The diagonals of a square with area 9 m² divide the square into four nonoverlapping triangles what is the sum of the perimeter of the four triangles...pls don't post irrelevant answers.If you know the answer pls explain the answer correctly..Irrelevant answers will be reported...
Answers
Answer:
Perimeter of each triangle=(3√2+3) metres
The sum of perimeters of all triangles=(12√2+12) metres
Explanation:
Area of the square=a²=9 m² [a= side of the square]
So, each side= 3 m
By Pythagoras theorem,
diagonal of the square=√(3²+3²)=3√2 m
see the next step in the file I have attached
We know that, all the triangles are equal (or congruent), so, perimeter of each triangle =(3√2+3) m
Thus, perimeter of all triangles added =(12√2+12) m
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ANSWER
MATHS
ABCD is a square of area of 4 square units which is divided into 4 non overlapping triangles as shown in figure, then sum of perimeters of the triangles so formed is
1213740
ANSWER
Sum of perimeter of Δlei
AO+OB+BA+BO+OC+CB+OC+CD+DO+DO+OA+AD
AO+OC OB+OD
↑ ↑
=AB+BC+CD+AD+2(AC+BD)
Length of diagonal =2
2
4(2)+2(2
2
)
4(2+
2
)
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