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zmey [24]
3 years ago
12

The area of the rhombus is 540 cm2; the length of one of its diagonals is 4.5 dm. What is the distance between the point of inte

rsection of the diagonals and the side of the rhombus?

Mathematics
2 answers:
NARA [144]3 years ago
8 0

Try this solution (this is not the shortest way!).

Note, the length of the diagonal is 45 cm.=4,5 dm.

Answer: the side is 22.5 cm., the EF is 12 cm.

klemol [59]3 years ago
6 0

Answer:

Step-by-step explanation:

area of rhombus = base x height or diagonal 1 x diagonal 2 / 2

one diagonal = 4.5dm = 45cm

the other diagonal = 540 / 45 * 2 = 24cm

the diagonals in a rhombus are perpendicular 2 each other

so half diagonals and one side of the rhombus form a right-angled triangle

side^2 = half-diagonal 1^2 + half-diagonal 2^2

= (45/2)^2 + (24/2)^2

= 650.25

side = sqrt (650.25) = 25.5cm

540 = height x side

height = 540 / 25.5 = 21.1765cm

distance between the point of intersection of the diagonals and the side of the rhombus = height / 2 = 10.59cm

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=============================================

How I got those values:

We have 12 hearts out of 60 cards total in our simulation or experiment. So 12/60 = (12*1)/(12*5) = 1/5 is the experimental probability. In the simulation, 1 in 5 cards were a heart.

Theoretically it should be 1 in 4, or 1/4, since we have 13 hearts out of 52 total leading to 13/52 = (13*1)/(13*4) = 1/4. This makes sense because there are four suits and each suit is equally likely.

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For example, let's say you flip a coin 20 times and get 8 heads. We see that 8/20 = 0.40 is close to 0.50 which is the theoretical probability of getting heads. If you flip that same coin 100 times and get 46 heads, then 46/100 = 0.46 is the experimental probability which is close to 0.50, and that probability is likely to get closer if you flipped it say 1000 times or 10000 times.

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