Answer to Question #149312 in Trigonometry for yui

Question

A pyramidal monument has an octagonal base, each side of which is 2 m. Each lateral edge of the pyramid is 3m. Find the total cost of painting the lateral surface of the monument at PhP (Philippine Pesos) 730.00 per square meter?
FINAL ANSWER: _______________________________________

Answer 1

A pyramidal monument has an octagonal base, each side is 2m. Hence BC=DC=DE=EF=FG=GH=HI=IB=2m

Lateral edge of pyramid = AB=AC=AD=AE=AF=AG=AH=AI=3m

Now in $\Delta$ ABC AB=AC=3m, AP _|_ BC

Hence BP = PC=2/2=1m

$AC^2 = AP^2+PC^2\\ 3^2 = AP^2+I^2 \implies 9 = AP^2+1 \implies 9-1 = AP^2$

Hence $AP^2 = 8 \implies AP = \sqrt8 = 2*\sqrt2$

Now ar of $\Delta$ ABC = 1/2*base*heihght = 1/2 * BC * AP = $1/2*2*2*\sqrt2 = 2\sqrt2m^2$

Now lateral surface area of pyramid = $8*2\sqrt2 = 16\sqrt2m^2$

Cost of painting the lateral surface of monument is: $16\sqrt2*730 = 16517.86$

Answer: PhP 16518