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
New: SSC CHSL tier 1 answer key 2024 out | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
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
.
Related Questions
Know More about
Staff Selection Commission Combined High ...
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Get Updates Brochure
Your Staff Selection Commission Combined Higher Secondary Level Exam brochure has been successfully mailed to your registered email id “”.
