Question : A right pyramid stands on a square base of diagonal $10\sqrt2$ cm. If the height of the pyramid is 12 cm, the area (in cm2) of its slant surface is:
Option 1: 520 cm2
Option 2: 420 cm2
Option 3: 360 cm2
Option 4: 260 cm2
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Correct Answer: 260 cm 2
Solution :
Use the formulas:
Area of the slant surface = $\frac{1}{2}×p×l$
Perimeter, $p$ = $4a$
Slant height, $l$ = $\sqrt{(\frac{a}{2})^{2}+h^{2}}$
where $l$ = slant height, $h$ = height, $p$ = perimeter and $a$ = side of square base.
Diagonal of square base $= 10\sqrt2$ cm
Height of the pyramid, $h = 12$ cm
Side of square base, $a = \frac{1}{\sqrt2}×10\sqrt2 = 10$ cm
Perimeter, $p = 4a = 4×10 = 40$ cm
Slant height, $l = \sqrt{(\frac{a}{2})^{2}+h^{2}} = \sqrt{5^{2}+12^{2}} = \sqrt{169} = 13$ cm
Area of the slant surface $= \frac{1}{2}×40×13 = 260$ cm
2
Hence, the correct answer is 260 cm
2
.
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