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What is the area of the two-dimensional cross section that is parallel to face ABC ?

 

 May 24, 2018
 #1
avatar+993 
+2

Hey oscar.a1551!

 

The area cross section of ABC is equal to the are of ABC.

 

Since ABC is a right triangle, we can calculate its area. 

 

We know that the hypotenuse is 17, and the height is 8, but we need to figure out the base. 

 

Using the Pythagorean Theorem: a^2 + b^2 = c^2 

 

In a right triangle, the sum of the square of the two legs is equal to the square of the hypotenuse. 

 

If the base is x, 8^2 + x^2 = 17^2 , x = 15

 

The area of the triangle is half of the base times height.

 

8 * 15 / 2 = 60.

 

I hope this helped,

 

Gavin

 May 24, 2018
 #2
avatar+8205 
0

Area of that cross section = Area of triangle ABC.

Using Pythagoras theorem, BC = \(\sqrt{17^2-8^2} = 15\)

Area of ABC = 8*15/2 = 60 sq ft

Area of the cross section = 60 sq ft :)

 May 26, 2018

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