Question 1
The cuboid ABCDEFGH has sides of length 5cm. Find the length of the line AH to 2 decimal places.
Select the correct answer from the list below:
A: 5 cm
B: 7.07 cm
C: 8.66 cm
D: 10 cm
CORRECT ANSWER: C: 8.66 cm
WORKED SOLUTION:
Adding in the line AH creates a triangle ACH.
The length of AC can be found by using Pythagoras’ Theorem on the triangle ABC.
AC^2=5^2+5^2
AC^2=25+25
AC^2=50
AC=\sqrt{50}
AC=7.07
We can now put this value into the triangle ACH and use Pythagoras’ Theorem again to find the length of AH.
AH^2=7.07^2+5^2
AH^2=49.99+25
AH^2=74.99
AH=\sqrt{74.99}
AH=8.66
Question 2
The cuboid ABCDEFG has volume 73.5m^3, AB=7m and HC=3m.
Find the length of EB to 2 decimal places.
Select the correct answer from the list below:
A: 8.38 m
B: 7.66 m
C: 9.38 m
D: 3.5 m
CORRECT ANSWER: A: 8.38 m
WORKED SOLUTION:
V=length\times width\times height 73.5=7\times width\times 3 73.5=21\times width width=73.5\div 21 width=3.5
Adding in the line EB creates a triangle BDE.
The length of BD can be found by using Pythagoras’ Theorem on the triangle ABD.
BD^2=7^2+3.5^2
BD^2=49+12.25
BD^2=61.25
BD=\sqrt{61.25}
BD=7.83
We can now put this value into the triangle BDE and use Pythagoras’ Theorem again to find the length of EB.
EB^2=3^2+7.83^2
EB ^2=9+61.31
EB ^2=70.31
EB =\sqrt{70.31}
EB =8.381527...
EB =8.38 m
Question 3
ABCDE is a square based pyramid, with E directly above the centre of the base.
The perpendicular height of the pyramid is 12cm and the base has area 225cm^2.
Find the length of EB to 2 decimal places.
Select the correct answer from the list below:
A: 14.15 cm
B: 19.21 cm
C: 16.02 cm
D: 13.84 cm
CORRECT ANSWER: C: 16.02 cm
WORKED SOLUTION:
Start by finding the lengths of the sides of the base. To find the lengths of a square from its area, square root.
AB=\sqrt{225} AB=15cm
We can now put in the lengths we know.
To find the length of EB we can create the triangle EFB.
To find the length of FB we can create a triangle from F to the midpoint of AB, call it x.
The shorter sides of this triangle are 7.5cm because F is in the centre of the square.
FB^2=7.5^2+7.5^2
FB ^2=56.25+56.25
FB ^2=112.5
FB =\sqrt{112.5}
FB =10.61
We can now find the length of EB.
EB^2=12^2+10.61^2
EB ^2=144+112.57
EB ^2=256.57
EB =\sqrt{256.57}
EB =16.02
Question 4
ABCDEFGH is a cuboid.
AB= 8m, BC= 6m, and CH= 5m.
Find the angle between AH and the plane ABCD.
Give your answer to 2 Significant figures.
Select the correct answer from the list below:
A: 27\degree
B: 44\degree
C: 62\degree
D: 71\degree
CORRECT ANSWER: A: 27\degree
WORKED SOLUTION:
To find the angle between AH and ABCD, we need to create the triangle ACH.
To find AC, use Pythagoras on the triangle ABC.
AC^2=8^2+6^2
AC^2=64+36
AC^2=100
AC=\sqrt{100}
AC =10
And now, we can just use trigonometry to find the angle HAC.
tan{(\theta)}=\frac{5}{10}
\theta =tan^{-1}\left(\frac{5}{10}\right)
\theta =30\degree
Question 5
ABCDEFGH is a cuboid.
AB=a, BC=b, CH=c.
Find the length of AH in terms of a, b, and c.
Select the correct answer from the list below:
A: \sqrt{c^2-(a^2+b^2)}
B: a^2+b^2+c^2
C: \sqrt{a^2+b^2}+c^2
D: \sqrt{a^2+b^2+c^2}
CORRECT ANSWER: D: \sqrt{a^2+b^2+c^2}
WORKED SOLUTION:
To find the length of AH we need to use the triangle ACH.
But, we don’t know AC so need to find that first using the triangle ABC.
AC^2=a^2+b^2 AC=\sqrt{a^2+b^2}
Now that we know AC, we can put that into the first triangle to find AH.
AH^2=(\sqrt{a^2+b^2})^2+c^2 AH^2=a^2+b^2+c^2
AH=\sqrt{a^2+b^2+c^2}Question 6
ABCDEFGH is a cuboid.
AB=a, BC=b, CH=c.
Given that the angle between AH and the planeABCD is \theta, find the value of \theta in terms of a, b, c, and tan.
Select the correct answer from the options below:
A: \theta=tan^{-1}\left(\frac{c}{\sqrt{a^2+b^2}}\right)
B: \theta=tan\left(\frac{c}{\sqrt{a^2+b^2}}\right)
C: \theta=tan^{-1}\left(\frac{\sqrt{a^2+b^2}}{c}\right)
D: \theta=tan^{-1}(a^2+b^2+c^2)
CORRECT ANSWER: A: \theta=tan^{-1}\left(\frac{c}{\sqrt{a^2+b^2}}\right)
WORKED SOLUTION:
To find the angle \theta, we need to use the triangle ACH.
But, we don’t know AC so need to find that first using the triangle ABC.
AC^2=a^2+b^2 AC=\sqrt{a^2+b^2}
Now that we know AC, we can put that into the first triangle to find AH.
We can now use trigonometry to find \theta.
tan(\theta)=\frac{opp}{adj}
tan(\theta)=\frac{c}{\sqrt{a^2+b^2}}
\theta=tan^{-1}\left(\frac{c}{\sqrt{a^2+b^2}}\right)