One way would be 49 = 21 + 7 + 7 +7 + 7
The area of a rectangle is length times width
I think that you should imagine what regular shape the figure would have been.
The fact that two sides are 9inches makes me think it would have been a square.
The area of a square is A= l² where l is the length of one side and A=area
A=9*9
A=81 in.²
Next, find the area of the part that is taken away,(it looks like a triangle). to find that, if the figure was a square, the length on top would be 9 not 5, so the length gone is 9-5=4, so the base of the triangle is 4,
If the figure was a square, the length on the other side would be 9 not 4 so the length gone is 9-4=5, so the height of the triangle is 5
the area of a triangle is A= 1/2bh, where A is area, b is base and h is height.
A=1/2*4*5 A=10 so the area of the part gone is 10 inches²
The area of the figure is the area of the square -the area of the triangle
A=81-10
A=71inches²
,
Answer:
x=−12 or x=16
Step-by-step explanation:
also go back to my question and explain if it is real or non-real
Answer:
20.6
Step-by-step explanation:
The problem can be solved by applyPythagoreanean theory
a^2+b^2= c^2
19^2 + b^2= 28^2
361 + b^2= 784
b^2= 784-361
= 423
b= √423
= 20.6