Three important properties of the diagonals of a rhombus that we need for this problem are:
1. the diagonals of a rhombus bisect each other
2. the diagonals form two perpendicular lines
3. the diagonals bisect the angles of the rhombus
First, we can let O be the point where the two diagonals intersect (as shown in the attached image). Using the properties listed above, we can conclude that ∠AOB is equal to 90° and ∠BAO = 60/2 = 30°.
Since a triangle's interior angles have a sum of 180°, then we have ∠ABO = 180 - 90 - 30 = 60°. This shows that the ΔAOB is a 30-60-90 triangle.
For a 30-60-90 triangle, the ratio of the sides facing the corresponding anges is 1:√3:2. So, since we know that AB = 10, we can compute for the rest of the sides.



Similarly, we have



Now, to find the lengths of the diagonals,


So, the lengths of the diagonals are 10 and 10√3.
Answer: 10 and 10√3 units
9x - 2y = 11 ... (i)
5x - 2y = 15 ... (ii)
Subtracting equation (ii) from (i) we get;
4x + 0 = -4
4x=-4 , x = -1
Replacing x = -1 in equation (i) we get;
9(-1) - 2y = 11
-9 - 2y = 11
-2y = 20
y = 20 ÷ -2 = -10
The solution to the system of equations is (-1,-10).
Answer:
80.48m2
Step-by-step explanation:
I think this is right but it may be wrong.
Answer: x=7
Step-by-step explanation:
Since AB is a bisector, it cuts the angle in half, and since we know that the angle is a right angle, we know it’s 90 degrees, so knowing that you use the equation 7x-4= 45 (since 45 is half of 90), add 4 to both sides so you have 7x=49, the divide 49 by 7 and you get 7, so x= 7
Answer:
.
Step-by-step explanation:
How many unique combinations are possible in total?
This question takes 5 objects randomly out of a bag of 50 objects. The order in which these objects come out doesn't matter. Therefore, the number of unique choices possible will the sames as the combination
.
How many out of that 2,118,760 combinations will satisfy the request?
Number of ways to choose 2 red candies out a batch of 28:
.
Number of ways to choose 3 green candies out of a batch of 8:
.
However, choosing two red candies out of a batch of 28 red candies does not influence the number of ways of choosing three green candies out of a batch of 8 green candies. The number of ways of choosing 2 red candies and 3 green candies will be the product of the two numbers of ways of choosing
.
The probability that the 5 candies chosen out of the 50 contain 2 red and 3 green will be:
.