Large Sides Means Large Area, Right?

Both the triangles are isosceles triangles, which means that median of the triangle on the non-equal side is also the normal. We have two pictures now:

Thus both the triangles have the same area now, which 12*5 = 60 sq units.

By Heron's formula:

Triangle A:

$S_{1}=\frac{13+13+10}{2}=\frac{36}{2}=18$

Area $(A_{1})=\sqrt{S(S-a)(S-b)(S-c)}=\sqrt{18(18-13)(18-13)(18-10)}$

$=\sqrt{18\times5\times5\times8}$

$=\sqrt{2\times3^{2}\times5^{2}\times2^{3}}$

$=\sqrt{2^{4}\times3^{2}\times5^{2}}$

$=2^{2}\times3\times5=4\times15=60$

Triangle B:

$S_{2}=\frac{13+13+24}{2}=\frac{50}{2}=25$

Area $(A_{2})=\sqrt{S(S-a)(S-b)(S-c)}=\sqrt{25(25-13)(25-13)(25-24)}$

$=\sqrt{25\times12\times12\times1}$

$=\sqrt{5^{2}\times12^{2}\times1^{2}}$

$=5\times12\times1=60$

$A_{1}=A_{2}=60$

Thus, the answer is $\boxed{Equal}$

The first method is easy the triangles are both right triangles with side lengths (5,12,13).But if you didn't realize that First find the length of the height and then you are going to realize that the base and the height are the same for the two triangles proving that the area would be same.The length and base should be be 12 and 5 in both triangles. In advance you could use herons formula

Check out the 5:12:13 right triangle in both cases.

Very easy use herons formula

Triangle A. (Semi perimeter=18) area=√18(18-13)(18-13)(18-10) =√18 5 5*8 =√3600 =60

Triangle B (semi perimeter =25) Area=√25(25-13)(25-13)(25-24) =√25 12 12*1 =√3600 =60

1. draw a construction line perp from mid of the base of the triangles,
2. find the height of a triangle and simply
3. then apply formula of area of triangle, as A= (Height x Base)/2
   for both the triangle the procedure used is same and then
   
4. campare the both Areas of a Traingle

The height of an isosceles triangle can be calculated by using pythagoras theorem. Consider the unequal side as the base. One of the equal sides and half of the length of the unequal side and the height form a right triangle. The equal side serves as the hypotenuse.

Hence, Square(Equal Side)= Square(Height) + Square(Half of Base or Unequal Side)

Thus, for each of the triangles, Height = 12 , Base = 10 and area = 60 (Triangle 1) [square(13)=square(12-height)+square(5-half of base)] Height = 5, Base=24 and area=60 (Triangle 2).[square(13)=square(5-height)+square(12-half of base)]

Therefore the triangles have equal area.

In an isosceles triangle the altitude bisects the dissimilar base.

Use c^2 = a^2 + b^2 to solve for the altitudes.

Then find the areas. They both equal 60.

only one property of an isosceles triangle can make this just a two lines sum........that is the perpendicular when dropped on the base divides the base into two equal parts.............in the first triangle.half of the base is 5 and its ht is 12.............in second triangle.half of the base is 12 and its ht is 5............now area of a triange is 1/2.base .altitude value of whose is same for both triangles ........thats it...........!!!!!!!!

In the figure below, the vertical to the base of the isosceles triangle, h = sqrt(a^2-(b/2)^2). The area of the triangle is (h * b) / 2 = (sqrt(a^2-(b/2)^2) * b) / 2. Replacing a with 13 and b with either 10 or 24, the result is always 60.

Think about Pythagorean triples! (5,12,13) is a triple so it immediately gives the answer.

Bisect them and you get two triangles, 5-12-13 on the sides.

Both are isosceles triangle. For triangle A find height =sqrt(13^2 - 5^2) =12, so area of triangle 'A'= 1/2 (base)(height)=1/2 (10)(12)=60; for triangle 'B' height= sqrt(13^2 - 12^2)=5, hence area of 'B'=1/2(24)(5)=60.

Use heron's formula and find out

Use Heron's Formula. Most Mathematical to Prove!!

it is very is problem with some calculation by determine area using heron's formula. s=\frac{a+b+c}{2}. A = \sqrt{s(s-a)(s-b)(s-c)},

Heron's Formula

Bashed it out with Heron's. I'm not sure exactly why, but somehow they are equal. :D