Answer:The claim is correct
Explanation:Assume the given triangle ABCperimeter of triangle ABC = AB + BC + CA ............> I
Now, we have:D is the midpoint of AB, this means that:
AD = DB = (1/2) AB ..........> 1E is the midpoint of AC, this means that:
AE = EC = (1/2) AC ...........> 2DE is the midsegment in triangle ABC, this means that:
DE = (1/2) BC ...........> 3perimeter 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) .........> IIFrom 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 :)
X=5 if that's what you were asking for
Answer:
Step-by-step explanation:
Let length and width of a rectangle be x, x+1
Area = 90 sq.cm
length * width = 90
x * (x + 1) = 90
x² + x = 90
x² + x - 90 = 0
x² + 10x - 9x - 9 * 10 = 0
x ( x + 10 ) - 9 (x + 10 ) = 0
(x + 10 ) (x - 9) = 0
x + 10 = 0 or x - 9 = 0
x = -10 { impossible . as length cannot be negative} ;
x = 9
Length = 9 , width = 10
Answer:
but dude what to answer............