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 \(B\).
Solution
Let \(A\) be the measure of angle \(A\) and \(B\) be the measure of angle \(B\). Hence \[ A = 2B \] Angles \(A\) and \(B\) are complementary; hence \[ A + B = 90^\circ \] But \(A = 2B\); hence \[ 2B + B = 90 \] \[ 3B = 90 \] \[ B = \frac{90}{3} = 30^\circ \] \[ A = 2B = 60^\circ \]
ABCD is a parallelogram such that \( AB \parallel DC \) and \( DA \parallel CB \). The length of side \( AB \) is \( 20 \, \text{cm} \). E is a point between A and B such that the length of \( AE \) is \( 3 \, \text{cm} \). F is a point between points D and C. Find the length of \( DF \) such that the segment \( EF \) divides the parallelogram into two regions with equal areas.
Solution
Let \( A_1 \) be the area of the trapezoid AEFD. Hence \[ A_1 = \frac{1}{2} h (AE + DF) = \frac{1}{2} h (3 + DF), \] where \( h \) is the height of the parallelogram.
Now let \( A_2 \) be the area of the trapezoid EBCF. Hence \[ A_2 = \frac{1}{2} h (EB + FC) \] We also have \[ EB = 20 - AE = 17, \quad FC = 20 - DF \] Substituting \( EB \) and \( FC \) into \( A_2 \): \[ A_2 = \frac{1}{2} h (17 + 20 - DF) = \frac{1}{2} h (37 - DF) \]
For \( EF \) to divide the parallelogram into two regions of equal area, we set \( A_1 = A_2 \): \[ \frac{1}{2} h (3 + DF) = \frac{1}{2} h (37 - DF) \] Multiply both sides by 2 and divide by \( h \): \[ 3 + DF = 37 - DF \] Solve for \( DF \): \[ 2DF = 37 - 3 \] \[ 2DF = 34 \] \[ DF = 17 \ \text{cm} \]
Find the measure of angle \( A \) in the figure below.
Solution
A first interior angle of the triangle is supplementary to the angle whose measure is \( 129^\circ \) and is equal to:
\[ 180^\circ - 129^\circ = 51^\circ \]A second interior angle of the triangle is supplementary to the angle whose measure is \( 138^\circ \) and is equal to:
\[ 180^\circ - 138^\circ = 42^\circ \]The sum of all three angles of the triangle is equal to \( 180^\circ \). Hence:
\[ A + 51^\circ + 42^\circ = 180^\circ \] \[ A = 180^\circ - 51^\circ - 42^\circ = 87^\circ \]
ABC is a right triangle. AM is perpendicular to BC. The size of angle \( \angle ABC \) is equal to \( 55^\circ \). Find the size of angle \( \angle MAC \).
Solution
The sum of all angles in triangle \(ABC\) is equal to \(180^\circ\). Hence \[ \angle ABC + \angle ACB + 90^\circ = 180^\circ \] Substitute \(\angle ABC = 55^\circ\) and solve for \(\angle ACB\): \[ \angle ACB = 180^\circ - 90^\circ - 55^\circ = 35^\circ \] The sum of all angles in triangle \(AMC\) is equal to \(180^\circ\). Hence \[ \angle MAC + \angle ACB + 90^\circ = 180^\circ \] Substitute \(\angle ACB = 35^\circ\) and solve for \(\angle MAC\): \[ \angle MAC = 180^\circ - 90^\circ - 35^\circ = 55^\circ \]Find the size of angle \( MBD \) in the figure below.
Solution
The sum of all angles in triangle \( AMC \) is equal to \( 180^\circ \). Hence
\[ 56 + 78 + \angle AMC = 180 \] \[ \angle AMC = 180 - 56 - 78 = 46^\circ \]Angles \( AMC \) and \( DMB \) are vertical angles and therefore equal in measure. Hence
\[ \angle DMB = 46^\circ \]The sum of angles of triangle \( DMB \) is equal to \( 180^\circ \). Hence
\[ \angle MBD + \angle DMB + 62 = 180 \]Substitute \(\angle DMB = 46^\circ\) and solve for \(\angle MBD\):
\[ \angle MBD + 46 + 62 = 180 \] \[ \angle MBD = 180 - 46 - 62 = 72^\circ \]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 \(DOB\).
Solution
Angle \(AOB = 132^\circ\) and is also the sum of angles \(AOD\) and \(DOB\). Hence \[ \angle AOD + \angle DOB = 132^\circ \quad \text{(I)} \] Angle \(COD = 141^\circ\) and is also the sum of angles \(COB\) and \(BOD\). Hence \[ \angle COB + \angle DOB = 141^\circ \quad \text{(II)} \] We now add the left sides together and the right sides together to obtain a new equation: \[ \angle AOD + \angle DOB + \angle COB + \angle DOB = 132 + 141 \quad \text{(III)} \] Note that \[ \angle AOD + \angle DOB + \angle COB = 180^\circ \] Substitute \(\angle AOD + \angle DOB + \angle COB\) in (III) by \(180^\circ\) and solve for \(\angle DOB\): \[ 180^\circ + \angle DOB = 132^\circ + 141^\circ \] \[ \angle DOB = 273^\circ - 180^\circ = 93^\circ \]
Find the size of angle \(x\) in the figure.
Solution
The interior angle of the quadrilateral on the left that is supplementary to \(x\) is equal to: \[ 180 - x \] The interior angle of the quadrilateral on the left that is supplementary to the angle of measure \(111^\circ\) is: \[ 180 - 111 = 69^\circ \] The sum of all interior angles of the quadrilateral is equal to \(360^\circ\). Hence: \[ 41 + 94 + (180 - x) + 69 = 360 \] Solving for \(x\): \[ 384 - x = 360 \] \[ x = 384 - 360 = 24^\circ \]
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 \( 432 \) square cm.
Solution
If the total area of the rectangle is \( 432 \ \text{cm}^2 \), the area of one square is \[ \frac{432}{12} = 36 \ \text{cm}^2 \] Let \( x \) be the side of one small square. The area of one small square is \[ x^2 = 36 \] Solving for \( x \): \[ x = 6 \ \text{cm} \] The length \( L \) of the rectangle is \( 4x \) and the width \( W \) is \( 3x \): \[ L = 4 \times 6 = 24 \ \text{cm}, \quad W = 3 \times 6 = 18 \ \text{cm} \] The perimeter \( P \) of the rectangle is \[ P = 2(L + W) = 2(24 + 18) = 84 \ \text{cm} \]
ABC is a right triangle with the size of angle \( ACB \) equal to \( 74^\circ \). The lengths of the sides \( AM \), \( MQ \), and \( QP \) are all equal. Find the measure of angle \( QPB \).
Solution
Angle \( CAB \) in the right triangle \( ACB \) is given by \[ 90^\circ - 74^\circ = 16^\circ \] Sides \( AM \) and \( MQ \) are equal in length, and therefore triangle \( AMQ \) is isosceles. Hence, \[ \angle AQM = \angle QAM = 16^\circ \] The sum of all interior angles in triangle \( AMQ \) is equal to \( 180^\circ \). Hence \[ 16^\circ + 16^\circ + \angle AMQ = 180^\circ \] Solving for \( \angle AMQ \), we get \[ \angle AMQ = 180^\circ - 32^\circ = 148^\circ \] Angle \( QMP \) is supplementary to angle \( AMQ \). Hence \[ \angle QMP = 180^\circ - \angle AMQ = 180^\circ - 148^\circ = 32^\circ \] Lengths \( QM \) and \( QP \) are equal; hence triangle \( QMP \) is isosceles, and therefore \[ \angle QPM = \angle QMP = 32^\circ \] Angle \( QPB \) is supplementary to angle \( QPM \). Hence \[ \angle QPB = 180^\circ - \angle QPM = 180^\circ - 32^\circ = 148^\circ \]Find the area of the given shape.
Solution
The area of the given shape may be found by subtracting the area of the right triangle (red) from the area of the large rectangle (see figure below).
Sides of the right triangle (red) are:
\[
15 - 10 = 5 \text{ cm}, \quad 20 - 8 = 12 \text{ cm}
\]
The area of the given shape is:
\[
\text{Area} = 20 \times 15 - \frac{1}{2} \times 12 \times 5 = 300 - 30 = 270 \text{ cm}^2
\]
Find the area of the shaded region.
Solution
The area of the shaded shape can be found by subtracting the area of the rectangle at the top left from the area of the large rectangle. \[ \text{Dimensions of the rectangle at top left:} \quad \text{length} = 30 - 8 = 22 \text{ cm}, \quad \text{width} = 15 - 4 = 11 \text{ cm} \] \[ \text{Area of the shaded shape} = 30 \times 15 - 22 \times 11 = 450 - 242 = 208 \text{ cm}^2 \]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.
Solution
Let \(2x\) be the side length of the large square (see figure below).
The area of the large square is
\[
(2x) \times (2x) = 4x^2
\]
The area of the inscribed square is
\[
y \times y = y^2
\]
Using the Pythagorean theorem for the right triangle formed by half the large square, we get
\[
y^2 = x^2 + x^2 = 2x^2
\]
The ratio \(R\) of the area of the outside square to the area of the inside square is
\[
R = \frac{4x^2}{y^2} = \frac{4x^2}{2x^2} = \frac{4}{2} = 2 : 1
\]