On each face of a cuboid,the sum of its perim | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the dimensions of the cuboid be $x, y,$ and $z$. The faces have dimensions $(x, y), (y, z),$ and $(x, z)$.The sum of the perimeter and area fo

Vedclass · Accepted Answer

$28$ and $35$

Vedclass · Answer

(D) Let the dimensions of the cuboid be $x, y,$ and $z$. The faces have dimensions $(x, y), (y, z),$ and $(x, z)$.The sum of the perimeter and area for each face is given by:$2(x+y) + xy = 16 \quad (i)$$2(y+z) + yz = 24 \quad (ii)$$2(x+z) + xz = 31 \quad (iii)$Adding $4$ to both sides of each equation to factorize:$xy + 2x + 2y + 4 = 16 + 4 \implies (x+2)(y+2) = 20 \quad (iv)$$yz + 2y + 2z + 4 = 24 + 4 \implies (y+2)(z+2) = 28 \quad (v)$$xz + 2x + 2z + 4 = 31 + 4 \implies (x+2)(z+2) = 35 \quad (vi)$Let $X = x+2, Y = y+2, Z = z+2$. Then $XY=20, YZ=28, XZ=35$.Multiplying these: $(XYZ)^2 = 20 	imes 28 	imes 35 = 19600 \implies XYZ = 140$.$Z = \frac{XYZ}{XY} = \frac{140}{20} = 7 \implies z+2 = 7 \implies z = 5$.$X = \frac{XYZ}{YZ} = \frac{140}{28} = 5 \implies x+2 = 5 \implies x = 3$.$Y = \frac{XYZ}{XZ} = \frac{140}{35} = 4 \implies y+2 = 4 \implies y = 2$.The volume $V = xyz = 3 	imes 2 	imes 5 = 30$.Since $30$ lies between $28$ and $35$,option $(d)$ is correct.

On each face of a cuboid,the sum of its perimeter and its area is written. Among the six numbers so written,there are three distinct numbers and they are $16, 24$ and $31$. The volume of the cuboid lies between

Similar Questions

If $\alpha, \beta$ are the real roots of $x^2+p x+q=0$ and $\alpha^4, \beta^4$ are the roots of $x^2-r x+s=0$,then the equation $x^2-4 q x+2 q^2-r=0$ has always

If $\tan A$ and $\tan B$ are the roots of the quadratic equation $3x^2 - 10x - 25 = 0$,then the value of $3 \sin^2 (A + B) - 10 \sin (A + B) \cos (A + B) - 25 \cos^2 (A + B)$ is

If the equation $x^8 - kx^2 + 3 = 0$ has a real solution,then the least integral value of $k$ is-

The number of solutions of the equation $x^2 + y^2 = a^2 + b^2 + c^2$,where $x, y, a, b, c$ are all prime numbers,is

$E_1: a+b+c=0$,if $1$ is a root of $ax^2+bx+c=0$. $E_2: b^2-a^2=2ac$,if $\sin \theta, \cos \theta$ are the roots of $ax^2+bx+c=0$. Which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module