how high is rhombus if diagonals are 36 cm and 12cm
e = one diagonal
f = the other diagonal
\(e = 12\ cm\) and \(f = 36\ cm\)
The four sides all have the same length \( a\)
\(h\) ist the height of the rhombus
cos-rule
\(\begin{array}{rcll} e^2 &=& 2a^2 - 2a^2\cdot \cos{(A)} \\ f^2 &=& 2a^2 - 2a^2\cdot \cos{(B)} \\\\ 2A+2B &=& 360^\circ \\ A+B &=& 180^\circ \\ B &=& 180^\circ - A \\ \cos{(B)} &=& \cos{(180^\circ - A)} = -\cos{(A)} \\\\ e^2 &=& 2a^2 - 2a^2\cdot \cos{(A)} \\ f^2 &=& 2a^2 + 2a^2\cdot \cos{(A)} \\\\ \cos{(A)} = \frac{2a^2-e^2}{2a^2} &=& \frac{f^2-2a^2}{2a^2}\\ 2a^2-e^2 &=& f^2-2a^2 \\ 4a^2 &=& e^2 + f^2\\ 2a &=& \sqrt{ e^2 + f^2 } \\ \mathbf{a} &\mathbf{=}& \mathbf{\frac{ \sqrt{ e^2 + f^2 } } {2} }\\ \end{array}\)
\(\begin{array}{rcll} \cos{(A)} &=& \frac{2a^2-e^2}{2a^2} \\ \cos^2{(A)} &=& \frac{(2a^2-e^2)^2}{4a^4} \\ 1-\cos^2{(A)} &=& 1-\frac{(2a^2-e^2)^2}{4a^4} \\ 1-\cos^2{(A)} &=& \frac{4a^4 - (2a^2-e^2)^2}{4a^4} \\ 1-\cos^2{(A)} &=& \frac{4a^4 - 4a^2 + 4a^2e^2-e^4 }{4a^4} \\ 1-\cos^2{(A)} &=& \frac{4a^2e^2-e^4 }{4a^4} \\ 1-\cos^2{(A)} &=&\frac{ e^2\cdot (4a^2-e^2) }{4a^4} \qquad | \qquad 4a^2 = e^2 + f^2\\ 1-\cos^2{(A)} &=&\frac{ e^2\cdot ( e^2 + f^2-e^2) }{4a^4} \\ 1-\cos^2{(A)} &=&\frac{ e^2\cdot f^2 }{4a^4} \qquad | \qquad \sin^2{(A)}=1-\cos^2{(A)}\\ \sin^2{(A)} &=&\frac{ e^2\cdot f^2 }{4a^4} \\ \mathbf{\sin{(A)}} &\mathbf{=}&\mathbf{\frac{ e\cdot f }{2a^2} } \\ \end{array}\)
\(\begin{array}{rcll} h &=& a\cdot \sin{(A)} \\ h &=& a\cdot \frac{ e\cdot f }{2a^2} \\ h &=&\frac{ e\cdot f }{2a} \qquad & | \qquad 2a = \sqrt{ e^2 + f^2 }\\ \end{array} \\ \boxed{~ \begin{array}{rcll} h &=&\frac{ e\cdot f }{\sqrt{ e^2 + f^2 }} \\ \end{array} ~}\\\)
\(\begin{array}{rcll} h &=&\frac{ 12\cdot 36 }{\sqrt{ 12^2 + 36^2 }} \\ h &=&\frac{ 432 }{\sqrt{ 1440 }} \\ h &=&\frac{ 432 }{37.9473319220} \\\\ \mathbf{h} &\mathbf{=}& \mathbf{11.3841995766\ cm }\\ \end{array}\)
