Question

A design is made on a rectangular tile of dimensions $50{\rm{ cm}} \times {\rm{70 cm}}$ as shown in the given figure. The design shows $8$triangle, each of sides $26{\rm{ cm}}$, $17{\rm{ cm}}$and $25{\rm{ cm}}$. Find the total area of the design and the remaining area of the tile.

Answer 1

Hint: To find the total area of the design. It is enough to find the area of one triangle since the area of all the triangles are equal. To find the area of the triangle we will use heron’s formula. Then the area of total design is equal to eight times the area of the triangle. To find the area of the remaining part. Find the area of rectangular tile. The area of the remaining triangle is equal to total rectangular tile minus total area of the design.


Complete step by step solution
Given:

The rectangle tile length is $l = 70{\rm{ cm}}$.

The rectangle tile width is $b = 50{\rm{ cm}}$.

The sides of the triangle are $17{\rm{ cm}}$, $25{\rm{ cm}}$and $26{\rm{ cm}}$.

We know the formula for the area of the rectangle is the product of length and width.
$A = l \times b$

On substituting the value of length and width in the above formula we get,
$70{\rm{ cm}} \times {\rm{50 cm}} = {\rm{3500 c}}{{\rm{m}}^2}$

Since, the design have $8$ congruent triangles with sides $17{\rm{ cm}}$, $25{\rm{ cm}}$and $26{\rm{ cm}}$. If we find the area of one triangle, then it would be enough. The formula to find the area of triangle is given as,
$A = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $

Where, $s$ is half of the perimeter of the triangle and $a$, $b$ and $c$ are the sides of the triangle.

Since, we know that the formula to find the perimeter is,
$s = \dfrac{{a + b + c}}{2}$

On substituting values of $a$, $b$ and $c$we get,
$\begin{array}{c}
s = \dfrac{{17 + 25 + 26}}{2}\\
 = \dfrac{{68}}{2}\\
 = 34
\end{array}$

Now, on putting the values of $s$, $a$, $b$ and $c$in $\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $ we get,

$\begin{array}{l}
\sqrt {34\left( {34 - 17} \right)\left( {34 - 25} \right)\left( {34 - 26} \right)} \\
\sqrt {34 \times 17 \times 9 \times 8} \\

\end{array}$$\begin{array}{l}
A = \sqrt {34\left( {34 - 17} \right)\left( {34 - 25} \right)\left( {34 - 26} \right)} \\
A = \sqrt {34 \times 17 \times 9 \times 8} \\
A = 17 \times 3 \times 4\\
A = 204
\end{array}$

Since, the area of all the triangle are equal then the area of all the 8 triangles will be,
$\begin{array}{l}
{A_t} = 8 \times A\\
{A_t} = 8 \times 204\\
{A_t} = 1632
\end{array}$

The area of design is $1632{\rm{ c}}{{\rm{m}}^2}$.

Now, the area of the remaining area is equal to the area rectangular tile minus area of the design.

On substituting the rectangular tile area and area of total design we get,
$3500 - 1632 = 1868{\rm{ c}}{{\rm{m}}^2}$

Hence, the area of remaining rectangular tile is $1868\;{\rm{c}}{{\rm{m}}^2}$.


Note: Since, it is given that all the $8$ triangles are congruent so we found area for only one triangle. If the triangles are not congruent then we need to find the areas for all the triangles separately.