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Mathematics

1 answer:

aleksley [76]2 years ago

4 0

Answer:

See brainly.com/question/23882263 for A and B parts of the first question

$HF = \sqrt{EH^2+EF^2} = \sqrt{2^2+3^2} = \sqrt{13}$

$d = \sqrt{2^2+5^2} = \sqrt{4+25} = \sqrt{29}$

Diagonals are perpendicular. Side length of the rhombus:

$a = \sqrt{(8/2)^2+(6/2)^2} = \sqrt{4^2+3^2} = \sqrt{25} = 5$

Perimeter:

P = 4a = 4(5) = 20

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