A Wire of length 1m 20cm is cut into two equal pieces. One piece is bent to form a square and the other one bent into form a rectangle of length 22 cm.
i. What is the length of each piece?
a. 10 cm b. 0.6 m
c. 60 cm
d. Both B and C
ii. What measure is breadth of rectangle?
a. 8 cm
b. 16 cm C. 22 cm d. 60 cm
iii. The side of square is
a. 10 cm b. 16 cm
C. 15 cm
d. 14 cm
iv. Area of rectangle is
a. 225 cm b. 225 m²
C. 176 cm?
d. 176 m?
v. Perimeter of square is
a. 60 cm b. 120 cm c. 60 m
d. 240 cm
Answers
Answered by
1
Answer:
Let the 2 pieces be x,y
Given,
x+y=20
Perimeter of square =x
Side=
4
x
Area of square =
16
x
2
Perimeter of triangle =y
Side=
3
y
Area of triangle =
4
3
(
3
y
)
2
=
36
3
y
2
z= area of square + area of triangle
z=
16
x
2
+
36
3
y
2
z=
16
x
2
+
36
3
(20−x)
2
differentiating the equation, we get,
dx
dz
=
8
x
−
18
3
(20−x)
equate,
dx
dz
=0
we have,
8
x
−
18
3
(20−x)
=0
8
x
=
18
3
(20−x)
solving the above equation, we get,
→x=
(9+4
3
)
80
3
y=20−
(9+4
3
)
80
3
→y=
(9+4
3
)
180
dx
2
d
2
z
=
8
1
+
18
3
>0
Thus z is minimum, when x=
(9+4
3
)
80
3
and y=
(9+4
3
)
180
Hence, wire of length 20cm should be cut into pieces of lengths
(9+4
3
)
80
3
,
(9+4
3
)
180
m
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