Question

7 work out the surface area of these triangular prisms. the triangular cross - section of each prism is a right - angled triangle. give your answer to part c correct to one decimal place (1 d.p.).

Explanation:

Step1: Find hypotenuse of prism a

First, calculate the hypotenuse of the right-angled triangular cross-section for prism a using the Pythagorean theorem:
$$c = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\ \text{cm}$$

Step2: Calculate surface area of prism a

Surface area of a triangular prism = $2\times(\text{area of triangular cross-section}) + \text{sum of (side length}\times\text{length of prism)}$
Area of triangular cross-section: $\frac{1}{2}\times5\times12 = 30\ \text{cm}^2$
Sum of rectangular faces: $5\times20 + 12\times20 + 13\times20 = 100 + 240 + 260 = 600\ \text{cm}^2$
Total surface area: $2\times30 + 600 = 60 + 600 = 660\ \text{cm}^2$

Step3: Find hypotenuse of prism b

Calculate the hypotenuse of the right-angled triangular cross-section for prism b:
$$c = \sqrt{12^2 + 15^2} = \sqrt{144 + 225} = \sqrt{369} \approx 19.209\ \text{mm}$$

Step4: Calculate surface area of prism b

Area of triangular cross-section: $\frac{1}{2}\times12\times15 = 90\ \text{mm}^2$
Sum of rectangular faces: $12\times30 + 15\times30 + 19.209\times30 = 360 + 450 + 576.27 = 1386.27\ \text{mm}^2$
Total surface area: $2\times90 + 1386.27 = 180 + 1386.27 = 1566.27\ \text{mm}^2$

Step5: Find hypotenuse of prism c

Calculate the hypotenuse of the right-angled triangular cross-section for prism c:
$$c = \sqrt{1.1^2 + 2^2} = \sqrt{1.21 + 4} = \sqrt{5.21} \approx 2.2825\ \text{m}$$

Step6: Calculate surface area of prism c

Area of triangular cross-section: $\frac{1}{2}\times1.1\times2 = 1.1\ \text{m}^2$
Sum of rectangular faces: $1.1\times4.5 + 2\times4.5 + 2.2825\times4.5 = 4.95 + 9 + 10.27125 = 24.22125\ \text{m}^2$
Total surface area: $2\times1.1 + 24.22125 = 2.2 + 24.22125 = 26.42125\ \text{m}^2$, rounded to 1 decimal place: $26.4\ \text{m}^2$

Answer:

a. $660\ \text{cm}^2$
b. $1566.3\ \text{mm}^2$
c. $26.4\ \text{m}^2$