Case study (4 Marks)

Question 14 Marks

Sara hold a Japanese folding fan in her hand as shown in the figure. It is shaped like a sector of a circle and made of a thin material such as paper or feather. 'The inner and outer radii are 3cm and 5cm respectively. 'The fan has three colours i.e., red, blue, and green.

Based on the above information, answer the following questions.

The region given in the figure represents.
If the region containing red color makes an angle of 20º at the center, then find the perimeter of the region containing red color.
If the region containing blue color makes an angle of 80º at the center, then find the area of the region having blue color.
Or
If the region containing green color makes an angle of 60º at the center, then find the area of the region having green color.

Answer

1. Minor sector2. Perimeter of the region containing red color = 2 + 2 + length of arc of sector having radius 3cm + length of arc of sector having radius 5cm.
$=4+2\times\frac{22}{7}\times3\times\frac{20^\circ}{360^\circ}+2\times\frac{22}{7}\times5\times\frac{20^\circ}{360^\circ}$
$=4+\frac{44}{7}\times\frac{1}{18}\times8=4+\frac{176}{63}=4+2.79=\text{6.79cm}$
3. Area of the region containing blue color.
$=\frac{22}{7}\times5\times5\times\frac{80^\circ}{360^\circ}-\frac{22}{7}\times3\times3\times\frac{80^\circ}{360^\circ}$
$=\frac{22}{7}\times\frac{2}{9}\times[25-9]=\frac{44}{63}(16)=\text{11.17cm}^2$
Or
Area of the region containing green color.
$=\frac{22}{7}\times\frac{60^\circ}{360^\circ}[5\times5\times-3\times3]=\frac{22}{7}\times\frac{1}{6}\times16=\text{8.38cm}^2$

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Question 24 Marks

Pookalam is the flower bed or flower pattern designed during Onam in Kerala. It is similar as Rangoli in North India and Kolam in Tamil Nadu. During the festival of Onam, your school is planning to conduct a Pookalam competition. Your friend who is a partner in competition, suggests two designs given below.

Design I: This design is made with a circle of radius 32cm leaving equilateral triangle ABC in the middle as shown in the given figure.
Design II: This Pookalam is made with 9 circular design each of radius 7cm.

The side of equilateral triangle is:
The altitude of the equilateral triangle is:
The area of square is:
Or
Area of each circular design is:

Answer

$1.32√3cm2.48cm$
$3.1764cm^2$
Or
$154 cm^2$

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Question 34 Marks

Gayatri have a triangular shaped grass field. At the three corners of the field, a cow, a buffalo, and a horse are tied separately to the pegs by means of ropes of 3.5m each to graze in the field, as shown in the figure. Sides of the triangular field are 25m, 24m and 7m. Based on the above information, answer the following questions.

Area of the region grazed by the cow is ?
Area of region grazed by the buffalo and the horse is:
Total area grazed by the cow, the buffalo and the horse is:
Or
Find the area of the field that cannot be grazed.

Answer

1. Area of region grazed by the cow
$\frac{\angle\text{A}}{360^\circ}\times\pi\times(3.5)^2$2. Area of the region grazed by the buffalo and the horse
$\frac{\angle\text{B}}{360^\circ}\times\pi\times(3.5)^2+\frac{\angle\text{C}}{360^\circ}\times\pi\times(3.5)^2$
$=\frac{(\angle\text{B}+\angle\text{C)}}{360^\circ}\times\pi\times(3.5)^2$
3. Total area grazed by the cow, the horse and the buffalo
$\frac{\angle\text{A}}{360^\circ}\times\pi\times(3.5)^2+\frac{(\angle\text{B}+\angle\text{C)}}{360^\circ}\times\pi\times(3.5)^2$
$=\frac{(\angle\text{A}+\angle\text{B}+\angle\text{C)}}{360^\circ}\times\pi\times(3.5)^2$
$=\frac{22}{7}\times\frac{108^\circ}{360^\circ}\times3.5\times3.5$
($\because$ sum of interior angles of a triangle is 180º)
$=\frac{77}{4}=\text{19.25m}^2$
Or
Area of the field that cannot be grazed = Area of $\triangle\text{ABC}$ - Area of region grazed by all the three animals.
$= 84 - 19.25 = 64.75m^2$

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Question 44 Marks

A builder of residential project have a vacant square land of side 21m. He wants to make a temple in the shape of semi-circle and a park in the shape of two quadrants of a circle as shown in the figure.

Based on the above information, answer the following questions.

Find the area of square.
Area of two quadrants, shown in figure, is?
Find the area of semi-circular temple.
Or
Find the area of unshaded region.

Answer

1.Area of square $ABCD = 21 \times 21 = 441m^2$2.Area of two quadrants $=2\Big(\pi\text{r}^2\times\frac{90^\circ}{360^\circ}\Big)$
$=\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2}\times\frac{1}{2}=\text{173.25m}^2$
3.Area of semi-circular temple $=\frac{1}{2}(\pi\text{r}^2)$
$=\frac{1}{2}\times\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2}=\text{173.25m}^2$
Or
Area of unshaded region = Area of semi-circle + Area of two quadrants.
$= 173.25 + 173.25 = 346.5m^2$

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Question 54 Marks

Principle of a school decided to give badges to students who are chosen for the post of Head boy, Head girl, Prefect, and Vice Prefect. Badges are circular in shape with two color area, red and silver, as shown in figure. The diameter of the region representing red color is 22cm and silver color is filled in l0.5 cm wide ring. Based on the above information, answer the following questions.

The radius of circle representing the red region is:
Find the area of the red region.
Find the radius of the circle formed by combining the red and silver region.
Or
Find the area of the silver region.

Answer

1.Radius of circle representing red region
$=\frac{22}{2}=\text{11cm}$ [$\because$ Diameter = 22cm (Given)]2.Area of red region $\pi\text{r}^2$
$=\frac{22}{7}\times11\times11=\text{380.28cm}^2$
3.Radius of circle formed by combining red and silver region = Radius of red region + width of silver sign.
= (11+ 10.5)cm = 21.5cm
Or
Area of silver region = Area of combined region - Area of red region.
$=\frac{22}{7}\times21.5\times21.5-\text{380.28}$
$= 1452.78 - 380.28 = 1072.50cm^2$

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MATHS Case study (4 Marks)

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