GMAT Geometry Solutions and Explanations
Sample 2

Below are detailed solutions to the geometry problems in GMAT Geometry Practice – Sample 2. Each solution follows GMAT-style reasoning.

Solution to Question 1

Let \(y\) be the vertical angle opposite \(x\). Since lines \(L_1\) and \(L_2\) are parallel, angle \(y\) and the \(130^\circ\) angle are supplementary:

\[ y + 130 = 180 \] \[ y = 50^\circ \]

Vertical angles are equal, therefore:

\[ x = 50^\circ \]

Parallel lines angle solution

Solution to Question 2

The sum of the angles in a triangle is \(180^\circ\):

\[ (4x + 10) + (4x + 5) + 3x = 180 \] \[ 11x = 165 \quad \Rightarrow \quad x = 15 \]

Solution to Question 3

The area of triangle \(ABC\) can be expressed using either altitude:

\[ \frac{1}{2}(AN)(BC) = \frac{1}{2}(BM)(AC) \]

Substitute the given values:

\[ \frac{1}{2}(8)(10) = \frac{1}{2}(BM)(12) \] \[ 80 = 12BM \] \[ BM = \frac{80}{12} \approx 6.7 \]

Solution to Question 4

Since \(BA = BC\), triangle \(ABC\) is isosceles and angles at \(A\) and \(C\) are equal.

\[ A + C + 40 = 180 \] \[ 2A = 140 \quad \Rightarrow \quad A = 70^\circ \]

Angle \(x\) is supplementary to angle \(A\):

\[ x + 70 = 180 \quad \Rightarrow \quad x = 110^\circ \]

Solution to Question 5

Let each leg of the right isosceles triangle be \(x\).

\[ x^2 + x^2 = 24^2 \] \[ 2x^2 = 576 \quad \Rightarrow \quad x^2 = 288 \]

The area is:

\[ A = \frac{1}{2}x^2 = \frac{1}{2}(288) = 144 \]

Solution to Question 6

The total area is the sum of the areas of two triangles sharing base \(AC\):

\[ \frac{1}{2}(20)(AC) + \frac{1}{2}(28)(AC) = 240 \] \[ 48AC = 480 \quad \Rightarrow \quad AC = 10 \]

Solution to Question 7

Let the side of the square be \(x\).

\[ x^2 = 144 \quad \Rightarrow \quad x = 12 \]

The diagonal \(d \) of the square equals the diameter of the circle:

\[ x^2 + x^2 = d^2 \Rightarrow d = 12\sqrt{2} \]

Radius:

\[ r = 6\sqrt{2} \]

Area of the circle:

\[ A = \pi r^2 = \pi(6\sqrt{2})^2 = 72\pi \]

Solution to Question 8

Consecutive angles in a parallelogram are supplementary:

\[ (2x + 10) + (x + 20) = 180 \] \[ 3x = 150 \quad \Rightarrow \quad x = 50 \] \[ M = 110^\circ, \quad N = 70^\circ \]

Solution to Question 9

Using the perimeter formula:

\[ 2L + 2W = 90 \] \[ 2L + 20 = 90 \Rightarrow L = 35 \]

Area:

\[ A = LW = 35 \times 10 = 350 \]

Solution to Question 10

Square area: \(x^2\) Rectangle area: \((2x)(3x) = 6x^2\)

\[ \text{Ratio} = \frac{x^2}{6x^2} = \frac{1}{6} \]

GMAT Geometry Solutions and Explanations Sample 2