Dimensions of a closed wooden box are in the ratio 5:4:3 . If cost of painting its outer surface excluding the base at the rate of Rs 5 per square decameter is Rs 9250 find the dimensions of the box.
Answers
Answer:
Outer dimensions of the box are 10,8,5 respectively.
The thickness of wood is 1cm.
∴ inner dimensions are 10−2,8−2 and 5−1 i.e., 8 cm., 6 cm. and 4 cm.
∴ outer surface area =10×8+2×5(10+8)
=80+180=260cm2
and inner surface area =8×6+2×4(8+6)
=48+112=160cm2
and Area of top =lb−(l−2)(b−2)=10×8−8×6=32
So total SA=260+160+32=452 sq. cm
Answer:
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Step-by-step explanation:
Given external dimensions of a closed wooden box are in the ratio 5:4:3.if the cost of painting its outer surface at the rate of Rs 5 per Rs 11750. we have to find the dimension of box.
Let length=5x, breadth=4x and height=3x
T.S.A=
2(lb+bh+hl)
=
2[(5x x 4x)+(4x x 3x)+(3x x 5x)]
=
94x^2
Also given outer surface at the rate of Rs 5 per dm^2= Rs 9250
TSA= 5(9250)= 46250dm^2
Hence, equating we get
94x^2=46250
x^2=46250/94
x^2=492.02
x=22.181
Length= 5x =5 x 22.181= 110.905dm
= 4x = 4 x 22.181 = 88.724dm
=3x = 3 x 22.181= 66.543dm
Hope it helps....