Question

1. in the paper airplane shown, ( abcdcong efgh ), ( mangle b = mangle bcd = 90 ), and ( mangle bad = 128 ). find ( mangle fgh ). 90 62 52 128

Explanation:

Step1: Recall the sum of interior angles of a quadrilateral

The sum of interior angles of a quadrilateral is \(360^{\circ}\). For quadrilateral \(ABCD\), we know \(m\angle B = 90^{\circ}\), \(m\angle BCD=90^{\circ}\), and \(m\angle BAD = 128^{\circ}\). Let \(m\angle ADC=x\). Then \(m\angle B+m\angle BCD + m\angle BAD+m\angle ADC=360^{\circ}\).

Step2: Calculate \(m\angle ADC\)

Substitute the known values into the formula: \(90 + 90+128+x = 360\).
Simplify the left - hand side: \(308+x = 360\).
Solve for \(x\): \(x=360 - 308=52^{\circ}\).

Step3: Use the property of congruent quadrilaterals

Since \(ABCD\cong EFGH\), corresponding angles are equal. \(\angle FGH\) and \(\angle ADC\) are corresponding angles.

Answer:

\(52\)