(a) The diagram shows a wooden structure in the form of a cone, mounted on a hemispherical base. The vertical height of the cone is 48 m and the base radius...

(a) Surface Area of the Wooden Structure

The structure consists of a cone mounted on a hemispherical base. We need to find the total surface area of this structure.

Given:

• Radius of both the cone and the hemisphere, $r=14r\; =\; 14$ m
• Height of the cone, $h=48h\; =\; 48$ m
• $\pi =\frac{22}{7}\backslash pi\; =\; \backslash frac\{22\}\{7\}$

Step 1: Calculate the Slant Height of the Cone

The slant height $ll$ of the cone can be found using the Pythagorean theorem: $l=\sqrt{h^2+r^2}=\sqrt{48^2+14^2}l\; =\; \backslash sqrt\{h^2\; +\; r^2\}\; =\; \backslash sqrt\{48^2\; +\; 14^2\}$

l=h2+r2​=482+142​ $l=\sqrt{2304+196}=\sqrt{2500}=50\text{ m}l\; =\; \backslash sqrt\{2304\; +\; 196\}\; =\; \backslash sqrt\{2500\}\; =\; 50\; \backslash text\{\; m\}$

Step 2: Calculate the Surface Area of the Cone

The surface area of the cone (excluding the base) is:

$A_{\text{cone}}=\pi rl=\frac{22}{7}\times 14\times 50A\_\{\backslash text\{cone\}\}\; =\; \backslash pi\; r\; l\; =\; \backslash frac\{22\}\{7\}\; \backslash times\; 14\; \backslash times\; 50$ $A_{\text{cone}}=22\times 50=1100{\text{ m}}^{2}A\_\{\backslash text\{cone\}\}\; =\; 22\; \backslash times\; 50\; =\; 1100\; \backslash text\{\; m\}^2$

Step 3: Calculate the Surface Area of the Hemisphere

The surface area of a hemisphere (excluding the base) is: $A_{\text{hemisphere}}=2\pi r^2=2\times \frac{22}{7}\times 14^{2}A\_\{\backslash text\{hemisphere\}\}\; =\; 2\; \backslash pi\; r^2\; =\; 2\; \backslash times\; \backslash frac\{22\}\{7\}\; \backslash times\; 14^2$

Ahemisphere​=2πr2=2×722​×142 $A_{\text{hemisphere}}=2\times \frac{22}{7}\times 196A\_\{\backslash text\{hemisphere\}\}\; =\; 2\; \backslash times\; \backslash frac\{22\}\{7\}\; \backslash times\; 196$ $A_{\text{hemisphere}}=2\times 22\times 28=1232{\text{ m}}^{2}A\_\{\backslash text\{hemisphere\}\}\; =\; 2\; \backslash times\; 22\; \backslash times\; 28\; =\; 1232\; \backslash text\{\; m\}^2$

Step 4: Calculate the Total Surface Area

The total surface area of the structure is the sum of the surface area of the cone and the surface area of the hemisphere: $A_{\text{total}}=A_{\text{cone}}+A_{\text{hemisphere}}=1100+1232=2332{\text{ m}}^{2}A\_\{\backslash text\{total\}\}\; =\; A\_\{\backslash text\{cone\}\}\; +\; A\_\{\backslash text\{hemisphere\}\}\; =\; 1100\; +\; 1232\; =\; 2332\; \backslash text\{\; m\}^2$

Atotal​=Acone​+Ahemisphere​=1100+1232=2332 m2

Thus, the surface area of the structure, correct to three significant figures, is: $2330{\text{ m}}^{2}2330\; \backslash text\{\; m\}^2$

(b) Finding Sesay's Present Age

Given:

• Five years ago, Musah was twice as old as Sesay.
• The sum of their current ages is 100.

Let Musah's present age be $MM$ and Sesay's present age be $SS$.

Five years ago:

• Musah's age was $M-5M\; -\; 5$
• Sesay's age was $S-5S\; -\; 5$

According to the problem, five years ago, Musah was twice as old as Sesay: $M-5=2(S-5)M\; -\; 5\; =\; 2(S\; -\; 5)$

$M-5=2(S-5)$ $M-5=2S-10M\; -\; 5\; =\; 2S\; -\; 10$ $M=2S-5M\; =\; 2S\; -\; 5$

We are also given that the sum of their current ages is 100: $M+S=100M\; +\; S\; =\; 100$

Substitute: $M=2S-5M\; =\; 2S\; -\; 5$

$M=2S-5$ into $M+S=100M\; +\; S\; =\; 100$: $(2S-5)+S=100(2S\; -\; 5)\; +\; S\; =\; 100$ $3S-5=1003S\; -\; 5\; =\; 100$ $3S=1053S\; =\; 105$ $S=35S\; =\; 35$

So, Sesay's present age is: $S=35S\; =\; 35$

$S=35$

To verify, calculate Musah's present age: $M=2S-5=2(35)-5=70-5=65M\; =\; 2S\; -\; 5\; =\; 2(35)\; -\; 5\; =\; 70\; -\; 5\; =\; 65$

$M=2S-5=2(35)-5=70-5=65$

Check the sum: $M+S=65+35=100M\; +\; S\; =\; 65\; +\; 35\; =\; 100$

Thus, Sesay's present age is $3535$ years.