PLS HELP I WILL GIVE BRAINIEST If triangle BCD were rotated over the y-axis from point (0, 0), what would be the coordinates of
the new triangle's vertices?
A. (0, -1) , (-4, -3), (-2, -1)
B. (-1, 0), (-5, -2), (-3, 4)
C. (1, 0), (5, 2), (3, 0)
D. (1, 0), (1, -2), (-3, -4)
A. (0, -1) , (-4, -3), (-2, -1)
B. (-1, 0), (-5, -2), (-3, 4)
C. (1, 0), (5, 2), (3, 0)
D. (1, 0), (1, -2), (-3, -4)
1 answer:
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Hello,
Reminders:
Reminders:
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xls
</span>
Answer:
6m
Step-by-step explanation:
the side of first square = 3 m
area of first square = 3^2= 9 m^2
let the side of second square = x m
since diagonal of first square = x m as per the question
in a square each angle is 90 degree
therefore by applying pythagorus theorem ,
diagonal ^2 = side^2 + side ^2
diagonal^2 = 3^2+3^2=9+9
DIAGONAL= square root of 18
diagonal= 3 m = side of second square
therefore in second square ,
one angle is 90 degree , therefore by applying pythagorue theorem
diaqonal^2 = side ^2 + side ^2 = (3)^2 + ( 3
)^2
diagonal ^2 = 18 + 18
diagonal = square root of 36
therefore diagonal of second square is 6m
hope it helps you... please mark me as the brainliest..