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mathematicsprinted 93 · PDF 94
165 pagesrevision 46935
printed 93 · PDF 94
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3W = 37.8 - 20.1 cm, so 3W = 17.7. This gives W as 5.9 cm. Substituting this
value back into the first equation, we can find 2H, and then H as 7.1 cm. Now
using this value of H, we can find 2L and thus L, and find it to be 11.8 cm.
b) i) - We can calculate the base area as the area of an 18*15 cm rectangle.
This yields 270 cm2. Since EDC is a right triangle, the height ED can be
calculated as the square root of the difference of the squares of 24 and 18.
This gives the height as root 252 cm. We can now use this into the given
formula to find the volume as approximately 1480 cubic centimetres. Density
is mass divided by volume, so in this case, it would be 800g/1480cm3, giving
0.541g/cm3.
b) ii) - We want to find angle EBD. We know that BD is the diagonal of the
base rectangle, so it would be the hypotenuse of a right triangle with sides 15
and 18. Thus, BD would be the root of the sum of the squares of 15 and 18,
yielding root 549 cm. We know the height of the right triangle, EBD, which is
the root of 26. Knowing the base and the height, we can use the trigonometric
ratio of tan, which is the opposite over adjacent, to calculate the value of the
tangent of the angle we want to find. This gives tan theta as root (252/549).
Using the tan inverse function, also known as the arctan, we can calculate
theta as 34.1 degrees.