Question

1. calculate the surface area and volume of the square based pyramids below.

a.
73 in
110 in
110 in
$sa=ab$
$ab=110×110$
$ab=12100$
b.
61 m
22 m
22 m

Explanation:

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Part a: Square Pyramid (110 in base, 73 in slant height)

Step1: Calculate base area

Base is a square: $A_{base} = s^2 = 110^2 = 12100$ in²

Step2: Calculate lateral surface area

4 congruent triangles: $A_{lateral} = 4 \times \frac{1}{2} \times s \times l = 2 \times 110 \times 73 = 16060$ in²

Step3: Total surface area

Sum base + lateral: $SA = 12100 + 16060 = 28160$ in²

Step4: Calculate pyramid height

Use Pythagoras: $h = \sqrt{l^2 - (\frac{s}{2})^2} = \sqrt{73^2 - 55^2} = \sqrt{5329 - 3025} = \sqrt{2304} = 48$ in

Step5: Calculate volume

Formula: $V = \frac{1}{3} \times A_{base} \times h = \frac{1}{3} \times 12100 \times 48 = 193600$ in³
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Part b: Square Pyramid (22 m base, 61 m slant height)

Step1: Calculate base area

Base is a square: $A_{base} = s^2 = 22^2 = 484$ m²

Step2: Calculate lateral surface area

4 congruent triangles: $A_{lateral} = 4 \times \frac{1}{2} \times s \times l = 2 \times 22 \times 61 = 2684$ m²

Step3: Total surface area

Sum base + lateral: $SA = 484 + 2684 = 3168$ m²

Step4: Calculate pyramid height

Use Pythagoras: $h = \sqrt{l^2 - (\frac{s}{2})^2} = \sqrt{61^2 - 11^2} = \sqrt{3721 - 121} = \sqrt{3600} = 60$ m

Step5: Calculate volume

Formula: $V = \frac{1}{3} \times A_{base} \times h = \frac{1}{3} \times 484 \times 60 = 9680$ m³

Answer:

a. Surface Area = 28160 square inches, Volume = 193600 cubic inches
b. Surface Area = 3168 square meters, Volume = 9680 cubic meters