ABCD is a square, $AC=BD=4\sqrt{2}cm,AE=DE=2.5cm$. Find the area of the adjoining figure ABCDE:
A. $19c{{m}^{2}}$
B. $22c{{m}^{2}}$
C. $17c{{m}^{2}}$
D. None of the above
Answer
683.4k+ views
- Hint-We will be using the concept of mensuration. We will be using Heron’s formula to find the area of the triangle.We know that the heron’s formula gives the area of the triangle when we know all the sides of the triangle.
Complete step-by-step solution -
We will be using the concept of mensuration. We will be using Heron’s formula to find the area of the triangle.We know that the heron’s formula gives the area of the triangle when we know all the sides of the triangle.
We have been given a figure and have to find its area.
So, the area of figure = area of square ABCD + area of triangle AED.
Now, to find the area of the square we need to find its side from the diagonal given to us as $4\sqrt{2}cm$. Let the side of the square be $xcm$. We know that diagonal is $x\sqrt{2}cm$.
So, according to question,
$\begin{align}
& x\sqrt{2}cm=4\sqrt{2}cm \\
& x=4cm \\
\end{align}$
Therefore, area of square $={{\left( side \right)}^{2}}$
$\begin{align}
& ={{x}^{2}} \\
& =16cm \\
\end{align}$
Now, to find area of $\Delta AED$ we will use Heron’s formula,
$area=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$
Where ‘s’ is semi – perimeter and a, b, c are sides of triangle AED.
Semi – perimeter of $\Delta AES=S=\dfrac{4+2.5+2.5}{2}$
$=\dfrac{9}{2}$
Therefore,
$\begin{align}
& area=\sqrt{\dfrac{9}{2}\left( \dfrac{9}{2}-4 \right)\left( \dfrac{9}{2}-\dfrac{5}{2} \right)\left( \dfrac{9}{2}-\dfrac{5}{2} \right)} \\
& =\sqrt{\dfrac{9}{2}\left( \dfrac{1}{2} \right)\left( \dfrac{4}{2} \right)\left( \dfrac{4}{2} \right)} \\
& =\sqrt{\dfrac{9}{2}\left( \dfrac{1}{2} \right)\left( 2 \right)\left( 2 \right)} \\
& ==\sqrt{\dfrac{9}{{{2}^{2}}}\times {{2}^{2}}} \\
& =\sqrt{9} \\
& =3c{{m}^{2}} \\
\end{align}$
Therefore, total area of figure ABCDE,
$\begin{align}
& \left( 16+3 \right)c{{m}^{2}} \\
& =19c{{m}^{2}} \\
\end{align}$
Hence, option (A) is correct.
Note: To solve these types of questions one must be familiar with all ways of finding the area of triangle, square, etc.
Complete step-by-step solution -
We will be using the concept of mensuration. We will be using Heron’s formula to find the area of the triangle.We know that the heron’s formula gives the area of the triangle when we know all the sides of the triangle.
We have been given a figure and have to find its area.
So, the area of figure = area of square ABCD + area of triangle AED.
Now, to find the area of the square we need to find its side from the diagonal given to us as $4\sqrt{2}cm$. Let the side of the square be $xcm$. We know that diagonal is $x\sqrt{2}cm$.
So, according to question,
$\begin{align}
& x\sqrt{2}cm=4\sqrt{2}cm \\
& x=4cm \\
\end{align}$
Therefore, area of square $={{\left( side \right)}^{2}}$
$\begin{align}
& ={{x}^{2}} \\
& =16cm \\
\end{align}$
Now, to find area of $\Delta AED$ we will use Heron’s formula,
$area=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$
Where ‘s’ is semi – perimeter and a, b, c are sides of triangle AED.
Semi – perimeter of $\Delta AES=S=\dfrac{4+2.5+2.5}{2}$
$=\dfrac{9}{2}$
Therefore,
$\begin{align}
& area=\sqrt{\dfrac{9}{2}\left( \dfrac{9}{2}-4 \right)\left( \dfrac{9}{2}-\dfrac{5}{2} \right)\left( \dfrac{9}{2}-\dfrac{5}{2} \right)} \\
& =\sqrt{\dfrac{9}{2}\left( \dfrac{1}{2} \right)\left( \dfrac{4}{2} \right)\left( \dfrac{4}{2} \right)} \\
& =\sqrt{\dfrac{9}{2}\left( \dfrac{1}{2} \right)\left( 2 \right)\left( 2 \right)} \\
& ==\sqrt{\dfrac{9}{{{2}^{2}}}\times {{2}^{2}}} \\
& =\sqrt{9} \\
& =3c{{m}^{2}} \\
\end{align}$
Therefore, total area of figure ABCDE,
$\begin{align}
& \left( 16+3 \right)c{{m}^{2}} \\
& =19c{{m}^{2}} \\
\end{align}$
Hence, option (A) is correct.
Note: To solve these types of questions one must be familiar with all ways of finding the area of triangle, square, etc.
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