Question

in the figure quadrilateral PQRS is a rectangle .if pq=14 cm qr=21 cm .find areas of parts x y and z​

Answer 1

x =theta \360*pie r square
90\360*22\7*14*14

=154cm square

y= qr =21
qm+qr=qr
14+mr=21

Mr=7

y =90/360°*22/7*7*7
=38.5cm square

z = l*b
=14*21
=488 cm square

488-154-38.5
=101.5


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Answer 2

Area of x is 154$(cm^{2})$, Area of y is 38.5$(cm^{2})$ and Area of z is 101.5$(cm^{2})$

Step-by-step explanation:

1. Here PQRS is a rectangle which

  Length QR= 21 (cm)

  Breadth = 14 (cm)

  $\mathbf{Area of PQRS = 21\times 14= 294 (cm^{2})}$    ...1)

2. Now taking large quadrant circle PQM, which radius is breadth of rectangle PQRS. It can be see from figure.

   Now radius of quadrant circle PQM= PQ = 14 (cm)

   So area of quadrant circle PQM(x)

   $\mathbf{(x)=\frac{\pi\times radius^{2}}{4}=\frac{22}{7}\times \frac{1}{4}\times PQ^{2}}$

   

  $\mathbf{(x)=\frac{22}{7}\times \frac{1}{4}\times 14^{2}=154(cm^{2})}$     ...2)

3. Now taking small quadrant circle MRN, which radius can be see from figure.

   Now radius of quadrant circle MRN=MR  =21-14=7 (cm)

   So area of quadrant circle MRN(y)

   $\mathbf{(y)=\frac{\pi\times radius^{2}}{4}=\frac{22}{7}\times \frac{1}{4}\times MR^{2}}$

   

  $\mathbf{(y)=\frac{22}{7}\times \frac{1}{4}\times 7^{2}=38.5(cm^{2})}$     ...3)

4. Now area of remaining part(z)= Area of rectangle PQRS - area of quadrant circle PQM(x)-area of quadrant circle MRN(y)

 

  Area of remaining part(z) = 294-154-38.5

  Area of remaining part(z) = 101.5 $(cm^{2})$