The net of a triangular pyramid is shown.
1 answer:
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An algebraic equation enables the expression of equality between variable expressions
The reason the above value is correct is given as follows:
The given parameters are;
The symbol for the area of a triangle ΔXYZ = [XYZ]
The side length of the given square ABCD = 1
The location of point <em>E</em> = Side on square ABCD
The location of point <em>F</em> = Side on square ABCD
∠EAF = 45°
The area of ΔCEF, [CEF] = 1/9 (corrected by using a similar online question)
Required:
To find the value of [AEF]
Solution:
The area of a triangle = (1/2) × Base length × Height
Let <em>x</em> = EC, represent the base length of ΔCEF, and let <em>y</em> = CF represent the height of triangle ΔCEF
We get;
The area of a triangle ΔCEF, [CEF] = (1/2)·x·y = x·y/2
The area of ΔCEF, [CEF] = 1/9 (given)
∴ x·y/2 = 1/9
ΔABE:
= BC - EC = 1 - x
The area of ΔABE, [ABE] = (1/2)×AB ×BE
AB = 1 = The length of the side of the square
The area of ΔABE, [ABE] = (1/2)× 1 × (1 - x) = (1 - x)/2
ΔADF:
= CD - CF = 1 - y
The area of ΔADF, [ADF] = (1/2)×AD ×DF
AD = 1 = The length of the side of the square
The area of ΔADF, [ADF] = (1/2)× 1 × (1 - y) = (1 - y)/2
The area of ΔAEF, [AEF] = [ABCD] - [ADF] - [ABE] - [CEF]
[ABCD] = Area of the square = 1 × 1
From , we have;
, which gives;
Area of a triangle = (1/2) × The product of the length of two sides × sin(included angle between the sides)
∴ [AEF] = (1/2) × ×
× sin(∠EAF)
= √((1 - x)² + 1),
= √((1 - y)² + 1)
[AEF] = (1/2) × √((1 - x)² + 1) × √((1 - y)² + 1) × sin(45°)
Which by using a graphing calculator, gives;
Squaring both sides and plugging in , gives;
Subtracting the right hand side from the equation from the left hand side gives;
36·y² - 40·y + 8 = 0
Plugging in and rationalizing surds gives;
Therefore;
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