ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and D(5, – 1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.

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Answers

2015-12-23T06:54:40+05:30
P is the mid point of A(-1,-1); B(-1,4)
P(x,y)=x₁+x₂/2 , y₁+y₂/2
here x₁=-1, x₂=-1
y₁=-1, y₂=4
P(x,y) = -1+(-1)/2 and -1+4/2
=-1-1/2 and 3/2
=-1 and 1.5
∴P(-1, 1.5)

Q is the mid point of B(-1,4); C(5,4)
Q(x,y)=x₁+x₂/2 , y₁+y₂/2
= -1+5/2 , 4+4/2
=4/2 , 8/2
=2,4
∴Q(2,4)

R is the mid point of C(5,4)D(5,-1)
R(x,y) = 5+5/2 , 4+(-1)/2
=10/2,4-1/2
=5,3/2
=5,1.5
∴R(5,1.5)

S is the mid point of A(-1,-1) ; D(5,-1)
S(x,y)= -1+(-1)/2+5+(-1)/2
=-1-1/2, 5-1/2
=-2/2 , 4/2
=-1,2
∴S(-1,2)

PQRS is a quadrilateral but to prove what type of quad. it is use the distance formula
P(-1, 1.5) ; Q(2,4)
PQ=√[(x₂-x₁)²+(y₂-y₁)²]
PQ = √(2-(-1))²+(4-1.5)²
= √(3)²+(2.5)²
=√9+6.25
=√15.25
∴ PQ = √15.25

Q(2,4); R(5,1.5)
QR = √[(5-2)²+(1.5-4)²]
=√(3)²+(-2.5)²
=√9+6.25
=√15.25
∴QR = √15.25

R(5,1.5) ; S(-1,2)
RS=√[(5+1)²+(2-1.5)²]
= √6²+(0.5)²
=√36+0.25
=√36.25
∴RS=√36.25

S(-1, 2) ; P(-1,1.5)
SP= √[(-1+1)²+(1.5-2)²]
=√(0)²+(0.5)²
=√0+0.25
=√0.25
=0.5
∴SP = 0.5

On comparing all sides joined by the midpoints of the rectangle we find that PQRS is just a quadrilateral not any type of quad.

Thank You,.....
Mark as brainliest if helpful.......
Yours, Jahnavi
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