if the sum length, breadth and height of is 21cm, and the length of the diagnol is 12cm, find the surface area of cuboid.
sanikapatil9:
given- lenght = 12 height = 21 cm cuboid = ???????? surface area of cuboid = 2(lb+bh+lh) = 2(12×b+b×21+12×21) = 2 (12b+21b+252) = 285×2 = 570..........is the cuboid of this equation
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diagonal = √l²+b²+h² = 13
⇒l²+b²+h² = 13² = 169
l+b+h=21
⇒(l+b+h)²=21² = 441
⇒l²+b²+h²+2(lb+bh+lh)=441
⇒169 + total surface area = 441
how did u get l^2+b^2+h^2 as 13^2?
diagonal = √l²+b²+h² = 12
⇒l²+b²+h² = 12= 169
l+b+h=21
⇒(l+b+h)²=21² = 441
⇒l²+b²+h²+2(lb+bh+lh)=441
⇒169 + total surface area = 441
⇒ TSA = 441-169 = 272 cm²
cost = 272×1.40 = Rs. 380.8
⇒l²+b²+h² = 12= 169
l+b+h=21
⇒(l+b+h)²=21² = 441
⇒l²+b²+h²+2(lb+bh+lh)=441
⇒169 + total surface area = 441
⇒ TSA = 441-169 = 272 cm²
cost = 272×1.40 = Rs. 380.8
sorry wait i will give you again ...above one is wrong...
orry
ok
can u pls send a picture of it... it would be more clear..
diagonal = √l²+b²+h² = 12
⇒l²+b²+h² = 12² = 144
l+b+h=21
⇒(l+b+h)²=21² = 441
⇒l²+b²+h²+2(lb+bh+lh)=441
⇒144 + total surface area = 441
⇒ TSA = 441-144= 297 cm²
⇒l²+b²+h² = 12² = 144
l+b+h=21
⇒(l+b+h)²=21² = 441
⇒l²+b²+h²+2(lb+bh+lh)=441
⇒144 + total surface area = 441
⇒ TSA = 441-144= 297 cm²
this is the correct answer
thank you
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