If K(0,5), L(-5, 0), M(3,0) and N(8, 5) are the vertices of a quadrilateral, then the quadrilateral KLMN is a _
Answers
Step-by-step explanation:
Given :-
K(0,5), L(-5, 0), M(3,0) and N(8, 5) are the vertices of a quadrilateral.
To find :-
Find the quadrilateral KLMN ?
Solution :-
Given vertices of the quadrilateral are K(0,5),
L(-5, 0), M(3,0) and N(8, 5).
Lenth of KL :-
Let (x1, y1) = K(0,5) => x1 = 0 and y1 = 5
Let (x2, y2) = L(-5,0) => x2 = -5 and y2 = 0
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> The distance between K and L
=> KL = √[(-5-0)²+(0-5)²]
=> KL = √[(-5)²+(-5)²]
=> KL = √(25+25)
=> KL = √50
=> KL = √(2×25)
=> KL = 5√2 units
Lenth of LM :-
Let (x1, y1) = L(-5,0) => x1 = -5 and y1 = 0
Let (x2, y2) = M(3,0) => x2 = 3 and y2 = 0
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> The distance between L and M
=> LM = √[(3-(-5))²+(0-0)²]
=> LM = √[(3+5)²+(0)²]
=> LM= √(8²+0)
=> LM = √64
=> LM= 8 units
Lenth of MN :-
Let (x1, y1) = M(3,0) => x1 = 3 and y1 = 0
Let (x2, y2) = N(8, 5) => x2 = 8 and y2 = 5
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> The distance between M and N
=> MN = √[(8-3)²+(5-0)²]
=> MN = √[(5)²+(5)²]
=> MN = √(25+25)
=> MN = √50
=> MN = √(2×25)
=> MN = 5√2 units
Lenth of KL :-
Let (x1, y1) = K(0,5) => x1 = 0 and y1 = 5
Let (x2, y2) = N(8, 5) => x2 = 8 and y2 = 5
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> The distance between K and N
=> KN = √[(8-0)²+(5-5)²]
=> KN = √[(8)²+(0)²]
=> KN = √64
=> KN = 8 units
Lenth of KM :-
Let (x1, y1) = K(0,5) => x1 = 0 and y1 = 5
Let (x2, y2) = M(3,0) => x2 = 3 and y2 = 0
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> The distance between K and M
=> KM = √[(3-0)²+(0-5)²]
=> KM = √[(3)²+(-5)²]
=> KM = √(9+25)
=> KM = √34
=> KM = √34 units
Lenth of LN :-
Let (x1, y1) = L(-5,0) => x1 = -5 and y1 = 0
Let (x2, y2) = N(8, 5) => x2 = 8 and y2 = 5
We know that
The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
=> The distance between L and N
=> LN = √[(8-(-5))²+(5-0)²]
=> LN = √[(8+5)²+(5)²]
=> LN = √(13²+5²)
=> LN = √169+25
=> LN = √194 units
We have ,
KL = MN
LM = KN
KM ≠ LN
We have , Opposite sides are equal.
and diagonals are not equal.
So , Given quadrilateral is a Parallelogram.
Answer :-
The quadrilateral KLMN is a Parallelogram .
Used Properties :-
→ In a Parallelogram, Two pairs of opposite sides are equal.
→ In a Parallelogram , The diagonals are not equal.
Used formulae :-
Distance formula :-
→The distance between two points (x1, y1) and
(x2, y2) is √[(x2-x1)²+(y2-y1)²] units
