Question

ABCD is a rectangle whose three Vertices are B (12,0), C(12,5) and
D(0,5) then find the length of one of its diagonal.​

Answer 1

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Step-by-step explanation:

Using Pythagoras theorem,

AC

2

=AB

2

+BC

2

=4

2

+3

2

AC

2

=16+9=25

Taking square root on both the sides, we get

AC =5 units.

Answer 2

Given:

• ABCD is a rectangle whose three vertices are B(12,0), C(12,5) and D(0,5)

Find:

• Length of any 1 Diagonal

Solution:

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\linethickness{0.7mm}\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,2.74){\framebox(0.25,0.25)}\put(4.74,0.01){\framebox(0.25,0.25)}\put(-0.5,-0.4){\bf D}\put(-0.5,3.2){\bf A }\put(5.3,-0.4){\bf C}\put(5.3,3.2){\bf B}\qbezier(0,0)(0,0)(5,3)\put(5.6, - 0.4){\bf (12,5)} \put(5.6,3.2){\bf (12,0)} \put( - 0.1, - 0.4){\bf (0,5)} \end{picture}$

Here, we know

$\displaystyle \boxed{ \sf Distance \: Formula,BD = \sqrt{ (x_2 - x_1)^2 + (y_2-y_1)^2} \: }$

where,

• $\red{\rm x_1 = 12}$
• $\pink{\rm x_2 = 0}$
• $\green{\rm y_1 = 0}$
• $\red{\rm y_2 = 5}$

So,

$\displaystyle \sf \dashrightarrow BD = \sqrt{ (x_2 - x_1)^2 + (y_2-y_1)^2}$

$\displaystyle \sf \dashrightarrow BD = \sqrt{ (0 - 12)^2 + (5-0)^2}$

$\displaystyle \sf \dashrightarrow BD = \sqrt{ (- 12)^2 + (5)^2}$

$\displaystyle \sf \dashrightarrow BD = \sqrt{ 144+ 25}$

$\displaystyle \sf \dashrightarrow BD = \sqrt{ 169}$

$\displaystyle \sf \dashrightarrow BD = \sqrt{ {13}^{2} }$

$\displaystyle \sf \dashrightarrow BD = 13units$

$\displaystyle \sf \therefore BD = 13units$

______________________________

Hence, Length of Diagonal BD = 13units