Question

please help me with this

Answer 1

Step-by-step explanation:

Perimeter of a rhombus =4×side length.= 64 m.

Length of the side of rhombus = 64/4. = 16 m.

Let ABCD is a rhombus in which AB=BC=CD=DA= 16 m. and diagonals meet at

point O . Diagonal AC=22 m.

In right angled triangle AOB:- OB^2+OA^2= AB^2

or. (BD/2)^2+(AC/2)^2= AB^2

or. (BD/2)^2+(22/2)^2 = 16^2. or. (BD/2)^2=256–121=135

or. BD/2=3√15. => BD= 6√15 m.

Area of the rhombus ABCD= (1/2).(AC×BD)= 22×6√15/2=66√15 m^2. Proved.

Answer 2

Step-by-step explanation:

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sorry about that my small brother type this sorry