Question

In rectangle ABCD, the diagonals intersect at E. If m angle∠AED= y+10 , and m angle∠AEB= 4y−30 , find m angle∠AED, m angle∠ADE and m angle∠EDC.

Answer 1

The sume of angle AED and angle AEB is 180° because they are supplements, so:

`\begin{gathered} \angle AED+\angle AEB=180 \\ (y+10)+(4y-30)=180 \\ 5y-20=180 \\ 5y=180+20=200 \\ y=(200)/(5)=40 \\ \angle AED=40+10=50 \\ \angle AEB=4\cdot40-30=130 \end{gathered}`

Angle AED is 50° and angle AEB is 130°.

The triangle AED is isosceles, so angle EAD and angle AED are equal, so:

`\begin{gathered} \angle EAD=\angle ADE=x \\ \angle AED+\angle EAD+\angle ADE=180 \\ 50+x+x=180 \\ 2x=180-50 \\ x=(130)/(2)=65 \\ \angle EAD=\angle ADE=65 \end{gathered}`

Also, angle ADE and angle EDC are complements, so:

`\begin{gathered} \angle ADE+\angle EDC=90 \\ 65+\angle EDC=90 \\ \angle EDC=90-65 \\ \angle EDC=25 \end{gathered}`

The angle EDC is 25°.