Home / IB MYP Year 4-5: Exntended Mathematics : Unit 4: Geometry -Volume and capacity (additional shapes) MYP Style Questions

IB MYP Year 4-5: Exntended Mathematics : Unit 4: Geometry -Volume and capacity (additional shapes) MYP Style Questions

IB myp 4-5 MATHEMATICS – Practice Questions- All Topics

Topic :GeometryVolume and capacity

Topic :Geometry- Weightage : 21 % 

All Questions for Topic : Volume and capacity (additional shapes),Enlargement around a given point,Enlargement by a rational factor,Gradients of perpendicular lines,Identical representation of transformations

Question

Question (a)

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 (b)

The whole sphere of ice melted in the cone, as shown in the diagram below.

Find the value of $h$.

▶️Answer/Explanation

Ans:

Given:
Radius of the melted sphere, $r = 2.80 \, \mathrm{cm}$
Radius of the cone, $R = 2.86 \, \mathrm{cm}$

We can still use the concept of equal volumes to find the height $h_{\text{cone}}$.

The volume of the cone is given by:

$V_{\text{cone}} = \frac{1}{3} \pi R^2 h_{\text{cone}}$

The volume of the sphere is given by:

$V_{\text{sphere}} = \frac{4}{3} \pi r^3$

Setting the volumes equal to each other:

$\frac{1}{3} \pi R^2 h_{\text{cone}} = \frac{4}{3} \pi r^3$

Canceling out $\pi$ and simplifying, we have:

$R^2 h_{\text{cone}} = 4r^3$

Substituting the values $r = 2.80 \, \mathrm{cm}$ and $R = 2.86 \, \mathrm{cm}$, we get:

$(2.86)^2 h_{\text{cone}} = 4(2.80)^3$

$8.1796 h_{\text{cone}} = 87.5456$

Dividing both sides by $8.1796$, we find:

$h_{\text{cone}} \approx \frac{87.5456}{8.1796} \approx 10.69$

Therefore, the height of the melted ice in the cone is approximately $10.7 \, \mathrm{cm}$.

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