IB myp 4-5 MATHEMATICS – Practice Questions- All Topics
Topic :Geometry-Circle Geometry
Topic :Geometry- Weightage : 21 %
All Questions for Topic : Metric conversions,Volume of regular polyhedra,Similarity and congruence,Coordinate geometry, including distance, midpoint and gradient formulae,Movement on a plane- isometric transformations, enlargements and tessellations,y=m x+c, gradients and intercepts (see also functions),Gradient of parallel lines,Circle geometry,Rotation around a given point
Question (3 marks)
Show that $r=2.80 \mathrm{~cm}$, correct to three significant figures.
▶️Answer/Explanation
Ans:
To solve the equation $\cos 15^\circ = \frac{2.7}{r}$ for $r$, we can rearrange the equation to isolate $r$ on one side:
$\cos 15^\circ = \frac{2.7}{r}$
Multiplying both sides by $r$:
$r \cos 15^\circ = 2.7$
Dividing both sides by $\cos 15^\circ$:
$r = \frac{2.7}{\cos 15^\circ}$
Now, let’s calculate the value of $r$:
$r = \frac{2.7}{\cos 15^\circ} \approx \frac{2.7}{0.965925}$ (using the cosine value of $15^\circ$)
$r \approx \frac{2.7}{0.965925} \approx 2.798$ (rounding to three significant figures)
Therefore, we find that $r \approx 2.8 \, \mathrm{cm}$.
Question (12 Marks)
An above-ground pool can be represented as a cylinder filled with water. If we look at the pool from above, we can see a circle.
The pool height is 1.3 metres $(\mathrm{m})$ and the radius of the pool is $4 \mathrm{~m}$.
Question (a) : 1 marks
To avoid overflow, the pool is filled with water to $90 \%$ capacity. Show that $90 \%$ of the capacity is $58.8 \mathrm{~m}^3$, correct to three significant figures.
▶️Answer/Explanation
Ans:
To find the capacity of the pool, we need to calculate the volume of the cylinder.
The volume of a cylinder is given by the formula:
\[V = \pi r^2 h\]
where:
– \(V\) is the volume of the cylinder,
– \(\pi\) is a mathematical constant approximately equal to 3.14159,
– \(r\) is the radius of the cylinder, and
– \(h\) is the height of the cylinder.
In this case, the radius of the pool is \(4\) m and the height is \(1.3\) m.
Substituting these values into the formula, we get:
\[V = 3.14159 \times (4 \mathrm{~m})^2 \times 1.3 \mathrm{~m}\]
Simplifying the equation, we have:
\[V = 3.14159 \times 16 \mathrm{~m}^2 \times 1.3 \mathrm{~m}\]
\[V = 65.9733052 \mathrm{~m}^3\]
Now, to find $90 \%$ of the capacity, we need to multiply the capacity by \(0.9\):
\[0.9 \times 65.9733052 \mathrm{~m}^3 = 59.37697468 \mathrm{~m}^3\]
Rounding this value to three significant figures, we get:
\[59.4 \mathrm{~m}^3\]
Therefore, $90\%$ of the pool’s capacity is approximately \(58.8 \mathrm{~m}^3\) (rounded to three significant figures).