Area of a Quadrilateral!

Geometry Level 4

In quadrilateral A B C D ABCD above, B A C = 3 x \angle{BAC} = 3x , B C A = x \angle{BCA} = x , C A D = 5 x \angle{CAD} = 5x , A C D = 3 x \angle{ACD} = 3x and B C = C D = 1 BC = CD = 1 . Find the area of quadrilateral A B C D ABCD .

Note: A variation of the above problem was taken from a problem I found online.


The answer is 0.43301270.

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3 solutions

Rocco Dalto
Oct 25, 2020

For A D C : \triangle{ADC} :

sin ( 5 x ) D C = sin ( 8 x ) A C sin ( 5 x ) sin ( 8 x ) = D C A C = B C A C = 1 A C \dfrac{\sin(5x)}{DC} = \dfrac{\sin(8x)}{AC} \implies \dfrac{\sin(5x)}{\sin(8x)} = \dfrac{DC}{AC} = \dfrac{BC}{AC} = \dfrac{1}{AC}

For A B C : \triangle{ABC} :

sin ( 3 x ) B C = sin ( 4 x ) A C sin ( 3 x ) sin ( 4 x ) = B C A C = = 1 A C \dfrac{\sin(3x)}{BC} = \dfrac{\sin(4x)}{AC} \implies \dfrac{\sin(3x)}{\sin(4x)} = \dfrac{BC}{AC} = = \dfrac{1}{AC}

\implies

sin ( 5 x ) sin ( 8 x ) = sin ( 3 x ) sin ( 4 x ) sin ( 5 x ) sin ( 4 x ) = sin ( 8 x ) sin ( 3 x ) = \dfrac{\sin(5x)}{\sin(8x)} = \dfrac{\sin(3x)}{\sin(4x)} \implies \sin(5x)\sin(4x) = \sin(8x)\sin(3x) =

2 sin ( 4 x ) cos ( 4 x ) sin ( 3 x ) sin ( 5 x ) = 2 cos ( 4 x ) sin ( 3 x ) 2\sin(4x)\cos(4x)\sin(3x) \implies \sin(5x) = 2\cos(4x)\sin(3x)

Using the identity 2 cos ( A ) sin ( B ) = sin ( A + B ) sin ( A B ) 2\cos(A)\sin(B) = \sin(A + B) - \sin(A - B) , with A = 4 x A = 4x and

B = 3 x B = 3x we obtain: sin ( 5 x ) = sin ( 7 x ) sin ( x ) sin ( 7 x ) sin ( 5 x ) = sin ( x ) \sin(5x) = \sin(7x) - \sin(x) \implies \sin(7x) - \sin(5x) = \sin(x)

or sin ( 7 x ) + sin ( 5 x ) = sin ( x ) \sin(7x) + \sin(-5x) = \sin(x)

Using the identity sin ( A ) + sin ( B ) = 2 sin ( A + B 2 ) cos ( A B 2 ) , \sin(A) + \sin(B) = 2\sin(\dfrac{A + B}{2})\cos(\dfrac{A - B}{2}), with A = 7 x A = 7x and

B = 5 x B = -5x we obtain: 2 sin ( x ) cos ( 6 x ) = sin ( x ) 2 cos ( 6 x ) = 1 cos ( 6 x ) = 1 2 2\sin(x)\cos(6x) = \sin(x) \implies 2\cos(6x) = 1 \implies \cos(6x) = \dfrac{1}{2}

6 x = 6 0 x = 1 0 \implies 6x = 60^{\circ} \implies \boxed{x = 10^{\circ}}

Using the above diagram with x = 1 0 h 2 = sin ( 3 0 ) = 1 2 x = 10^{\circ} \implies h_{2} = \sin(30^{\circ}) = \dfrac{1}{2} and h 1 = sin ( 1 0 ) h_{1} = \sin(10^{\circ})

For A C AC we have: A C sin ( 14 0 ) = 1 sin ( 3 0 ) = 2 \dfrac{AC}{\sin(140^{\circ})} = \dfrac{1}{\sin(30^{\circ})} = 2 A C = 2 sin ( 14 0 ) \implies AC = 2\sin(140^{\circ})

\implies The total area A A B C D = 1 2 ( 2 sin ( 14 0 ) sin ( 1 0 ) ) + 1 2 ( 2 sin ( 14 0 ) ) ( 1 2 ) = A_{ABCD} = \dfrac{1}{2}(2\sin(140^{\circ})\sin(10^{\circ})) + \dfrac{1}{2}(2\sin(140^{\circ}))(\dfrac{1}{2}) =

sin ( 14 0 ) ( sin ( 1 0 ) + 1 2 ) = 1 2 sin ( 14 0 ) ( 2 sin ( 1 0 ) + 1 ) \sin(140^{\circ})(\sin(10^{\circ}) + \dfrac{1}{2}) = \dfrac{1}{2}\sin(140^{\circ})(2\sin(10^{\circ}) + 1) 0.43301270 \approx \boxed{0.43301270}

... = 3 4 =\frac{\sqrt3}{4}

which is the area of an equilateral triangle with side 1 1 . I wonder if there's a direct geometric way to show that?

Chris Lewis - 7 months, 2 weeks ago

There is an excellent geometric proof of the fact that x = 10 x=10{}^\circ by @Maria Kozlowska here .

Thanos Petropoulos - 7 months, 2 weeks ago

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Very nice and simple solution.

Rocco Dalto - 7 months, 2 weeks ago
ChengYiin Ong
Nov 5, 2020

From the angle conditions (solutions above showed that x = 1 0 x=10^{\circ} ), we know that A B D C AB||DC and so the quadrilateral is a trapezium, the height of the trapezium is 1 × sin ( 18 0 A B C ) = sin 4 0 1\times \sin (180^{\circ}-\angle ABC)=\sin 40^{\circ} .

By sine rule, A B = sin 1 0 × 1 sin 3 0 = 2 sin 1 0 . AB=\sin 10^{\circ}\times \frac{1}{\sin 30^{\circ}}=2\sin 10^{\circ}. So the area of quadrilateral A B C D ABCD is 1 2 ( 1 + 2 sin 1 0 ) ( sin 4 0 ) 0.43301270 \frac{1}{2}(1+2\sin 10^{\circ})(\sin 40^{\circ})\approx \boxed{0.43301270}

Chew-Seong Cheong
Oct 26, 2020

By sine rule :

A C B C = A C C D = A C 1 sin ( 18 0 4 x ) sin 3 x = sin ( 18 0 8 x ) sin 5 x SInce sin ( 18 0 θ ) = sin θ sin 4 x sin 3 x = sin 8 x sin 5 x And sin 2 θ = 2 sin θ cos θ 1 sin 3 x = 2 cos 4 x sin 5 x sin 5 x sin 3 x = 2 cos 4 x Also sin θ = cos ( 9 0 θ ) sin 5 x sin 3 x = sin ( 9 0 4 x ) 1 2 5 x = 9 0 4 x x = 1 0 3 x = 3 0 x = 1 0 \begin{aligned} \frac {AC}{BC} & = \frac {AC}{CD} = \frac {AC}1 \\ \frac \blue{\sin (180^\circ - 4x)}{\sin 3x} & = \frac \blue{\sin (180^\circ - 8x)}{\sin 5x} & \small \blue{\text{SInce } \sin(180^\circ - \theta) = \sin \theta} \\ \frac \blue{\sin 4x}{\sin 3x} & = \frac \blue{\sin 8x}{\sin 5x} & \small \blue{\text{And } \sin 2\theta = 2 \sin \theta \cos \theta} \\ \frac \blue 1{\sin 3x} & = \frac \blue{2\cos 4x}{\sin 5x} \\ \frac {\sin 5x}{\sin 3x} & = 2 \blue{\cos 4x} & \small \blue{\text{Also }\sin \theta = \cos (90^\circ - \theta)} \\ \frac \blue{\sin 5x}\red{\sin 3x} & = \frac \blue{\sin (90^\circ - 4x)}\red{\frac 12} \\ \implies \blue{5x} & = \blue{90^\circ - 4x} \implies x = 10^\circ \\ \implies \red{3x} & = \red{30^\circ} \implies x = 10^\circ \end{aligned}

Fill in the measures of all the angles. Extend B C BC and D A DA to meet at E E . Then we note that E = 4 0 \angle E = 40^\circ and C D E \triangle CDE and A B E \triangle ABE are isosceles and similar. Therefore the area of quadrilateral A B C D ABCD :

[ A B C D ] = [ C D E ] [ A B E ] = 1 2 1 2 sin 8 0 1 2 A B 2 sin 8 0 By sine rule, A B = sin 1 0 sin 3 0 = sin 8 0 2 ( 1 sin 2 1 0 sin 2 3 0 ) 0.433 \begin{aligned} [ABCD] & = [CDE]-[ABE] \\ & = \frac 12 \cdot 1^2 \cdot \sin 80^\circ - \frac 12 \cdot AB^2 \cdot \sin 80^\circ & \small \blue{\text{By sine rule, }AB = \frac {\sin 10^\circ}{\sin 30^\circ}} \\ & = \frac {\sin 80^\circ}2 \left(1 - \blue{\frac {\sin^2 10^\circ}{\sin^2 30^\circ}} \right) \\ & \approx \boxed{0.433} \end{aligned}

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