Question

Construct a rectangle ABCD of length 6 cm and breadth 4 cm. Construct angle bisectors of $\angle {\text{A}}$ and $\angle {\text{B}}$. Let them meet at O. What is the distance of O from length AB?
$
  {\text{A}}{\text{. 3}} \\
  {\text{B}}{\text{. 10}} \\
  {\text{C}}{\text{. 2}} \\
  {\text{D}}{\text{. 8}} \\
 $

Answer 1

Hint: Here, we will proceed by using the scale and protractor to draw the rectangle ABCD and the angle bisectors of $\angle {\text{A}}$ and $\angle {\text{B}}$. Then, we will apply the formula for tangent trigonometric function i.e., $\tan \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}$ in order to obtain the distance of O from length AB.

Complete step-by-step answer:

For construction of a rectangle ABCD with length AB=CD=6 cm and breadth BC=AD=4 cm.
Draw a straight line having a measure of 6 cm with the help of a scale. Mark the end points of this drawn line as A and B. Since, we know that all the interior angles of any rectangle is equal to ${90^0}$.
So, we will draw two perpendiculars using protectors at the ends A and B such that the length of these drawn perpendiculars is equal to 4 cm. Now, mark these other ends of the perpendiculars (apart from the ends marked as A and B) as D and C respectively such that the perpendicular drawn from point A is AD and that drawn from point B is BC each having a measure of 4 cm. Now, join the end points by drawing a straight line from point D to point C. Now, this ABCD is a rectangle having length AB=CD=6 cm and breadth BC=AD=4 cm.
For construction of angle bisectors of angles $\angle {\text{A}}$ and $\angle {\text{B}}$
Since, we know that the angle bisector of any angle divides that angle equally. So, the angle bisector of angle $\angle {\text{A}}$ which is equal to ${90^0}$ divides this angle into two equal angles which are each equal to ${45^0}$ and the angle bisector of angle $\angle {\text{B}}$ which is also equal to ${90^0}$ divides this angle into two equal angles which are each equal to ${45^0}$.
Here, we will draw two lines from the vertices A and B which are ${45^0}$ inclined to the length AB using a protector and mark the point where these two lines are intersecting as point O. So, AO and BO are the required two angle bisectors of $\angle {\text{A}}$ and $\angle {\text{B}}$.
Now, let us take point P as the midpoint of line AB i.e., AP=PB=$\dfrac{{{\text{AB}}}}{2} = \dfrac{6}{2}$=3 cm.
As we know that in any right angled triangle, $\tan \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}$
In right angled triangle AOP, we can write
$
  \tan {45^0} = \dfrac{{{\text{OP}}}}{{{\text{AP}}}} = \dfrac{{{\text{OP}}}}{3} \\
   \Rightarrow {\text{OP}} = 3\tan {45^0} \\
 $
According to trigonometric table, $\tan {45^0} = 1$
$ \Rightarrow {\text{OP}} = 3 \times 1 = 3$ cm
Therefore, the distance of point O from length AB is OP which is equal to 3 cm.
Hence, option A is correct.

Note: In this particular problem, we have taken point P and constructed the line segment OP just to make a right angled triangle (i.e., right angled triangle AOP) so that we can evaluate the value of OP. Also, in any right angled triangle the side opposite to the right angle is hypotenuse, the side opposite to the considered angle (here it is angle ${45^0}$) and the remaining side is base.