer is 8+6^2*\sqrt(3)/4+(2*\sqrt(2))^2*\sqrt(3)/4.

2. TRUE. We proved that c is always odd in a primitive Pythagorean triple, and c=108 is even.

3. 39^2=(40-1)^2=1600-80+1=1521, 80^2=6400, and 89^2=(90-1)^2=8100-180+1=7921, and 1521+6400=7921. It is primitive because 39=3*13 (primes), and 3 and 13 do not divide 80.

4. A circle of radius 2 has area 4*\pi. A square of side \sqrt(12) has area 12. But 4*pi is greater than 12 since \pi is greater than 3. So the circle has larger area.

5. A regular hexagon consists of 6 equilateral triangles, each of side length s. The area of an equilateral triangle is s^2*\sqrt(3)/4, so six of them have area 6*s^2*\sqrt(3)/4.