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What is the area of triangle ABC if AB is tangent to the circle?
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How can it be proved that the circumradius of a triangle is the product of three sides divided by four times the area of the triangle?
We know that R = a/2sinA area of triangle = 1/2 bc sinA sin A = 2(area of triangle)/bc R = (a/2)*2(area of triangle)/bc R = abc/4*(area of triangle)
The radius of a incircle is 14cm what is the side of the equilateral triangle abc?
Where the side of the equilateral triangle is s and the radius of the inscribed circle is r:s = 2r * tan 30° = 48.50 cm
Which side lengths form a triangle that is similar to triangle ABC?
Any triangle whose sides are in the same ratio with the corresponding sides of ABC.
Triangle ABC has two sides that measure 12 inches and 16 inches. which could be the measures of the corresponding sides of a triangle that is similar to triangle ABC?
One pair of corresponding sides of a triangle that are similar to triangle ABC are 6 inches and 8 inches. However, there are multiple answers to this question.
Can triangle abc be a right triangle if side ac2 side bc3 and side ab4?
No.
Prove that the tangent at a to the circumcircle of triangle abc is parallel to bc where triangle abc is a isosceles triangle?
The circumcircle of a triangle is the circle that passes through the three vertices. Its center is at the circumcenter, which is the point O, at which the perpendicular bisectors of the sides of the triangle are concurrent. Since our triangle ABC is an isosceles triangle, the perpendicular line to the base BC of the triangle passes through the vertex A, so that OA (the part of the bisector perpendicular line to BC) is a radius of the circle O. Since the tangent line at A is perpendicular to the radius OA, and the extension of OA is perpendicular to BC, then the given tangent line must be parallel to BC (because two or more lines are parallel if they are perpendicular to the same line).
What is the area enclosed by a chord of a circle?
The area enclosed by a chord is equal to the area enclosed by a segment minus the area enclosed by the triangle with the same corners as the segment. To visualise it, draw a circle and put a chord on it. Label the chord AB and the centre of the circle C. The area of sector AB equal to the area of sector ABC minus the area of triangle ABC.
In a circle chord AC has a length of 6 If the length of arc ABC is equal to one-sixth of the circumference of the circle what is the area of the circle?
The triangle ABC is an equallateral triangle since angle ABC is one sixth of 360 degress of the circle and the angles BAC and BCA are equal of the remaining 180-60=120 degrees. With radius BC (or BA) being 6; the areaof the circle is pi (r)squared; 36 piArea of the circle is 36piMalcolm Lowe
How do you find the area of a triangle using cotangent?
(c2) / (2 cot A + cot B) = Area of Triangle ABC
How do you work out the area of an ABC triangle?
you just have to measure it in inches all the sidesAnother Answer:-Area of a triangle = 0.5*base*height
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
C program to find area of a triangle?
find the area of abc a[2,3] c[6,0]
Find the ratio of area of the triangles ABC and PQR such that AB = BC=CA = 6m each and PQ, QR, RP are half of the sides AB BC and CA respectively?
Since the sides of triangle are equal, the triangles are equilateral. Just for your information, in this question, we do not require the length of sides. It is just additional information. :) The area of equilateral triangle is: (√3)/4 × a², where a is the side of the equilateral triangle. For triangle ABC, area will be = (√3)/4 × a² (Let 'a' is the side of triangle ABC) Since, side of triangle PQR is half that of ABC, it will be = a/2 Therefore, area of triangle PQR = (√3)/4 × (a/2)² = (√3)/16 × a² Take the ratio of areas of triangle ABC and PQR: [(√3)/4 × a²] / [(√3)/16 × a²] = 4:1
How can it be proved that the circumradius of a triangle is the product of three sides divided by four times the area of the triangle?
We know that R = a/2sinA area of triangle = 1/2 bc sinA sin A = 2(area of triangle)/bc R = (a/2)*2(area of triangle)/bc R = abc/4*(area of triangle)
What type of triangle is angle ABC?
ABC angle is an angle,not a triangle!
If is both the altitude and median of triangle ABC then triangle ABC is?
It is isosceles.
If is both the altitude and median of triangle ABC then triangle ABC is .?
It is isosceles.
