An equilateral triangle is inscribed in a circle of radius 4r. Express the area A within the circle but outside the triangle as a function of the length 5x of the side of the triangle.

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Malena · Accepted Answer

In equilateral triangle:  Γ radius =side3  4γ=5x3  4γ=5x43  A=π×γ2=π(5x43)2  A=25x2π48  A=34×(side)2=34×(5x)2=253×x24  A=25π×x248–2534×x2

RizerMix · Accepted Answer

Result:A=16π·r2−2534·x2Solution:The radius of the circle is given as 4r. Since the triangle is inscribed in the circle, each vertex of the triangle lies on the circumference of the circle. Therefore, the distance from the center of the circle to any vertex of the triangle is equal to the radius.Using this information, we can draw a line segment from the center of the circle to one of the vertices of the equilateral triangle, creating a right triangle. The hypotenuse of this right triangle is the radius of the circle (4r), and one of the legs is half the side length of the equilateral triangle (5x2).We can use the Pythagorean theorem to find the length of the other leg of the right triangle. Let&#039;s denote this length as y.By applying the Pythagorean theorem, we have:(5x2)2+y2=(4r)2Simplifying the equation, we get:25x24+y2=16r2Now, let&#039;s find the area of the equilateral triangle. The formula for the area of an equilateral triangle is:Area=34·(side&amp;nbsp;length)2Substituting the given side length of 5x, we get:Area=34·(5x)2=2534·x2Next, we need to find the area within the circle but outside the equilateral triangle. This can be done by subtracting the area of the equilateral triangle from the area of the circle.The formula for the area of a circle is:Area=π·(radius)2Substituting the given radius of 4r, we get:Area=π·(4r)2=16π·r2Finally, we can express the area A within the circle but outside the triangle as a function of the side length 5x of the triangle:A=16π·r2−2534·x2Thus, the area within the circle but outside the triangle is given by A=16π·r2−2534·x2.

Vasquez · Accepted Answer

Step 1:The formula for the area of an equilateral triangle with side length s is given by:Atriangle=34·s2In this case, the side length of the triangle is 5x. Therefore, the area of the equilateral triangle is:Atriangle=34·(5x)2Step 2:Now, let&#039;s find the area within the circle but outside the triangle. This area can be obtained by subtracting the area of the triangle from the total area of the circle.The formula for the area of a circle with radius r is given by:Acircle=π·r2In this case, the radius of the circle is 4r. Therefore, the area of the circle is:Acircle=π·(4r)2To find the area within the circle but outside the triangle (A), we subtract the area of the triangle from the area of the circle:A=Acircle−AtriangleSubstituting the formulas for Acircle and Atriangle:A=π·(4r)2−34·(5x)2Hence, the area within the circle but outside the triangle (A) as a function of the length 5x of the side of the triangle is given by:A=π·(4r)2−34·(5x)2Please note that this solution assumes that the center of the circle coincides with the centroid of the equilateral triangle.

Don Sumner · Accepted Answer

The radius of the circle is given as 4r, which means the diameter is 2(4r)=8r. Since the triangle is inscribed in the circle, each side of the equilateral triangle is equal to the diameter, which is 8r.The area of an equilateral triangle with side length 5x can be calculated using the formula: Atriangle=34(5x)2.In this case, we have Atriangle=34(5x)2=34(25x2)=253x24.Now, we can find the area within the circle but outside the triangle. This region can be calculated by subtracting the area of the triangle from the area of the entire circle.The area of the circle is given by the formula: Acircle=πr2.Since the radius of the circle is 4r, the area of the circle becomes Acircle=π(4r)2=π(16r2)=16πr2.To find the area within the circle but outside the triangle, we subtract the area of the triangle from the area of the circle: A=Acircle−Atriangle.Substituting the previously calculated values, we have A=16πr2−253x24.Therefore, the area A within the circle but outside the triangle is given by the equation: A=16πr2−253x24.

An equilateral triangle is inscribed in a circle of radius 4r. Express the area A within the circle but outside the triangle as a function of the length 5x of the side of the triangle.

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