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A square-based pyramid has a base side length of 20 cm and a slant edge of 26 cm. Calculate the total number of surfaces and the total surface area.

Mathematics

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2 parts
a

5 surfaces\text{5 surfaces}

b

1360 cm2\text{1360 cm}^2

Step-by-step solution

5 steps

Here's the solution to question 5:

Given:
Length of a side of the base of the square-based pyramid, a=20a = 20 cm.
Length of the slant edge, s=26s = 26 cm.

Part a

A square-based pyramid has one square base and four triangular lateral faces.

Therefore, the total number of surfaces is 1+4=5 surfaces1 + 4 = \boxed{\text{5 surfaces}}.

Part b

1360 cm2\text{1360 cm}^2

To find the total surface area (TSA) of the pyramid, we need the base area and the lateral surface area.

The formula for TSA is TSA=Base Area+Lateral Surface Area\text{TSA} = \text{Base Area} + \text{Lateral Surface Area}.
Base Area =a2= a^2.
Lateral Surface Area =2al= 2al, where ll is the slant height.

  1. Calculate the vertical height (hh) of the pyramid.

    We use the relationship between the slant edge (ss), vertical height (hh), and half of the base diagonal. Alternatively, we can use the relationship s2=h2+(a22)2=h2+a22s^2 = h^2 + \left(\frac{a\sqrt{2}}{2}\right)^2 = h^2 + \frac{a^2}{2}.

    s2=h2+a22s^2 = h^2 + \frac{a^2}{2}

    Substitute the given values:

    262=h2+202226^2 = h^2 + \frac{20^2}{2} 676=h2+4002676 = h^2 + \frac{400}{2} 676=h2+200676 = h^2 + 200 h2=676−200h^2 = 676 - 200 h2=476h^2 = 476 h=476 cmh = \sqrt{476} \text{ cm}

  2. Calculate the slant height (ll) of the pyramid.

    We use the relationship between the slant height (ll), vertical height (hh), and half of the base side length (a2\frac{a}{2}).

    l2=h2+(a2)2l^2 = h^2 + \left(\frac{a}{2}\right)^2

    Substitute the values for h2h^2 and aa:

    l2=476+(202)2l^2 = 476 + \left(\frac{20}{2}\right)^2 l2=476+102l^2 = 476 + 10^2 l2=476+100l^2 = 476 + 100 l2=576l^2 = 576 l=576l = \sqrt{576} l=24 cml = 24 \text{ cm}

  3. Calculate the total surface area (TSA).

    TSA=a2+2al\text{TSA} = a^2 + 2al TSA=202+2×20×24\text{TSA} = 20^2 + 2 \times 20 \times 24 TSA=400+960\text{TSA} = 400 + 960 TSA=1360 cm2\text{TSA} = 1360 \text{ cm}^2

    The total surface area of the pyramid is 1360 cm2\boxed{\text{1360 cm}^2}.

Part c· 2 steps

To compare the vertical height (hh) and slant height (ll)

  1. State the values of hh and ll.

    Vertical height h=476h = \sqrt{476} cm.
    Slant height l=24l = 24 cm.

  2. Compare the values.

    We know that 212=44121^2 = 441 and 222=48422^2 = 484. So, 476\sqrt{476} is between 21 and 22.
    h≈21.82h \approx 21.82 cm.
    Since 24>21.8224 > 21.82, the slant height is greater than the vertical height.
    Comparison: The slant height (l=24l = 24 cm) is greater than the vertical height (h=476≈21.82h = \sqrt{476} \approx 21.82 cm).

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