Question

In the paper airplane,ABCD is congruent to EFGH, m<B=m<BCD=90 and m<BAD= 131.Find m<GHE.

Answer 1

Answer:

$m\angle GHE=49^{\circ}$

Step-by-step explanation:

We have been given a paper airplane in which $ABCD\cong EFGH$.

Since $ABCD\cong EFGH$, therefore, by the definition of congruence corresponding angles of both quadrilaterals will be equal.

$m\angle BAD=m\angle FEH$

$m\angle B=m\angle F$

$m\angle BCD=m\angle FGC$

$m\angle CDA=m\angle GHE$

We know that all interior angles of quadrilateral add up-to 360 degree, so we can set an equation as:

$m\angle BAD+m\angle B+m\angle BCD+m\angle CDA=360^{\circ}$

$m\angle 131^{\circ}+90^{\circ}+90^{\circ}+m\angle CDA=360^{\circ}$

$m\angle 311^{\circ}+m\angle CDA=360^{\circ}$

$m\angle 311^{\circ}-311^{\circ}+m\angle CDA=360^{\circ}-311^{\circ}$

$m\angle CDA=49^{\circ}$

Therefore, the measure of angle GHE is 49 degrees.

Answer 2

I hope this helps you