A math problem in the second day of junior high school

Solution 1:

Do DE‖AC after point D, extend the line from DE to BC at point E, and do DF⊥BC at point F.

ADEC is a parallelogram with CE=AD=3 cm and DE=AC=BD.

So △BDE is an isosceles right triangle, and DF is also the midline on the hypotenuse BE, so BF =1/2 * BE =1/2× (BC+CE) = 5.

That is, the height of trapezoidal ABCD is 5.

So the area is: S= 1/2*(AD+BC)*DF=25.

Solution 2:

Let the intersection of AC and BD be H, after which HP⊥BC and P are vertical feet, and extend the PH to AD in Q, because AD‖BC and HQ are the height of △AHD.

Because ABCD is an isosceles trapezoid, BH=CH, that is, △BHC is an isosceles right triangle, and HP on the side of BC is also the midline, so HP= 1/2*BC.

Similarly, HQ= 1/2*AD,

So PQ=HQ+HP= 1/2*(AD+BC)=5, that is, the height of the trapezoid is 5, so the area is: S= 1/2*(AD+BC)*DF=25.

Solution 3: (Algebraic Calculation)

Let the intersection of AC and BD be h,

In terms of symmetry, that is, △BHC is an isosceles right triangle, so BH=CH=BC/√2,

Similarly, AH=DH=AD/√2 cm,

AC=AH+CH=(AD+BC)/√2=5√2 cm，

Similarly, BD=5√2 cm,

The area of a quadrilateral (whether concave or convex) whose diagonals are perpendicular to each other is equal to half of the diagonal product (easily obtained from the area formula of a diamond), so,

S_ABCD= 1/2*AC*BD=25 .

If you are not familiar with this conclusion, you can consider it this way.

s△AHD = 1/2 * AH * DH；

s△AHB = 1/2 * AH * BH；

s△CHB = 1/2 * CH * BH；

s△CHD = 1/2 * CH * DH；

S_ABCD=S△AHD+S△AHB+S△CHB+S△CHD

= 1/2*(AH+CH)*(BH+DH)

= 1/2*AC*BD

=25 。 )

Solution 4: (or algebraic calculation)

Let the intersection of AC and BD be h,