Geometry problems in the sixth grade of primary school mathematics

The height of parallelogram ABCD =90/ 10=9,

Triangle BEF is similar to triangle DAF, [AAA]

Area of triangle BEF: area of triangle DAF =(5: 10)? = 1/4

The area of triangular DAF =4* the area of triangular BEF,

The area of the triangle is BAD = 10*9/2=90/2,

That is, the area of triangle ABF+the area of triangle DAF =90/2,

Area of triangle ABF +4* area of triangle BEF =90/2, ... 1)

The area of the triangle ABE =5*9/2=45/2,

The area of triangle ABF+ the area of triangle BEF =45/2,

Area of triangle ABF =45/2- area of triangle BEF, ... 2)

2) Substitute 1):

The area of the triangle BEF =(90/2-45/2)/3= 15/2.

Triangular area DEF = triangular bed area-triangular bed area BEF

=5*9/2- 15/2= 15.