Question #144907
A regular square pyramid has an altitude of 15 m and a volume of 125 m3.

1. What is the base edge of a square pyramid?
2. What is the slant height of the pyramid?
3. What is the lateral area in m2?
1
Expert's answer
2020-11-17T17:28:05-0500

We have h=15 m,  V=125 m3.h=15\ \text{m}, \ \ V=125 \ \text{m}^3.

1 . Let aa be the base edge of this square pyramid.

Volume of the pyramid is V=13BhV=\frac{1}{3} B h , where B=a2B=a^2 is area of base.

V=13a2hV=\frac{1}{3}a^2h

a2=3Vha^2=\frac{3V}{h}

a=3Vh=3×12515=25=5a=\sqrt{\frac{3V}{h}}= \sqrt{\frac{3\times 125}{15}}=\sqrt{25}=5

2.Let ll be the slant height of the pyramid.

Then l=h2+(a2)2=152+2.52=925/4=537/215.2 ml=\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}

3.We can find the lateral area using formula L=12PlL=\frac{1}{2}Pl , where P=4aP=4a is the perimeter of the base, ll is the slight height.

L=4al2=4×5×5372×2=2537152 m2L=\frac{4al}{2}=\frac{4\times 5\times 5\sqrt{37}}{2\times 2}=25\sqrt{37}\approx 152\ \text{m}^2 .




Answer: 1. a=5 ma=5\ \text{m} , 2. l=537/215.2 ml=5\sqrt{37}/{2}\approx 15.2\ \text{m} , 3. L=2537152 m2L=25\sqrt{37}\approx 152\ \text{m}^2 .



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