# Finding the diagonals of a rhombus with side length $13$, where the sum of the diagonals is $34$ [closed]

How can we find diagonals of the rhombus with side length $$a=13$$ cm and sum of diagonals $$d_1+d_2=34$$ cm?

Anything doesn't seem to work...

I would really appreciate it, if anyone could help me / solve it :) Since I have only basic knowledge in geometry, please post your EASIEST answer possible.

## closed as off-topic by Henrik, Abcd, metamorphy, Cesareo, IuliJan 12 at 9:25

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Results on quadratic equations make this problem easy to solve:

• As the diagonals in a rhombus are perpendicular, applying Pythagoras results in $$d_1^2+d_2^2=4\cdot13^2.$$
• On the other, the hypothesis $$\;d_1+d_2=34$$ implies $$4\cdot 17^2=(d_1+d_2)^2=d_1^2+d_2^2+2d_1d_2=4\cdot13^2+2d_1d_2,$$ so that $$\;d_1d_2=240$$. Therefore, it comes down to the standard problem of finding two numbers , given their sum $$s$$ and their product $$p$$. We know they're the roots (if any) of the quadratic equation $$x^2-sx+p=0.$$
• Yes! Thank you for "dumbing" it down for me :) – Jaanis Soosaar Jan 11 at 20:10

Hint: The diagonals of a rhombus are perpendicular. Set up a system of equations using the given information and the Pythagorean theorem.

You can also use the theorem of cosines: $$d_1^2=13^2+13^2-2\cdot 13^2\cos(180^{\circ}-\alpha)$$

$$d_2^2=13^2+13^2-2\cdot 13^2\cos(\alpha)$$ and use that $$d_1+d_2=34$$ and note that $$\cos(\pi-x)=-\cos(x)$$

Here's how to do this problem the wrong way. Be suspicious of the number $$13$$ and remember that "$$5$$-$$12$$-$$13$$" forms a right triangle. Guess that the diagonals split the rhombus into $$4$$ seperate "$$5$$-$$12$$-$$13$$" right triangles, which implies diagonals of lengths $$10$$ and $$24$$, which do actually add to $$34$$, so that must be the correct answer.

Perpendicular lengths of rhombus semi diagonals $$(x,y)$$ can be mentally solved

$$2x+2y=34\quad x+y=17 \tag1$$

Perimeter

$$4 \sqrt{x^2+y^2} = 4 (13)\quad \rightarrow {x^2+y^2}=13^2$$

Using identity

$$2xy= (x+y)^2-(x^2+y^2) =17^2-13^2= 30(4)=120 \rightarrow xy=60 \tag2$$

Since the sum and product of $$x$$ and $$y$$ are known we can factorize

$$(x-5)(y-12)=0 \rightarrow d_1= 2x =10, d_2=2y= 24. \tag3$$