Question

prove that the diagonals of a rectangle ABCD with vertices a(2,-1) b(5,-1) c(5,6) d(2,6) are equal and bisect each other

Answer 1

$\textbf{Given:}$

$\text{Points A(2,-1), B(5,-1), C(5,6) and D(2,6)}$

$\textbf{To prove:}$

$\text{diagonals of rectangle ABCD are equal and}$

$\text{bisect each other}$

$\textbf{Solution:}$

$\textbf{Length of diangonal AC}$

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

$=\sqrt{(2-5)^2+(-1-6)^2}$

$=\sqrt{9+49}$

$=\sqrt{55}$

$\textbf{Length of diangonal BD}$

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

$=\sqrt{(5-2)^2+(-1-6)^2}$

$=\sqrt{9+49}$

$=\sqrt{55}$

$\therefore\textbf{Diagonals are equal}$

$\textbf{Midpoint of AC}$

$=(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})$

$=(\dfrac{2+5}{2},\dfrac{-1+6}{2})$

$=(\dfrac{7}{2},\dfrac{5}{2})$

$=(\dfrac{7}{2},\dfrac{5}{2})$

$\textbf{Midpoint of BD}$

$=(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})$

$=(\dfrac{2+5}{2},\dfrac{-1+6}{2})$

$=(\dfrac{7}{2},\dfrac{5}{2})$

$=(\dfrac{7}{2},\dfrac{5}{2})$

$\implies\text{Midpoint of diagonal AC=Midpoint of diagonal BD}$

$\therefore\textbf{Diagonals bisect each other}$

Find more:

What is the side of a rhombus whose diagonal is D1 and D2​

https://brainly.in/question/16865762

Answer 2

Step-by-step explanation:

ΔADC and ΔBDC are right angled triangle with AD and BC are hypotenuse.

AC

2

=AB

2

+DC

2

AC

2

=(5−2)

2

+(6+1)

2

=9+48=58 sq.unit

BD

2

=DC

2

+CB

2

BD

2

=(5−2)

2

+(−1−6)

2

=9+49=58 sq.unit

Hence, both the diagonals are equal in length.

In ΔABO and ΔCDO

Since, ∠OAB=∠OCD, ∠OBA=∠ODC (Both are alternate interior angles of parallel lines)

and AB=CD

Therefore ΔABO≅ΔCDO

⇒AO=CO and BO=DO

Therefore, Both diaginals bisects each other.

solution