the diagonals of a rhombus have lengths 4 and 12. what is the length of a side in simplest form.

Answer 1

a=sqroot(12^2+4^2)

a=sqroot(144+16)

a=sqroot(160)

a=sqroot(16*10)

a=4*sqroot(10)=12.64911
Becouse sqroot(16)=4

a=4*sqroot(10)=12.64911

Answer 2

That is the wrong answer.

p=first diagonal=4
q=second diagonal=12

a=sqroot[(p/2)^2+(q/2)^2]

a=sqroot[(4/2)^2+(12/2)^2]

a=sqroot(2^2+6^2)

a=sqroot(4+36)

a=sqroot(40)

a=sqroot(4*10)

a=2*sqroot(10)

OR:

p^2+q^2=4a^2 Divide with 4

a^2=(p^2+q^2)/4

a=sqroot[(p^2+q^2)/4]

a=sqroot(p^2+q^2)/sqroot(4)

a=sqroot(p^2+q^2)/2

a=sqroot(4^2+12^2)/2

a=sqroot(16+144)/2

a=sqroot(160)/2

a=sqroot(16*10)/2

a=4*sqroot(10)/2

a=2*sqroot(10)

Check this.

In google type:
mathworld.wolfram rhombus

When page be open click on :
Rhombus-from Wolfram MathWorld