1. The complement of an angle is 10 degrees less than the angle itself. Find the angle and it's complement.
Find the perimeter and area of the following figures with the vertices given:
a) A (3, 5) B (3, 1) C (0, 1)
b) Q (-3, 2) R (1, 2) S (1, -4) T (-3,-4)
c) G (-4, 1) H (4,1) I (0, -2)
d) K (-4, -2) L (-1,2) M (8,2) N (5, -2)
e) O (3,2) P (1,2) Q(1,-4) R(3,-4)
f) S (0,0) T(3,-2) U (8,0)
Please help. I already graphed the figures but I have no idea how to find the perimeters and areas. Don't just give me the answers, but also tell me how you found it. Thank you.
To find the angle and its complement, we can set up an equation. Let the angle be x degrees. Then the complement of the angle will be (x - 10) degrees.
According to the problem, the complement of the angle is 10 degrees less than the angle itself. We can write this as:
(x - 10) = x - 10
Simplifying the equation, we get:
x - 10 = x - 10
0 = 0
Since the equation is true for any value of x, it means that there are infinitely many solutions. Therefore, we cannot find a unique value for the angle and its complement.
Now, let's find the perimeter and area of the given figures:
a) A (3, 5) B (3, 1) C (0, 1)
To find the perimeter, we need to find the lengths of all sides AB, BC, and AC and add them up.
AB = |3 - 3| + |5 - 1| = 0 + 4 = 4
BC = |3 - 0| + |1 - 1| = 3 + 0 = 3
AC = |0 - 3| + |1 - 5| = 3 + 4 = 7
Perimeter of ABC = AB + BC + AC = 4 + 3 + 7 = 14
To find the area, we can use the Shoelace formula or the formula for the area of a triangle given its coordinates (Heron's formula). Let's use the Shoelace formula:
Area of ABC = 1/2 * |(3*1 + 3*1 + 0*5) - (5*3 + 1*3 + 1*0)| = 1/2 * |(3 + 3 + 0) - (15 + 3 + 0)| = 1/2 * |-6| = 3
b) Q (-3, 2) R (1, 2) S (1, -4) T (-3,-4)
Perimeter of QRST = QR + RS + ST + TQ
QR = |1 - (-3)| + |2 - 2| = 4 + 0 = 4
RS = |1 - 1| + |-4 - 2| = 0 + 6 = 6
ST = |-3 - 1| + |-4 - (-4)| = 4 + 0 = 4
TQ = |-3 - (-3)| + |-4 - 2| = 0 + 6 = 6
Perimeter of QRST = 4 + 6 + 4 + 6 = 20
To find the area of QRST, we can split it into two triangles: QRS and STQ. Then use the Shoelace formula to find the area of each triangle and sum them up:
Area of QRS = 1/2 * |(-3*2 + 1*2 + 1*(-4)) - (2*1 + 1*(-4) + (-3)*1)| = 1/2 * |(-6 + 2 - 4) - (2 - 4 - 3)| = 1/2 * |-10| = 5
Area of STQ = 1/2 * |(-3*(-4) + (-3)*1 + 1*1) - ((-4)*1 + 1*(-3) + 1*(-3))| = 1/2 * |(12 - 3 + 1) - ((-4) + (-3) + (-3))| = 1/2 *|16| = 8
Area of QRST = Area of QRS + Area of STQ = 5 + 8 = 13
c) G (-4, 1) H (4,1) I (0, -2)
Perimeter of GHI = GH + HI + IG
GH = |4 - (-4)| + |1 - 1| = 8 + 0 = 8
HI = |4 - 0| + |1 - (-2)| = 4 + 3 = 7
IG = |0 - 4| + |-2 - 1| = 4 + 3 = 7
Perimeter of GHI = 8 + 7 + 7 = 22
To find the area, we can split it into two triangles: GHI and GII'. The line segment II' is the y-axis.
Area of GHI = 1/2 * |(-4*1 + 4*1 + 0*(-2)) - (1*4 + 1*0 + (-4)*(-2))| = 1/2 * |(-4 + 4 + 0) - (4 + 0 + 8)| = 1/2 * |-12| = 6
Area of GII' = 1/2 * |(-4*(-2) + (-4)*0 + 0*1) - (1*(-4) + 1*0 + (-4)*1)| = 1/2 * |(8 + 0 + 0) - (-4 + 0 - 4)| = 1/2 * |16| = 8
Area of GHI = Area of GHI + Area of GII' = 6 + 8 = 14
Continued...
Sure! Let's start with the first question.
1. The complement of an angle is 10 degrees less than the angle itself. Find the angle and its complement.
Let's assume the angle is represented by 'x' degrees. According to the given information, the complement of the angle is 10 degrees less than the angle itself. Mathematically, this can be represented as (x - 10) degrees.
Since the sum of the angle and its complement is 90 degrees (as they form a right angle), we have the equation:
x + (x - 10) = 90
Simplifying the equation:
2x - 10 = 90
Adding 10 to both sides:
2x = 100
Dividing both sides by 2:
x = 50
Therefore, the angle is 50 degrees, and its complement is (50 - 10) = 40 degrees.
Now let's move on to the second part of your question:
2. Find the perimeter and area of the given figures with the vertices provided.
To find the perimeter of a polygon with given vertices, you need to calculate the sum of the lengths of all the sides. For the area, you can use the formula specific to each shape.
a) A (3, 5) B (3, 1) C (0, 1)
To find the perimeter, calculate the distance between each pair of consecutive vertices and sum them up:
Distance between A and B: √((3-3)^2 + (5-1)^2) = √(0^2 + 4^2) = 4 units
Distance between B and C: √((3-0)^2 + (1-1)^2) = √(3^2 + 0^2) = 3 units
Distance between C and A: √((0-3)^2 + (1-5)^2) = √((-3)^2 + (-4)^2) = √(9 + 16) = √25 = 5 units
Perimeter = 4 + 3 + 5 = 12 units
To find the area, you can divide the polygon into two triangles. You can calculate the area of each triangle using the formula: (1/2) * base * height.
Triangle ABC:
Base = Distance between A and B = 4 units
Height = Distance between A and C along the y-axis = |5 - 1| = 4 units
Area of Triangle ABC = (1/2) * 4 * 4 = 8 square units
Now, you can continue using the above process to find the perimeter and area for the remaining figures (b, c, d, e, and f) using the given vertices.