Question Video: Using the Pythagorean Theorem in 3D to Find the Area of a Shape in a Cube | Nagwa Question Video: Using the Pythagorean Theorem in 3D to Find the Area of a Shape in a Cube | Nagwa

Question Video: Using the Pythagorean Theorem in 3D to Find the Area of a Shape in a Cube Mathematics • Second Year of Secondary School

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Given that 𝐴𝐡𝐢𝐷𝐸𝐹𝐺𝐻 is a cube whose edge length is 6√2 cm and 𝑋 is the midpoint of line segment 𝐴𝐡, find the area of the rectangle π·π‘‹π‘ŒπΈ.

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Video Transcript

Given that 𝐴𝐡𝐢𝐷𝐸𝐹𝐺𝐻 is a cube whose edge length is six root two centimeters and 𝑋 is the midpoint of line segment 𝐴𝐡, find the area of the rectangle π·π‘‹π‘ŒπΈ.

In the given cube, we are told that the length, width, and height will all be equal to six root two centimeters, and we are asked to find the area of the rectangle π·π‘‹π‘ŒπΈ. We recall that the area of any rectangle is equal to the length multiplied by the width or the base multiplied by the height. In this question, this is equal to the length of the line segment 𝐷𝑋 multiplied by the length of the line segment π‘‹π‘Œ. We know that π‘‹π‘Œ is the height of the cube and is therefore equal to six root two centimeters.

In order to calculate the length of 𝐷𝑋, let’s consider the right triangle 𝐷𝐴𝑋. We know that the length 𝐷𝐴 is equal to six root two centimeters. And since 𝑋 is the midpoint of line segment 𝐴𝐡, then the length 𝐴𝑋 is equal to half of this. It is equal to three root two centimeters. We can now use the Pythagorean theorem to calculate the missing length. This states that π‘Ž squared plus 𝑏 squared is equal to 𝑐 squared, where 𝑐 is the length of the hypotenuse of the right triangle and π‘Ž and 𝑏 are the lengths of the two shorter sides. Substituting in our values, 𝐷𝑋 squared is equal to six root two squared plus three root two squared. Squaring six root two gives us 72 as six squared is 36 and root two squared is two.

In the same way, three root two squared is equal to 18. Our equation simplifies to 𝐷𝑋 squared is equal to 72 plus 18. Square rooting both sides of this equation and since 𝐷𝑋 must be positive, we have 𝐷𝑋 is equal to the square root of 90. Using our laws of radicals or surds, we can rewrite this as root nine multiplied by root 10. And since the square root of nine is three, the length of 𝐷𝑋 is three root 10 centimeters.

We are now in a position to calculate the area of the rectangle. We multiply three root 10 by six root two. Three multiplied by six is 18, and root 10 multiplied by root two is root 20. The area of the rectangle is therefore equal to 18 root 20. We can then write root 20 as root four multiplied by root five, which is equal to two root five. And multiplying this by 18 gives us 36 root five. The area of rectangle π·π‘‹π‘ŒπΈ is therefore equal to 36 root five square centimetres.

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