Answer:
Length of the base = 15 cm
Volume = 1125√2 cm³ = 1590.99 cm³ (2 d.p.)
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{6.6cm}\underline{Lateral Surface Area of a square pyramid}\\\\$LSA=a \sqrt{a^2+4h^2}$\\\\where:\\\phantom{ww}$\bullet$ $a$ is the base edge \\\phantom{ww}$\bullet$ $h$ is the height\\\end{minipage}}[/tex]
[tex]\boxed{\begin{minipage}{6.6cm}\underline{Base Area of a square pyramid}\\\\$BA=a^2$\\\\where:\\\phantom{ww}$\bullet$ $a$ is the base edge \\\end{minipage}}[/tex]
If the lateral surface area (LSA) of the pyramid is three times the base area (BA):
[tex]\implies a \sqrt{a^2+4h^2}=3a^2[/tex]
Rearrange the equation to isolate h²:
[tex]\implies \sqrt{a^2+4h^2}=3a[/tex]
[tex]\implies (\sqrt{a^2+4h^2})^2=(3a)^2[/tex]
[tex]\implies a^2+4h^2=9a^2[/tex]
[tex]\implies 4h^2=8a^2[/tex]
[tex]\implies h^2=2a^2[/tex]
[tex]\boxed{\begin{minipage}{6.6cm}\underline{Total Surface Area of a square pyramid}\\\\$SA=a^2+a \sqrt{a^2+4h^2}$\\\\where:\\\phantom{ww}$\bullet$ $a$ is the base edge \\\phantom{ww}$\bullet$ $h$ is the height\\\end{minipage}}[/tex]
Given:
Substitute these values into the formula for the total surface area and solve for a:
[tex]\implies a^2+a\sqrt{a^2+4(2a^2)}=900[/tex]
[tex]\implies a^2+a\sqrt{a^2+8a^2}=900[/tex]
[tex]\implies a^2+a\sqrt{9a^2}=900[/tex]
[tex]\implies a^2+a(3a)=900[/tex]
[tex]\implies a^2+3a^2=900[/tex]
[tex]\implies 4a^2=900[/tex]
[tex]\implies a^2=225[/tex]
[tex]\implies a=15[/tex]
Therefore, the length of the base of the square based pyramid is:
To find the height (h), substitute the found value of a into the formula for h² and solve for h:
[tex]\implies h^2=2(15)^2[/tex]
[tex]\implies \sqrt{h^2}=\sqrt{2(15)^2}[/tex]
[tex]\implies h=\sqrt{2}\sqrt{(15)^2}[/tex]
[tex]\implies h=15\sqrt{2}[/tex]
Therefore, the height of the square based pyramid is:
[tex]\boxed{\begin{minipage}{5cm}\underline{Volume of a square pyramid}\\\\$V=\dfrac{1}{3}a^2h$\\\\where:\\\phantom{ww}$\bullet$ $a$ is the base edge \\\phantom{ww}$\bullet$ $h$ is the height\\\end{minipage}}[/tex]
Given:
Substitute the values into the formula and solve for volume:
[tex]\implies V=\dfrac{1}{3}(15)^2 \cdot 15\sqrt{2}[/tex]
[tex]\implies V=\dfrac{225}{3} \cdot 15\sqrt{2}[/tex]
[tex]\implies V=75\cdot 15\sqrt{2}[/tex]
[tex]\implies V=1125\sqrt{2}[/tex]
[tex]\implies V=1590.99[/tex]
Therefore, the volume of the square based pyramid is:
