Question 14 Marks
Use of mobile screen for long hours makes your eye sight weak and give you headaches. Children who are addicted to play $"\text{PUBG}"$ can get easily stressed out. To raise social awareness about ill effects of playing $\text{PUBG},$ a school decided to start $'\text{BAN PUBG}\ '$ campaign, in which student are asked to prepare campaign board in the shape of a rectangle. One such campaign board made by class $X$ student of the school is shown in the figure.
Based on the above information, answer the following questions:
$(i)$ Find the coordinates of the point of intersection of diagonals $AC$ and $BD$ .
$(ii)$ Find the length of the diagonal $AC$ .
$(iii)\ (a$) Find the area of the campaign Board $\text{ABCD}$ .
OR
$(b)$ Find the ratio of the length of side $A B$ to the length of the diagonal $AC$.
Based on the above information, answer the following questions:
$(i)$ Find the coordinates of the point of intersection of diagonals $AC$ and $BD$ .
$(ii)$ Find the length of the diagonal $AC$ .
$(iii)\ (a$) Find the area of the campaign Board $\text{ABCD}$ .
OR
$(b)$ Find the ratio of the length of side $A B$ to the length of the diagonal $AC$.
Answer
View full question & answer→$(i)$ Since $\text{ABCD}$ is a rectangle,
its diagonals bisect each other
Using mid-point formula, we can find the point of intersection of diagonals $AC$ and $BD$.
$(x, y)=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$
Let the point be $O ( x , y )$.
The midpoint of diagonal $BD$ will be:
$(x, y)=\left(\frac{7+1}{2}, \frac{1+5}{2}\right)$
$(x, y)=\left(\frac{8}{2}, \frac{6}{2}\right)$
Or, $(x, y)=(4,3)$
Therefore, the coordinates of the point of intersection of diagonals $=(4,3)$
$(ii)$ If the distance between two points $A \left( x _1, y _1\right)$ and $C \left( x _2, y _2\right)$ is $AC$
By using the distance formula, we get
$AC =\sqrt{\left(x_1-x_1\right)^2+\left(y_2-y_1\right)^2}$
$ =\sqrt{(7-1)^2+(5-1)^2}$
$ =\sqrt{(6)^2+(4)^2}$
$ =\sqrt{(36+16)}$
$ =\sqrt{52}$
So, the length of the diagonal $AC =52$ units.
$(iii)\ (a$) Area of a rectangle $=$ Length $\times$ Breadth.
Let $\overline{ AB }$ and $\overline{ AD }$ be the measure of the vertical and horizontal side, correspondingly.
By the formula of the distance between two points on a plain:
$\overline{AB}=\sqrt{\left(x_{B}-x_{A}\right)^2+\left(y_{B}-y_{A}\right)^2}$
$\overline{AB}=\sqrt{(7-1)^2+(1-1)^2}=6$
$\overline{AD}=\sqrt{(1-1)^2+(5-1)^2}=4$
The area of the rectangle is given by:
$\text{ABCD} =\overline{AB} \times \overline{AD}$
$ =6 \times 4$
$ =24 \text { units square }$
OR
$(b$) Length of side $A B=6\text { units }$
Length of diagonal $AC$
$=\sqrt{52} \text { units }$
Therefore, the ratio between them
$=\frac{6}{\sqrt{52}}$
its diagonals bisect each other
Using mid-point formula, we can find the point of intersection of diagonals $AC$ and $BD$.
$(x, y)=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$
Let the point be $O ( x , y )$.
The midpoint of diagonal $BD$ will be:
$(x, y)=\left(\frac{7+1}{2}, \frac{1+5}{2}\right)$
$(x, y)=\left(\frac{8}{2}, \frac{6}{2}\right)$
Or, $(x, y)=(4,3)$
Therefore, the coordinates of the point of intersection of diagonals $=(4,3)$
$(ii)$ If the distance between two points $A \left( x _1, y _1\right)$ and $C \left( x _2, y _2\right)$ is $AC$
By using the distance formula, we get
$AC =\sqrt{\left(x_1-x_1\right)^2+\left(y_2-y_1\right)^2}$
$ =\sqrt{(7-1)^2+(5-1)^2}$
$ =\sqrt{(6)^2+(4)^2}$
$ =\sqrt{(36+16)}$
$ =\sqrt{52}$
So, the length of the diagonal $AC =52$ units.
$(iii)\ (a$) Area of a rectangle $=$ Length $\times$ Breadth.
Let $\overline{ AB }$ and $\overline{ AD }$ be the measure of the vertical and horizontal side, correspondingly.
By the formula of the distance between two points on a plain:
$\overline{AB}=\sqrt{\left(x_{B}-x_{A}\right)^2+\left(y_{B}-y_{A}\right)^2}$
$\overline{AB}=\sqrt{(7-1)^2+(1-1)^2}=6$
$\overline{AD}=\sqrt{(1-1)^2+(5-1)^2}=4$
The area of the rectangle is given by:
$\text{ABCD} =\overline{AB} \times \overline{AD}$
$ =6 \times 4$
$ =24 \text { units square }$
OR
$(b$) Length of side $A B=6\text { units }$
Length of diagonal $AC$
$=\sqrt{52} \text { units }$
Therefore, the ratio between them
$=\frac{6}{\sqrt{52}}$