GRE Math: Two More Example Geometry Problems and Solutions

Example Geometry Problem #3

Example Geometry Problem #3

The problem states that the triangle depicted is an isosceles triangle. Remember, there are two qualities that define an isosceles triangle:

1) two identical sides
2) two identical angles

If you have one, you will automatically have the other. The other important fact about isosceles triangles that you need to remember for this question is that the sides that are identical are the two sides that are opposite the two angles that are the same (and vice-versa).

OK, so how does this information help us solve the problem? Well, we know that the two bottom angles are equal, because they are both labeled y. This means that the two sides of the triangle (not the bottom side) are equal because they are opposite the two identical angles. Since they’re equal, we can say

2x – 2 = 3x – 8 [subtract 2x from both sides]
-2 = x – 8 [add 8 to both sides]
6 = x

Now we know that x is 6.

The question is asking about the area of the triangle, which is given by the equation

Atriangle = ½bh

where b is the length of the base, and h is the height. From the diagram, we can see that the height is simply x, which we know to be 6. The base is defined to be

3x – 2

so we simply substitute 6 for x, and evaluate.

3x – 2 = 3(6) – 2
3x – 2 = 16

So the base of the triangle is 16, and the height is 6; thus, the area is

Atriangle = ½(16)(6) = 48

The answer is (C).

Example Geometry Problem #4

Example Geometry Problem #4

The area of a circle is given by the equation

Acircle = πr2

So the area of a semicircle is half of this

Asemicircle = ½πr2

The areas of the three semicircles are given in the problem:

A1 = 8π = ½πr12
A2 = 25π/2 = ½πr22
A3 = 18π = ½πr32

So if we solve for the radius in each of these cases, we get

8π = ½πr12
25π/2 = ½πr22
18π = ½πr32
[multiply all three equations by 2 on both sides]

16π = πr12
25π = πr22
36π = πr32
[divide all three equations by π on both sides]

16 = r12
25 = r22
36 = r32
[take the square root of both sides on all three equations]

4 = r1
5 = r2
6 = r3

Now remember, these are the radii, not the diameters! In order to find the perimeter of the triangle, which is what the question is asking for, we need to multiply each of these by 2 to get the diameters of the semicircles (which are also the sides of the triangle). Thus, the perimeter is

Ptriangle = 2(4) + 2(5) + 2(6) = 30

The answer is (D).

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4 Responses to GRE Math: Two More Example Geometry Problems and Solutions

  1. yannah kliner says:

    GEOMETRY is so hard . my gosh .

  2. yannah kliner says:

    sabrina michell beale you know YANNAH KLINER geometry is not hard if you have brain . duh .

  3. steve haverberg says:

    An isosceles triangle, base length equal 1, base angles equal 30 degrees. What is the length of the two sides?

    • Calvin says:

      Dear Steve,

      This may be solved either with trigonometry or by recognizing that half of the isosceles will form a 30-60-90 triangle. Either way, first you divide the triangle in half by drawing a line from the midpoint of the side with length 1 to the opposite vertex. You may then either calculate the length of the hypotenuse of that triangle like this:

      cos(x) = adjacent/hypotenuse
      cos(30) = .5/hypotenuse
      hypotenuse = .5/cos(30)

      Or like this:

      The relationship between the sides of a 30:60:90 triangle is a:a*3^(1/2):2a, where a is the length of the shorter leg, a*3^(1/2) is the length of the longer leg, and 2a is the length of the hypotenuse. We know the longer leg is .5, so:

      .5 = a*3^(1/2)
      a = .5/(3^(1/2))
      2a = 1/(3^(1/2))

      So, either way it comes out to radical 3 over 3 (r3/3), which is length of each of the two equal sides of the original isosceles triangle.


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