VERY SHORT ANSWER TYPEQUESTIONS
Q1. PQRS is a diameter of a circle of radius 6 cm. The equal lengths PQ, QR and RS are drawn on PQ and QS as diameters, as shown in figure. Find the perimeter of the shaded region.
Sol. Diameter PS = 12 cm [∵ Radius OS = 6 cm]
Since PQ, QR and RS are three equal parts of diameter,
Now, the total required perimeter
Q2. A sheet of paper is in the form of a rectangle ABCD in which AB = 40 cm, and BC = 28 cm. A semicirclular portion with BC as diameter is cut off. Find the area of the remaining paper.
Sol. Length of the paper = 40 cm, width of the paper = 28 cm.
∴ Area of the rectangle =length × breadth = 40 × 28 cm^{2} = 1120 cm^{2}
Again, diameter of semicircle = 28 cm.
⇒ Radius of the semicircle = 28/2 = 14 cm.
∴ Area of the semicircle
∴ Area of the remaining paper = 1120 cm^{2} − 308 cm^{2} = 812 cm^{2}.
Q3. Find the area of the shaded region in the figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
Sol. Side of the square = 14 cm
∴ Area of the square = 14 × 14 cm^{2} = 196 cm^{2}
Also, diameter of each semicircle = side of the square = 14 cm.
⇒ Radius = 14/2 = = 7 cm.
Area of 1 semicircle = 1/2 πr^{2}
⇒ Area of both semicircles = 2 × 77 cm2
= 154 cm^{2}.
Now, the area of the shaded region = [Area of the square ABCD] − [Area of both the semicircles]
= (196 − 154) cm^{2} = 42 cm^{2}.
Q4. Find the perimeter of the shaded region, if ABCD is a square of side 21 cm and APB and CPD are semicircles. (use π = 22/7)
Sol. Here diameter = 21 cm ⇒ r = 21/2
Thus the required perimeter of shaded region is 108 cm.
Q5. A park is in the form of a rectangle 120 m long and 100 m wide. At the centre, there is a circular lawn of radius Find the area of the park excluding the lawn.
[Take π = 22/7]
Sol. Length of the park l = 120 m
Breadth of the park b = 100 m
∴ Area of the park excluding the central park
= 12000 − 3300 m^{2}
= 8700 m^{2}.
Q6. In the given figure, OAPB is a sector of a circle of radius 3.5 cm with the centre at O and ∠AOB = 120°. Find the length of OAPBO.
Sol. Here, the major sector angle is given by θ = 360° − 120° = 240°
∴ Circumference of the sector APB
∴ Perimeter of OAPBO = [Circumference of sector AOB] + OA + OB]
Q7. Find the area of the shaded region of the following figure, if the diameter of the circle with centre O is 28 cm and AQ = 1/4 AB.
Sol. We have AB = 28 cm
∴
⇒ BQ = 28 − 7 = 21 cm
∴ Area of the semicircle having diameter as 21 cm
∴ Also area of the semicircle having diameter as 7 cm
Thus the area of the shaded region
= 192.5 cm^{2}.
Q8. PQRS is a square land of side 28 m. Two semicircular grass covered postions are to be made on two of its opposite sides as shown in the figure. How much area will be left uncovered? [Take π = 22/7]
Sol. Side of the square = 28 m
∴ Area of the square PQRS = 28 × 28 m^{2}
Diameter of a semicircle = 28 m
⇒ Radius of a semicircle = 28/2 = 14m
∴ Area of 1 semicircle = 1/2πr^{2} =
⇒ Area of both the semicircles = 2 × 308 m^{2} = 616 m^{2}
∴ Area of the square left uncovered = (28 × 28) − 616 m^{2} = 784 − 616 m^{2} = 168 m^{2}.
Q9. Find the area of a square inscribed in a circle of radius 10 cm.
Sol. Let ABCD be the square such that
AB = BC = 10 cm
∴ AC^{2} = AB^{2} + BC^{2}
AB^{2} + BC^{2} = (10 × 2)
⇒ x^{2} + x^{2} = (20)2
[Let AB = BC = x]
⇒ 2x^{2} = 400
⇒
∴ Area of the square = 200 cm^{2}.
Q10. In the given figure, O is the centre of a circular arc and AOB is a straight line. Find the perimeter of the shaded region.
Sol. O is the centre of the circle.
∴ AB is its diameter.
In right Δ ABC,
AC^{2} + BC^{2} = AB^{2}
⇒ 12^{2} + 16^{2} = AB^{2}
⇒ 144 + 256 = AB^{2} ⇒ AB^{2} = 400
⇒
∴ Circumference of semicircle ACB
∴ Perimeter of the shaded region = 22/7 cm + 12 cm + 16 cm
= 31.43 cm + 12cm + 16cm
= 59.43 cm
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