Perimeter: P=62 feet
P=2(b+h)
62=2(b+h)
Dividing both sides of the equation by 2:
62/2=2(b+h)/2
31=b+h
b+h=31 (1)
Area: A=bh (2)
Isolating h in equation (1)
(1) b+h=31→b+h-b=31-b→h=31-b (3)
Replacing h by 31-b in equation (2)
(2) A=bh
A=b(31-b)
A=31b-b^2
To maximize the area:
A'=0
A'=(A)'=(31b-b^2)'=(31b)'-(b^2)'=31-2b^(2-1)→A'=31-2b
A'=0→31-2b=0
Solving for b:
31-2b+2b=0+2b
31=2b
Dividing both sides by 2:
31/2=2b/2
31/2=b
b=31/2=15.5
Replacing b by 31/2 in equation (3)
h=31-b
h=31-31/2
h=(2*31-31)/2
h=(62-31)/2
h=31/2
The dimensions are 31/2 ft x 31/2 ft = 15.5 ft x 15.5 ft
The area with these dimensions is: A=(15.5 ft)(15.5 ft)→A=240.25 ft^2
These dimensions are not in the options
1) The first option has an area of: A=(18 ft)(13 ft)→A=234 ft^2
2) The second option has an area of: A=(19 ft)(12 ft)→A=228 ft^2
3) The third option has an area of: A=(17 ft)(14 ft)→A=238 ft^2
The third option has the largest area.
Answer: Third option
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It is given that, B ≅ BC and AD ≅ CD
We need BD perpendicular to AC, then only we can say triangles AXB and CXB are congruent using the HL theorem.
If BD perpendicular to AC, means that AB and CB are the hypotenuse of triangles AXB and CXB respectively.
from the given information ABCD is a square
If BD and AC bisect each other then AX = CX
Then only we can immediately possible to prove that triangles AXD and CXD are congruent by SSS congruence theorem
Answer:
11
Step-by-step explanation:
x3+x-5
3*4=12
12+4-5
16-5=11
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Answer:
32
Step-by-step explanation:
7m + 7m = 14m
46m - 14m = 32m
