Question

Diagram of the adjacent picture frame has outer dimensions$ = 24\,cm \times 28\,cm$ and inner dimensions $ = 16\,cm \times 20\,cm$. Find the area of each section of the frame, if the width of each section is the same.

Answer 1

Hint:In order to find the area of each section of the frame with the mentioned diagrams, first draw a rough diagram in which the sections are distributed. The section obtained would be in the form of a trapezium, so find the area of the trapezium using its formula and get the results. Similarly, it is to be done for other sections.

Formula used:
Area of Trapezium $ = \dfrac{1}{2}\left( {{\text{Sum of Parallel Sides}}} \right) \times height$

Complete step by step answer:
We are given that the outer dimension of the frame is $ = 24cm \times 28cm$, whereas the inner dimension is $ = 16cm \times 20cm$ and the width for each section is the same.Considering the width to be ‘w’, and naming all the parts of the frame obtained between the dimensions. From the above information the diagram obtained is:

Now, from the figure, we got that the frame has been divided into 4 parts, namely as 1,2,3,4:
We need to find the width first;
For that, from figure we can see $AB = w + 16 + w$ and $AB = 24$
So, substituting AB in the equation and solving it, we get:
$AB = w + 16 + w$
$ \Rightarrow 24 = 2w + 16$
Subtracting both the sides by $16$:
$ \Rightarrow 24 - 16 = 2w + 16 - 16$
$ \Rightarrow 8 = 2w$
Dividing both the sides by $2$:
$ \Rightarrow \dfrac{8}{2} = \dfrac{{2w}}{2}$
$ \Rightarrow 4 = w$
Hence, we get that the width of the section of the frame is $4{\text{ cm}}$.
From the symmetry of the frame, we can see that:
Area of section 1=Area of section 3
Area of section 2= Area of section 4
So, we need to find only two sections, and the rest will be the same.

Area of Section 1: In Section 1 from the figure, we can see that AB and EF are parallel and the width ${\text{ = 4 cm}}$ as obtained. The shape of the parts obtained are all in the forms of trapezium. That means Area of section 1= Area of trapezium
The formula for Area of Trapezium $ = \dfrac{1}{2}\left( {{\text{Sum of Parallel Sides}}} \right) \times height$
So, substituting the values of parallel sides AB and EF as ${\text{24 cm}}$ and ${\text{16 cm}}$, height as ${\text{4 cm}}$ in the above equation, and we get:
${\text{Area of Section 1}} = \dfrac{1}{2}\left( {{\text{Sum of Parallel Sides}}} \right) \times height \\
\Rightarrow {\text{Area of Section 1}} =\dfrac{1}{2}\left( {AB + EF} \right)w \\
\Rightarrow {\text{Area of Section 1}} =\dfrac{1}{2}\left( {24 + 16} \right)4 \\
\Rightarrow {\text{Area of Section 1}} =\dfrac{1}{2}\left( {40} \right)4 \\
\Rightarrow {\text{Area of Section 1}} =\dfrac{1}{2}.160 \\
\Rightarrow {\text{Area of Section 1}} =80{\text{ c}}{{\text{m}}^2} $
Therefore, area of Section 1$ = 80{\text{ c}}{{\text{m}}^2}$

Area of section 2: Area of section 2 is also in the form of trapezium where EH and AD are the parallel sides of length ${\text{28 cm}}$ and ${\text{20 cm}}$. So,
${\text{Area of Section 2}} = \dfrac{1}{2}\left( {{\text{Sum of Parallel Sides}}} \right) \times height \\
\Rightarrow {\text{Area of Section 2}} =\dfrac{1}{2}\left( {AD + EH} \right)w \\
\Rightarrow {\text{Area of Section 2}} =\dfrac{1}{2}\left( {28 + 20} \right)4 \\
\Rightarrow {\text{Area of Section 2}} =\dfrac{1}{2}\left( {48} \right)4 \\
\Rightarrow {\text{Area of Section 2}} =\dfrac{1}{2}.192 \\
\therefore {\text{Area of Section 2}} =96{\text{ c}}{{\text{m}}^2} $
Therefore, ${\text{area of Section 2}} = 96{\text{ c}}{{\text{m}}^2}$

Hence, area of Section 1= area of section 3$ = 80{\text{ c}}{{\text{m}}^2}$ and area of Section 2= area of section 4$ = 96{\text{ c}}{{\text{m}}^2}$.

Note:It’s recommended to draw the diagram of the given information, as it will be helpful to figure out the shapes of the sections, their height and width. If asked to find the area between the outer and the inner frame, then simply add all the sections 1, 2,3 and 4.