Question

An object moves from point A to C along the rectangle . Find the magnitude of the displacement of the object
{ A-B=12 cm
B-C=5 cm
C-D=12 cm
D-A=5 cm
it is a rectangle AB and CD length are same (12cm) & BC and DA breadth are same (5cm)
​

Answer 1

Answer:

13 cm

Step-by-step explanation:

As per the provided information in the given question, we have :

An object moves from point A to C along the rectangle.

Length [ AB & CD ] is 12 cm and breadth [ BC & DA ] is 5 cm.

We are asked to calculate his displacement.

Displacement is the shortest distance from initial to the final position of the body. The shortest distance between two points is always a straight line. Here, object's initial position is A and final position is C. Thus, as the shortest distance between two points is always a straight line, so the diagonal of the rectangle (AC) will be the displacement.

Now, as angles of a rectangle measure 90° each. So, by using the Pythagoras property,

\begin{gathered}\longmapsto \rm { \ell^2 + b^2 = (Diagonal)^2} \\ \end{gathered}⟼ℓ2+b2=(Diagonal)2

\begin{gathered}\longmapsto \rm { (AB)^2 + (BC)^2 = (AC)^2} \\ \end{gathered}⟼(AB)2+(BC)2=(AC)2

\begin{gathered}\longmapsto \rm { (12 \; cm)^2 + (5 \; cm)^2 = (AC)^2} \\ \end{gathered}⟼(12cm)2+(5cm)2=(AC)2

\begin{gathered}\longmapsto \rm { 144 \; cm^2 + 25 \; cm^2 = AC^2} \\ \end{gathered}⟼144cm2+25cm2=AC2

\begin{gathered}\longmapsto \rm { 169 \; cm^2 = AC^2} \\ \end{gathered}⟼169cm2=AC2

\begin{gathered}\longmapsto \rm { \sqrt{169 \; cm^2} = AC} \\ \end{gathered}⟼169cm2=AC

\begin{gathered}\longmapsto \bf { 13 \; cm= AC} \\ \end{gathered}⟼13cm=AC

∴ Magnitude of the displacement is 13 cm.

\rule{200}2

Answer 2

Answer:

An object moves from point A to C along the rectangle . Find the magnitude of the displacement of the object

{ A-B=12 cm

B-C=5 cm

C-D=12 cm

D-A=5 cm

it is a rectangle AB and CD length are same (12cm) & BC and DA breadth are same (5cm)