P, Q, R and S tye midpoints of the sides AB, BC, CD, AD respectively of a quadrilateral ABCD. Prove that area of PQRS = ½ area of ABCD.
Answers
Answer:
Here, we are joining A and C.
In ΔABC
P is the mid point of AB
Q is the mid point of BC
PQ∣∣AC [Line segments joining the mid points of two sides of a triangle is parallel to AC(third side) and also is half of it]
PQ=
2
1
AC
In ΔADC
R is mid point of CD
S is mid point of AD
RS∣∣AC [Line segments joining the mid points of two sides of a triangle is parallel to third side and also is half of it]
RS=
2
1
AC
So, PQ∣∣RS and PQ=RS [one pair of opposite side is parallel and equal]
In ΔAPS & ΔBPQ
AP=BP [P is the mid point of AB)
∠PAS=∠PBQ(All the angles of rectangle are 90
o
)
AS=BQ
∴ΔAPS≅ΔBPQ(SAS congruency)
∴PS=PQ
BS=PQ & PQ=RS (opposite sides of parallelogram is equal)
∴ PQ=RS=PS=RQ[All sides are equal]
∴ PQRS is a parallelogram with all sides equal
∴ So PQRS is a rhombus.
Step-by-step explanation:
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