Q: What is the area of a triangle 4 in by 2 in?

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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)

3 units2

The area of a right-angle triangle is half of the area of a rectangle 8cm x 4cm. Find the rectangle area and divide by half to find the area of the right-angle triangle. Therefore: 8 x 4 / 2 = 16cm2 * * * * * The above answer assumes that the two given lengths are the shorter legs of the triangle. It is, however, possible that these are the hypotenuse and one leg. In that case, the second leg is 4*sqrt(2) cm and the area of the triangle is 4 * 4*sqrt(2) / 2 = 8*sqrt(2) = 11.31 cm2.

Area of triangle = 1/2 base times height1/2 * 12 * 4 = 24

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.

Related questions

area of triangle =(1/2)(base)(height) 16=(1/2)(4)h h=8

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)

A triangle is half a square The area of a square is height × width(base) So, the area of a triangle is height × base ÷ 2 5×4÷2 = 10

area of a triangle = (base x height)/2 = (4 units x 14 units)/2 =28 square units

3 units2

The area of a right-angle triangle is half of the area of a rectangle 8cm x 4cm. Find the rectangle area and divide by half to find the area of the right-angle triangle. Therefore: 8 x 4 / 2 = 16cm2 * * * * * The above answer assumes that the two given lengths are the shorter legs of the triangle. It is, however, possible that these are the hypotenuse and one leg. In that case, the second leg is 4*sqrt(2) cm and the area of the triangle is 4 * 4*sqrt(2) / 2 = 8*sqrt(2) = 11.31 cm2.

The formula for finding the Area of a Triangle is:Base X Height / 2= (9 X 4) / 2= 36 / 2= 18 u2

Area = side^2 x (√3)/4

Area of triangle = 1/2 base times height1/2 * 12 * 4 = 24

area of triangle = half (base*height) i.e. 1/2*(4*4)= 8 square inch.

The formula for the area of a triangle is A = ½bh. We can use this formula to find the height, h, of the triangle: 16 = ½*4*h 16=2h /2 /2 8=h The height of the triangle is 8 units.

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.