What is the greatest common factor needed to simplify 60/96
1 answer:
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The missing measures are listed below:
In this question we must determine the measures of the <em>missing</em> angles by both <em>geometric</em> and <em>algebraic</em> means, whose procedure is presented below:
-
Given
Alternate internal angles
Complementary angles
Alternate internal angles
Isosceles triangle
Sum of internal angles in a triangle
Supplementary angles
,
Vertical angles
Sum of internal angles in a triangle
Sum of internal angles in a triangle
Sum of internal angles in a triangle/Result
The missing measures are listed below:
To learn more on angles, we kindly invite to check this verified question: brainly.com/question/15767203
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 :)