Given EG = 16 and FH = 12, what is the length of one side of the rhombus?
1 answer:
Length of one side of the rhombus is 10.
If there is a rhombus EFGH, such as yours, with points E, F, G, H (such as in the .jpeg image in attachment), and if the notation of rhombus side is a (rhombus has 4 sides and each of them is of the same length):
FH/2 = 12/2 = 6 (shorter diagonal of rhombus, it's half)
EG/2 = 16/2 = 8 (longer diagonal of rhombus, it's half)
-----------------------------------------
a^2 = (FH/2)^2 + (EG/2)^2
a^2 = 6^2 + 8^2
a^2 = 36 + 64
a^2 = 100
a = square root of 100
a = 10
If there is a rhombus EFGH, such as yours, with points E, F, G, H (such as in the .jpeg image in attachment), and if the notation of rhombus side is a (rhombus has 4 sides and each of them is of the same length):
FH/2 = 12/2 = 6 (shorter diagonal of rhombus, it's half)
EG/2 = 16/2 = 8 (longer diagonal of rhombus, it's half)
-----------------------------------------
a^2 = (FH/2)^2 + (EG/2)^2
a^2 = 6^2 + 8^2
a^2 = 36 + 64
a^2 = 100
a = square root of 100
a = 10
You might be interested in
ANSWER
3) b
EXPLANATION
Given that:
and
Recall that the sine and cosine functions are equal for complementary angles.
This implies that,