Answer :
Answer:
Step-by-step explanation:
It is given that A pyramid has a rectangular base with edges of length 10 and 24. The vertex of the pyramid is directly 13 units above the center of the base.. then The total surface area = area of rectangular base + area of 2 isosceles triangles with a base of 24 units + area of 2 isosceles triangles with a base of 10 units.
Area of rectangular base = [tex]24{\times}10=240 sq units[/tex]
The slant height of isosceles triangles with a base of 24 units =[tex](\frac{10}{2})^{2}+(13)^{2})^\frac{1}{2}=(25+169)^{\frac{1}{2}}=13.928units[/tex].
The area of 2 isosceles triangles with a base of 24 units=[tex]\frac{2{\times}24{\times}13.928}{2}=334.281 sq units[/tex]
The slant height of isosceles triangles with a base of 10 units =[tex]((12)^{2}+(13)^{2})^\frac{1}{2}=(144+169)^\frac{1}{2}=17.691 units[/tex]
The area of 2 isosceles triangles with a base of 10 units=[tex]\frac{2{\times}10{\times}17.691}{2}=176.918 sq units[/tex]
The total surface area of the pyramid = 240 + 334.281 + 176.918 = 591.97 sq units.
Answer:
10 square root of 313+24 square root of 194+240
Step-by-step explanation:
its right
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