Find the missing measure to the nearest tenth. Use 3.14 for \( \pi \). 5. rectangular pyramid: base length \( =15 \mathrm{~m} \) base width \( = \) ? height \( =21 \mathrm{~m} \) volume \( =2415 \mathrm{~m}^{3} \) 6. triangular pyramid: base width \( =8 \mathrm{~cm} \) base height \( =18 \mathrm{~cm} \) height \( = \) ? volume \( =624 \mathrm{~cm}^{3} \) \( \qquad \)
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To find the missing measures for the rectangular pyramid, we start with the formula for the volume of a rectangular pyramid: \[ V = \frac{1}{3} \times \text{Base Area} \times \text{Height} \] The base area (A) is calculated as: \[ A = \text{base length} \times \text{base width} = 15 \times b \] Setting this equal to the given volume: \[ 2415 = \frac{1}{3} \times (15 \times b) \times 21 \] Solving for b gives: \[ b = \frac{2415 \times 3}{15 \times 21} \] \[ b \approx 23.0 \mathrm{~m} \] For the triangular pyramid, we use the formula for volume as well: \[ V = \frac{1}{3} \times \text{Base Area} \times \text{Height} \] Here, the base area is: \[ A = \frac{1}{2} \times \text{base width} \times \text{base height} = \frac{1}{2} \times 8 \times 18 \] \[ A = 72 \mathrm{~cm}^2 \] Using the volume formula: \[ 624 = \frac{1}{3} \times 72 \times h \] Solving for h provides: \[ h = \frac{624 \times 3}{72} \] \[ h \approx 26.0 \mathrm{~cm} \]
