A pyramid has an equilateral triangle as its base of which each side is 1 m. Its slant edge is 3 m. The whole surface are of the pyramid is equal to :
A. $$\frac{{\sqrt 3 + 2\sqrt {13} }}{4}sq.m$$
B. $$\frac{{\sqrt 3 + 3\sqrt {13} }}{4}sq.m$$
C. $$\frac{{\sqrt 3 + 3\sqrt {35} }}{4}sq.m$$
D. $$\frac{{\sqrt 3 + 2\sqrt {35} }}{4}sq.m$$
Answer: Option C
Solution(By Examveda Team)
Area of base :$$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} \times {1^2}} \right){m^2} \cr & = \frac{{\sqrt 3 }}{4}{m^2} \cr} $$
Clearly, the pyramid has 3 triangular faces each with sides 3m, 3m and 1 m
So, area of each lateral face :
$$\eqalign{ & = \sqrt {\frac{7}{2} \times \left( {\frac{7}{2} - 3} \right)\left( {\frac{7}{2} - 3} \right)\left( {\frac{7}{2} - 1} \right)} {m^2} \cr & \left[ {\because s = \frac{{3 + 3 + 1}}{2} = \frac{7}{2}} \right] \cr & = \sqrt {\frac{7}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{5}{2}} {m^2} \cr & = \frac{{\sqrt {35} }}{4}{m^2} \cr} $$
∴ Whole surface area of the pyramid :
$$\eqalign{ & = \left( {\frac{{\sqrt 3 }}{4} + 3 \times \frac{{\sqrt {35} }}{4}} \right){m^2} \cr & = \frac{{\sqrt 3 + 3\sqrt {35} }}{4}{m^2} \cr} $$
Related Questions on Volume and Surface Area
A. 12$$\pi$$ cm3
B. 15$$\pi$$ cm3
C. 16$$\pi$$ cm3
D. 20$$\pi$$ cm3
In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of ground is:
A. 75 cu. m
B. 750 cu. m
C. 7500 cu. m
D. 75000 cu. m
A. 84 meters
B. 90 meters
C. 168 meters
D. 336 meters
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