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 absoulutely risiculous wuestion i can't even

Guest Mar 21, 2018

Best Answer 

 #1
avatar+6943 
+3

Look at triangle  ABE .  We know..

 

∠ABE  =  81°

AB  =  4 cm       and

BE  =  6 cm

 

So using the law of cosines...

 

AE2  =  62 + 42 - 2(6)(4) cos 81°

AE2  =  36 + 16 - 48 cos 81°

AE2  =  52 - 48 cos 81°          Take the positive square root of both sides.

AE  =  √[  52 - 48 cos 81°  ]    Plug this into a calculator to get...

AE  ≈   6.67

AE  ≈   6.7     cm

 

h, AE, and part of AC form a right triangle, and we know...

 

∠EAC  =  46°    and

AE  ≈  6.67       so...

 

sin 46°  =  h / AE

h   =   AE sin 46°

h   ≈   6.67sin 46°

h   ≈   4.8    cm

 

The base is a square with a side length of  4 cm  , so...

 

area of base =  (4 cm)2  =  16 cm2

height  =  h  ≈  4.8 cm

 

volume of pyramid  ≈  (1/3)(16 cm2)(4.8 cm)

volume of pyramid  ≈  25.6  cm3

hectictar  Mar 21, 2018
Sort: 

3+0 Answers

 #1
avatar+6943 
+3
Best Answer

Look at triangle  ABE .  We know..

 

∠ABE  =  81°

AB  =  4 cm       and

BE  =  6 cm

 

So using the law of cosines...

 

AE2  =  62 + 42 - 2(6)(4) cos 81°

AE2  =  36 + 16 - 48 cos 81°

AE2  =  52 - 48 cos 81°          Take the positive square root of both sides.

AE  =  √[  52 - 48 cos 81°  ]    Plug this into a calculator to get...

AE  ≈   6.67

AE  ≈   6.7     cm

 

h, AE, and part of AC form a right triangle, and we know...

 

∠EAC  =  46°    and

AE  ≈  6.67       so...

 

sin 46°  =  h / AE

h   =   AE sin 46°

h   ≈   6.67sin 46°

h   ≈   4.8    cm

 

The base is a square with a side length of  4 cm  , so...

 

area of base =  (4 cm)2  =  16 cm2

height  =  h  ≈  4.8 cm

 

volume of pyramid  ≈  (1/3)(16 cm2)(4.8 cm)

volume of pyramid  ≈  25.6  cm3

hectictar  Mar 21, 2018
 #2
avatar
+1

you are a GENIUS! LOVE YOU XX

Guest Mar 21, 2018
 #3
avatar+85766 
+1

Very nice, hectictar  !!!

 

cool cool cool

CPhill  Mar 21, 2018

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