Triangle EFG is an isosceles triangle with EG=EF. What is the approximate length of EF, and what is the approximate perimeter of
the triangle EFG? Round your answers to the nearest hundredth
The possible answers are
A. The approximate length of EF is 4.47 units, and the approximate perimeter of triangle EFG is 12.47 units
B. The approximate length of EF is 4.47 units, and the approximate perimeter of triangle EFG is 12.94 units.
C. The approximate length of EF is 4.58 units, and the approximate perimeter of the triangle EFG is 12.58 units.
D. The approximate length of EF is 4.58 units, and the approximate perimeter of triangle EFG is 13.16 units.
The possible answers are
A. The approximate length of EF is 4.47 units, and the approximate perimeter of triangle EFG is 12.47 units
B. The approximate length of EF is 4.47 units, and the approximate perimeter of triangle EFG is 12.94 units.
C. The approximate length of EF is 4.58 units, and the approximate perimeter of the triangle EFG is 12.58 units.
D. The approximate length of EF is 4.58 units, and the approximate perimeter of triangle EFG is 13.16 units.
2 answers:
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Answer:
The claim is correct
Explanation:
Assume the given triangle ABC
perimeter of triangle ABC = AB + BC + CA ............> I
Now, we have:
D is the midpoint of AB, this means that:
AD = DB = (1/2) AB ..........> 1
E is the midpoint of AC, this means that:
AE = EC = (1/2) AC ...........> 2
DE is the midsegment in triangle ABC, this means that:
DE = (1/2) BC ...........> 3
perimeter of triangle ADE = AD + DE + EA
Substitute in this equation with the corresponding lengths in 1,2 and 3:
perimeter of triangle ADE = (1/2) AB + (1/2) BC = (1/2) AC
perimeter of triangle ADE = (1/2)(AB+BC+AC) .........> II
From I and II, we can prove that:
perimeter of triangle ADE = (1/2) perimeter of triangle ABC
Which means that:
perimeter of midsegment triangle is half the perimeter of the original triangle.
Hope this helps :)
The claim is correct
Explanation:
Assume the given triangle ABC
perimeter of triangle ABC = AB + BC + CA ............> I
Now, we have:
D is the midpoint of AB, this means that:
AD = DB = (1/2) AB ..........> 1
E is the midpoint of AC, this means that:
AE = EC = (1/2) AC ...........> 2
DE is the midsegment in triangle ABC, this means that:
DE = (1/2) BC ...........> 3
perimeter of triangle ADE = AD + DE + EA
Substitute in this equation with the corresponding lengths in 1,2 and 3:
perimeter of triangle ADE = (1/2) AB + (1/2) BC = (1/2) AC
perimeter of triangle ADE = (1/2)(AB+BC+AC) .........> II
From I and II, we can prove that:
perimeter of triangle ADE = (1/2) perimeter of triangle ABC
Which means that:
perimeter of midsegment triangle is half the perimeter of the original triangle.
Hope this helps :)