The Statue of Unity situated in Gujarat is the world’s largest

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Case Study question maths class 10

Case Study 1:- The Statue of Unity situated in Gujarat is the world’s largest Statue which stands over a 58 m high base. As part of the project, a student constructed an inclinometer and wishes to find the height of Statue of Unity using it.                   [CBSE  2025]

He noted following observations from places :

Situation – I :

The angle of elevation of the top of Statue from Place A which is 80√3 m away from the base of the Statue is found to be 60º.

Situation – II :

The angle of elevation of the top of Statue from a Place B which is 40 m above the ground is found to be 30º and entire height of the Statue including the base is found to be 240 m.

Based on given information, answer the following questions :

(i) Represent the Situation – I with the help of a diagram.            (1)

(ii) Represent the Situation -II with the help of a diagram.           (1)

(iii) (a) Calculate the height of Statue excluding the base and also find the height including the base with the help of Situation -I.               (2)

OR

(iii) (b) Find the horizontal distance of point B (Situation – II) from the Statue and the value of \tan \alpha, where α is the angle of elevation of top of base of the Statue from point B.         (2)

Solution:- (i) the Situation – I with the help of a diagram

Base of Statue of Unity = 58 m

AR = 80√3 m

The Statue of Unity situated in Gujarat is the world's largest

(ii) the Situation -II with the help of a diagram.

The Statue of Unity situated in Gujarat is the world's largest

(iii) (a) From the diagram of situtaion-I

AR = 80√3 m

In ΔAPR

\tan 60 = \dfrac{PR}{AR}

⇒ \sqrt{3} = \dfrac{PR}{80\sqrt{3}}

⇒ PR = 240 m

Height of Statue  of Unity including base = 240 m

Height of Statue  of Unity excluding base = (240 – 58) m = 182 m

      OR

(iii) (b) From the diagram of situtaion-II

Height of Statue of unity including base = 240 m

Height of of point B = 40 m

In ΔPSB

PS = PR – SR = 240 – 40=200m

\tan \angle PBS = \dfrac{PS}{SB}

⇒ \tan 30 = \dfrac{200}{SB}

⇒ \dfrac{1}{\sqrt{3}} = \dfrac{200}{SB}

⇒ SB = 200\sqrt{3} m

Horizontal distance of Point B from Base = 200√3 m

In ΔQSB

QS = QR – SR = 58 – 40 = 18 m

\tan \alpha = \dfrac{QS}{SB}

⇒ \tan \alpha = \dfrac{18}{200\sqrt{3}}

⇒ \tan \alpha = \dfrac{3\sqrt{3}}{100}

Case Study 2:- Anurag purchased a farmhouse which is in the form of a semicircle of diameter 70 m. He divides it into three parts by taking a point P on the semicircle in such a way that ∠PAB = 30º as shown in the following figure, where O is the centre of semicircle.

In part I, he planted saplings of Mango tree, in part II, he grew tomatoes and in part III, he grew oranges.

    [CBSE    2025]

Based on given information, answer the following questions.

(i) What is the measure of ∠POA ?                       (1)

(ii) Find the length of wire needed to fence entire piece of land.          (1)

(iii) (a) Find the area of region in which saplings of Mango tree are planted.                     (2)

OR

(iii) (b) Find the length of wire neede to fence the region III.             (2)

Case Study 3:- In order to organise, Annual Sports day, a school prepared an eight lane running track with an integrated football field inside the track area as shown below :

The length of innermost lane of the track is 400 m and each subsequent lane is 7.6 m longer than the preceding lane.                  [CBSE  2025]

Based on given information, answer the following questions, using concept on Arithmetic Progression.

(i) What is the length of the 6^{th} lane ?            (1)

(ii) How long is the 8^{th} lane than that of 4^{th} lane ?                   (1)

(iii) (a) While practicing for a race, a student took distance covered by the student.            (2)

OR

(iii) (b) A student took one round each in lane 4 to 8. Find the total distance covered by the student.            (2)

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