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The formula to calculate the area of a triangle is 1/2 * base * height. To understand this, think of a rectangle or a square. To calculate the area of this object you would use length * width (which is the same as base * height). If you cut this object in half, you get a triangle. So that area of any triangle is 1/2 * base * height. I cannot answer your question because you are missing the triangle's height but you should be able to use the formula above to calculate the answer on your own.

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Q: What is the area of a triangle with base 12cm and height?
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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


Consider an isosceles triangle whose two equal sides are of length 4 What is the largest possible area for such a triangle?

An isosceles triangle with side length 4 has an altitude x. By the Pythagorean theorem, the base of the triangle is 2*SQRT(16-x2). The area of the triangle is 1/2 base times height, so A=x*(16-x2)1/2. the derivative, dA/dx=(16-x2)1/2 - x2/(16-x2)1/2. This is found with the product rule and chain rule. This shows the rate which the area of the triangle changes with respect to the altitute. At the x value of the maximum, the area will have stopped increasing and begun to decrease, so the rate of increase wil be zero. We just need to solve for x. (16-x2)1/2 - x2/(16-x2)1/2=0 (16-x2)1/2=x2/(16-x2)1/2 (16-x2)=x2 16=2x2 8=x2 SQRT(8)=x. Now we can solve the original equation for the maximum are. SQRT(8)*SQRT(16-8) SQRT(8)*SQRT(8)=8 So 8 is the largest possible area.


Find the area of each what A circle of radius 3cm A rectangle that has a length of 14 inches and a width of 2 inches A triangle with a base of 0.4 feet and a height of 0.7 feet?

1. The formula for finding the Area of a circle is Pie X R-squared. R 3 cm squared is 9cm X 3,14 (value of Pie) = 28.26 square inches. 2. The formula for finding the Area of a rectangle is Length X Width (LXW): 14 X 2 = 28 square inches. 3.The answer is...900 ins. are in 75 ft. 24


What are the impact of center of mass in a real life application?

Balance and stability. The centre of mass of an object must lie within the area of the object's base otherwise the object is unstable.


What is log 100 base e?

log 100 base e = log 100 base 10 / log e base 10 log 100 base 10 = 10g 10^2 base 10 = 2 log 10 base 10 = 2 log e base 10 = 0.434294 (calculator) log 100 base e = 2/0.434294 = 4.605175