IB DP Math AA Topic 3: Geometry and trigonometry -3D Geometry-IB Style Questionbank- SL Paper 2

 Question:

A solid metal ornament is in the shape of a right pyramid, with vertex V and square base ABCD. The centre of the base is X. Point V has coordinates (1, 5, 0) and point A has coordinates (-1, 1, 6).

(a) Find AV.
(b) Given that \(A\hat{V}B =40^{0}\), find AB.
The volume of the pyramid is 57.2 cm3, correct to three significant figures.
(c) Find the height of the pyramid, VX.
A second ornament is in the shape of a cuboid with a rectangular base of length 2x cm, width x cm and height y cm. The cuboid has the same volume as the pyramid.

(d) The cuboid has a minimum surface area of S cm2. Find the value of S.

Answer/Explanation

Ans:

(a) attempt to use the distance formula to find AV

(b) METHOD 1
attempt to apply cosine rule OR sine rule to find AB

METHOD 2
Let M be the midpoint of [AB]
attempt to apply right-angled trigonometry on triangle AVM
= 2× 7.48…× sin(200)
= 5.11888…
= 5.12 (cm)

(c) METHOD 1
equating volume of pyramid formula to 57.2
\(\frac{1}{3}\times 5.11….^{2}\times h=57.2\)
h = 6.54886…
h = 6.55 (cm)

METHOD 2
Let M be the midpoint of [AB]

(d) V = x × 2x × y= 57.2
       S = 2(2x2+xy+2xy)

Note: Condone use of A.

attempt to substitute \(y = \frac{57.2}{2x^{2}}\)   into their expression for surface area 

\(\left ( S(x) =\right )4x^{2}+6x\left ( \frac{57.2}{2x^{2}} \right )\)

EITHER
attempt to find minimum turning point on graph of area function

OR

\(\frac{dS}{dx}= 8x – 171.6x^{2}= 0\) OR x = 2.77849….

THEN
92.6401…

minimum surface area = 92.6(cm2)

Question:

A company is designing a new logo. The logo is created by removing two equal segments from a rectangle, as shown in the following diagram.

The rectangle measures 5cm by 4cm. The points A and B lie on a circle, with centre O and radius 2cm, such that AÔB = θ, where 0 < θ < π. This information is shown in the following diagram.

(a) Find the area of one of the shaded segments in terms of θ.
(b) Given that the area of the logo is 13.4cm2, find the value of θ.

Answer/Explanation

Ans:

(a) valid approach to find area of segment by finding area of sector – area of triangle

(b) EITHER
area of logo = area of rectangle – area of segments
5 × 4 – 2 × (2θ-2sinθ) = 13.4

OR

area of one segment \(= \frac{20-13.4}{2}(=3.3)\)
2θ-2sinθ = 3.3
THEN
θ = 2.35672…
θ = 2.36 (do not accept an answer in degrees)

Note: Award (M1)(A1)A0 if there is more than one solution. Award (M1)(A1FT)A0 if the candidate works in degrees and obtains a final answer of 135.030…

Question

The diameter of a spherical planet is 6 x 104 km .

(a) Write down the radius of the planet.[1]

The volume of the planet can be expressed in the form π(a × 10k) km3 where 1 ≤ a <10  and k ∈  \(\mathbb{Z}\)

(b) Find the value of a and the value of k . [3]

Answer/Explanation

Ans

Question

Let A(2, -3, 5) and B(-1, 1 , 5). Find
(a) the distance between A and B
(b) the distance between O and B
(c) the coordinates of the midpoint M of the line segment [AB]
(d) the coordinates of point C given that B is the midpoint of [AC].

Answer/Explanation

Ans
(a) \(d_{AB}=\sqrt{3^2+4^2+0^2}=5\)
(b) \(d_{OB}=\sqrt{1^2+1^2+5^2}=\sqrt{27}=3\sqrt{3}\)
(c) M(1/2, -1,5)
(d) C(-4,5,5)

 

Question

Complete the table

Answer/Explanation

Ans

 

Question

For a right pyramid of square base of slide 8 and vertical height 3 find
(a) the volume
(b) the surface are

Answer/Explanation

Ans
(a) \(V=\frac{1}{3}8^23=64\)
(b) \(AM^2=4^2+3^2 \Rightarrow AM =5\)
\(S=8^2+4 \times (\frac{1}{2} \times 8 \times 5)=64+80=144\)

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