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for #21 - 23, find the values of x and y which make the quadrilateral a…

Question

for #21 - 23, find the values of x and y which make the quadrilateral a parallelogram. show your work.

  1. ( m angle a = 3 y ); ( m angle d = 2 x - 5 ); ( m angle c = 123 ^ { circ } )
  2. ( x = )

( y = )

Explanation:

Step1: Use the property of parallelogram (opposite angles are equal)

In a parallelogram, \(m\angle A=m\angle C\) and \(m\angle D = m\angle B\), and consecutive - angles are supplementary (\(m\angle A+m\angle D=180^{\circ}\)).
Since \(m\angle C = 123^{\circ}\), and \(m\angle A=m\angle C\) (opposite - angles of a parallelogram are equal), but also \(m\angle A + m\angle D=180^{\circ}\) (consecutive - angles of a parallelogram are supplementary).
We know \(m\angle D=2x - 5\).
Substitute into the consecutive - angle formula: \(3y+(2x - 5)=180\).
Also, since \(m\angle C = 123^{\circ}\) and \(m\angle D=m\angle B\), and \(m\angle A+m\angle B = 180^{\circ}\) (another pair of consecutive angles), but using \(m\angle A+m\angle D=180^{\circ}\) is more straightforward.
First, use the property that \(m\angle D=m\angle B\) and \(m\angle A+m\angle D = 180^{\circ}\).
Since \(m\angle D=2x - 5\) and \(m\angle A = 3y\), and we know that \(m\angle D\) and \(m\angle C\) are consecutive angles (\(m\angle D+m\angle C=180^{\circ}\) for a parallelogram).
Set up the equation for \(\angle D\):
\(2x−5 + 123=180\)

Step2: Solve the equation for \(x\)

$$ LATEXBLOCK0 $$

Step3: Solve the equation for \(y\)

Substitute \(x = 31\) into \(m\angle A+m\angle D=180^{\circ}\) (where \(m\angle A = 3y\) and \(m\angle D=2x - 5\)).
First, find \(m\angle D\) when \(x = 31\): \(m\angle D=2\times31-5=62 - 5=57^{\circ}\).
Then, since \(m\angle A+m\angle D=180^{\circ}\), and \(m\angle A = 3y\), we have \(3y+57 = 180\).

$$ LATEXBLOCK1 $$

Answer:

\(x = 31\), \(y = 41\)