# GMAT Geometry Problems with Solutions and Explanations

Sample 2

Solutions with detailed explanations to problems in sample 2.
## Solution to Question 1Let y be the vertical angle to x. Since L1 and L2 are parallel, angle y and the angle with measure 130° are supplementary. Hence. y + 130 = 180 or y = 180 - 130 = 50° x and y are vertical angles and therefore have equal measures x = 50° ## Solution to Question 2The 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 3The area A of the given triangle may be calculated using the two altitutdes as followsA = (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 4Since 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 followsA + 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 5The 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 byA = (1/2) x * x = (1/2)x ^{2}We now use Pythagorean theorem to find x as follows x ^{2} + x^{2} = 24^{2}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 6Let 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 7Let x be the side of the square. The area is equal to x^{2}. Hencex ^{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} + x^{2} = d^{2}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 8Any 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 9Let L and W be the length and width of the rectangle. Using the formula of the perimeter, we write2L + 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 10The area of a square of side x is x^{2} and the area of a rectangle of width 2x and length 3x is (2 x)(3 x) = 6 x^{2}. The ratio of the area of the square to the area of a rectangle is given byx ^{2} / (6 x^{2}) = 1 / 6
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