**12.2.1 Solid Geometry (II), PT3 Focus Practice**

Question 1:

Question 1:

Diagram below shows closed right cylinder.

Calculate the total surface area, in cm

^{2}, of the cylinder.

$\left(\pi =\frac{22}{7}\right)$

SolutionSolution

**:**

Total surface area

= 2(π

$\begin{array}{l}=\left(2\times \frac{22}{7}\times {7}^{2}\right)+\left(2\times \frac{22}{7}\times 7\times 20\right)\\ =308+880\\ =1188c{m}^{2}\end{array}$
*r*^{2}) + 2π*rh***Question 2:**

Diagram below shows a right prism with right-angled triangle

*ABC*as its uniform cross section.Calculate the total surface area, in cm

^{2}, of the prism.

SolutionSolution

**:**

$\begin{array}{l}AB=\sqrt{{5}^{2}-{3}^{2}}\\ =\sqrt{25-9}\\ =\sqrt{16}\\ =4cm\end{array}$

Total surface area

= 2 (½× 3 × 4) + (3 × 10) + (4 × 10) + (5 × 10)

= 12 + 30 + 40 + 50

= 132 cm^{2}

^{ }

**Question 3:**

Diagram below shows a right pyramid with a square base.

Calculate the total surface area, in cm

^{2}, of the right pyramid.

SolutionSolution

**:**

*h*= 10

^{2}^{2}– 6

^{2}

= 100 – 36

= 64

*h*= √64

= 8cm

Total surface area of the right pyramid

= (12 × 12) + 4 × (½× 12 × 8)

= 144 + 192

= 336 cm

^{2 }