Name the type of quadrilateral formed by the following points A(3, 5) , B(6, 0) , C(1, - 3) and (- 2, 2) and explain, please solve fast how square is formed, don’t leave it like that
Answers
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0)
(ii) (- 3, 5), (3, 1), (0, 3), (- 1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Answer :
i) Let A = (–1, –2), B = (1, 0), C= (–1, 2) and D = (–3, 0)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = \(\sqrt{(1 + 1)^2 + (0 + 2)^2} \)
\( = \sqrt{4 + 4} \)
\(= 2 \sqrt{2}\)
BC = \(\sqrt{(-1 - 1)^2 + (2 - 0)^2} \)
\( = \sqrt{4 + 4} \)
\( = 2 \sqrt{2}\)
CD = \(\sqrt{(-3 + 1)^2 + (0 - 2)^2} \)
\( = \sqrt{4 + 4} \)
\( = 2 \sqrt{2}\)
DA = \(\sqrt{(-3 + 1)^2 + (0 - 2)^2} \)
\( = \sqrt{4 + 4} \)
\( = 2 \sqrt{2}\)
Therefore, all four sides of quadrilateral are equal. … (1)
Now, we will check the length of diagonals.
AC = \(\sqrt{(-1 + 1)^2 + (2 + 2)^2} \)
\( = \sqrt{0 + 16} \)
\( = 4\)
BD = \(\sqrt{(-3 - 1)^2 + (0 + 0)^2} \)
\( = \sqrt{16 + 0} \)
\( = 4\)
Therefore, diagonals of quadrilateral ABCD are also equal. … (2)
From (1) and (2), we can say that ABCD is a square.
(ii) Let A = (–3, 5), B= (3, 1), C= (0, 3) and D= (–1, –4)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = \(\sqrt{(-3 - 3)^2 + (1 - 5)^2} \)
\( = \sqrt{36 + 16} \)
\( = 2 \sqrt{13}\)
BC = \(\sqrt{(0 - 3)^2 + (3 - 1)^2} \)
\( = \sqrt{9 + 4} \)
\( = \sqrt{13}\)
CD = \(\sqrt{(-1 - 0)^2 + (-4 - 3)^2} \)
\( = \sqrt{1 + 49} \)
\( = 5\sqrt{2}\)
DA = \(\sqrt{(-1 + 3)^2 + (-4 - 5)^2} \)
\( = \sqrt{4 + 81} \)
\( = \sqrt{85}\)
We cannot find any relation between the lengths of different sides.
Therefore, we cannot give any name to the figure ABCD.
(iii) Let A = (4, 5), B= (7, 6), C= (4, 3) and D= (1, 2)
Using Distance Formula to find distances AB, BC, CD and DA, we get
AB = \(\sqrt{(7 - 4)^2 + (6 - 5)^2} \)
\( = \sqrt{9 + 1} \)
\( = \sqrt{10}\)
BC = \(\sqrt{(4 - 7)^2 + (3 - 6)^2} \)
\( = \sqrt{9 + 9} \)
\( = \sqrt{18}\)
CD = \(\sqrt{(1 - 4)^2 + (2 - 3)^2} \)
\( = \sqrt{9 + 1} \)
\( = \sqrt{10}\)
DA = \(\sqrt{(1 - 4)^2 + (2 - 5)^2} \)
\( = \sqrt{9 + 9} \)
\( = \sqrt{18}\)
Here opposite sides of quadrilateral ABCD are equal. … (1)
We can now find out the lengths of diagonals.
AC = \(\sqrt{(4 - 4)^2 + (3 - 5)^2} \)
\( = \sqrt{0 + 4} \)
\( = 2 \)
BD = \(\sqrt{(1 - 7)^2 + (2 - 6)^2} \)
\( = \sqrt{36 + 16} = \)
\( 2 \sqrt{13}\)
Here diagonals of ABCD are not equal. … (2)
From (1) and (2), we can say that ABCD is not a rectangle therefore it is a parallelogram.
Step-by-step explanation:
first find lengths of AB,BC,CD,DA
given:
points are A(3, 5) , B(6, 0) , C(1, - 3) and (- 2, 2).
now, AB = √[(6-3)²+(0-5)²]
= √[9+25] = √34
BC = √[(6-1)²+(0+3)²] = √[25+9] = √34
CD = √[(1-(-2))²+(2-(-3))²] = √[9+25] =√34
DA = √[(3-(-2))²+(2-5)²] = √[25+9] =√34
so, here AB=BC=CD=DA.
quadrilateral in which all sides are equal is either a Square or a Rhombus.
now if AB and BC are making an angle of 90, then it's square. if not, then it's rhombus.
to check perpendicularity it should follow below one.
slope of AB *Slope of BC = -1
we need to find slopes of AB,BC.
slope of AB = (Y2-Y1)/(X2-X1)
= (0-5)/(6-3)
= -5/3
slope of BC = (-3-0)/(1-6)
= -3/-5
3/5
now,
= slope of AB *Slope of BC
= (-5/3) * (3/5)
= -1
it is satisfying the condition,so AB and BC is perpendicular to each other.
So, the quadrilateral is Square.