TheCATExam
Sign in

CAT 2024 Slot 1 — QA Question 2

SphereEasy

The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq. cm., and the sum of the lengths of all its edges is 144 cm. The volume, in cubic cm, of the sphere is

Answer & solution

  • A

    1125π

  • 1125π√2

  • C

    750π√2

  • D

    750π

Solution

Medium

The sphere that circumscribes the box has the box's space diagonal as its diameter. So we only need l2+w2+h2l^2+w^2+h^2, which comes straight from the surface area and the sum of the edges via the identity (l+w+h)2=(l2+w2+h2)+2(lw+lh+wh)(l+w+h)^2=(l^2+w^2+h^2)+2(lw+lh+wh).

1

Translate the two given facts. Surface area =2(lw+lh+wh)=2(lw+lh+wh) and the sum of all 1212 edges =4(l+w+h)=4(l+w+h).

2(lw+lh+wh)=846⇒ lw+lh+wh=423(divide by 2)4(l+w+h)=144⇒ l+w+h=36(divide by 4)\begin{aligned} &2(lw+lh+wh)=846\\ &\Rightarrow\ lw+lh+wh=423 \quad\text{(divide by 2)}\\ &4(l+w+h)=144\\ &\Rightarrow\ l+w+h=36 \quad\text{(divide by 4)} \end{aligned}
2

Get the squared diagonal. Use the square-of-a-sum identity with the values from step 1.

(l+w+h)2=(l2+w2+h2)+2(lw+lh+wh)⇒ 362=(l2+w2+h2)+2⋅423(substitute step 1)⇒ 1296=(l2+w2+h2)+846⇒ l2+w2+h2=450\begin{aligned} &(l+w+h)^2=(l^2+w^2+h^2)+2(lw+lh+wh)\\ &\Rightarrow\ 36^2=(l^2+w^2+h^2)+2\cdot 423 \quad\text{(substitute step 1)}\\ &\Rightarrow\ 1296=(l^2+w^2+h^2)+846\\ &\Rightarrow\ l^2+w^2+h^2=450 \end{aligned}
3

Diagonal = diameter. For a box inscribed in a sphere the space diagonal equals the diameter 2r2r.

(2r)2=l2+w2+h2=450⇒ r2=4504=2252(from step 2)⇒ r=152\begin{aligned} &(2r)^2=l^2+w^2+h^2=450\\ &\Rightarrow\ r^2=\dfrac{450}{4}=\dfrac{225}{2} \quad\text{(from step 2)}\\ &\Rightarrow\ r=\dfrac{15}{\sqrt2} \end{aligned}
4

Volume of the sphere.

V=43πr3=43π(152)3⇒ V=43π⋅337522(cube the radius)⇒ V=4500π22=2250π2=1125π2\begin{aligned} &V=\tfrac43\pi r^3=\tfrac43\pi\left(\dfrac{15}{\sqrt2}\right)^{3}\\ &\Rightarrow\ V=\tfrac43\pi\cdot\dfrac{3375}{2\sqrt2} \quad\text{(cube the radius)}\\ &\Rightarrow\ V=\dfrac{4500\pi}{2\sqrt2}=\dfrac{2250\pi}{\sqrt2}=1125\pi\sqrt2 \end{aligned}
V=1125π2 cubic cmV=1125\pi\sqrt2\ \text{cubic cm}

Don't solve for l,w,hl,w,h individually — the volume needs only l2+w2+h2l^2+w^2+h^2, so combine surface area and edge-sum directly through (l+w+h)2(l+w+h)^2.