Geometry Practice Problems
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Problem 1: Complementary Angles
Angles \( A \) and \( B \) are complementary, and the measure of angle \( A \) is twice the measure of angle \( B \). Find the measures of angles \( A \) and \( B \).
View Step-by-Step Solution
Let \( A \) be the measure of angle \( A \) and \( B \) be the measure of angle \( B \). We are told that:
\[ A = 2B \]
Because angles \( A \) and \( B \) are complementary, their sum is \( 90^\circ \):
\[ A + B = 90^\circ \]
Substitute \( 2B \) for \( A \) in the equation:
\[ 2B + B = 90^\circ \]
\[ 3B = 90^\circ \]
\[ B = \dfrac{90^\circ}{3} = 30^\circ \]
Now, substitute \( B \) back to find \( A \):
\[ A = 2(30^\circ) = 60^\circ \]
Answer: Measure of \( A = 60^\circ \), Measure of \( B = 30^\circ \).
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Problem 2: Parallelogram Area
\( ABCD \) is a parallelogram such that \( AB \) is parallel to \( DC \) and \( DA \) is parallel to \( CB \). The length of side \( AB \) is \( 20 \text{ cm} \). Point \( E \) is between \( A \) and \( B \) such that the length of \( AE \) is \( 3 \text{ cm} \). Point \( F \) is between points \( D \) and \( C \).
Find the length of \( DF \) such that the segment \( EF \) splits the parallelogram into two regions with equal areas.
View Step-by-Step Solution
Let \( A_1 \) be the area of the trapezoid \( AEFD \). Its area is given by:
\[ A_1 = \dfrac{1}{2} h (AE + DF) = \dfrac{1}{2} h (3 + DF) \]
where \( h \) is the height of the parallelogram.
Let \( A_2 \) be the area of the trapezoid \( EBCF \). Its area is given by:
\[ A_2 = \dfrac{1}{2} h (EB + FC) \]
We can find the lengths of the remaining bases because the total length of the parallelogram sides is \( 20 \):
\[ EB = 20 - AE = 20 - 3 = 17 \]
\[ FC = 20 - DF \]
Substituting \( EB \) and \( FC \) into the equation for \( A_2 \):
\[ A_2 = \dfrac{1}{2} h (17 + 20 - DF) = \dfrac{1}{2} h (37 - DF) \]
For segment \( EF \) to divide the parallelogram into two equal-area regions, set \( A_1 = A_2 \):
\[ \dfrac{1}{2} h (3 + DF) = \dfrac{1}{2} h (37 - DF) \]
Multiply both sides by 2 and divide by \( h \):
\[ 3 + DF = 37 - DF \]
\[ 2DF = 34 \]
\[ DF = 17 \text{ cm} \]
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Problem 3: Triangle Interior Angles
Find the measure of angle \( A \) in the figure below.
View Step-by-Step Solution
To find angle \( A \), we must first find the other two interior angles of the triangle using the supplementary exterior angles.
The first interior angle (bottom left) is supplementary to the \( 129^\circ \) angle. A straight line is \( 180^\circ \), so:
\[ 180^\circ - 129^\circ = 51^\circ \]
The second interior angle (bottom right) is supplementary to the \( 138^\circ \) angle:
\[ 180^\circ - 138^\circ = 42^\circ \]
The sum of all three interior angles of a triangle is always \( 180^\circ \). Therefore:
\[ A + 51^\circ + 42^\circ = 180^\circ \]
\[ A = 180^\circ - 51^\circ - 42^\circ = 87^\circ \]
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Problem 4: Angles in a Right Triangle
\( ABC \) is a right triangle. \( AM \) is perpendicular to \( BC \). The size of angle \( ABC \) is equal to \( 55^\circ \). Find the size of angle \( MAC \).
View Step-by-Step Solution
The sum of all angles in the large triangle \( \triangle ABC \) is \( 180^\circ \). Because it is a right triangle at \( A \), we know the angles are \( 90^\circ \), \( 55^\circ \) (given for \( \angle ABC \)), and \( \angle ACB \):
\[ \angle ABC + \angle ACB + 90^\circ = 180^\circ \]
Substitute \( \angle ABC = 55^\circ \) and solve for \( \angle ACB \):
\[ 55^\circ + \angle ACB + 90^\circ = 180^\circ \]
\[ \angle ACB = 180^\circ - 145^\circ = 35^\circ \]
Now look at the smaller triangle \( \triangle AMC \). The line \( AM \) is perpendicular to \( BC \), meaning \( \angle AMC = 90^\circ \). The sum of angles in \( \triangle AMC \) is \( 180^\circ \):
\[ \angle MAC + \angle ACM + 90^\circ = 180^\circ \]
Note that \( \angle ACM \) is the exact same angle as \( \angle ACB \). Substitute \( 35^\circ \) into the equation:
\[ \angle MAC + 35^\circ + 90^\circ = 180^\circ \]
\[ \angle MAC = 180^\circ - 125^\circ = 55^\circ \]
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Problem 5: Vertical Angles
Find the size of angle \( MBD \) in the figure below.
View Step-by-Step Solution
First, find the missing angle in the left triangle, \( \triangle AMC \). The sum of its angles is \( 180^\circ \):
\[ 56^\circ + 78^\circ + \angle AMC = 180^\circ \]
\[ \angle AMC = 180^\circ - 134^\circ = 46^\circ \]
Angles \( \angle AMC \) and \( \angle DMB \) are vertical angles, which means they are equal in measure:
\[ \angle DMB = \angle AMC = 46^\circ \]
Now, use the sum of angles for the right triangle, \( \triangle DMB \), which is also \( 180^\circ \):
\[ \angle MBD + \angle DMB + 62^\circ = 180^\circ \]
Substitute \( \angle DMB = 46^\circ \):
\[ \angle MBD + 46^\circ + 62^\circ = 180^\circ \]
\[ \angle MBD = 180^\circ - 108^\circ = 72^\circ \]
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Problem 6: Intersecting Lines & Angles
The size of angle \( AOB \) is equal to \( 132^\circ \) and the size of angle \( COD \) is equal to \( 141^\circ \). Find the size of angle \( DOB \).
View Step-by-Step Solution
Angle \( \angle AOB = 132^\circ \) and is composed of the sum of angles \( \angle AOD \) and \( \angle DOB \):
\[ \angle AOD + \angle DOB = 132^\circ \quad \text{(Equation 1)} \]
Angle \( \angle COD = 141^\circ \) and is composed of the sum of angles \( \angle COB \) and \( \angle DOB \):
\[ \angle COB + \angle DOB = 141^\circ \quad \text{(Equation 2)} \]
Add Equation 1 and Equation 2 together:
\[ \angle AOD + \angle DOB + \angle COB + \angle DOB = 132^\circ + 141^\circ \]
Because \( A, O, \) and \( C \) form a straight line, the angles \( \angle AOD, \angle DOB, \) and \( \angle COB \) form a straight angle, meaning:
\[ \angle AOD + \angle DOB + \angle COB = 180^\circ \]
Substitute \( 180^\circ \) into our combined equation:
\[ 180^\circ + \angle DOB = 273^\circ \]
\[ \angle DOB = 273^\circ - 180^\circ = 93^\circ \]
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Problem 7: Quadrilateral Angles
Find the size of angle \( x \) in the figure.
View Step-by-Step Solution
To find \( x \), we first need to determine all the interior angles of the quadrilateral.
Two of the interior angles are given as \( 41^\circ \) and \( 94^\circ \).
The interior angle supplementary to the exterior angle \( x \) is: \( 180^\circ - x \).
The interior angle supplementary to the exterior angle \( 111^\circ \) is: \( 180^\circ - 111^\circ = 69^\circ \).
The sum of all interior angles of a quadrilateral is \( 360^\circ \). Therefore:
\[ 41^\circ + 94^\circ + (180^\circ - x) + 69^\circ = 360^\circ \]
Combine the constant values:
\[ 384^\circ - x = 360^\circ \]
\[ x = 384^\circ - 360^\circ = 24^\circ \]
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Problem 8: Grid Perimeter & Area
The rectangle below is made up of 12 congruent (same size) squares. Find the perimeter of the rectangle if the area of the rectangle is equal to \( 432 \text{ cm}^2 \).
View Step-by-Step Solution
If the total area of the rectangle is \( 432 \text{ cm}^2 \), the area of a single small square is:
\[ \dfrac{432}{12} = 36 \text{ cm}^2 \]
Let \( x \) be the side length of one small square. The area of one square is \( x^2 \):
\[ x^2 = 36 \implies x = 6 \text{ cm} \]
The grid shows the rectangle is 4 squares long and 3 squares wide. The dimensions of the large rectangle are:
\[ \text{Length } (L) = 4 \times 6 = 24 \text{ cm} \]
\[ \text{Width } (W) = 3 \times 6 = 18 \text{ cm} \]
The perimeter \( P \) of the rectangle is:
\[ P = 2(L + W) = 2(24 + 18) = 2(42) = 84 \text{ cm} \]
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Problem 9: Isosceles Triangles within a Right Triangle
\( ABC \) is a right triangle with the size of angle \( ACB \) equal to \( 74^\circ \). The lengths of the side segments \( AM \), \( MQ \), and \( QP \) are all equal. Find the measure of angle \( QPB \).
View Step-by-Step Solution
Assume the right angle is at \( B \) (which makes \( \angle ABC = 90^\circ \)). Angle \( \angle CAB \) in the right triangle \( \triangle ACB \) is:
\[ \angle CAB = 90^\circ - 74^\circ = 16^\circ \]
Because the segments \( AM \) and \( MQ \) are equal in length, triangle \( \triangle AMQ \) is an isosceles triangle. Therefore, its base angles are equal:
\[ \angle AQM = \angle QAM = 16^\circ \]
The sum of all interior angles in \( \triangle AMQ \) is \( 180^\circ \):
\[ 16^\circ + 16^\circ + \angle AMQ = 180^\circ \implies \angle AMQ = 148^\circ \]
Angle \( \angle QMP \) is supplementary to angle \( \angle AMQ \) (they form a straight line on the base):
\[ \angle QMP = 180^\circ - 148^\circ = 32^\circ \]
Because the lengths \( QM \) and \( QP \) are equal, triangle \( \triangle QMP \) is also an isosceles triangle. Therefore:
\[ \angle QPM = \angle QMP = 32^\circ \]
Angle \( \angle QPB \) is supplementary to angle \( \angle QPM \) (forming a straight line):
\[ \angle QPB = 180^\circ - 32^\circ = 148^\circ \]
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Problem 10: Area Subtraction
Find the area of the given composite shape.
View Step-by-Step Solution
The area of the given shape can be found easily by calculating the area of the full bounding rectangle and subtracting the "missing" right triangle in the top right corner.
The full bounding rectangle has dimensions of \( 20 \text{ cm} \) by \( 15 \text{ cm} \).
The sides of the "missing" right triangle (red in the diagram) are:
\[ \text{Base} = 20 - 8 = 12 \text{ cm} \]
\[ \text{Height} = 15 - 10 = 5 \text{ cm} \]
Calculate the Area:
\[ \text{Area} = \text{Area of Rectangle} - \text{Area of Triangle} \]
\[ \text{Area} = (20 \times 15) - \left( \dfrac{1}{2} \times 12 \times 5 \right) \]
\[ \text{Area} = 300 - 30 = 270 \text{ cm}^2 \]
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Problem 11: Shaded Area
Find the area of the shaded region.
View Step-by-Step Solution
The area of the shaded shape can be found by subtracting the area of the unshaded rectangle at the top left from the total area of the large outer rectangle.
First, find the dimensions of the unshaded rectangle at the top left:
\[ \text{Length} = 30 - 8 = 22 \text{ cm} \]
\[ \text{Width} = 15 - 4 = 11 \text{ cm} \]
Now, perform the area subtraction:
\[ \text{Area} = \text{Area of Large Rectangle} - \text{Area of Unshaded Rectangle} \]
\[ \text{Area} = (30 \times 15) - (22 \times 11) \]
\[ \text{Area} = 450 - 242 = 208 \text{ cm}^2 \]
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Problem 12: Inscribed Square Ratio
The vertices of the inscribed (inside) square bisect the sides of the second (outside) square. Find the ratio of the area of the outside square to the area of the inscribed square.
View Step-by-Step Solution
Let \( 2x \) be the side length of the large outside square. This means the midpoint of the side divides it into segments of length \( x \).
The area of the large outside square is:
\[ \text{Area}_{\text{outer}} = (2x) \times (2x) = 4x^2 \]
Let \( y \) be the side length of the inscribed square. Its area is \( y^2 \).
Because the vertices bisect the outer square, they create four right-angled triangles in the corners with base \( x \) and height \( x \). We can use the Pythagorean theorem to find \( y^2 \):
\[ y^2 = x^2 + x^2 = 2x^2 \]
The ratio \( R \) of the area of the outside square to the area of the inside square is:
\[ R = \dfrac{\text{Area}_{\text{outer}}}{\text{Area}_{\text{inner}}} = \dfrac{4x^2}{2x^2} = \dfrac{4}{2} = \dfrac{2}{1} \]
Answer: The ratio is 2:1.