Area of square (ABCD)=64cm²
area of square formed by joining mid points
[tex]area \: of \: square = {(side)}^{2} [/tex]
➦ 64 = (side)²
➦side = √64
➦side = 8cm
length of side of given square ABCD is 8cm
LMNO is the square obtained by joining the midpoints of square ABCD
Now,look at ∆ LBM
∆LBM is a right angled ∆
:. using Pythagoras theorem
LM²=LB²+BM²
→LM² = 4²+4²
→LM² = 16 +16
→LM = √32
→LM = 4√2
Now,area of square LMNO= (side)²
➦ area of square LMNO= (4√2)²
➦ area of square LMNO= 32cm²
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Verified answer
GIVEN:
Area of square (ABCD)=64cm²
TO FIND:
area of square formed by joining mid points
SOLUTION:
[tex]area \: of \: square = {(side)}^{2} [/tex]
➦ 64 = (side)²
➦side = √64
➦side = 8cm
length of side of given square ABCD is 8cm
LMNO is the square obtained by joining the midpoints of square ABCD
Now,look at ∆ LBM
∆LBM is a right angled ∆
:. using Pythagoras theorem
LM²=LB²+BM²
→LM² = 4²+4²
→LM² = 16 +16
→LM = √32
→LM = 4√2
Now,area of square LMNO= (side)²
➦ area of square LMNO= (4√2)²
➦ area of square LMNO= 32cm²
HENCE, REQUIRED AREA OF SQUARE LMNO IS 32CM²