
In square $ABCD$;$A = (3,4)$, $B = ( - 2,4)$, and $C = ( - 2, - 1)$ By plotting these points on a graph paper, find the area of the square.
Answer
474.3k+ views
Hint: First we have to plot the given points into a graph. Because we must know the distance between all the four pairs of points. Then we will start with the use of the distance formula to find the distance between all the four pairs of the points. After calculating the vertex of the point $D$ then we will find the area of the square $ABCD$.
Formula used: Distance between points formula $D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
Where $\left( {{x_1},{x_2}} \right)$ and $\left( {{y_1},{y_2}} \right)$ be the two points.
Area of the square $ = $ $side \times side$ unit sq.
Complete step-by-step answer:
Given points are $A = (3,4)$, $B = ( - 2,4)$, and $C = ( - 2, - 1)$
By plotting the points into graph we get,
First we have to find the fourth vertex $D$ and the area of the given square $ABCD$.
In the square $ABCD$, we know all the sides are equal thus, the distance of each of the vertices are same
Hence we can write,
$AB = BC = CD = DA$
The points plotted on the graph below.
From the graph, the coordinates of point $D$ are $(3, - 1)$
Hence,
Now using the distance formula $D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} ......(1)$
Using equation $(1)$ we can find out the side $AB$, so it will be
Here, $A = (3,4)$ where ${x_1} = 3,{y_1} = 4$ and $B = ( - 2,4)$ where ${x_2} = - 2,{y_2} = 4$
$\Rightarrow$$AB = \sqrt {{{( - 2 - 3)}^2} + {{(4 - 4)}^2}} $ $ = \sqrt {{{( - 5)}^2} + {0^2}} $$ = \sqrt {( - 5)( - 5) + 0} $$ = \sqrt {25} $$ = 5$ Units
Here,
$B = ( - 2,4)$ where ${x_1} = - 2$, ${y_1} = 4$ and
$C = ( - 2, - 1)$ where ${x_2} = - 2,{y_2} = - 1$
$\Rightarrow$$BC = \sqrt {{{( - 2 - ( - 2))}^2} + {{( - 1 - 4)}^2}} $$ = \sqrt {0 + {{( - 5)}^2}} = \sqrt {25} = 5$ Units
Now,
$C = ( - 2, - 1)$ where ${x_1} = - 2,{y_1} = - 1$ and
$D = (3, - 1)$ where ${x_2} = 3,{y_2} = - 1$
$\Rightarrow$$CD = \sqrt {{{(3 - ( - 2))}^2} + {{( - 1 - ( - 1))}^2}} = \sqrt {{5^2} + {0^2}} $$ = \sqrt {25} = 5$ Units
Now,
$D = (3, - 1)$where ${x_1} = 3,{y_1} = - 1$ and
$A = (3,4)$where ${x_2} = 3,{y_2} = 4$
$\Rightarrow$$DA = \sqrt {{{(3 - 3)}^2} + {{( - 1 - 4)}^2}} = \sqrt {{0^2} + {{( - 5)}^2}} = \sqrt {25} = 5$ Units
$D$ are $(3, - 1)$
Here we got the side $AB = 5$ unit sq.
Area of the square $ = $ side ×side
Area of the square $ABCD = AB \times AB$
Substituting the side value,
$ \Rightarrow 5 \times 5$
Multiplying both the numbers, we get
Hence,
Area of the square \[ABCD = 25\] Unit sq.
$\therefore $ The area of the square is $25$ unit sq.
Note: Whenever we face such types of problems the key behind the solution lies in the diagrammatic representation of the data provided then awareness about the basic formula can help in getting to the right track to reach the answer.
Formula used: Distance between points formula $D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
Where $\left( {{x_1},{x_2}} \right)$ and $\left( {{y_1},{y_2}} \right)$ be the two points.
Area of the square $ = $ $side \times side$ unit sq.
Complete step-by-step answer:
Given points are $A = (3,4)$, $B = ( - 2,4)$, and $C = ( - 2, - 1)$
By plotting the points into graph we get,

First we have to find the fourth vertex $D$ and the area of the given square $ABCD$.
In the square $ABCD$, we know all the sides are equal thus, the distance of each of the vertices are same
Hence we can write,
$AB = BC = CD = DA$
The points plotted on the graph below.
From the graph, the coordinates of point $D$ are $(3, - 1)$
Hence,

Now using the distance formula $D = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} ......(1)$
Using equation $(1)$ we can find out the side $AB$, so it will be
Here, $A = (3,4)$ where ${x_1} = 3,{y_1} = 4$ and $B = ( - 2,4)$ where ${x_2} = - 2,{y_2} = 4$
$\Rightarrow$$AB = \sqrt {{{( - 2 - 3)}^2} + {{(4 - 4)}^2}} $ $ = \sqrt {{{( - 5)}^2} + {0^2}} $$ = \sqrt {( - 5)( - 5) + 0} $$ = \sqrt {25} $$ = 5$ Units
Here,
$B = ( - 2,4)$ where ${x_1} = - 2$, ${y_1} = 4$ and
$C = ( - 2, - 1)$ where ${x_2} = - 2,{y_2} = - 1$
$\Rightarrow$$BC = \sqrt {{{( - 2 - ( - 2))}^2} + {{( - 1 - 4)}^2}} $$ = \sqrt {0 + {{( - 5)}^2}} = \sqrt {25} = 5$ Units
Now,
$C = ( - 2, - 1)$ where ${x_1} = - 2,{y_1} = - 1$ and
$D = (3, - 1)$ where ${x_2} = 3,{y_2} = - 1$
$\Rightarrow$$CD = \sqrt {{{(3 - ( - 2))}^2} + {{( - 1 - ( - 1))}^2}} = \sqrt {{5^2} + {0^2}} $$ = \sqrt {25} = 5$ Units
Now,
$D = (3, - 1)$where ${x_1} = 3,{y_1} = - 1$ and
$A = (3,4)$where ${x_2} = 3,{y_2} = 4$
$\Rightarrow$$DA = \sqrt {{{(3 - 3)}^2} + {{( - 1 - 4)}^2}} = \sqrt {{0^2} + {{( - 5)}^2}} = \sqrt {25} = 5$ Units
$D$ are $(3, - 1)$
Here we got the side $AB = 5$ unit sq.
Area of the square $ = $ side ×side
Area of the square $ABCD = AB \times AB$
Substituting the side value,
$ \Rightarrow 5 \times 5$
Multiplying both the numbers, we get
Hence,
Area of the square \[ABCD = 25\] Unit sq.
$\therefore $ The area of the square is $25$ unit sq.
Note: Whenever we face such types of problems the key behind the solution lies in the diagrammatic representation of the data provided then awareness about the basic formula can help in getting to the right track to reach the answer.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Express the following as a fraction and simplify a class 7 maths CBSE

The length and width of a rectangle are in ratio of class 7 maths CBSE

The ratio of the income to the expenditure of a family class 7 maths CBSE

How do you write 025 million in scientific notatio class 7 maths CBSE

How do you convert 295 meters per second to kilometers class 7 maths CBSE

Trending doubts
What constitutes the central nervous system How are class 10 biology CBSE

What are five examples of facts and opinions class 10 english CBSE

Which state has the longest coastline in India A Tamil class 10 social science CBSE

Distinguish between coming together federations and class 10 social science CBSE

10 examples of evaporation in daily life with explanations

What is the difference between anaerobic aerobic respiration class 10 biology CBSE
