What is the area of the two-dimensional cross section that is parallel to face ABC ?
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? ft²
The two-dimensional cross-section that is parallel to face ABC is just the triangle ABC.
We need to find the area of ABC.
Since it is a right triangle, we can use the Pythagorean theorem.
5^2+x^2=13^2
25+x^2=169
x^2=144
x=12
Therefore, the area of ABC is 5*12/2=30 ft^2.
You are very welcome!
:P
The two-dimensional cross-section that is parallel to face ABC is just the triangle ABC.
We need to find the area of ABC.
Since it is a right triangle, we can use the Pythagorean theorem.
5^2+x^2=13^2
25+x^2=169
x^2=144
x=12
Therefore, the area of ABC is 5*12/2=30 ft^2.
You are very welcome!
:P
The area of the two-dimensional cross section that is parallel to face ABC.that is Area of Δ DEF = 84 ft²
Step-by-step explanation:
Given : A triangular prism with some side measurement.
We have to find the area of the two-dimensional cross section that is parallel to face ABC.
Since, the cross section that is parallel to face ABC.
Since, face parallel to ABC is DEF .
And DEF is a triangle with ∠ E = 90°
So, Area of right angled triangle is 1/2 times base times height.
Base = 24 ft
and height is 7 ft
So, Area of Δ DEF = 1/2 times 24 times 7
Simplify , we have,
Area of Δ DEF = 84 ft²
Thus, The area of the two-dimensional cross section that is parallel to face ABC.that is Area of Δ DEF = 84 ft²