if the areas of two similar triangles are equal,prove that they are equal.?

Question

therom

Answer 1

Let's see if we can prove this with algebra.

One of the formulas for the area of a triangle involves only one side:

Area = a^2*sin(B)*sin(C)/(2*sin(B+C))

Since the given triangles are similar, angles B and C are the same in the two triangles, and thus the trigonometric functions of those angles are equal. Now if you say that the areas of the two triangles are equal, the side "a" must be the same in the two triangles. Thus the two triangles are equal by angle-side-angle.

Sam's solution, below, is longer than mine, but it is nicer in the sense that it uses the simpler and well known formula for the area of a triangle.

Answer 2

By the term "equal", if you mean "congruent",

We already know that similar triangles have the same corresponding angles.

Also, The area is equal implies that 1/2*h*b values of both triangles are equal. Since one of the sides (the base b) is equal in both triangles, along with the two anges formed on its ends, the triangles are congruent.

Answer 3

A similar triangle is a triangle that has its lengths in proportion to the other triangle. If the areas are the same, i.e. bxh/2=bxh/2, and the triangles are similar, then the triangles are simply reflections or rotations of each other. A triangle of length 1,2,3 cm would be similar to a triangle of lengths 2,4,6, but have a different area.

Answer 4

if triangles 1 and 2 are similar, then we can pick a base and height for each such that B1 = B2 * k and H1 = H2 * k, where k is the scale factor relating their absolute size.

If area is computed as (B * H) / 2, and if they have equal area, then:

(B1 * H1)/2 = (B2 * H2)/2

substituing we get

(B2 * K * H2 * k)/2 = (B2 * H2)/2

or

k*k*( B2 * H2 ) /2 = (B2 * H2)/2

and dividing both sides by (B2 * H2)/2 leaves k*k = 1, or k = 1 (or -1) but I claim we care only about magnitude.

hence B2 == B1 and H2 == H1

hence they are not only similar, but identical.

Answer 5

sure they're similar no longer because they have a basic aspect yet because they have each and each and every of the angles equivalent. If 2 triangles have each and each and every of the angles equivalent then they're similar. i became no longer waiting to appreciate your added ingredient so ought to you please write it in a proper way.

Answer 6

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