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For a triangle, the formula for the area is: A = (BH)/2

7*5/2 = 35/2 = 17.5

17.5 m

(If anyone else knows what a "sub" of a triangle is, please edit!)

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Q: What is the sub and area if the triangle is 5m Height and 7m base?
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What is the area of an equilateral triangle inscribed in a circle of radius 4cm?

Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.And so the area of the inscribed triangle is 12 cm2.Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.And so the area of the inscribed triangle is 12 cm2.Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.And so the area of the inscribed triangle is 12 cm2.Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.And so the area of the inscribed triangle is 12 cm2.


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Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.And so the area of the inscribed triangle is 12 cm2.Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.And so the area of the inscribed triangle is 12 cm2.Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.And so the area of the inscribed triangle is 12 cm2.Two vertices of the triangle and the centre of the circle make a smaller equilateral triangle with legs of 4 cm and the included angle is 360/3 = 120 degrees.Therefore the area of each of these sub-triangles = 1/2*ab*sin(C) = 1/2*4*4*sin(120) = 1/2*4*4*1/2 = 4 cm2.And so the area of the inscribed triangle is 12 cm2.


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