GMAT Geometry Problems with Solutions and Explanations
Sample 2

Solutions with detailed explanations to problems in sample 2.

Solution to Question 1

Let y be the vertical angle to x. Since L1 and L2 are parallel, angle y and the angle with measure 130 are supplementary. Hence

gmat problem 1 .


y + 130 = 180 or y = 180 - 130 = 50
x and y are vertical angles and therefore have equal measures
x = 50

Solution to Question 2

The sum of all 3 angles of any triangle is equal to 180 degrees. Hence
(4x + 10) + (4x + 5) + (3x) = 180
Group like terms and solve for x
11x = 165 or x = 15

Solution to Question 3

The area A of the given triangle may be calculated using the two altitutdes as follows
A = (1/2)(AN)(BC) or A = (1/2)(BM)(AC)
Hence
(1/2)(AN)(BC) = (1/2)(BM)(AC)
Multiply both sides by 2 and substitute known lengths
8 * 10 = MB * 12
Multiply both sides by 2 and substitute known lengths
BM = 80 / 12 = 6.7 (approximated to one decimal place)

Solution to Question 4

Since BA and BC have equal lengths, then the triangle is isosceles and the interior angles at A and C have equal measues which may be calculated as follows
A + C + 40 = 180 or 2A = 140 or A = 70
The interior angle at A and angle x are supplementary. Hence
70 + x = 180 or x = 110

Solution to Question 5

The two legs of a right isosceles triangle have equal lengths; let x be one of these lengths. The area A of the triangle is given by
A = (1/2) x * x = (1/2)x
2
We now use Pythagorean theorem to find x as follows
x
2 + x2 = 242
Simplify
2 x
2 = 576
x
2 = 288
We now calculate the area A as follows
A = (1/2)x
2 = (1/2) 288 = 144

Solution to Question 6

Let us express the total area as the area of the upper and lower triangles with commom base AC
(1/2)(20)(AC) + (1/2)(28)(AC) = 240
Multiply all terms by 2, simplify and solve for AC
(20)(AC) + (28)(AC) = 480
48 AC = 480
AC = 480 / 48 = 10

Solution to Question 7

Let x be the side of the square. The area is equal to x2. Hence
x
2 = 144 , solve for x, x = 12
The diameter d of the circle is equal to the length of the diagonal of the square. Using Pythagora's theorem, we obtain
x
2 + x2 = d2
Solve for d
144 + 144 = d
2
d = 12 √2
The radius r of the circle is equal to d/2
r = 12 √2 / 2 = 6 √2
The area A of the circle is given by
A = Pi r
2 = 72 Pi

Solution to Question 8

Any two consecutive angle of a parallelogram are supplementary. Hence
(2x + 10) + (x + 20) = 180
Solve for x
3x + 30 = 180
x = 50
We now evaluate M and N as follows
M = 2x + 10 = 2(50) + 10 = 110
N = x + 20 = 50 + 20 = 70

Solution to Question 9

Let L and W be the length and width of the rectangle. Using the formula of the perimeter, we write
2L + 2W = 90
W is given; hence
2L + 2(10) = 90
Solve for L
2L = 70 , L = 35
The area A of the rectangle is given by
A = L * W = 35 * 10 = 350

Solution to Question 10

The area of a square of side x is x2 and the area of a rectangle of width 2x and length 3x is (2 x)(3 x) = 6 x2. The ratio of the area of the square to the area of a rectangle is given by
x
2 / (6 x2) = 1 / 6

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