Happy diamond

$\text{Solution 1:}$

By Pythagorean theorem $a^2+b^2=13^2$ and the problem $2(a+b)=34$ , so $2ab=(a+b)^2-(a^2+b^2)=17^2-13^2=120$ where $ab=60$ . $\therefore$ rhombus area is $4\frac{ab}{2}=4(30)=\boxed{120}$

$\text{Solution 2:}$

Also we can get diagonal length by Pythagorean theorem. Make $b=17-a$ then $(17-a)^2+a^2=13^2 \Rightarrow 60-17a+a^2=(a-12)(a-5)=0$ so $a=12,5$ and $b=5,12$ respectively. $\therefore$ rhombus area is $\frac{2a\times2b}{2}=\frac{10\times240}{2}=\boxed{120}$

Consider the diagram. From the problem, $2a+2b=34$ or $a+b=17$ or $a=17-b$ . By pythagorean theorem, we have

$13^2=a^2+b^2$

$169=(17-b)^2+b^2$

$b^2-17b+60=0$

$b-5=0$ $\implies$ $b=5$

$b-12=0$ $\implies$ $b=12$

Based form the diagram, $b>a$ , therefore $b=12$ and $a=5$ . It follows that $2a=2(5)=10$ and $2b=2(12)=24$ .

The area of a rhombus is half the product of its diagonals. Therefore

$A=\dfrac{1}{2}(2a)(2b)=\dfrac{1}{2}(10)(24)=$ $\boxed{120}$

Una solución fácil y rápida es buscar una terna pitagórica con C=13, la cual seria 5^2 + 12^2=13^2, 5 y 12 serian la mitad de las diagonales, asi que las multiplicamos X2, ya solo nos queda aplicar la formula (Dxd/2)(24x10/2)=120

Let $a$ and $b$ be the diagonals of rhombus.

Thus, $Area = \dfrac{ab}{2}$ , so we need to manipulate the value of $\dfrac{ab}{2}$

We know that $4s^2 = a^2+b^2$ and $a+b = 34$ , and we have $s = 13$

Since $a^2+b^2 = (a+b)^2 - 2ab$

So $4s^2 = (a+b)^2 - 2ab$

Substitute all the given values to get,

$4\times13^2 = 34^2 - 2ab$

$\dfrac{ab}{2}=\boxed{120}$

it's simple with picture so that you can save this..... :-)