There are two wires of length 40 cm each .One of them has to be bend into a rectangle of length 12 cm and the other wire into a square of side 10 cm
Answers
Question:
There are two wires of length 40 cm each .One of them has to be bend into a rectangle of length 12 cm and the other wire into a square of side 10 cm
To Find: We have to find length if rectangle.
Perimeter of rectangle= 44cm
Answer:
36cm^(2)
Answer : Area of rectangle= length×breadth
17cm×5cm= 85cm^(2)
121cm^(2)-85cm^(2)= 36cm^(2)
The square has more area by 36cm^(2)
Solution:
The sides of square are equal.
Area of square= 11cm×11cm
The sides of square are equal. So, 44/4= 11cm
Area of square= 11cm×11cm
121cm^(2)
Area of rectangle= length×breadth
17cm×5cm= 85cm^(2)
121cm^(2)-85cm^(2)= 36cm^(2)
The square has more area by 36cm^(2)
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★ To Find: We have to find length if rectangle.
Perimeter of rectangle= 44cm
Answer:
36cm^(2)
★ Answer : Area of rectangle= length×breadth
17cm×5cm= 85cm^(2)
121cm^(2)-85cm^(2)= 36cm^(2)
The square has more area by 36cm^(2)
★★ Solution:
Area of square=a^2
The sides of square are equal.
So,44/4=11cmSo, 44/4= 11cmSo,44/4=11cm
Area of square= 11cm×11cm
121cm2121cm^2121cm
2
Areaofsquare=a2Area \: \: of \: \: square= a^2Areaofsquare=a
2
The sides of square are equal. So, 44/4= 11cm
Area of square= 11cm×11cm
121cm^(2)
44cm=2(l+b)44cm= 2(l+b)44cm=2(l+b)
44cm=2l+2b44cm= 2l+2b44cm=2l+2b
44cm=2l+2(5)cm44cm= 2l+2(5)cm44cm=2l+2(5)cm
44cm=2l+10cm44cm= 2l+10cm44cm=2l+10cm
44cm−10cm=2l44cm-10cm= 2l44cm−10cm=2l
34cm=2l34cm= 2l34cm=2l
34cm/2=l34cm/2= l34cm/2=l
17cm=l17cm= l17cm=l
Area of rectangle= length×breadth
17cm×5cm= 85cm^(2)
121cm^(2)-85cm^(2)= 36cm^(2)
The square has more area by 36cm^(2)
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