Given a right triangle with leg a = 6 and hypotenuse c = 10. Find the three heights, the three medians, the three bisectors, and the radii of the inscribed and circumscribed circles.

Step-by-step explanation:

a) The other leg of the right triangle is ...

  b = √(10² -6²) = √64 = 8

The area of the triangle is ...

  A = 1/2bh = 1/2(6)(8) = 24

Then the height to the long side is ...

  h = 2A/c = 48/10 = 4.8

The three heights are 4.8, 6, 8.

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b) The medians to the hypotenuse is half its length, because that is also the radius of the circumcircle: 10/2 = 5.

The medians to the legs are computed from the Pythagorean theorem:

  median to "b" = √(4² +6²) = √52 = 2√13

  median to "a" = √(3² +8²) = √73

The three medians are 5, 2√13, √73.

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c) The angle bisectors to the legs can be found in a fashion similar to the medians, using the angle bisector theorem to locate the point of intersection on the leg.

The angle bisector to "b" divides it into segments in the ratio 6:10, so the point  of intersection is 6/(6+10)×8 = 3 units from the right angle. Then its length is ...

  bisector to "b" = √(6² +3²) = √45 = 3√5

The angle bisector to "a" divides it into segments in the ratio 8:10, so the point of intersection is 8/(8+10)×6 = 8/3 units from the right angle. Then its length is ...

  bisector to "a" = √(8² +(8/3)²) = (8/3)√10

I find it easier to compute the bisector to the hypotenuse by considering the hypotenuse to be a line with x- and y-intercepts at 8 and 6. Then its equation can be 3x+4y=24, and its intersection with the line y=x will be x = y = 24/7. The  length of the segment to that point is ...

  bisector to "c" = (24/7)√2

The three angle bisectors are (24/7)√2, 3√5, (8/3)√10.

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d) The radius of the inscribed circle can be found from the formula ...

  r = √((s-a)(s-b)(s-c)/s) . . . .  where s is the semiperimeter = 24/2 = 12

  r = √(2×4×6/12) = 2

The hypotenuse is the diameter of the circumscribed circle, so the radius of that circle is c/2 = 5.

The radii of the incircle and circumcircle are 2 and 5, respectively.

b. an equilateral triangle inscribed in a circle

step-by-step explanation:

we are given the figure of two circles having a triangle inside.

now, we know that,

"when a figure is lying inside an another figure, then the outside figure is known as circumscribed figure and the figure lying inside is referred to as the inscribed figure".

as we see that, the circles are containing the triangle completely.

thus, the circles are the circumscribed figures and the triangle is the inscribed figure.

hence, we get, 'an equilateral triangle is inscribed in a circle'.

think about it deeply and passionately about this property

i am sorry