Question

ABCD and PQRC are rectangles and Q is the mid point of AC. show that P is the mid point of DC and R is the midpoint of BC. also the ratio of area( ABCD) and area( PQRC)?

Answer 1

Given,

AD=BC

AB=CD

AQ=QC

Now,

$AC^{2}=AD^{2}+DC^{2}\\(2*QC)^{2}= AD^{2}+DC^{2} \\4QC^{2}= AD^{2}+DC^{2} \\ 4(RC^{2}+QR^{2})= AD^{2}+DC^{2}$

=> 2RC=AD=BC=> R is mid point of BC

=>2QR=2PC=DC=>P is the mid point of DC.

Area of ABCD=BC*CD=2RC*2PC=4*RC*PC=4*area of PQRC

HENCE,

ratio of area(ABCD) AND area (PQRC)=4:1

Answer 2

Given :

ABCD and PQRC are rectangles .

Q is the mid point of AC.

To prove :

P is  mid point of DC and R is  mid point of BC.

To Find :

Ratio of area of ABCD and area PQRC =?

Solution :

Construction :

Join diagonal BD.

∴We know that in a rectangle length of diagonals are equal .

Now consider Δ BQR and Δ CQR :

$BQ = CQ= \frac{1}{2} AC=\frac{1}{2} BD$

QR = QR   (Common )

∴ Δ BQR  ≈ Δ CQR   (RHS congruence )

And BR= CR   ( cpct )

⇒ R is mid point of BC

$BR = CR +\frac{1}{2}BC$      - (1)

Similarly Δ QDP  and Δ CDP  are congruent .

∴ $DP = CP =\frac{1}{2} CD$    (cpct)     - (2)

⇒P is mid point of DC.

Now area (ABCD) = $l\times b= CD\times BC$

And area (PQRC) =$l'\times b'= PC\times RC$

Now using equation 1 and 2 :

$\frac{area(ABCD)}{area(PQRC)} =\frac{CD\times BC}{PC \times RC}$

                 =$\frac{CD \times BC}{\frac{1}{2}CD\times \frac{1}{2}BC }$

                  = 4:1

So the ratio of area of given two rectangles is 4:1.