How can I find the diagonal of a quadrilateral?

Question

Given a quadrilateral $MNPQ$ for which $MN=26$, $NP=30$, $PQ=17$, $QM=25$ and $MP=28$ how do I find the length of $NQ$?

Answer 1

For the answer to be unique we must assume the quadrilateral is convex and/or simple, which you said in a comment that we can assume.

One way to solve the problem is to place your quadrilateral in a Cartesian coordinate system, perhaps placing $M$ at the origin and $P$ at $(28,0)$, and assume that point $N$ is in the upper half-plane. This fixes the quadrilateral, with point $Q$ in the lower half-plane (by convexity).

The law of cosines in triangle $\triangle MNP$ tells us

$$\cos(\angle NMP)=\frac{26^2+28^2-30^2}{2\cdot 26\cdot 28}=\frac 5{13}$$

and we get

$$\sin(\angle NMP)=\sqrt{1-[\cos(\angle NMP)]^2}=\frac{12}{13}$$

So point $N$ is

$$\left(26\cos(\angle NMP),\ 26\sin(\angle NMP) \right)=\left(26\cdot\frac5{13},\ 26\cdot\frac{12}{13}\right)=(10,24)$$

We can similarly find that point $Q$ is $(20,-15)$. Then use the standard distance formula to get

$$NQ=\sqrt{1621}\approx 40.2616$$

I'm sure there are shorter ways, but this way makes each step easy and checkable.

Answer 2

Use the cosine law to find the two partial angles at $M$. Then you know the total angle and you can find $NQ$.

$$\mu_0=\arccos\left(\frac{MQ^2+MP^2-PQ^2}{2MQ\cdot MP}\right)\\ \mu_1=\arccos\left(\frac{MP^2+MN^2-NP^2}{2MP\cdot MN}\right)\\ NQ=\sqrt{MQ^2+MN^2-2MQ\cdot MN\cos\left(\mu_0+\mu_1\right)}.$$

Answer 3

Hint: use Heron's Formula for the area of a triangle $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ $$s=\frac{a+b+c}{2}$$ Firstly find the total area by using the two triangles $MNP$ and $MQP$, then you can use same procedures for triangles $NMQ$ and $NPQ$ but now the area is known and the length of $NQ$ unknown.