Question

# The figure formed by joining the mid-points of the adjacent sides of a square is a

A

rhombus

B

square

C

rectangle

D

parallelogram

Solution

## The correct option is A. squareLet ABCD be a square. Let P,Q,R and S be the mid points of the side AB,BC,CD and AD respectively.Consider ΔADC, since S and R are the mis points of AD and DC, therefore, by Mid Point Theorem,SR∥AC and SR=AC2...(i)Similarly, in ΔABC, since P and Q are the mis points of AB and BC, therefore, by Mid Point Theorem,PQ∥AC and PQ=AC2...(ii)From (i) and (ii)SR∥PQ and SR=PQ ...(iii)Therefore, PQRS is a parallelogram.Now, consider ΔAPS and ΔPBQAS=BQ [As AS and BQ are the sides of the square]AP=BP [P is the mid point of AB]∠SAP=∠QBP=90∘ [Each angle is 90∘ of a square]Therefore, ΔSAP and ΔBQP by SAS congruence rule.Hence, SP=PQ (By CPCT) ...(iv)From (iii) and (iv),⇒PQ=QR=RS=PS...(v)Now consider, quadrilateral ERFO,Since, EO∥RF and ER∥OF, therefore,⇒ Quadrilateral ERFO is a parallelogram.Now, we know that diagonals of a rhombus bisect each other at 90∘.∴∠EOF=90∘Hence, ∠SRQ=∠ERF=∠EOF=90∘ [Opposite angles of parallelogram are equal]...(vi)From (v) and (vi), PQRS is a square.MathematicsRD SharmaStandard IX

Suggest Corrections

3

Similar questions
View More

Same exercise questions
View More

People also searched for
View More