3.11.1 Pythagoras in 3D
3.11.1 Pythagoras in 3D
3.11.1 Pythagoras in 3D
The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In \( 3D \), this theorem can be extended to relate the lengths of the three sides of a right-angled solid shape called a “right rectangular prism”.
A right rectangular prism is a solid shape that has six faces, all of which are rectangles.
It is called “right” because the angles between adjacent faces are all right angles, and it is called “rectangular” because all of its faces are rectangles.
The Pythagorean theorem can be applied to the right rectangular prism as follows:
If a, b, and c are the lengths of the three edges of the right rectangular prism, where c is the length of the hypotenuse of a right-angled triangle that is formed by two of the edges, then:
\( c^2=a^2+b^2 \)
This equation can be used to calculate the length of any edge of a right rectangular prism if the lengths of the other two edges are known.
It can also be used to check if a given solid shape is a right rectangular prism or not.
Worked example:
The diagram shows a cuboid \( PQRSTUVW \).
\( PV=17.2 \) cm
The angle between the line \( PV \) and the base \( TUVW \) of the cuboid is \( 43° \).
Calculate \( PT \).
The lines \( PV, PT \)and angle \( PVT \) form a right triangle.
To make it clear, let’s draw the lines on to the diagram.
The side opposite the angle is needed and hypotenuse is given so use
\( sin (θ) = \frac{Opposite}{Hypotenuse} \)
\( sin (43°) = \frac{PT}{17.2} \)
Solve for \( PT \)
\( PT = 17.2\times sin (43°) \)
\( PT = 11.73037179… \)
Rounding to one decimal place
\( PT = 11.7 \)
\( PT = 11.7 \)
Test yourself
Question 1:
The diagram shows a triangular prism.
\( AF = 10 \) cm, \( AB = 24 \)cm and \( BC = 8 \)cm.
Angle \( FAB = \)Angle \( ADC = \)Angle \( BCD = 90° \).
Work out the size of the angle between the line \( BE \) and the plane \( ABCD \).
Give your answer correct to \( 1 \) decimal place.
[3]
Question 2:
The diagram shows cuboid \( ABCDEFGH \).
For this cuboid
the length of \( AB \) : the length of \( BC \) : the length of \( CF = 4\) : \( 2 \) : \( 3 \)
Calculate the size of the angle between \( AF \) and the plane \( ABCD \).
Give your answer correct to one decimal place.
[3]
Question 3:
The diagram shows a pyramid with a square base \( ABCD \).
The diagonal \( AC \) and \( BD \) intersect at \( M \).
The vertex \( V \) is vertically above \( M \).
\( AB=11 \)cm and \( AV=18.6 \)cm.
Calculate the angle that \( AV \) makes with the base.
[4]
Question 4:
The diagram shows a cuboid with dimensions \( 5.5 \) cm, \( 8 \) cm and \( 16.2 \) cm.
Calculate the angle between the line \( AB \) and the horizontal base of the cuboid.
[4]
Question 5:
The diagram shows the prism \( ABCDEF \) with cross section triangle \( ABC \).
Angle \( BEC=40° \) and angle \( ACB \) is obtuse.
\( AC=6 \) cm and \( CE=13 \)cm
The area of triangle \( ABC \) is \( 22 \) cm2.
Calculate the length of \( AB \).
Give your answer correct to one decimal place.
[6]