Making aeroplanes

Geometry Level 3

Given that the rectangle A B C D ABCD is folded along the diagonal B D BD , where A B = 4 AB= 4 and B C = 6 BC = 6 , find the area of the triangle P B D PBD .

Note: Figure not drawn to scale.

7 3 13 13 \frac { 3\sqrt { 13 } }{ 13 } 14 5 \frac { 14 }{ 5 } 26 3 \frac { 26 }{ 3 }

View wiki
Justin Tuazon by Justin Tuazon , 20, Philippines

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Lee Cho
Jul 4, 2015

We define x and y as the following: From the Pythagorean Theorem, we get x 2 + 4 2 = ( 6 − x ) 2 x 2 + 16 = 36 − 12 x + x 2 x = 5 3 { x }^{ 2 }+{ 4 }^{ 2 }={ (6-x) }^{ 2 }\\ { x }^{ 2 }+16=36-12x+{ x }^{ 2 }\\ x=\frac { 5 }{ 3 }

To find the length of the side B D ˉ \bar { BD } , we apply the same theorem and get 6 2 + 4 2 = ( B D ˉ ) 2 ( B D ˉ ) 2 = 52 B D ˉ = 2 13 { 6 }^{ 2 }+{ 4 }^{ 2 }={ (\bar { BD } ) }^{ 2 }\\ { (\bar { BD } ) }^{ 2 }=52\\ \bar { BD } =2\sqrt { 13 }

To find the value of y,

[ 1 2 ( 2 13 ) ] 2 + y 2 = ( 6 − 5 3 ) 2 13 + y 2 = 169 9 y 2 = 52 9 y = 2 13 3 { [\frac { 1 }{ 2 } (2\sqrt { 13 } )] }^{ 2 }+{ y }^{ 2 }={ (6-\frac { 5 }{ 3 } ) }^{ 2 }\\ 13+{ y }^{ 2 }=\frac { 169 }{ 9 } \\ { y }^{ 2 }=\frac { 52 }{ 9 } \\ y=\frac { 2\sqrt { 13 } }{ 3 }

Finally, we use A = b h 2 A=\frac { bh }{ 2 } to find the area of P B D PBD

A = 2 13 ( 2 13 3 ) 2 A = 52 3 ( 1 2 ) A = 26 3 A=\frac { 2\sqrt { 13 } (\frac { 2\sqrt { 13 } }{ 3 } ) }{ 2 } \\ A=\frac { 52 }{ 3 } (\frac { 1 }{ 2 } )\\ A=\boxed { \frac { 26 }{ 3 } }

This is my first solution, so any feedback is appreciated! :)

5 Helpful 0 Interesting 0 Brilliant 0 Confused

You can simplify this solution by realizing that a r e a o f t r i a n g l e P B D = a r e a o f t r i a n g l e A B D − a r e a o f t r i a n g l e P A B area of triangle PBD=area of triangle ABD-area of triangle PAB and that the triangle ABD have the areas of 12, via a r e a = 1 2 b a s e ∗ h e i g h t area=\frac{1}{2}base*height . The area of triangle PAB can be solved vy solving for the side, x = 5 3 x=\frac{5}{3} as shown above and reapplying the area formula gives the area of 10 3 \frac{10}{3} for the triangle PAB. Thus the area of triangle PBD is 12 − 10 3 = 26 3 12-\frac{10}{3}=\frac{26}{3}

Scott Ripperda - 5 years, 10 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...