Square \(ABCD\) has \(K\) on \(BC\), \(L\) on \(DC\), \(M\) on \(AD\), and \(N\) on \(AB\) such that \(KLMN\) forms a rectangle, \(\triangle AMN\) and \(\triangle LKC\) are congruent isosceles triangles, and also \(\triangle MDL\) and \(\triangle BNK\) are congruent isosceles triangles. If the total area of the four triangles is \(50\) cm\(^2\), what is the length of \(MK\)?
Let \(x\) represent the lengths of the equal sides of \(\triangle AMN\) and \(\triangle LKC\), and let \(y\) represent the lengths of the equal sides of \(\triangle MDL\) and \(\triangle BNK\).
Thus, area \(\triangle AMN =\) area \(\triangle LKC= \frac{1}{2} x^2\), and area \(\triangle MDL =\) area \(\triangle BNK = \frac{1}{2} y^2\).
Therefore, the total area of the four triangles is equal to \(\frac{1}{2} x^2+\frac{1}{2} x^2+\frac{1}{2} y^2+\frac{1}{2} y^2= x^2 + y^2\). Since we’re given that this area is \(50\) cm\(^2\), we have \(x^2 + y^2 = 50\).
Three different solutions to find the length of \(MK\) are provided.
Solution 1
In \(\triangle AMN\), \(MN^2=AM^2+AN^2 = x^2+ x^2\), and in \(\triangle BNK\), \(NK^2=BN^2+BK^2= y^2+y^2\).
Since \(MK\) is a diagonal of rectangle \(KLMN\), then by the Pythagorean Theorem we have \[\begin{aligned} MK^2 &=MN^2+NK^2\\ &=x^2+x^2+y^2+y^2\\ &=x^2+y^2+x^2+y^2\\ &=50+50\\ &=100 \end{aligned}\] Since \(MK>0\), we have \(MK= 10\) cm.
Solution 2
In \(\triangle AMN\), \(MN^2= x^2 + x^2=2x^2\). Therefore, \(MN=\sqrt{2}x\), since \(x >0\).
In \(\triangle BNK\), \(NK^2= y^2 + y^2= 2y^2\). Therefore, \(NK=\sqrt{2}y\), since \(y > 0\).
Since \(MK\) is a diagonal of rectangle \(KLMN\), then by the Pythagorean Theorem we have \[\begin{aligned} MK^2&=MN^2+NK^2\\ &=(\sqrt{2}x)^2+(\sqrt{2}y)^2\\ &=2x^2+2y^2\\ &=2(x^2+y^2)\\ &=2(50)\\ &=100 \end{aligned}\] Since \(MK>0\), we have \(MK= 10\) cm.
Solution 3
We construct the line segment \(KP\), where \(P\) lies on \(AD\) such that \(KP\) is perpendicular to \(AD\).
Then \(APKB\) is a rectangle. Furthermore, \(AP = BK = y\), \(PK = AB = x + y\), and \(PM = AM - AP = x - y\).
Since \(\triangle PKM\) is a right-angled triangle, by the Pythagorean Theorem we have \[\begin{aligned} MK^2&=PM^2+PK^2\\ &=(x-y)^2+(x+y)^2\\ &=x^2 - 2xy + y^2 + x^2 + 2xy + y^2\\ &=2x^2+2y^2\\ &=2(x^2+y^2)\\ &=2(50)\\ &=100 \end{aligned}\] Since \(MK>0\), we have \(MK= 10\) cm.