# Ex.13.1 Q6 Surface Areas and Volumes - NCERT Maths Class 9

## Question

A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is \(30\,\rm{ cm}\) long, \(25\,\rm cm\) wide and \(25\,\rm{ cm}\) high.

(i) What is the area of the glass?

(ii) How much of tape is needed for all the \(12\) edges?

## Text Solution

**Reasoning:**

A cuboid is enclosed by six rectangle regions called faces and it has \(12\) edges. The total surface area is the sum of the areas of the six faces which is nothing but the area of the glass.

(i) Area of the glass?

**What is the known?**

Measurements of herbarium.

**What is the unknown?**

The area of the glass.

**Steps:**

The area of the herbarium is the total surface area of the cuboid.

\[\begin{align}&\text{length(l) = 30}\,\rm{cm}\\&\text{breadth = 25}\,\rm{cm}\\&\text{height = 25}\,\rm{cm}\end{align}\]

Total surface area of the cuboid

\[\begin{align}&= 2(lb + bh + hl)\\ &= 2[25\!\times\!30\!+\!25 \!\times\! 25\!+\!30\!\times\!25]\\ &= 2[750 + 625 + 750]\\ &= 4250\,\rm{cm^2} \end{align}\]

Area of the glass \(\begin{align} = 4250\,\rm{cm^2}. \end{align}\)

(ii) How much of tape is needed for all the \(12\) edges?

**What is the unknown?**

Tape needed for all the \(12\) edge.

**Steps:**

Total length of the tape needed for \(12\) edges

\[\begin{align} & = 4(l + b + h)\\ &= 4(30 + 25 + 25)\\ &= 320\,\rm{cm}\end{align}\]

Tape required to cover all the \(12\) edges is \(320\,\rm{ cm}.\)