Answer:
10. To find the area of rhombus ABCD, we can use the formula:
Area = (diagonal 1 x diagonal 2) / 2
We need to find the length of both diagonals. Since the diagonals of a rhombus are perpendicular and bisect each other, we can use the Pythagorean theorem to find the length of each diagonal.
AC is one diagonal, AB is a side of the rhombus, and BD is the other diagonal divided by 2 (since BD bisects AC):
BD = AC/2 = 10/2 = 5 cm
Using the Pythagorean theorem with AC and BD:
AC^2 = AB^2 + BD^2
AC^2 = 13^2 + 5^2
AC^2 = 169 + 25
AC^2 = 194
AC = sqrt(194) ≈ 13.93 cm
Now that we have the lengths of both diagonals:
Area = (AC x BD) / 2
= (13.93 x 5) / 2
≈ 34.83 cm^2
Therefore, the area of rhombus ABCD is approximately 34.83 cm^2.
11. We know that the area of rhombus ABCD is 96 cm^2, and that BD is 8 cm. To find the length of AC, we can use the formula:
Area = (diagonal 1 x diagonal 2) / 2
Solving for diagonal 1 (AC):
AC = (2 x Area) / BD
= (2 x 96) / 8
= 24 cm
Therefore, the length of AC is 24 cm.
12. We know that AB is a side of the rhombus and that AB = 16 m. We also know that mZABD = 60 degrees. We can use trigonometry to find the length of the diagonals.
First, we can find the length of AD using the law of cosines:
AD^2 = AB^2 + BD^2 - 2(AB)(BD)cos(mZABD)
AD^2 = 16^2 + (2x)^2 - 2(16)(2x)cos(60)
AD^2 = 256 + 4 - 32x
AD^2 = 260 - 32x
AD = sqrt(260 - 32x)
Then, we can find the length of AC using the law of sines:
AC / sin(mZBAD) = AD / sin(mZABD)
AC / sin(120) = AD / sin(60)
AC = (AD x sin(120)) / sin(60)
AC = (sqrt(260 - 32x) x sqrt(3)) / 2
Now that we have the lengths of both diagonals:
Area = (AC x BD) / 2
= [(sqrt(260 - 32x) x sqrt(3)) / 2] x 8 / 2
= 2(sqrt(260 - 32x) x sqrt(3))
≈ 67.29 m^2
Therefore, the area of rhombus ABCD is approximately 67.29 square meters.
13. We know that the perimeter of the rhombus is 20 mm, so each side is 5 mm. We also know that AC is one of the diagonals. To find the length of the other diagonal, we can use the Pythagorean theorem.
Let x be half the length of the other diagonal:
AC^2 = (2x)^2 + 5^2
x^2 = (AC^2 - 25) / 4
x = sqrt((AC^2 - 25) / 4)
Now that we have the lengths of both diagonals:
Area = (AC x BD) / 2
= (AC x 2x) / 2
= xAC
= sqrt((AC^2 - 25) / 4) x AC
≈ 19.80 mm^2
Therefore, the area of rhombus ABCD is approximately 19.80 square millimeters.