We have h = 15 m , V = 125 m 3 . h=15\ \text{m}, \ \ V=125 \ \text{m}^3. h = 15 m , V = 125 m 3 .
1 . Let a a a be the base edge of this square pyramid.
Volume of the pyramid is V = 1 3 B h V=\frac{1}{3} B h V = 3 1 B h , where B = a 2 B=a^2 B = a 2 is area of base.
V = 1 3 a 2 h V=\frac{1}{3}a^2h V = 3 1 a 2 h
a 2 = 3 V h a^2=\frac{3V}{h} a 2 = h 3 V
a = 3 V h = 3 × 125 15 = 25 = 5 a=\sqrt{\frac{3V}{h}}= \sqrt{\frac{3\times 125}{15}}=\sqrt{25}=5 a = h 3 V = 15 3 × 125 = 25 = 5
2.Let l l l be the slant height of the pyramid.
Then l = h 2 + ( a 2 ) 2 = 1 5 2 + 2. 5 2 = 925 / 4 = 5 37 / 2 ≈ 15.2 m l=\sqrt{h^2+\left(\frac{a}{2}\right)^2}=\sqrt{15^2+2.5^2}=\sqrt{925/4}=5\sqrt{37}/2\approx 15.2\ \text{m} l = h 2 + ( 2 a ) 2 = 1 5 2 + 2. 5 2 = 925/4 = 5 37 /2 ≈ 15.2 m
3.We can find the lateral area using formula L = 1 2 P l L=\frac{1}{2}Pl L = 2 1 Pl , where P = 4 a P=4a P = 4 a is the perimeter of the base, l l l is the slight height.
L = 4 a l 2 = 4 × 5 × 5 37 2 × 2 = 25 37 ≈ 152 m 2 L=\frac{4al}{2}=\frac{4\times 5\times 5\sqrt{37}}{2\times 2}=25\sqrt{37}\approx 152\ \text{m}^2 L = 2 4 a l = 2 × 2 4 × 5 × 5 37 = 25 37 ≈ 152 m 2 .
Answer: 1. a = 5 m a=5\ \text{m} a = 5 m , 2. l = 5 37 / 2 ≈ 15.2 m l=5\sqrt{37}/{2}\approx 15.2\ \text{m} l = 5 37 / 2 ≈ 15.2 m , 3. L = 25 37 ≈ 152 m 2 L=25\sqrt{37}\approx 152\ \text{m}^2 L = 25 37 ≈ 152 m 2 .
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