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TEN CLASSIC GEOMETRY PROBLEMS -- Solutions
1.
When a transversal crosses parallel lines, the four acute angles formed are all equal, the four obtuse
angles formed are all equal, and any angles that are not equal are supplementary. The angle marked
x is obtuse, so it’s supplementary to the given 40o angle. 180 – 40 = 140. The answer is E.
2. The three angles of any triangle add up to 180o. Furthermore, the two angles opposite the equal
sides of an isosceles triangle are equal. The given angle measures 50o, so the other two angles split
the remaining 130o. Since those other two angles are equal, they each measure 65o. The angle
marked x is adjacent and supplementary to a 65o angle, so x = 180 – 65 = 115. The answer is J.
3. The formula for the area of a triangle is A = ½ bh. To apply this formula, you need the base and the
height. Here you can use CD for the base and AB for the height. So,
Area = ½ * CD * AB = ½ * 5 * 4 = 10.
The answer is A.
4. In similar triangles, corresponding sides are proportional. DE corresponds to AB, and DF
corresponds to AC, so we can set up this proportion:
AB = AC => 6 = 8 => 6x = 24 => x = 4
DE DF
3 x
The answer is H.
5. The Pythagorean Theorem says A2 + B2 = C2. Here, the legs are lengths 1 and 2, so just plug them in:
(1)2 + (2)2 = (Hypotenuse)2 => 1 + 4 = x2 => x2 = 5 => x = √5
6. The indicated angles tell you that triangle ABC is a 45-45-90 triangle and that triangle ACD is a 3060-90 triangle. AD is the hypotenuse and AC is the shorter leg of the 30-60-90 triangle, so AC = ½
AD = 3. AC is also the shorter leg of the 45-45-90 triangle, and AB—the side we’re looking for, is the
hypotenuse. Therefore, AB = AC√2 = 3√2
7. The formula for the area of a trapezoid is A = b1 + b2 *h
2
Where b1 and b2 are the lengths of the parallel sides. You could think of it as the height times the
average of the bases. You’re given the height (3 inches), one base (8 inches) and enough information
to figure out the other base. Notice that triangle ABE is a 3-4-5 triangle, so AE = 4 inches. And if you
were to drop an altitude down from point C, you’d get another 3-4-5 triangle on the right. From
there you can see the base is 16 inches, then plug the numbers into the formula:
8 + 16 * 3 => 12 x 3 = 36
2
The answer is E.
8. The formula for the area of a circle is πr2, where r is the radius. If the diameter is 8 inches, then the
radius is 4 inches, which we plug into the formula:
π(4)2 => 16π
The answer is H.
9. The central angle of minor arc AB is 40o, which is 1/9 of the whole circle’s 360o. The length of minor
arc AB, therefore, is 1/9 of the whole circle’s circumference:
C = 2πr = 2π9 = 18π 1/9 * 18π = 2π
You could also solve using the formula for the length of an arc: 2πr * angle
360
The answer is B.
10. The shaded region is equal to the area of the square minus the area of the semi-circle. The area of the
square is 10 x 10 = 100. The radius of the circle is half of 10, or 5, so the area of the whole circle is
π(5)2 = 25π, and the area of the semicircle is 25π. The square minus the circle is:
2
100 - 25π
2
The answer is H.