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How would you construct a right triangle given the length of a leg and the radius of the circumscribed circle?
Wiki User
∙ 14y ago
To construct a right triangle given the radius of the circumscribed circle and the length of a leg, begin with two ideas. First, the diameter of the circle is equal to twice the radius. That's pretty easy. Second, the diameter of the circle is the length of the hypotenuse. The latter is a key to construction. Draw your circle, and draw in a diameter, which is the hypotenuse of the right triangle, as was stated. Now set you compass for the length of the leg of the triangle. With this set, place the point of the compass on one end of the diameter (the hypotenuse of your triangle), and draw an arc through the circumference of the circle. The point on the curve of the circle where the arc intersects it will be a vertex of your right triangle. All that remains is to add the two legs or sides of the triangle. Draw in line segments from each end of the hypotenuse (that diameter) to the point where your arc intersected the curve of the circle. You've constructed your right triangle. Note that any pair of lines that is drawn from the ends of the diameter of a circle to a point on the curve of the circle will create a right triangle.
Wiki User
∙ 14y ago
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How do you construct a circle that has radius with an equal length to the given line segment?
Adjust the compass to the given line segment then construct the circle.
How do you use a straightedge and a compass to construct a triangle with 2 sides having the same length?
first draw your baseline, then set your compasses to say twice that length, then draw a circle of the same radius from each end of the baseline, use either intersection as the apex of your isosceles triangle.
The side of a square is 20 cmfind the areas of the circumscribed and inscribed circles?
The radius length r of the inscribed circle equals to one half of the length side of the square, 10 cm. The area A of the inscribed circle: A = pir2 = 102pi ≈ 314 cm2 The radius length r of the circumscribed circle equals to one half of the length diagonal of the square. Since the diagonals of the square are congruent and perpendicular to each other, and bisect the angles of the square, we have sin 45⁰ = length of one half of the diagonal/length of the square side sin 45⁰ = r/20 cm r = (20 cm)(sin 45⁰) The area A of the circumscribed circle: A = pir2 = [(20 cm)(sin 45⁰)]2pi ≈ 628 cm2.
How do you construct a 3 4 5 triangle Read?
It will be a right angled triangle with sides of 3 and 4 units with an hypotenuse of 5 units in length.
How do you construct a triangle when the only measurements given are the length of the sides?
Draw a horizontal line AB equal to one of the side lengths. From A draw an arc of a circle of radius one of the remaining lengths. From B draw an arc of a circle of radius the third length. Where the arcs intersect is point C. Join AC and BC. Voila!
How do you construct a circle that has radius with an equal length to the given line segment?
Adjust the compass to the given line segment then construct the circle.
How do you use a straightedge and a compass to construct a triangle with 2 sides having the same length?
first draw your baseline, then set your compasses to say twice that length, then draw a circle of the same radius from each end of the baseline, use either intersection as the apex of your isosceles triangle.
The side of a square is 20 cmfind the areas of the circumscribed and inscribed circles?
The radius length r of the inscribed circle equals to one half of the length side of the square, 10 cm. The area A of the inscribed circle: A = pir2 = 102pi ≈ 314 cm2 The radius length r of the circumscribed circle equals to one half of the length diagonal of the square. Since the diagonals of the square are congruent and perpendicular to each other, and bisect the angles of the square, we have sin 45⁰ = length of one half of the diagonal/length of the square side sin 45⁰ = r/20 cm r = (20 cm)(sin 45⁰) The area A of the circumscribed circle: A = pir2 = [(20 cm)(sin 45⁰)]2pi ≈ 628 cm2.
What are the dimensions of an isosceles triangle of least area that can be circumscribed about a circle of radius r?
The isosceles triangle of least area that can be circumscribed about a circle of radius r turns out to be not just isosceles, but also equilateral. Each side has length 2r x ( 3 )0.5 . The area is r2 x (27)0.5 . Thanks are due to litotes for pointing out that the original answer did not actually answer the question ! tpm Since the equilateral triangle is also an isosceles triangle, we can say that at least area that can be circumscribed to a circle is the area of an equilateral triangle.If we are talking only for isosceles triangle where base has different length than two congruent sides, we can say that at least area circumscribed to a circle with radius r, is the area of an isosceles triangle whose base angles are very close to 60 degrees. Solution: Let say that the isosceles triangle ABC is circumscribed to a circle with radius r, where BA = BC. We know that the center of the circle inscribed to a triangle is the point of the intersection of the three angle bisectors of the triangle. Let draw these angle bisectors, and denote with D the point where the bisector drawn from the vertex, B, of the triangle, intersects the base AC. Since the triangle is an isosceles triangle, then BD bisects the base and it is perpendicular to the base. So that AD = DC, OD = r, and the triangles ADB and AOD are right triangles (O is the center of the circle). In the triangle ADB, we have:tan A = BD/AD, so that AD = BD/tan A In the triangle AOD, we have:tan A/2 = OD/AD, so that AD = r/tan A/2, and AC = 2r/tan A/2 Therefore,BD/tan A = r/tan A/2, andBD = (r tan A)/tan A/2 Area of triangle ABC = (1/2)(AC)(BD) = (1/2)(2r/ tan A/2)[(r tan A)/tan A/2] = (r2 tan A)/tan2 A/2 After we try different acute angles measure, we see that the smallest area would be: If the angle A= 60⁰,then the Area of the triangle ABC = r2 tan 60⁰/tan2 30⁰ ≈ 5.1961r2 If the angle A= 59.8⁰,then the Area of the triangle ABC = (r2 tan 59.8⁰)/tan2 29.9⁰ ≈ 5.1962r2
How do you construct a 3 4 5 triangle Read?
It will be a right angled triangle with sides of 3 and 4 units with an hypotenuse of 5 units in length.
How do you construct a triangle when the only measurements given are the length of the sides?
Draw a horizontal line AB equal to one of the side lengths. From A draw an arc of a circle of radius one of the remaining lengths. From B draw an arc of a circle of radius the third length. Where the arcs intersect is point C. Join AC and BC. Voila!
When a triangle with one side length the diameter of a circle is inscribed in that circle does it have to be a right triangle?
Yes. It follows from one of the circle theorems which states that the angle subtended in a semicircle is a right angle.
When a circle is drawn through each vertex of a right triangle the triangles hypotenuse will be equal to what?
The length of the circle's diameter
Is it true or false that the radius of the circle inscribed an equilateral triangle equals one-half the length of the length of the triangles median?
This is true. The answer is obvious if you think about it the following way: an equilateral triangle has three equal sides, and every point on the circumference of a circle is the same distance from the center of the circle. Therefore, it is safe to assume that the circle will touch the midpoint of each side of the triangle. It also means that the center of the circle will be in the center of the triangle. Therefore, the radius of the circle will travel from the center of the triangle to the midpoint of one of the sides. This will cover the distance of half the triangle's median.
What is the largest angle of a triangle with sides of 5.8cm by 14.1cm by 8.3cm showing work?
Such a triangle would be impossible to construct with the given 3 dimensions because in order to construct a triangle the sum of its 2 shortest sides must be greater than the length of its longest side.
How do you find the area of a hexagon that is inscribed in a circle?
The area of any hexagon is 6(0.5)(L)(L sin 60o) = 3L2 sin 60o, where L is the length of one side and is also the radius of the circumscribed circle.
What side length can make a triangle with side lengths 8 and 12?
In order to construct a triangle the sum of its 2 smallest sides must be greater than its longest side.
