Answer
Verified
411.9k+ views
Hint: Here, will use the formula of the area of a triangle and substitute the given vertices in that formula to find the required area. A triangle is a two-dimensional figure which has three sides and three vertices.
Formula Used:
Area of a triangle $ = \left| {\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]} \right|$
Complete step-by-step answer:
According to the question,
We are given the three vertices of a triangle.
Let the vertices of the triangle be $A = \left( {1,1} \right)$, $B = \left( {3,1} \right)$ and $C = \left( {5,7} \right)$.
Hence, we have to find the area of $\vartriangle ABC$ whose three vertices are given.
Now, we will use the formula:
Area of a triangle $ = \left| {\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]} \right|$………………….$\left( 1 \right)$
Now, substituting $\left( {{x_1},{y_1}} \right) = \left( {1,1} \right)$, $\left( {{x_2},{y_2}} \right) = \left( {3,1} \right)$ and $\left( {{x_3},{y_3}} \right) = \left( {5,7} \right)$, in equation $\left( 1 \right)$, we get
Area of $\vartriangle ABC = \left| {\dfrac{1}{2}\left[ {1\left( {1 - 7} \right) + 3\left( {7 - 1} \right) + 5\left( {1 - 1} \right)} \right]} \right|$
$ \Rightarrow $ Area of $\vartriangle ABC = \left| {\dfrac{1}{2}\left[ {1\left( { - 6} \right) + 3\left( 6 \right) + 5\left( 0 \right)} \right]} \right|$
Solving this further, we get,
$ \Rightarrow $ Area of $\vartriangle ABC = \left| {\dfrac{1}{2}\left[ { - 6 + 18 + 0} \right]} \right| = \left| {\dfrac{1}{2}\left( {12} \right)} \right| = \left| 6 \right|$
Now, we have used the modulus sign because the area of a triangle cannot be negative.
Hence, Area of $\vartriangle ABC = 6$ square units
Therefore, the area of the triangle with vertices $\left( {1,1} \right),\left( {3,1} \right)$and $\left( {5,7} \right)$ is 6 square units.
Hence, option A is the correct answer.
Note:
We can also find the area of the triangle using the help of determinants.
We will use the formula:
Area of triangle
$ = \left| {\dfrac{1}{2}\left| \begin{gathered}
{x_1}{\text{ }}{y_1}{\text{ }}1 \\
{x_2}{\text{ }}{y_2}{\text{ }}1 \\
{x_3}{\text{ }}{y_3}{\text{ }}1 \\
\end{gathered} \right|} \right|$
Now, Substituting $\left( {{x_1},{y_1}} \right) = \left( {1,1} \right)$, $\left( {{x_2},{y_2}} \right) = \left( {3,1} \right)$ and $\left( {{x_3},{y_3}} \right) = \left( {5,7} \right)$ we get,
Area of $\vartriangle ABC = \left| {\dfrac{1}{2}\left| {\begin{array}{*{20}{l}}
1&1&1 \\
3&1&1 \\
5&7&1
\end{array}} \right|} \right|$
Now, solving the determinant,
$ \Rightarrow ar\vartriangle ABC = \left| {\dfrac{1}{2}\left[ {1\left( {1 - 7} \right) - 1\left( {3 - 5} \right) + 1\left( {21 - 5} \right)} \right]} \right|$
$ \Rightarrow ar\vartriangle ABC = \left| {\dfrac{1}{2}\left[ { - 6 + 2 + 16} \right]} \right|$
Solving further,
$ \Rightarrow ar\vartriangle ABC = \left| {\dfrac{1}{2} \times 12} \right| = \left| 6 \right|$
Therefore, area of $\vartriangle ABC = 6$square units
Hence, option A is the correct answer.
Also, we have used the ‘modulus sign’ while finding the area of the triangle because it means that we have to take the absolute value of the terms present inside it, i.e. we will only take the non-negative values of the terms present inside the modulus when we will remove it. Hence, we have used Modulus, keeping in mind that area of a triangle can never be negative.
Formula Used:
Area of a triangle $ = \left| {\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]} \right|$
Complete step-by-step answer:
According to the question,
We are given the three vertices of a triangle.
Let the vertices of the triangle be $A = \left( {1,1} \right)$, $B = \left( {3,1} \right)$ and $C = \left( {5,7} \right)$.
Hence, we have to find the area of $\vartriangle ABC$ whose three vertices are given.
Now, we will use the formula:
Area of a triangle $ = \left| {\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]} \right|$………………….$\left( 1 \right)$
Now, substituting $\left( {{x_1},{y_1}} \right) = \left( {1,1} \right)$, $\left( {{x_2},{y_2}} \right) = \left( {3,1} \right)$ and $\left( {{x_3},{y_3}} \right) = \left( {5,7} \right)$, in equation $\left( 1 \right)$, we get
Area of $\vartriangle ABC = \left| {\dfrac{1}{2}\left[ {1\left( {1 - 7} \right) + 3\left( {7 - 1} \right) + 5\left( {1 - 1} \right)} \right]} \right|$
$ \Rightarrow $ Area of $\vartriangle ABC = \left| {\dfrac{1}{2}\left[ {1\left( { - 6} \right) + 3\left( 6 \right) + 5\left( 0 \right)} \right]} \right|$
Solving this further, we get,
$ \Rightarrow $ Area of $\vartriangle ABC = \left| {\dfrac{1}{2}\left[ { - 6 + 18 + 0} \right]} \right| = \left| {\dfrac{1}{2}\left( {12} \right)} \right| = \left| 6 \right|$
Now, we have used the modulus sign because the area of a triangle cannot be negative.
Hence, Area of $\vartriangle ABC = 6$ square units
Therefore, the area of the triangle with vertices $\left( {1,1} \right),\left( {3,1} \right)$and $\left( {5,7} \right)$ is 6 square units.
Hence, option A is the correct answer.
Note:
We can also find the area of the triangle using the help of determinants.
We will use the formula:
Area of triangle
$ = \left| {\dfrac{1}{2}\left| \begin{gathered}
{x_1}{\text{ }}{y_1}{\text{ }}1 \\
{x_2}{\text{ }}{y_2}{\text{ }}1 \\
{x_3}{\text{ }}{y_3}{\text{ }}1 \\
\end{gathered} \right|} \right|$
Now, Substituting $\left( {{x_1},{y_1}} \right) = \left( {1,1} \right)$, $\left( {{x_2},{y_2}} \right) = \left( {3,1} \right)$ and $\left( {{x_3},{y_3}} \right) = \left( {5,7} \right)$ we get,
Area of $\vartriangle ABC = \left| {\dfrac{1}{2}\left| {\begin{array}{*{20}{l}}
1&1&1 \\
3&1&1 \\
5&7&1
\end{array}} \right|} \right|$
Now, solving the determinant,
$ \Rightarrow ar\vartriangle ABC = \left| {\dfrac{1}{2}\left[ {1\left( {1 - 7} \right) - 1\left( {3 - 5} \right) + 1\left( {21 - 5} \right)} \right]} \right|$
$ \Rightarrow ar\vartriangle ABC = \left| {\dfrac{1}{2}\left[ { - 6 + 2 + 16} \right]} \right|$
Solving further,
$ \Rightarrow ar\vartriangle ABC = \left| {\dfrac{1}{2} \times 12} \right| = \left| 6 \right|$
Therefore, area of $\vartriangle ABC = 6$square units
Hence, option A is the correct answer.
Also, we have used the ‘modulus sign’ while finding the area of the triangle because it means that we have to take the absolute value of the terms present inside it, i.e. we will only take the non-negative values of the terms present inside the modulus when we will remove it. Hence, we have used Modulus, keeping in mind that area of a triangle can never be negative.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Change the following sentences into negative and interrogative class 10 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE