Asked by K
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
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
Answers
Answered by
Steve
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.
(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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.