19√π5÷{16∆4}÷96+64√29
Answers
Answer:
Circumference = 39.6 cm
We know:
Circumference of a circle = 2πr
⇒39.6 =2×
22
7
×r⇒
39.6×7
2×22
=r⇒r=6.3 cm
Also,
Area of the circle = πr2
=
22
7
×6.3×6.3=124.74 cm2
Page No 705:
Question 2:
The area of a circle is 98.56 cm2. Find its circumference.
ANSWER:
Let the radius of the circle be r.
Now,
Area=98.56⇒πr2=98.56⇒
22
7
×r2=98.56⇒r=5.6
Now,
Circumference = 2πr=2×
22
7
×5.6=35.2 cm
Hence, the circumference of the circle is 35.2 cm.
Page No 705:
Question 3:
The circumference of a circle exceeds its diameter by 45 cm. Find the circumference of the circle.
ANSWER:
Let the radius of the circle be r.
Now,
Circumference = Diameter + 45
⇒2πr=2r+45⇒2πr-2r=45⇒2r(
22
7
-1)=45⇒2r×
15
7
=45⇒r=10.5 cm
∴ Circumference = Diameter + 45 = 2(10.5) + 45 = 66 cm
Hence, the circumference of the circle is 66 cm.
Page No 705:
Question 4:
A copper wire when bent in the form of a square encloses an area of 484 cm2. The same wire is not bent in the form of a circle. Find the area enclosed by the circle.
ANSWER:
Area of the circle = 484 cm2
Area of the square = Side2
⇒484=Side2⇒222=Side2⇒Side=22 cm
Perimeter of the square = 4×Side
Perimeter of the square = 4×22
= 88 cm
Length of the wire = 88 cm
Circumference of the circle = Length of the wire = 88 cm
Now, let the radius of the circle be r cm.
Thus, we have:
2πr=88⇒2×
22
7
×r=88⇒r=14
Area of the circle = πr2
=
22
7
×14×14=616 cm2
Thus, the area enclosed by the circle is 616 cm2.
Page No 705:
Question 5:
A wire when bent in the form of an equilateral triangle encloses an area of 121
√
3
cm2. The same wire is bent to form a circle. Find the area enclosed by the circle.
ANSWER:
Area of an equilateral triangle=
√
3
4
×(Side)2⇒121
√
3
=
√
3
4
×(Side)2⇒121×4=(Side)2⇒Side=22 cm
Perimeter of an equilateral triangle = 3×Side
=3×22=66 cm
Length of the wire=66 cm
Now, let the radius of the circle be r cm.
We know:
Circumference of the circle = Length of the wire
2πr =66⇒2×
22
7
×r=66⇒r=
66×7
2×22
⇒r=
21
2
⇒r =10.5
Step-by-step explanation:
all seprate qst plz find urs