last math question…?

solve for AC first then to find the area multiply AC byAB.

by phytagorean theorem:

BC^2 = AB^2 + AC^2

AC^2 = BC^2 – AB^2

AC = sqrt(BC^2 – AB^2)

The diagonal AC divides the parallelogram into two congruent right triangles–CAB and ACD.

CAB is a right triangle with BC the hypotenuse.

(AC)² = (BC)² – (AB)² = 13² – 12² = 169 – 144 = 25

AC = 5

Calculate the area of triangle CAB.

Area = ½bh = ½*AC*AB = ½*5*12 = 30

Triangle ACD is congruent to triangle CAB so the area of the parallelogram ABCD is:

2*30 = 60 cm²

enable the width be x metre. Then length is x + 4 section is x (x + 4) squarem. = x^2 + 4x = 77 = x^2 + 4x – 77 = 0 x^2 +11x – 7x – 77 = 0 x(x + 11) – 7( x-11) = 0 (x+11) (x – 7) = 0 x = -11 or 117. -11 isn’t perfect contained interior the on the spot case. as a result x = 7 as a result the width is 7 and length is 11m.

60cm

ab^2 = 144

bc^2 =169

169 – 144 =25

sqrt of 25 = 5

AC = 5

AB = 12

5*12 = 60

Area = 60cm

60cm

ab^2 = 144

bc^2 =169

169 – 144 =25

sqrt of 25 = 5

AC = 5

AB = 12

5*12 = 60

Area = 60cm

60cm

ab^2 = 144

bc^2 =169

169 – 144 =25

sqrt of 25 = 5

AC = 5

AB = 12

5*12 = 60

Area = 60cm

simple! just multiply 12 by 13!

I think you can do Area equals .5 base times height cosine(90),then multiply by four i think

i think the answer is 60cm..