Question

ABCD is a rectangle in which AB=30cm AD=16cm as shown in fig a quadrilateral DBPQ with all integers side is drawn inside the rectangle find the minimum intregl perimeter of quad ?

Answer 1

Given : ABCD is a rectangle in which AB = 30cm, AD = 16cm as shown in fig. a quadrilateral DPBQ with all integers sides is drawn inside the rectangle (AP = 5cm & Qc = 7cm)​

To find  : the minimum integral perimeter of quad. DPBQ?

Solution:

Here concept to be used is sum of two sides of triangle > third side

in ΔADP  minimum DP  = 12

so that 12 + 5 > 16

in ΔABP  minimum BP  = 26

so that 26 + 5 > 30

in ΔBCQ  minimum BQ  = 10

so that 10 + 7 > 16

in ΔCDQ  minimum DQ  = 24

so that 24 + 7 > 30

Now lets Check  Triangle formed by these sides with diagonal of rectangle

Diagonal of rectangle = √30² + 16² = 34

DB = 34

DP + PB > DB  

=> 12 + 26 > 34  ( satisfied)

DQ + QB > DB

=> 10 + 24 > 34  ( not satisfied)

Hence one of the side out of DQ & QB has to be greater

minimum integral perimeter > 12 + 26 + 10 + 24  

minimum integral perimeter >72

minimum integral perimeter of quad. DPBQ >72 cm

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Answer 2

Answer:

Let x,y,z,w be the side of quadrilateral respectively

then

16 +5 >x>16-5 (by triangle law) equation 1

similarity for

y,z,w

we have

35>y>25 eq 2,

23>z>9 eq 3,

37>w>23 eq 4,

now adding above all equations 1,2,3,4 we get

116>x+y+z+w>68

so minimum possible value for the perimeter(sum of all the sides x+y+z+w) in integral form will be 69.(I.e greater than 68)