Question

In a rectangle ABCD, the lengths of the sides AB, BC, CD, DA are \[\left( {5x + 2y + 2} \right)\] cm, \[\left( {x + y + 4} \right)\] cm, \[\left( {2x + 5y - 7} \right)\] cm, and \[\left( {3x + 2y - 11} \right)\] cm respectively. Which of the following statements is/are true?
(a) One of the sides of the rectangle is 15 cm long
(b) Each diagonal of the rectangle is 39 cm long
(c) Perimeter of the rectangle is 102 cm
(d) Area of the rectangle is 560 \[{\rm{c}}{{\rm{m}}^2}\]

Answer 1

Hint: Here, we need to find the correct statements. We will use the property of rectangles to form two linear equations in two variables. Then, we will solve them to find the values of the variables, and thus, obtain the length of the sides of the rectangle. Then, we will use Pythagoras’s theorem to find the length of the diagonals. We will use formulas for perimeter and area of the rectangle to find their value. Finally, we will check which of the given statements are correct.

Complete step-by-step answer:
First, will form two linear equations in two variables using the given information.
We know that opposite sides of a rectangle are equal.
Therefore, in rectangle ABCD, we get
\[\begin{array}{l}AB = CD\\BC = DA\end{array}\]
Substituting \[AB = \left( {5x + 2y + 2} \right)\] cm and \[CD = \left( {2x + 5y - 7} \right)\] cm in the equation \[AB = CD\], we get
\[ \Rightarrow 5x + 2y + 2 = 2x + 5y - 7\]
Rewriting the equation, we get
\[ \Rightarrow 5x + 2y + 2 - 2x - 5y + 7 = 0\]
Adding and subtracting the like terms, we get
\[ \Rightarrow 3x - 3y + 9 = 0\]
Dividing both sides by 3, we get
\[ \Rightarrow x - y + 3 = 0\]…………………..\[\left( 1 \right)\]
Substituting \[BC = \left( {x + y + 4} \right)\] cm and \[DA = \left( {3x + 2y - 11} \right)\] cm in the equation \[BC = DA\], we get
\[ \Rightarrow x + y + 4 = 3x + 2y - 11\]
Rewriting the equation, we get
\[ \Rightarrow 3x + 2y - 11 - x - y - 4 = 0\]
Adding and subtracting the like terms, we get
\[ \Rightarrow 2x + y - 15 = 0\]………………….\[\left( 2 \right)\]
We can observe that the equations \[\left( 1 \right)\] and \[\left( 2 \right)\] are a pair of linear equations in two variables.
We will solve the equations to find the values of \[x\] and \[y\].
Rewriting equation \[\left( 1 \right)\], we get
\[ \Rightarrow x = y - 3\]
Substituting \[x = y - 3\] in equation \[\left( 2 \right)\], we get
\[ \Rightarrow 2\left( {y - 3} \right) + y - 15 = 0\]
Multiplying the terms using the distributive law of multiplication, we get
\[ \Rightarrow 2y - 6 + y - 15 = 0\]
Adding and subtracting the like terms, we get
\[ \Rightarrow 3y - 21 = 0\]
Adding 21 on both the sides, we get
\[ \Rightarrow 3y = 21\]
Dividing both sides by 3, we get
\[ \Rightarrow y = 7\]
Substituting \[y = 7\] in the equation \[x = y - 3\], we get
\[ \Rightarrow x = 7 - 3\]
Subtracting the terms in the expression, we get
\[ \Rightarrow x = 4\]
Now, we will find the lengths of the sides of the rectangle.
Substituting \[x = 4\] and \[y = 7\] in the equation \[AB = \left( {5x + 2y + 2} \right)\] cm, we get
\[ \Rightarrow AB = 5\left( 4 \right) + 2\left( 7 \right) + 2\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow AB = 20 + 14 + 2\\ \Rightarrow AB = 36{\rm{ cm}}\end{array}\]
Substituting \[x = 4\] and \[y = 7\] in the equation \[BC = \left( {x + y + 4} \right)\] cm, we get
\[ \Rightarrow BC = 4 + 7 + 4\]
Simplifying the expression, we get
\[ \Rightarrow BC = 15{\rm{ cm}}\]
Thus, we get
\[ \Rightarrow AB = CD = 36{\rm{ cm}}\]
\[ \Rightarrow BC = DA = 15{\rm{ cm}}\]
We can observe that one of the pairs of sides of the rectangle is of length 15 cm.
Therefore, option (a) is correct.
Now, we will find the length of the diagonal AC using Pythagoras’s theorem.
We know that the interior angles of a rectangle are right angles.
Therefore, the triangle ABC is a right angled triangle.
Using Pythagoras’s theorem in right triangle ABC, we get
Using the Pythagoras’s theorem in right angled triangle ADB, we get
\[ \Rightarrow A{C^2} = A{B^2} + B{C^2}\]
Substituting \[AB = 36\] cm and \[BC = 15\] cm in the equation, we get
\[ \Rightarrow A{C^2} = {36^2} + {15^2}\]
Applying the exponents on the bases, we get
\[ \Rightarrow A{C^2} = 1296 + 225\]
Adding the terms in the expression, we get
\[ \Rightarrow A{C^2} = 1521\]
Taking the square root of both sides, we get
\[ \Rightarrow AC = 39\] cm
The diagonals of a rectangle are equal.
Therefore, we get
\[ \Rightarrow AC = BD = 39\] cm
We can observe that each of the diagonal of the rectangle measures 39 cm.
Thus, option (b) is correct.
Now, we will find the perimeter of the rectangle.
The perimeter of a rectangle is the sum of its four sides.
Perimeter of rectangle ABCD \[ = \left( {AB + BC + CD + DA} \right)\] cm
Therefore, we get
Perimeter of rectangle ABCD \[ = \left( {36 + 15 + 36 + 15} \right)\] cm
Adding the terms, we get
Perimeter of rectangle ABCD \[ = 102\] cm
We can observe that the perimeter of rectangle ABCD is 102 cm.
Thus, option (c) is correct.
Now, we will find the area of the rectangle ABCD.
The area of a rectangle is the product of its length and its breadth, that is the product of any pair of adjacent sides.
Area of rectangle ABCD \[ = \left( {AB \times BC} \right){\rm{ c}}{{\rm{m}}^2}\]
Therefore, we get
\[ \Rightarrow \] Area of rectangle ABCD \[ = \left( {36 \times 15} \right){\rm{ c}}{{\rm{m}}^2}\]
Multiplying the terms, we get
\[ \Rightarrow \]Area of rectangle ABCD \[ = 540{\rm{ c}}{{\rm{m}}^2}\]
We can observe that the area of rectangle ABCD is 540 \[{\rm{c}}{{\rm{m}}^2}\].
Thus, option (d) is incorrect.
Therefore, the correct statements are options (a), (b), and (c).

Note: We have formed two linear equations in two variables and simplified them to find the number. A linear equation in two variables is an equation of the form \[ax + by + c = 0\], where \[a\] and \[b\] are not equal to 0. For example, \[2x - 7y = 4\] is a linear equation in two variables.
We used the Pythagoras’s theorem in the solution to find AB. The Pythagoras’s theorem states that the square of the hypotenuse of a right angled triangle is equal to the sum of squares of the other two sides, that is \[{\rm{Hypotenus}}{{\rm{e}}^2} = {\rm{Perpendicula}}{{\rm{r}}^2} + {\rm{Bas}}{{\rm{e}}^2}\].
We have used the distributive law of multiplication to find the product of 2 and \[y - 3\]. The distributive law of multiplication states that \[a\left( {b + c} \right) = a \cdot b + a \cdot c\].