2 Answers

Danaley · Answer 1 · 2017-06-15T14:58:22+0000

Answer:

$m\angle GHE=49^(\circ)$

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.

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