Question

is a rhombus, Is it a square?
(C) Prove that the quadrilateral ABCD is a rectangle if A =(-4,-1), B =(-2,-4), C =(4,0) and
D=(2, 3).
(d) Prove that the quadrilateral ABCD is a square if A =(4, 3), B=(6,4), C =(5, 6) and D =(3,5).​

Answer 1

Step-by-step explanation:

(C)

$distance \: between \: two \: points = \\ \sqrt{( {x2 - x1)}^{ 2} + {(y2 - y1)}^{2} }$

AB=

$\sqrt{13}$

BC=

$2 \sqrt{13}$

CD=

$\sqrt{13}$

DA=

$2 \sqrt{13}$

AC=

$\sqrt{65}$

BD=

$\sqrt{65}$

AB=CD ,BC=DA and AC=BD

Opposite sides are equal.Diagonals are also equal.

Hence it is a rectangle.

(D)

AB=

$\sqrt{5}$

BC=

$\sqrt{5}$

CD=

$\sqrt{5}$

DA=

$\sqrt{5}$

AC=

$\sqrt{10}$

BD=

$\sqrt{10}$

All sides are equal and diagonals are also equal.

Hence it is a square.