1) The base angles of an isosceles triangle are (50-x)^6 and (30x-12)^6 what is the vertex angle?
2) The area of a trapezoid is 52cm^2 if it's bases are 8 cm apart find the sum of their heights
3) What is the area of a circle inscribed inside an equilateral triangle whose area is 4√3 cm^2
Please show solution
1) the base angles are equal, so
(50-x)^6 = (30x-12)^6
50-x = 30x-12
...
Find x, evaluate the angles, and subtract them from 180
2) I assume you mean lengths. If they are b and B,
8(b+B)/2 = 52
3)The medians of a triangle meet 2/3 of the way from a vertex to the opposite side. This triangle's medians are also its altitude. The side length is 4, so its altitude is 2√3. Thus, the radius of the circle is 2/3 of that.
1) To find the vertex angle of an isosceles triangle, we need to find the value of x that makes the base angles equal.
Given: Base angles are (50 - x)^6 and (30x - 12)^6.
Since the base angles of an isosceles triangle are equal, we can set them equal to each other:
(50 - x)^6 = (30x - 12)^6
Now, we can solve this equation. However, finding the value of x for this equation might require numerical methods or complex computational techniques. It is not feasible to solve this equation step-by-step using simple algebraic methods.
2) The area of a trapezoid is given by the formula: Area = (1/2)(sum of bases)(height)
Given: Area = 52 cm^2 and the bases are 8 cm apart.
Let the lengths of the two bases be b1 and b2, and let h1 and h2 be the heights corresponding to the bases b1 and b2, respectively.
We are given that the bases are 8 cm apart, so b2 - b1 = 8.
The sum of the heights (h1 + h2) can be determined by rearranging the formula for the area:
Area = (1/2)(b1 + b2)(h1 + h2)
Plugging in the given values, we have:
52 = (1/2)(b1 + b2)(h1 + h2)
We also have the relationship: b2 - b1 = 8
Solving these two equations simultaneously will give us the values of h1 and h2, and we can then determine their sum.
3) The area of an equilateral triangle can be calculated using the formula: Area = (sqrt(3)/4)(side length)^2
Given: The area of the equilateral triangle is 4√3 cm^2.
We can set up an equation using the area formula for the equilateral triangle:
4√3 = (sqrt(3)/4)(side length)^2
To find the side length, we need to solve this equation for side length:
4√3 = (sqrt(3)/4)(side length)^2
Multiplying both sides by 4/√3, we have:
4(4/√3) = (side length)^2
16/√3 = (side length)^2
Taking the square root of both sides, we get:
side length = sqrt(16/√3)
Finally, we can calculate the area of the inscribed circle in the equilateral triangle. The radius of the inscribed circle is equal to 1/3 of the height of the equilateral triangle. Since the area of the equilateral triangle is given as 4√3 cm^2, we can determine the height of the equilateral triangle using the formula: area = (√3/4)(side length)^2.
Substituting the given area value, we have:
4√3 = (√3/4)(side length)^2
Solving for side length, we get:
(side length)^2 = (4√3)/(√3/4)
Taking the square root of both sides, we get:
side length = √(4)/(√(√3/4))
Finally, the radius of the inscribed circle is (1/3) times the height of the equilateral triangle, which is equal to (√3/2)(side length). Plugging in the value of side length, we can calculate the radius. Then, we can calculate the area of the inscribed circle using the formula: Area = π(radius)^2.
1) To find the vertex angle of an isosceles triangle, we can use the fact that the sum of the measures of the angles in a triangle is always 180 degrees. In an isosceles triangle, the base angles are equal.
Given that the base angles of the triangle are (50-x)^6 and (30x-12)^6, we can set up the equation:
(50-x)^6 + (30x-12)^6 = 180
Then we solve this equation for x, and substitute the value of x back into (50-x)^6 to find the measure of the base angle. Finally, we can subtract the base angle from 180 to find the vertex angle.
2) The area (A) of a trapezoid is given by the formula: A = 1/2 * (b1 + b2) * h, where b1 and b2 are the lengths of the bases, and h is the height of the trapezoid.
In this case, we are given the area (A) as 52 cm^2 and the distance between the bases as 8 cm. Let's denote the height of the trapezoid as h1 and h2, where h1 is the height corresponding to b1, and h2 is the height corresponding to b2.
We can use the formula for the area of a trapezoid and the given information to set up the equation:
52 = 1/2 * (b1 + b2) * h
We know that the distance between the bases, b2 - b1, is 8 cm. So we can re-write the equation as:
52 = 1/2 * (b1 + b1 + 8) * h
Now, our equation depends on the sum of the bases. To simplify the equation, we need to express b1 in terms of h. We can use the fact that similar triangles have proportional sides.
Let's denote the height of the trapezoid as h1 and h2, and let's denote the shorter base as b1. We can set up a proportion between the height of the trapezoid and the corresponding segment on the longer base:
h1 / x = h2 / (x + 8)
We can cross-multiply and solve this proportion for h1 in terms of h2:
h1 = (h2 * x) / (x + 8)
Substitute this expression for h1 into the equation for the area of the trapezoid:
52 = 1/2 * (b1 + b1 + 8) * ((h2 * x) / (x + 8))
We can simplify this equation and solve for h2.
Once we have the values of h1 and h2, we can find their sum to get the sum of the heights of the bases.
3) To find the area of a circle inscribed inside an equilateral triangle, we can use the fact that the radius of the circle is equal to 1/3 times the height of the equilateral triangle.
Let A be the area of the equilateral triangle, and h be the height of the equilateral triangle. We have the equation:
A = (sqrt(3) / 4) * h^2
Given that the area of the equilateral triangle is 4√3 cm^2, we can substitute this value into the equation:
4√3 = (sqrt(3) / 4) * h^2
Simplifying this equation, we find:
h^2 = 16
Taking the square root of both sides, we have:
h = 4
The height of the equilateral triangle is 4 cm, and the radius of the inscribed circle is 1/3 * 4 = 4/3 cm.
To find the area of the circle, we can use the formula:
A = π * r^2
Substituting the value of the radius, we get:
A = π * (4/3)^2
Simplifying this expression, we find the area of the circle.