# GMAT Geometry Problems with Solutions and Explanations Sample 2

 Solutions and detailed explanations to problems in sample 2. What is the measure of angle x if L1 and L2 are parallel lines? . Solution 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 . y + 130 = 180 or y = 180 - 130 = 50° x and y are vertical angles and therefore have equal measures x = 50° Find x. . Solution 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 ABC is a triangle where AN is perpendicular to CB and BM perpendicular to AC. The length of BC is 10, that of AC is 12 and that of AN is 8. Find the length of BM. . Solution 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) The lengths of sides BA and BC of triangle ABC are equal. Find the measure of angle x. . Solution 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° Find the area of a right isosceles triangle with hypotenuse equal to 24. Solution 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)x2 We now use Pythagora's theorem to find x as follows x2 + x2 = 242 Simplify 2 x2 = 576 x2 = 288 We now calculate the area A as follows A = (1/2)x2 = (1/2) 288 = 144 The total area of the shape below is equal to 240. Find the length of segment AC. . Solution 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 The circle in the figure has all its vertices on the circle. The area of the square is 144. What is the area of the circle? . Solution Let x be the side of the square. The area is equal to x2. Hence x2 = 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 x2 + x2 = d2 Solve for d 144 + 144 = d2 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 r2 = 72 Pi The measures, in degrees, of two consecutive interior angles M and N of a parallelogram are given by 2x + 10 and x + 20 respectively. Find the measures of the two angles. Solution 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° The perimeter of a rectangle is 90 and its width is 10. What is its area? Solution 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 What is the ratio of the area of a square of side x to the area of a rectangle of width 2 x and length 3 x? Solution 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 x2 / (6 x2) = 1 / 6 More Math Practice Tests Free Practice for GAMT Math tests Free Compass Math tests Practice Free Practice for SAT, ACT Math tests Free GRE Quantitative for Practice Free AP Calculus Questions (AB and BC) with Answers