In square abcd, the centre o and the midpoint m of side ab are connected with vertices c and d. prove that areas of △amd, △bmc, △cdo and area of mcod are equal.

The area of a triangle is given by the formula ...

... A = (1/2)bh

where b is the base length and h is the height perpendicular to the base.

If we let the length of the sides of the square be represented by s, then we have ...

... AM = MB = MO = s/2

and the areas of the triangles are ...

... AAMD = (1/2)(s)(s/2) = s²/4

... ABMC = (1/2)(s)(s/2) = s²/4

... ACDO = (1/2)(s)(s/2) = s²/4

and the area of the quadrilateral is the sum of the areas of triangles MOC and MOD, so is ...

... AMCOD = (1/2)(s/2)(s/2) +(1/2)(s/2)(s/2) = s²/8 +s²/8 = s²/4

Hence the described areas are equal to each other. (All are s²/4.)

answer: 1, 3 and 5 are correct

step-by-step explanation:

1 x=22.52

2: 65/39=604+39x

65x-39x=604

26x=604

x=604/26

×=23.23