Perimeter= 2(l+b)

Let the perimeter of 1st room be => 2(L1+B1) = 60
=> (L1+B1) = 30 -----(1)

Let the perimeter of 2nd room be => 2(L2+B2) = 140
=> (L2+B2) = 70 -----(2)

Let the perimeter of 3rd room be => 2(L3+B3) = 100
=> (L3+B3) = 50 -----(3)

Let the perimeter of 4th room be => 2(L4+B4) = 80
=> (L4+B4) = 40 -----(4)

Now (L1+B1)+(L2+B2)+(L3+B3)+(L4+B4) Are the corners or the sum of edge of first floor.
30+70+50+40=190

Answer is E

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When you add the perimeter of all the rooms, we add the internal wall twice, so we will have to subtract the length of internal wall from the total, and that is nothing but length and breadth of complete floor.
Length=x+y=x+p+q
Breadth=a+b
So the total perimeter =\( I+II+III+IV=2(x+a)+2(b+(x+p)+2(a+y)+2(q+b)\)
\(=2(x+a+b+(x+p)+a+y+q+b)\)
\(=2(x+y+b+q+p+a+b+a+b)=2(2L+2B)=2(2(L+B))\)
=> \(60+140+100+80=380 =2(2(L+B))....2(L+B)=190\)

E

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The perimeter of each rectangular room consists of \(2(l+w)\)

Out of each \(2(l+w)\), \((l+w)\) is part of the larger rectangle.

Therefore to find the perimeter of the larger triangle, we need to divide each smaller rectangle perimeter by 2 and add them up:

\(60 + 140 + 100 + 80 = \frac{380 }{ 2} = 190\)

Answer is D.
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FOR SUCH QUESTION, SEE HOW MANY TIMES IS THE PERIMETER INCLUDED IN THE GIVEN THE FIGURE.

HERE IF YOU SEE CLEARLY, YOU WILL FIND THAT 1/2 OF EACH ROOM'S PERIMETER GIVES A PART OF PERIMETER ASKED FOR.

HENCE, YOU WILL:
1/2 THE PERIMETER OF THE 1ST ROOM +
1/2 PERIMIEERT OF THE 2ND ROOM +
1/2 PERIMETER OF 3RD ROOM +
1/2 PERIMETER OF THE 4TH ROOM TO GET THE FINAL PERIMETER OF THE FLOOR.

official answer:E

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