Method 1: Half of diagonal product
Solution: Because the quadrilateral ABCD is an isosceles trapezoid.
So AC=BD
Parallel BC
Because DE is in parallel with AC
So the quadrilateral is a parallelogram.
So AC=DE
AD=CE=3
BD vertical DE
So the angle BDE=90 degrees
So the triangle BDE is a right triangle.
From Pythagorean Theorem:
BE^2=BD^2+DE^2
Because BE=BC+CE
BC=7
BD=DE
So BD=DE=5 times the root number 2.
So S- trapezoidal ABCD= 1/2*AC*BD=25.
Method 2: Cutting repair method
Solution: Because the quadrilateral ABCD is an isosceles trapezoid.
So AC=BD
Parallel BC
Because DE is in parallel with AC
So AD=CE
AC=DE
S triangle ACD=S triangle DCE=S triangle ABD
BD vertical DE
So the angle BDE=90 degrees
From Pythagorean Theorem:
BE^2=BD^2+DE^2
Because BE=BC+CE
BC=7
AD=3
So BD=DE=5 times the root number 2.
Because s trapezoid ABCD=S triangle ABD+S triangle BDC.
S triangle BDE=S triangle BDC+S triangle DCE
So s trapezoid ABCD=S triangle BDE
Because the S triangle BDE= 1/2*BD*DE=25.
So the S-trapezoid ABCD=25.
Method 3: (upper bottom+lower bottom) times the height and divides by 2.
Solution: the intersection d is DF, perpendicular to BC and F.
So BFD angle =90 degrees.
From Pythagorean Theorem:
BD^2=DF^2+BF^2
Because the quadrilateral ABCD is an isosceles trapezoid
So AC=BD
Parallel BC
Because DE is in parallel with AC
So the quadrilateral is a parallelogram.
So AD=CE
AC=DE
Because AC vertical BD
So BD is perpendicular to DE.
So the angle BDE=90 degrees
BD=DE
So the triangle BDE is an isosceles triangle.
So DBF angle =45 degrees.
Derived from Pythagorean theorem;
BE^2=BD^2+DE^2
Because BE=BC+CE
AD=3
BC=7
So BD=DE=5 times the root number 2.
Because DBF angle +BFD angle +BDF angle = 180 degrees.
So DBF angle = BDF angle =45 degrees.
So DF=BF
So DF=5
Because S- trapezoid ABCD= 1/2*(AD+BC)*DF=25.
So the S-trapezoid ABCD=25.