\displaystyle \text{SOME APPLICATION OF TRIGONOMETRY }

\displaystyle \textbf{Case Study - 1}

\displaystyle \text{A group of Class X students visited India Gate on an educational trip. Their teacher explained}
\displaystyle \text{that India Gate is a famous monument in New Delhi and is approximately }42\text{ m high. The}
\displaystyle \text{students decided to use their knowledge of trigonometry to study heights, distances and angles}
\displaystyle \text{related to the monument.} \displaystyle \\

\displaystyle \textbf{Question 1: }\text{What is the angle of elevation of the top of India Gate if the students are}
\displaystyle \text{standing at a distance of }42\text{ m from the monument?}
\displaystyle \text{(a) }30^\circ\qquad\text{(b) }45^\circ\qquad\text{(c) }60^\circ\qquad\text{(d) }0^\circ
\displaystyle \text{Answer:}
\displaystyle \text{Let the angle of elevation be }\theta.
\displaystyle \text{Height of India Gate}=42\text{ m}
\displaystyle \text{Horizontal distance from India Gate}=42\text{ m}
\displaystyle \tan\theta=\frac{\text{Opposite side}}{\text{Adjacent side}}
\displaystyle =\frac{42}{42}=1
\displaystyle \therefore \theta=45^\circ
\displaystyle \therefore \text{The angle of elevation is }45^\circ.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The students want to view the top of India Gate at an angle of elevation of}
\displaystyle 60^\circ.\text{ At what distance from the monument should they stand?}
\displaystyle \text{(a) }24.25\text{ m}\qquad\text{(b) }20.12\text{ m}\qquad\text{(c) }42\text{ m}\qquad\text{(d) }24.64\text{ m}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required distance from the monument}=x\text{ m.}
\displaystyle \tan60^\circ=\frac{42}{x}
\displaystyle \sqrt3=\frac{42}{x}
\displaystyle x=\frac{42}{\sqrt3}
\displaystyle =\frac{42\sqrt3}{3}=14\sqrt3\text{ m}
\displaystyle \approx24.25\text{ m}
\displaystyle \therefore \text{They should stand approximately }24.25\text{ m from the monument.}
\displaystyle \therefore \text{The corrected option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If the altitude of the Sun is }60^\circ,\text{ find the height of a vertical tower that}
\displaystyle \text{casts a shadow of length }20\text{ m.}
\displaystyle \text{(a) }20\sqrt3\text{ m}\qquad\text{(b) }\frac{20}{\sqrt3}\text{ m}\qquad\text{(c) }\frac{15}{\sqrt3}\text{ m}\qquad\text{(d) }15\sqrt3\text{ m}
\displaystyle \text{Answer:}
\displaystyle \text{Let the height of the tower}=h\text{ m.}
\displaystyle \text{Length of shadow}=20\text{ m}
\displaystyle \tan60^\circ=\frac{h}{20}
\displaystyle \sqrt3=\frac{h}{20}
\displaystyle h=20\sqrt3\text{ m}
\displaystyle \therefore \text{The height of the tower is }20\sqrt3\text{ m.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The ratio of the height of a vertical rod to the length of its shadow is }1:1.
\displaystyle \text{Find the angle of elevation of the Sun.}
\displaystyle \text{(a) }30^\circ\qquad\text{(b) }45^\circ\qquad\text{(c) }60^\circ\qquad\text{(d) }90^\circ
\displaystyle \text{Answer:}
\displaystyle \text{Let the angle of elevation of the Sun be }\theta.
\displaystyle \frac{\text{Height of rod}}{\text{Length of shadow}}=\frac11
\displaystyle \tan\theta=1
\displaystyle \therefore \theta=45^\circ
\displaystyle \therefore \text{The angle of elevation of the Sun is }45^\circ.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The angle formed by the line of sight with the horizontal when the object viewed}
\displaystyle \text{is below the horizontal level is called:}
\displaystyle \text{(a) Corresponding angle}\qquad\text{(b) Angle of elevation}
\displaystyle \text{(c) Angle of depression}\qquad\text{(d) Complete angle}
\displaystyle \text{Answer:}
\displaystyle \text{When an object is below the horizontal level of the observer, the angle between the}
\displaystyle \text{horizontal line and the downward line of sight is called the angle of depression.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

 

\displaystyle \text{AREAS RELATED TO CIRCLES }

\displaystyle \textbf{Case Study - 1}

\displaystyle \text{Pookalam is a flower bed or floral pattern designed during Onam in Kerala. It is similar to}
\displaystyle \text{Rangoli in North India and Kolam in Tamil Nadu. During the festival of Onam, a school is}
\displaystyle \text{planning to conduct a Pookalam competition. A student participating in the competition}
\displaystyle \text{suggests the following two designs. Observe the designs carefully and answer the questions.}
\displaystyle \\

\displaystyle \text{Design I: An equilateral triangle }ABC\text{ is inscribed in a circle of radius }32\text{ cm, as shown}
\displaystyle \text{in the figure.}
\displaystyle \text{Design II: This Pookalam consists of }9\text{ equal circular designs, each of radius }7\text{ cm,}
\displaystyle \text{arranged in a square }ABCD\text{ as shown in the figure. Take }\pi=\frac{22}{7}.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Refer to Design I. The side of the equilateral triangle is:}
\displaystyle \text{(a) }12\sqrt3\text{ cm}\qquad\text{(b) }32\sqrt3\text{ cm}\qquad\text{(c) }48\text{ cm}\qquad\text{(d) }64\text{ cm}
\displaystyle \text{Answer:}
\displaystyle \text{For an equilateral triangle of side }a\text{ and circumradius }R,
\displaystyle R=\frac{a}{\sqrt3}
\displaystyle 32=\frac{a}{\sqrt3}
\displaystyle a=32\sqrt3\text{ cm}
\displaystyle \therefore \text{The side of the equilateral triangle is }32\sqrt3\text{ cm.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Refer to Design I. The altitude of the equilateral triangle is:}
\displaystyle \text{(a) }8\text{ cm}\qquad\text{(b) }12\text{ cm}\qquad\text{(c) }48\text{ cm}\qquad\text{(d) }52\text{ cm}
\displaystyle \text{Answer:}
\displaystyle \text{Side of the equilateral triangle}=32\sqrt3\text{ cm}
\displaystyle \text{Altitude of an equilateral triangle}=\frac{\sqrt3}{2}\times\text{side}
\displaystyle =\frac{\sqrt3}{2}\times32\sqrt3
\displaystyle =48\text{ cm}
\displaystyle \therefore \text{The altitude of the equilateral triangle is }48\text{ cm.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Refer to Design II. The area of square }ABCD\text{ is:}
\displaystyle \text{(a) }1264\text{ cm}^2\qquad\text{(b) }1764\text{ cm}^2\qquad\text{(c) }1830\text{ cm}^2\qquad\text{(d) }1944\text{ cm}^2
\displaystyle \text{Answer:}
\displaystyle \text{Radius of each circular design}=7\text{ cm}
\displaystyle \therefore \text{Diameter of each circular design}=14\text{ cm}
\displaystyle \text{There are }3\text{ circular designs along each side of the square.}
\displaystyle \therefore \text{Side of square}=3\times14=42\text{ cm}
\displaystyle \text{Area of square}=42^2
\displaystyle =1764\text{ cm}^2
\displaystyle \therefore \text{The area of square }ABCD\text{ is }1764\text{ cm}^2.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Refer to Design II. The area of each circular design is:}
\displaystyle \text{(a) }124\text{ cm}^2\qquad\text{(b) }132\text{ cm}^2\qquad\text{(c) }144\text{ cm}^2\qquad\text{(d) }154\text{ cm}^2
\displaystyle \text{Answer:}
\displaystyle \text{Radius of each circular design}=7\text{ cm}
\displaystyle \text{Area of each circular design}=\pi r^2
\displaystyle =\frac{22}{7}\times7\times7
\displaystyle =154\text{ cm}^2
\displaystyle \therefore \text{The area of each circular design is }154\text{ cm}^2.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Refer to Design II. The area of the remaining portion of square }ABCD\text{ is:}
\displaystyle \text{(a) }378\text{ cm}^2\qquad\text{(b) }260\text{ cm}^2\qquad\text{(c) }340\text{ cm}^2\qquad\text{(d) }278\text{ cm}^2
\displaystyle \text{Answer:}
\displaystyle \text{Area of square }ABCD=1764\text{ cm}^2
\displaystyle \text{Area of }9\text{ circular designs}=9\times154
\displaystyle =1386\text{ cm}^2
\displaystyle \text{Area of remaining portion}=1764-1386
\displaystyle =378\text{ cm}^2
\displaystyle \therefore \text{The area of the remaining portion of square }ABCD\text{ is }378\text{ cm}^2.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

 

\displaystyle \textbf{Case Study - 2}

\displaystyle \text{A brooch is a small piece of jewellery with a pin at the back so that it can be fastened to a}
\displaystyle \text{dress, blouse or coat. Two brooch designs are shown below. Observe them carefully and answer}
\displaystyle \text{the questions that follow.} \displaystyle \text{Design A: Brooch A is made of silver wire in the form of a circle of diameter }28\text{ mm. Silver}
\displaystyle \text{wire is also used to make }4\text{ diameters that divide the circle into }8\text{ equal sectors.}
\displaystyle \text{Design B: Brooch B consists of an inner circular silver part surrounded by a golden ring. The}
\displaystyle \text{circumference of the silver part is }44\text{ mm, and the golden ring is }3\text{ mm wide throughout.}
\displaystyle \text{Take }\pi=\frac{22}{7}.
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Refer to Design A. The total length of silver wire required is:}
\displaystyle \text{(a) }180\text{ mm}\qquad\text{(b) }200\text{ mm}\qquad\text{(c) }250\text{ mm}\qquad\text{(d) }280\text{ mm}
\displaystyle \text{Answer:}
\displaystyle \text{Diameter of the circle}=28\text{ mm}
\displaystyle \text{Circumference of the circle}=\pi d
\displaystyle =\frac{22}{7}\times28=88\text{ mm}
\displaystyle \text{Length of wire required for }4\text{ diameters}=4\times28=112\text{ mm}
\displaystyle \text{Total length of silver wire}=88+112=200\text{ mm}
\displaystyle \therefore \text{The total length of silver wire required is }200\text{ mm.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Refer to Design A. The area of each sector of the brooch is:}
\displaystyle \text{(a) }44\text{ mm}^2\qquad\text{(b) }52\text{ mm}^2\qquad\text{(c) }77\text{ mm}^2\qquad\text{(d) }68\text{ mm}^2
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the circle}=\frac{28}{2}=14\text{ mm}
\displaystyle \text{Area of the circle}=\pi r^2
\displaystyle =\frac{22}{7}\times14\times14=616\text{ mm}^2
\displaystyle \text{The }4\text{ diameters divide the circle into }8\text{ equal sectors.}
\displaystyle \therefore \text{Area of each sector}=\frac{616}{8}=77\text{ mm}^2
\displaystyle \therefore \text{The area of each sector is }77\text{ mm}^2.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Refer to Design B. The circumference of the outer edge of the golden part is:}
\displaystyle \text{(a) }48.49\text{ mm}\qquad\text{(b) }82.2\text{ mm}\qquad\text{(c) }72.50\text{ mm}\qquad\text{(d) }62.86\text{ mm}
\displaystyle \text{Answer:}
\displaystyle \text{Circumference of the inner silver circle}=44\text{ mm}
\displaystyle 2\pi r=44
\displaystyle 2\times\frac{22}{7}\times r=44
\displaystyle r=7\text{ mm}
\displaystyle \text{Width of the golden ring}=3\text{ mm}
\displaystyle \therefore \text{Outer radius}=7+3=10\text{ mm}
\displaystyle \text{Outer circumference}=2\pi(10)=20\pi
\displaystyle =\frac{440}{7}\text{ mm}\approx62.86\text{ mm}
\displaystyle \therefore \text{The circumference of the outer edge is approximately }62.86\text{ mm.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Refer to Design B. The area of the golden part is:}
\displaystyle \text{(a) }18\pi\text{ mm}^2\qquad\text{(b) }44\pi\text{ mm}^2\qquad\text{(c) }51\pi\text{ mm}^2\qquad\text{(d) }64\pi\text{ mm}^2
\displaystyle \text{Answer:}
\displaystyle \text{Inner radius}=7\text{ mm},\qquad\text{Outer radius}=10\text{ mm}
\displaystyle \text{Area of the golden part}=\pi(R^2-r^2)
\displaystyle =\pi(10^2-7^2)
\displaystyle =\pi(100-49)
\displaystyle =51\pi\text{ mm}^2
\displaystyle \therefore \text{The area of the golden part is }51\pi\text{ mm}^2.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{A boy rolls Brooch B along its outer edge. How many complete revolutions will it}
\displaystyle \text{make to cover a distance of }80\pi\text{ mm?}
\displaystyle \text{(a) }2\qquad\text{(b) }3\qquad\text{(c) }4\qquad\text{(d) }5
\displaystyle \text{Answer:}
\displaystyle \text{Outer radius of Brooch B}=10\text{ mm}
\displaystyle \text{Distance covered in one revolution}=2\pi r
\displaystyle =2\pi(10)=20\pi\text{ mm}
\displaystyle \text{Number of revolutions}=\frac{80\pi}{20\pi}=4
\displaystyle \therefore \text{The brooch will make }4\text{ complete revolutions.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

 

\displaystyle \text{SURFACE AREAS AND VOLUMES }

\displaystyle \textbf{Case Study - 1}

\displaystyle \text{Adventure camps provide children with opportunities to practise decision-making and become}
\displaystyle \text{more independent. Some students of a school went to an adventure camp at Sakleshpur. At the}
\displaystyle \text{camp, some students were served a welcome drink in a cylindrical glass and others in a}
\displaystyle \text{hemispherical cup. The dimensions of the containers are shown in the figure. After enjoying}
\displaystyle \text{the jungle trek, the students had to make tents for shelter. Each group of four students was}
\displaystyle \text{given }551\text{ m}^2\text{ of canvas. Of this, }1\text{ m}^2\text{ was used for stitching and wastage. The}
\displaystyle \text{remaining canvas was used to make the curved surface of a conical tent of radius }7\text{ m.}
\displaystyle \text{Take }\pi=\frac{22}{7}. \displaystyle \\

\displaystyle \textbf{Question 1: }\text{The volume of the cylindrical glass is:}
\displaystyle \text{(a) }295.75\text{ cm}^3\qquad\text{(b) }7415.5\text{ cm}^3\qquad\text{(c) }384.88\text{ cm}^3
\displaystyle \text{(d) }404.25\text{ cm}^3
\displaystyle \text{Answer:}
\displaystyle \text{Diameter of the cylindrical glass}=7\text{ cm}
\displaystyle \therefore r=\frac{7}{2}=3.5\text{ cm},\qquad h=10.5\text{ cm}
\displaystyle \text{Volume of cylinder}=\pi r^2h
\displaystyle =\frac{22}{7}\times3.5\times3.5\times10.5
\displaystyle =404.25\text{ cm}^3
\displaystyle \therefore \text{The volume of the cylindrical glass is }404.25\text{ cm}^3.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The volume of the hemispherical cup is:}
\displaystyle \text{(a) }179.67\text{ cm}^3\qquad\text{(b) }89.83\text{ cm}^3\qquad\text{(c) }172.25\text{ cm}^3
\displaystyle \text{(d) }210.60\text{ cm}^3
\displaystyle \text{Answer:}
\displaystyle \text{Diameter of the hemispherical cup}=7\text{ cm}
\displaystyle \therefore r=3.5\text{ cm}
\displaystyle \text{Volume of hemisphere}=\frac{2}{3}\pi r^3
\displaystyle =\frac{2}{3}\times\frac{22}{7}\times(3.5)^3
\displaystyle \approx89.83\text{ cm}^3
\displaystyle \therefore \text{The volume of the hemispherical cup is approximately }89.83\text{ cm}^3.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Which container holds more juice and by how much?}
\displaystyle \text{(a) Hemispherical cup, }195\text{ cm}^3\qquad\text{(b) Cylindrical glass, }207\text{ cm}^3
\displaystyle \text{(c) Hemispherical cup, }280.85\text{ cm}^3\qquad\text{(d) Cylindrical glass, }314.42\text{ cm}^3
\displaystyle \text{Answer:}
\displaystyle \text{Volume of cylindrical glass}=404.25\text{ cm}^3
\displaystyle \text{Volume of hemispherical cup}\approx89.83\text{ cm}^3
\displaystyle \text{Difference}=404.25-89.83
\displaystyle =314.42\text{ cm}^3
\displaystyle \therefore \text{The cylindrical glass holds approximately }314.42\text{ cm}^3\text{ more juice.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The height of the conical tent prepared to accommodate four students is:}
\displaystyle \text{(a) }18\text{ m}\qquad\text{(b) }10\text{ m}\qquad\text{(c) }24\text{ m}\qquad\text{(d) }14\text{ m}
\displaystyle \text{Answer:}
\displaystyle \text{Canvas available for making the tent}=551-1=550\text{ m}^2
\displaystyle \text{Radius of the conical tent}=7\text{ m}
\displaystyle \text{Curved surface area of cone}=\pi rl
\displaystyle 550=\frac{22}{7}\times7\times l
\displaystyle 550=22l
\displaystyle l=25\text{ m}
\displaystyle \text{Using }l^2=r^2+h^2,
\displaystyle 25^2=7^2+h^2
\displaystyle h^2=625-49=576
\displaystyle h=24\text{ m}
\displaystyle \therefore \text{The height of the conical tent is }24\text{ m.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{How much space on the ground is occupied by each student in the conical tent?}
\displaystyle \text{(a) }54\text{ m}^2\qquad\text{(b) }38.5\text{ m}^2\qquad\text{(c) }86\text{ m}^2\qquad\text{(d) }24\text{ m}^2
\displaystyle \text{Answer:}
\displaystyle \text{Area of the ground covered by the tent}=\pi r^2
\displaystyle =\frac{22}{7}\times7\times7
\displaystyle =154\text{ m}^2
\displaystyle \text{Number of students}=4
\displaystyle \text{Space occupied by each student}=\frac{154}{4}
\displaystyle =38.5\text{ m}^2
\displaystyle \therefore \text{Each student occupies }38.5\text{ m}^2\text{ of space on the ground.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

 

\displaystyle \textbf{Case Study - 2}

\displaystyle \text{The Great Stupa at Sanchi is one of the oldest stone structures in India and an important}
\displaystyle \text{monument of Indian architecture. It was originally commissioned by Emperor Ashoka in the}
\displaystyle 3\text{rd century BCE. Its nucleus was a simple hemispherical brick structure built over the relics}
\displaystyle \text{of the Buddha. It is a good example of a combination of solid figures. Consider a large}
\displaystyle \text{hemispherical dome with a cuboidal structure mounted on top. Take }\pi=\frac{22}{7}. \displaystyle \\

\displaystyle \textbf{Question 1: }\text{Calculate the volume of the hemispherical dome if its height is }21\text{ m.}
\displaystyle \text{(a) }19404\text{ m}^3\qquad\text{(b) }2000\text{ m}^3\qquad\text{(c) }15000\text{ m}^3\qquad\text{(d) }19000\text{ m}^3
\displaystyle \text{Answer:}
\displaystyle \text{For a hemisphere, height}=\text{radius.}
\displaystyle \therefore r=21\text{ m}
\displaystyle \text{Volume of hemisphere}=\frac{2}{3}\pi r^3
\displaystyle =\frac{2}{3}\times\frac{22}{7}\times21^3
\displaystyle =19404\text{ m}^3
\displaystyle \therefore \text{The volume of the hemispherical dome is }19404\text{ m}^3.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The formula for the volume of a sphere is:}
\displaystyle \text{(a) }\frac{2}{3}\pi r^3\qquad\text{(b) }\frac{4}{3}\pi r^3\qquad\text{(c) }4\pi r^2\qquad\text{(d) }2\pi r^2
\displaystyle \text{Answer:}
\displaystyle \text{Volume of a sphere}=\frac{4}{3}\pi r^3
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The area of cloth required to cover the curved surface of the hemispherical}
\displaystyle \text{dome of radius }14\text{ m is:}
\displaystyle \text{(a) }1222\text{ m}^2\qquad\text{(b) }1232\text{ m}^2\qquad\text{(c) }1200\text{ m}^2\qquad\text{(d) }1400\text{ m}^2
\displaystyle \text{Answer:}
\displaystyle \text{Curved surface area of a hemisphere}=2\pi r^2
\displaystyle =2\times\frac{22}{7}\times14\times14
\displaystyle =1232\text{ m}^2
\displaystyle \therefore \text{The area of cloth required is }1232\text{ m}^2.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{A cuboidal structure of dimensions }8\text{ m}\times6\text{ m}\times4\text{ m is mounted on a}
\displaystyle \text{hemispherical dome of radius }14\text{ m. If the bottom of the cuboid is not exposed, find}
\displaystyle \text{the total exposed surface area of the combined structure.}
\displaystyle \text{(a) }1344\text{ m}^2\qquad\text{(b) }1232\text{ m}^2\qquad\text{(c) }1392\text{ m}^2\qquad\text{(d) }1932\text{ m}^2
\displaystyle \text{Answer:}
\displaystyle \text{Curved surface area of hemispherical dome}=2\pi r^2
\displaystyle =2\times\frac{22}{7}\times14^2=1232\text{ m}^2
\displaystyle \text{Lateral surface area of cuboid}=2h(l+b)
\displaystyle =2\times4(8+6)=112\text{ m}^2
\displaystyle \text{Area of the exposed top face}=8\times6=48\text{ m}^2
\displaystyle \text{Exposed surface area of cuboid}=112+48=160\text{ m}^2
\displaystyle \text{Total exposed surface area}=1232+160=1392\text{ m}^2
\displaystyle \therefore \text{The total exposed surface area is }1392\text{ m}^2.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The volume of the cuboidal structure of dimensions }8\text{ m}\times6\text{ m}\times4\text{ m is:}
\displaystyle \text{(a) }182.45\text{ m}^3\qquad\text{(b) }282.45\text{ m}^3\qquad\text{(c) }292\text{ m}^3\qquad\text{(d) }192\text{ m}^3
\displaystyle \text{Answer:}
\displaystyle \text{Volume of cuboid}=l\times b\times h
\displaystyle =8\times6\times4
\displaystyle =192\text{ m}^3
\displaystyle \therefore \text{The volume of the cuboidal structure is }192\text{ m}^3.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

 

\displaystyle \textbf{Case Study - 3}

\displaystyle \text{On a Sunday, your parents took you to a fair. You saw many toys on display and wanted them to}
\displaystyle \text{buy you a Rubik's Cube and a strawberry ice cream. Observe the figures and answer the}
\displaystyle \text{questions that follow. Take }\pi=\frac{22}{7}. \displaystyle \\

\displaystyle \textbf{Question 1: }\text{The length of the space diagonal of a cube whose edge measures }6\text{ cm is:}
\displaystyle \text{(a) }3\sqrt3\text{ cm}\qquad\text{(b) }3\sqrt6\text{ cm}\qquad\text{(c) }\sqrt{12}\text{ cm}\qquad\text{(d) }6\sqrt3\text{ cm}
\displaystyle \text{Answer:}
\displaystyle \text{Space diagonal of a cube}=a\sqrt3
\displaystyle =6\sqrt3\text{ cm}
\displaystyle \therefore \text{The length of the space diagonal is }6\sqrt3\text{ cm.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The volume of a cube whose edge measures }7\text{ cm is:}
\displaystyle \text{(a) }256\text{ cm}^3\qquad\text{(b) }196\text{ cm}^3\qquad\text{(c) }343\text{ cm}^3\qquad\text{(d) }434\text{ cm}^3
\displaystyle \text{Answer:}
\displaystyle \text{Volume of a cube}=a^3
\displaystyle =7^3=343\text{ cm}^3
\displaystyle \therefore \text{The volume of the cube is }343\text{ cm}^3.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{What is the curved surface area of the hemispherical ice cream if its base}
\displaystyle \text{radius is }7\text{ cm?}
\displaystyle \text{(a) }309\text{ cm}^2\qquad\text{(b) }308\text{ cm}^2\qquad\text{(c) }803\text{ cm}^2\qquad\text{(d) }903\text{ cm}^2
\displaystyle \text{Answer:}
\displaystyle \text{Curved surface area of a hemisphere}=2\pi r^2
\displaystyle =2\times\frac{22}{7}\times7\times7
\displaystyle =308\text{ cm}^2
\displaystyle \therefore \text{The curved surface area of the hemisphere is }308\text{ cm}^2.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the slant height of a cone whose radius is }7\text{ cm and height is }24\text{ cm.}
\displaystyle \text{(a) }26\text{ cm}\qquad\text{(b) }25\text{ cm}\qquad\text{(c) }52\text{ cm}\qquad\text{(d) }62\text{ cm}
\displaystyle \text{Answer:}
\displaystyle l=\sqrt{r^2+h^2}
\displaystyle =\sqrt{7^2+24^2}
\displaystyle =\sqrt{49+576}
\displaystyle =\sqrt{625}=25\text{ cm}
\displaystyle \therefore \text{The slant height of the cone is }25\text{ cm.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the exposed surface area of the cone with a hemispherical ice cream}
\displaystyle \text{placed on it, if their common radius is }7\text{ cm and the slant height of the cone is }25\text{ cm.}
\displaystyle \text{(a) }858\text{ cm}^2\qquad\text{(b) }885\text{ cm}^2\qquad\text{(c) }588\text{ cm}^2\qquad\text{(d) }855\text{ cm}^2
\displaystyle \text{Answer:}
\displaystyle \text{Exposed surface area}=\text{CSA of cone}+\text{CSA of hemisphere}
\displaystyle =\pi rl+2\pi r^2
\displaystyle =\frac{22}{7}\times7\times25+2\times\frac{22}{7}\times7^2
\displaystyle =550+308
\displaystyle =858\text{ cm}^2
\displaystyle \therefore \text{The exposed surface area is }858\text{ cm}^2.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

 

\displaystyle \text{STATISTICS}

\displaystyle \textbf{Case Study - 1}

\displaystyle \text{COVID-19 is an infectious disease caused by the SARS-CoV-2 virus. The following tables show}
\displaystyle \text{the age distribution of cases admitted during a particular day in two different hospitals.} \displaystyle \\

\displaystyle \text{Table 1:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Age (in years)}&5-15&15-25&25-35&35-45&45-55&55-65\\ \hline \text{No. of cases}&6&11&21&23&14&5\\ \hline \end{array}
\displaystyle \\

\displaystyle \text{Table 2:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Age (in years)}&5-15&15-25&25-35&35-45&45-55&55-65\\ \hline \text{No. of cases}&8&16&10&42&24&12\\ \hline \end{array}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Refer to Table 1. The mode of the given data is:}
\displaystyle \text{(a) }32.24\qquad\text{(b) }34.36\qquad\text{(c) }36.82\qquad\text{(d) }42.24
\displaystyle \text{Answer:}
\displaystyle \text{The highest frequency is }23,\text{ so the modal class is }35-45.
\displaystyle l=35,\quad h=10,\quad f_1=23,\quad f_0=21,\quad f_2=14
\displaystyle \text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h
\displaystyle =35+\frac{23-21}{2(23)-21-14}\times10
\displaystyle =35+\frac{2}{11}\times10
\displaystyle =36.82\text{ years (approx.).}
\displaystyle \therefore \text{The mode of the given data is }36.82\text{ years (approx.).}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Refer to Table 1. The upper limit of the modal class is:}
\displaystyle \text{(a) }15\qquad\text{(b) }25\qquad\text{(c) }35\qquad\text{(d) }45
\displaystyle \text{Answer:}
\displaystyle \text{The highest frequency is }23,\text{ corresponding to the class }35-45.
\displaystyle \therefore \text{Modal class}=35-45
\displaystyle \therefore \text{Upper limit of the modal class}=45\text{ years.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Refer to Table 1. The mean of the given data is:}
\displaystyle \text{(a) }26.2\qquad\text{(b) }32.4\qquad\text{(c) }33.5\qquad\text{(d) }35.4
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|c|}\hline \text{Class}&f_i&x_i&f_ix_i\\ \hline 5-15&6&10&60\\ 15-25&11&20&220\\ 25-35&21&30&630\\ 35-45&23&40&920\\ 45-55&14&50&700\\ 55-65&5&60&300\\ \hline \text{Total}&80&&2830\\ \hline \end{array}
\displaystyle \text{Mean}=\frac{\sum f_ix_i}{\sum f_i}
\displaystyle =\frac{2830}{80}=35.375
\displaystyle \approx35.4\text{ years}
\displaystyle \therefore \text{The mean of the given data is approximately }35.4\text{ years.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Refer to Table 2. The mode of the given data is:}
\displaystyle \text{(a) }41.4\qquad\text{(b) }48.2\qquad\text{(c) }55.3\qquad\text{(d) }64.6
\displaystyle \text{Answer:}
\displaystyle \text{The highest frequency is }42,\text{ so the modal class is }35-45.
\displaystyle l=35,\quad h=10,\quad f_1=42,\quad f_0=10,\quad f_2=24
\displaystyle \text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h
\displaystyle =35+\frac{42-10}{2(42)-10-24}\times10
\displaystyle =35+\frac{32}{50}\times10
\displaystyle =41.4\text{ years}
\displaystyle \therefore \text{The mode of the given data is }41.4\text{ years.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Refer to Table 2. The median of the given data is:}
\displaystyle \text{(a) }32.7\qquad\text{(b) }40.2\qquad\text{(c) }42.3\qquad\text{(d) }48.6
\displaystyle \text{Answer:}
\displaystyle N=8+16+10+42+24+12=112
\displaystyle \frac{N}{2}=56
\displaystyle \text{Cumulative frequencies are }8,\ 24,\ 34,\ 76,\ 100,\ 112.
\displaystyle \therefore \text{Median class}=35-45
\displaystyle l=35,\quad cf=34,\quad f=42,\quad h=10
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-cf}{f}\times h
\displaystyle =35+\frac{56-34}{42}\times10
\displaystyle =35+\frac{220}{42}
\displaystyle \approx40.24\text{ years}
\displaystyle \approx40.2\text{ years}
\displaystyle \therefore \text{The median of the given data is approximately }40.2\text{ years.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

 

\displaystyle \textbf{Case Study - 2}

\displaystyle \text{Electricity consumption refers to the amount of electrical energy used by consumers. A survey}
\displaystyle \text{was conducted among }56\text{ families of Colony A to record their weekly electricity consumption.}
\displaystyle \text{The following table shows the distribution of their weekly electricity consumption.} \displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Weekly consumption (in units)}&0-10&10-20&20-30&30-40&40-50&50-60\\ \hline \text{No. of families}&16&12&18&6&4&0\\ \hline \end{array}
\displaystyle \text{A similar survey was conducted among }80\text{ families of Colony B, and the data were recorded}
\displaystyle \text{as follows.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Weekly consumption (in units)}&0-10&10-20&20-30&30-40&40-50&50-60\\ \hline \text{No. of families}&0&5&10&20&40&5\\ \hline \end{array}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{Refer to the data for Colony A. The median weekly consumption is:}
\displaystyle \text{(a) }12\text{ units}\qquad\text{(b) }16\text{ units}\qquad\text{(c) }20\text{ units}\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle N=16+12+18+6+4=56
\displaystyle \frac{N}{2}=28
\displaystyle \text{Cumulative frequencies are }16,\ 28,\ 46,\ 52,\ 56.
\displaystyle \text{Taking the median class as }10-20,
\displaystyle l=10,\quad cf=16,\quad f=12,\quad h=10
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-cf}{f}\times h
\displaystyle =10+\frac{28-16}{12}\times10
\displaystyle =10+10=20\text{ units}
\displaystyle \therefore \text{The median weekly consumption is }20\text{ units.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Refer to the data for Colony A. The mean weekly consumption is:}
\displaystyle \text{(a) }19.64\text{ units}\qquad\text{(b) }22.5\text{ units}\qquad\text{(c) }26\text{ units}
\displaystyle \text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|c|}\hline \text{Class}&f_i&x_i&f_ix_i\\ \hline 0-10&16&5&80\\ 10-20&12&15&180\\ 20-30&18&25&450\\ 30-40&6&35&210\\ 40-50&4&45&180\\ 50-60&0&55&0\\ \hline \text{Total}&56&&1100\\ \hline \end{array}
\displaystyle \text{Mean}=\frac{\sum f_ix_i}{\sum f_i}
\displaystyle =\frac{1100}{56}
\displaystyle \approx19.64\text{ units}
\displaystyle \therefore \text{The mean weekly consumption is approximately }19.64\text{ units.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Refer to the data for Colony A. The modal class of the given data is:}
\displaystyle \text{(a) }0-10\qquad\text{(b) }10-20\qquad\text{(c) }20-30\qquad\text{(d) }30-40
\displaystyle \text{Answer:}
\displaystyle \text{The highest frequency is }18.
\displaystyle \text{This frequency corresponds to the class interval }20-30.
\displaystyle \therefore \text{The modal class is }20-30.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Refer to the data for Colony B. The modal weekly consumption is:}
\displaystyle \text{(a) }38.2\text{ units}\qquad\text{(b) }43.6\text{ units}\qquad\text{(c) }26\text{ units}\qquad\text{(d) }32\text{ units}
\displaystyle \text{Answer:}
\displaystyle \text{The highest frequency is }40,\text{ so the modal class is }40-50.
\displaystyle l=40,\quad h=10,\quad f_1=40,\quad f_0=20,\quad f_2=5
\displaystyle \text{Mode}=l+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h
\displaystyle =40+\frac{40-20}{2(40)-20-5}\times10
\displaystyle =40+\frac{20}{55}\times10
\displaystyle \approx43.64\text{ units}
\displaystyle \approx43.6\text{ units}
\displaystyle \therefore \text{The modal weekly consumption is approximately }43.6\text{ units.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Refer to the data for Colony B. The mean weekly consumption is:}
\displaystyle \text{(a) }15.65\text{ units}\qquad\text{(b) }32.8\text{ units}\qquad\text{(c) }38.75\text{ units}\qquad\text{(d) }48\text{ units}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|c|}\hline \text{Class}&f_i&x_i&f_ix_i\\ \hline 0-10&0&5&0\\ 10-20&5&15&75\\ 20-30&10&25&250\\ 30-40&20&35&700\\ 40-50&40&45&1800\\ 50-60&5&55&275\\ \hline \text{Total}&80&&3100\\ \hline \end{array}
\displaystyle \text{Mean}=\frac{\sum f_ix_i}{\sum f_i}
\displaystyle =\frac{3100}{80}
\displaystyle =38.75\text{ units}
\displaystyle \therefore \text{The mean weekly consumption is }38.75\text{ units.}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

 

\displaystyle \text{PROBABILITY}

\displaystyle \textbf{Case Study - 1}

\displaystyle \text{On a weekend, Rani was playing cards with her family. A standard deck has }52\text{ cards.}
\displaystyle \text{Her brother draws one card at random from the deck. Based on this information, answer the}
\displaystyle \text{following questions.} \displaystyle \\

\displaystyle \textbf{Question 1: }\text{Find the probability of getting a king of red colour.}
\displaystyle \text{(a) }\frac{1}{26}\qquad\text{(b) }\frac{1}{13}\qquad\text{(c) }\frac{1}{52}\qquad\text{(d) }\frac{1}{4}
\displaystyle \text{Answer:}
\displaystyle \text{There are }2\text{ red kings, one of hearts and one of diamonds.}
\displaystyle \text{Total number of cards}=52
\displaystyle P(\text{a red king})=\frac{2}{52}=\frac{1}{26}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the probability of getting a face card.}
\displaystyle \text{(a) }\frac{1}{26}\qquad\text{(b) }\frac{1}{13}\qquad\text{(c) }\frac{2}{13}\qquad\text{(d) }\frac{3}{13}
\displaystyle \text{Answer:}
\displaystyle \text{Each suit has }3\text{ face cards: Jack, Queen and King.}
\displaystyle \text{Number of face cards}=3\times4=12
\displaystyle P(\text{a face card})=\frac{12}{52}=\frac{3}{13}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the probability of getting a jack of hearts.}
\displaystyle \text{(a) }\frac{1}{26}\qquad\text{(b) }\frac{1}{52}\qquad\text{(c) }\frac{3}{52}\qquad\text{(d) }\frac{3}{26}
\displaystyle \text{Answer:}
\displaystyle \text{There is only one jack of hearts in a deck of }52\text{ cards.}
\displaystyle P(\text{jack of hearts})=\frac{1}{52}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the probability of getting a red face card.}
\displaystyle \text{(a) }\frac{3}{13}\qquad\text{(b) }\frac{1}{13}\qquad\text{(c) }\frac{1}{52}\qquad\text{(d) }\frac{1}{4}
\displaystyle \text{Answer:}
\displaystyle \text{The red suits are hearts and diamonds.}
\displaystyle \text{Each red suit has }3\text{ face cards.}
\displaystyle \text{Number of red face cards}=2\times3=6
\displaystyle P(\text{a red face card})=\frac{6}{52}=\frac{3}{26}
\displaystyle \text{None of the given options is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the probability of getting a spade.}
\displaystyle \text{(a) }\frac{1}{26}\qquad\text{(b) }\frac{1}{13}\qquad\text{(c) }\frac{1}{52}\qquad\text{(d) }\frac{1}{4}
\displaystyle \text{Answer:}
\displaystyle \text{There are }13\text{ spade cards in a deck of }52\text{ cards.}
\displaystyle P(\text{a spade})=\frac{13}{52}=\frac{1}{4}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

 

\displaystyle \textbf{Case Study - 2}

\displaystyle \text{Rahul and Ravi planned to play Business, a board game in which they were supposed to use}
\displaystyle \text{two dice. They take turns rolling the two dice. Based on this information, answer the following}
\displaystyle \text{questions.} \displaystyle \\

\displaystyle \textbf{Question 1: }\text{Ravi got the first chance to roll the two dice. What is the probability that the sum}
\displaystyle \text{of the numbers appearing on the top faces of the dice is }8\text{?}
\displaystyle \text{(a) }\frac{1}{26}\qquad\text{(b) }\frac{5}{36}\qquad\text{(c) }\frac{1}{18}\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \text{Total number of possible outcomes}=6\times6=36
\displaystyle \text{Outcomes with sum }8\text{ are }(2,6),(3,5),(4,4),(5,3),(6,2).
\displaystyle \text{Number of favourable outcomes}=5
\displaystyle P(\text{sum }8)=\frac{5}{36}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Rahul got the next chance. What is the probability that the sum of the numbers}
\displaystyle \text{appearing on the top faces of the dice is }13\text{?}
\displaystyle \text{(a) }1\qquad\text{(b) }\frac{5}{36}\qquad\text{(c) }\frac{1}{18}\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \text{The maximum possible sum on two dice is }6+6=12.
\displaystyle \text{Therefore, getting a sum of }13\text{ is an impossible event.}
\displaystyle P(\text{sum }13)=0
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Now it was Ravi's turn. What is the probability that the sum of the numbers}
\displaystyle \text{appearing on the top faces of the dice is less than or equal to }12\text{?}
\displaystyle \text{(a) }1\qquad\text{(b) }\frac{5}{36}\qquad\text{(c) }\frac{1}{18}\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \text{The possible sums on two dice range from }2\text{ to }12.
\displaystyle \text{Hence, the sum is always less than or equal to }12.
\displaystyle P(\text{sum}\leq12)=\frac{36}{36}=1
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Rahul got the next chance. What is the probability that the sum of the numbers}
\displaystyle \text{appearing on the top faces of the dice is equal to }7\text{?}
\displaystyle \text{(a) }\frac{5}{9}\qquad\text{(b) }\frac{5}{36}\qquad\text{(c) }\frac{1}{6}\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \text{Outcomes with sum }7\text{ are }(1,6),(2,5),(3,4),(4,3),(5,2),(6,1).
\displaystyle \text{Number of favourable outcomes}=6
\displaystyle P(\text{sum }7)=\frac{6}{36}=\frac{1}{6}
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Now it was Ravi's turn. What is the probability that the sum of the numbers}
\displaystyle \text{appearing on the top faces of the dice is greater than }8\text{?}
\displaystyle \text{(a) }1\qquad\text{(b) }\frac{5}{36}\qquad\text{(c) }\frac{1}{18}\qquad\text{(d) }\frac{5}{18}
\displaystyle \text{Answer:}
\displaystyle \text{For a sum greater than }8,\text{ the possible sums are }9,10,11\text{ and }12.
\displaystyle \text{Number of outcomes for these sums}=4+3+2+1=10
\displaystyle P(\text{sum}>8)=\frac{10}{36}=\frac{5}{18}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

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