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# rhombus

+12
389
3

how high is rhombus if diagonals are 36 cm and 12cm

Guest Jan 24, 2016

#1
+19207
+10

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

$$\begin{array}{rcll} e^2 &=& 2a^2 - 2a^2\cdot \cos{(A)} \\ f^2 &=& 2a^2 - 2a^2\cdot \cos{(B)} \\\\ 2A+2B &=& 360^\circ \\ A+B &=& 180^\circ \\ B &=& 180^\circ - A \\ \cos{(B)} &=& \cos{(180^\circ - A)} = -\cos{(A)} \\\\ e^2 &=& 2a^2 - 2a^2\cdot \cos{(A)} \\ f^2 &=& 2a^2 + 2a^2\cdot \cos{(A)} \\\\ \cos{(A)} = \frac{2a^2-e^2}{2a^2} &=& \frac{f^2-2a^2}{2a^2}\\ 2a^2-e^2 &=& f^2-2a^2 \\ 4a^2 &=& e^2 + f^2\\ 2a &=& \sqrt{ e^2 + f^2 } \\ \mathbf{a} &\mathbf{=}& \mathbf{\frac{ \sqrt{ e^2 + f^2 } } {2} }\\ \end{array}$$

$$\begin{array}{rcll} \cos{(A)} &=& \frac{2a^2-e^2}{2a^2} \\ \cos^2{(A)} &=& \frac{(2a^2-e^2)^2}{4a^4} \\ 1-\cos^2{(A)} &=& 1-\frac{(2a^2-e^2)^2}{4a^4} \\ 1-\cos^2{(A)} &=& \frac{4a^4 - (2a^2-e^2)^2}{4a^4} \\ 1-\cos^2{(A)} &=& \frac{4a^4 - 4a^2 + 4a^2e^2-e^4 }{4a^4} \\ 1-\cos^2{(A)} &=& \frac{4a^2e^2-e^4 }{4a^4} \\ 1-\cos^2{(A)} &=&\frac{ e^2\cdot (4a^2-e^2) }{4a^4} \qquad | \qquad 4a^2 = e^2 + f^2\\ 1-\cos^2{(A)} &=&\frac{ e^2\cdot ( e^2 + f^2-e^2) }{4a^4} \\ 1-\cos^2{(A)} &=&\frac{ e^2\cdot f^2 }{4a^4} \qquad | \qquad \sin^2{(A)}=1-\cos^2{(A)}\\ \sin^2{(A)} &=&\frac{ e^2\cdot f^2 }{4a^4} \\ \mathbf{\sin{(A)}} &\mathbf{=}&\mathbf{\frac{ e\cdot f }{2a^2} } \\ \end{array}$$

$$\begin{array}{rcll} h &=& a\cdot \sin{(A)} \\ h &=& a\cdot \frac{ e\cdot f }{2a^2} \\ h &=&\frac{ e\cdot f }{2a} \qquad & | \qquad 2a = \sqrt{ e^2 + f^2 }\\ \end{array} \\ \boxed{~ \begin{array}{rcll} h &=&\frac{ e\cdot f }{\sqrt{ e^2 + f^2 }} \\ \end{array} ~}\\$$

$$\begin{array}{rcll} h &=&\frac{ 12\cdot 36 }{\sqrt{ 12^2 + 36^2 }} \\ h &=&\frac{ 432 }{\sqrt{ 1440 }} \\ h &=&\frac{ 432 }{37.9473319220} \\\\ \mathbf{h} &\mathbf{=}& \mathbf{11.3841995766\ cm }\\ \end{array}$$

heureka  Jan 25, 2016
edited by heureka  Jan 25, 2016
edited by heureka  Jan 25, 2016
Sort:

#1
+19207
+10

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

$$\begin{array}{rcll} e^2 &=& 2a^2 - 2a^2\cdot \cos{(A)} \\ f^2 &=& 2a^2 - 2a^2\cdot \cos{(B)} \\\\ 2A+2B &=& 360^\circ \\ A+B &=& 180^\circ \\ B &=& 180^\circ - A \\ \cos{(B)} &=& \cos{(180^\circ - A)} = -\cos{(A)} \\\\ e^2 &=& 2a^2 - 2a^2\cdot \cos{(A)} \\ f^2 &=& 2a^2 + 2a^2\cdot \cos{(A)} \\\\ \cos{(A)} = \frac{2a^2-e^2}{2a^2} &=& \frac{f^2-2a^2}{2a^2}\\ 2a^2-e^2 &=& f^2-2a^2 \\ 4a^2 &=& e^2 + f^2\\ 2a &=& \sqrt{ e^2 + f^2 } \\ \mathbf{a} &\mathbf{=}& \mathbf{\frac{ \sqrt{ e^2 + f^2 } } {2} }\\ \end{array}$$

$$\begin{array}{rcll} \cos{(A)} &=& \frac{2a^2-e^2}{2a^2} \\ \cos^2{(A)} &=& \frac{(2a^2-e^2)^2}{4a^4} \\ 1-\cos^2{(A)} &=& 1-\frac{(2a^2-e^2)^2}{4a^4} \\ 1-\cos^2{(A)} &=& \frac{4a^4 - (2a^2-e^2)^2}{4a^4} \\ 1-\cos^2{(A)} &=& \frac{4a^4 - 4a^2 + 4a^2e^2-e^4 }{4a^4} \\ 1-\cos^2{(A)} &=& \frac{4a^2e^2-e^4 }{4a^4} \\ 1-\cos^2{(A)} &=&\frac{ e^2\cdot (4a^2-e^2) }{4a^4} \qquad | \qquad 4a^2 = e^2 + f^2\\ 1-\cos^2{(A)} &=&\frac{ e^2\cdot ( e^2 + f^2-e^2) }{4a^4} \\ 1-\cos^2{(A)} &=&\frac{ e^2\cdot f^2 }{4a^4} \qquad | \qquad \sin^2{(A)}=1-\cos^2{(A)}\\ \sin^2{(A)} &=&\frac{ e^2\cdot f^2 }{4a^4} \\ \mathbf{\sin{(A)}} &\mathbf{=}&\mathbf{\frac{ e\cdot f }{2a^2} } \\ \end{array}$$

$$\begin{array}{rcll} h &=& a\cdot \sin{(A)} \\ h &=& a\cdot \frac{ e\cdot f }{2a^2} \\ h &=&\frac{ e\cdot f }{2a} \qquad & | \qquad 2a = \sqrt{ e^2 + f^2 }\\ \end{array} \\ \boxed{~ \begin{array}{rcll} h &=&\frac{ e\cdot f }{\sqrt{ e^2 + f^2 }} \\ \end{array} ~}\\$$

$$\begin{array}{rcll} h &=&\frac{ 12\cdot 36 }{\sqrt{ 12^2 + 36^2 }} \\ h &=&\frac{ 432 }{\sqrt{ 1440 }} \\ h &=&\frac{ 432 }{37.9473319220} \\\\ \mathbf{h} &\mathbf{=}& \mathbf{11.3841995766\ cm }\\ \end{array}$$

heureka  Jan 25, 2016
edited by heureka  Jan 25, 2016
edited by heureka  Jan 25, 2016
#2
+85918
+5

Here's another method :

Length of a side  = sqrt(6^2 + 18^2)  = sqrt (36 + 324)  = sqrt (360)

And using the Law of Cosines, we can find the smaller interior angle of the rhombus

12^2  = 360 + 360 - 2(360)cos(theta)

[144 - 360 - 360]  / [ -2(360)]  = cos (theta)

cos-1  [ [144 - 360 - 360]  / [ -2(360] ]  = theta  = 36.869897645844°

Using the Law of Sines.....we can find the height - h - as follows :

sqrt(360)   = h / sin(36.869897645844°)

h = sqrt(360)* sin (36.869897645844°)  =  11.3841995766061656  cm

Here's a pic :  [DE  is the height]

CPhill  Jan 26, 2016
#3
+19207
+10

e = one diagonal

f = the other diagonal

h ist the height of the rhombus

$$\boxed{~ \begin{array}{rcll} h &=&\frac{ e\cdot f }{\sqrt{ e^2 + f^2 }} \\ \end{array} ~}\\$$

or

$$\begin{array}{rcll} h &=&\frac{ e\cdot f }{\sqrt{ e^2 + f^2 }}\\ h &=&\frac{ 1 }{ \frac { \sqrt{ e^2 + f^2 }}{ e\cdot f}} \\ h &=&\frac{ 1 }{ \sqrt{ \frac{e^2}{e^2\cdot f^2} + \frac{f^2}{e^2\cdot f^2} }} \\ h &=&\frac{ 1 }{ \sqrt{ \frac{1}{f^2} + \frac{1}{e^2} }} \\ \frac{1}{h} &=&\sqrt{ \frac{1}{f^2} + \frac{1}{e^2} } \\ \frac{1}{h^2} &=&\frac{1}{f^2} + \frac{1}{e^2} \\ \frac{1}{h^2} &=&\frac{1}{e^2} + \frac{1}{f^2} \\ \end{array}$$

$$\boxed{~ \begin{array}{rcll} \frac{1}{h^2} &=&\frac{1}{e^2} + \frac{1}{f^2} \end{array} ~}$$

heureka  Jan 26, 2016
edited by heureka  Jan 26, 2016

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