T r i a n g l e s Triangles form by a c r e a s e crease

Geometry Level 3

Given that A B C D ABCD is a rectangle and E F EF & G H GH are the crease. Given that A B = 12 AB = 12 & A D = 5 AD = 5 and if they intersect at entire point O O , Find the A r e a ( S h a d e d R e g i o n ) Area (Shaded Region) .

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N o t e : Note:

Round your answer to the nearest h u n d r e d t h s hundredths .


The answer is 5.21.

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Christian Daang by Christian Daang , 20, Philippines

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1 solution

Christian Daang
Jan 19, 2015

S o l u t i o n : Solution:

L e n g t h Length of crease = W L ∗ L 2 + W 2 \cfrac{W}{L} * \sqrt{L^{2} + W^{2}}

-> = 5 12 ∗ 5 2 + 1 2 2 = \cfrac{5}{12} * \sqrt{5^{2} + 12^{2}}

-> = 5 12 ∗ 169 = \cfrac{5}{12} * \sqrt{169}

-> = 65 12 = \cfrac{65}{12}

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Since if we fold the paper like this,

A E + E B D = A B AE + EBD = AB

-> a + 25 + a 2 = 12 a + \sqrt{25 + a^{2}} = 12

-> a − 12 = − 25 + a 2 a - 12 = -\sqrt{25 + a^{2}}

-> a = 119 24 a = \cfrac{119}{24}

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Then, A E = H D = C F = B G = 119 24 AE = HD = CF = BG = \cfrac{119}{24}

Draw a imaginary line E H EH and G F GF so that, E G F H EGFH is a rectangle where in, E H = G F = 5 EH = GF = 5 and G E = F H = 12 − 2 ∗ 119 24 = 25 12 GE = FH = 12 - 2*\cfrac{119}{24} = \cfrac{25}{12}

∴ \therefore

A r e a ( S h a d e d R e g i o n ) = 1 2 A r e a ( E F G H ) = 5 ∗ 25 12 2 = 125 24 t h a t i s a p p r o x i m a t e l y e q u a l t o 5.21 s q . u n i t s Area(Shaded Region) = \cfrac{1}{2} Area(EFGH) = \cfrac{5*\cfrac{25}{12}}{2} = \cfrac{125}{24} that is approximately equal to \boxed{5.21 sq. units}

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You asked at the same time solved'

Cabanting Perez Francis - 6 years, 4 months ago

What is a crease by the way?

Cabanting Perez Francis - 6 years, 4 months ago

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