ABCD is a parallelogram in which cd = 7cm Ad = 5cm ND adc = 125.. Find correct to one decimal place the area of the parallelogram.

28.7cm²

The sketch of the parallelogram is attached to this response.

As shown in the sketch,

h is the height of the parallelogram

ADC is 125°

AD = 5cm

CD = 7cm

Also

Since the adjacent angles of a parallelogram are supplementary (add up to 180°), it means that;

=> ADC + BCD = 180°

=> 125° + BCD = 180°

=> BCD = 180° - 125°

=> BCD = 55°

The area (A) of a parallelogram is given by;

A = b x h            ----------(i)

Where

b = base of the parallelogram

h = height of the parallelogram

From the sketch,

b = CD = 7cm

h can be found by using the sine trigonometric function as follows

sin BCD =

=> sin 55° =

=> h = 5 sin 55°

=> h = 5 x 0.8192

=> h = 4.096cm

Now, substitute the values of h and b into equation (i) as follows;

A = 7 x 4.096

A = 28.672

A ≅ 28.7 (to one decimal place)

Therefore, the area of the parallelogram is 28.7cm²

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length

10.416 m/s

step-by-step explantion:

distance=100m

time= 9.6s

average speed=100/9.6

=10.416 m/s