MCQ Test: Heights and Distances - 2 - SSC CGL MCQ

Given, the distance walking, CD = 138 m
Let, The height of the monument, AB = h metre
BD = x metre, ∠ACB = α and ∠ADB = β

∴ BC = CD + BD = (138 + x) m
We know that,

x = 5h – 138 ...(i)
Now, in ΔABD,

⇒ 7x = 12h
⇒ 7(5h – 138) = 12h [From eq. (i)]
⇒ 35h – 966 = 12h ⇒ 23h = 966 ⇒ h = 42 m
∴ The height of the monument is 42 metre.
Hence, option C is correct.

Given, AB = 48 m
Let, the height of the tower, CD = h metre
And, BC = x metre
∴ AC = AB + BC = (48 + x) m
In ΔBCD,

x = h √3 ...(i)
Now, in ACD,


⇒ h = 96 + 2h √3 – 48 √3 – 3h
⇒ 4h – 2h √3 = 48(2 – √3)
⇒ 2h(2 –√3) = 48(2 – √3)
⇒ h = 24 m
∴ The height of the tower is 24 metre.
Hence, option B is correct.

Let, the height of the pillar, AB = h metre.
When the sun's angle of elevation was 30°, then the length of shadow of the pillar is BD.
And, when the sun's angle of elevation is 45°, then the length of shadow of the pillar is BC = x metre (let).
When the sun changes from 30° to 45°, then the length of shadow of the pillar decreases CD = 20 (given)
∴ BD = BC + CD = (x + 20) m
In ΔABC,

⇒ h = x ...(i)
Now, in ΔABD,

⇒ h 3 = x + 20
⇒ h √3 = h + 20 [From eq. (i)]
⇒ h ( √3 – 1) = 20

∴ The height of the pillar is 10 (√3 + 1) metre.
Hence, option D is correct.

AC = Flag, AB = building = 10 m
∠APB = 30°; ∠CPB = 45°
In Δ APB,

⇒ PB = 10 √3 m
In ΔPBC,

⇒ PB = AB + AC ⇒ 10 √3 = 10 + AC
⇒ AC = 10 √3 – 10
⇒ 10(√3 – 1) m = 10(1.732 – 1) m
= 10 × 0.732 = 7.32 m.
Hence, option D is correct.

DE = 36 – 24 = 12 m
From Δ ADE,

Hence, option B is correct.

Given, the height of the aeroplane, BC = 1500 √3 meter
Let, an aeroplane travelled distance BD = x metre in 15 seconds.
And, AC = y metre
∴ AE = CE + AC = (x + y) m
In ΔABC,

y = 1500 m
Now, in ΔADE,

x + y = 4500
x + 1500 = 4500 x = 3000 m
∴ The aeroplane travelled 3000 m distance in 15 seconds.
∴ Speed = 3000/15 = 200 m/sec
Hence, option A is correct.

Let, the height of the tower, AB = h metre
And, the angle of elevation = Θ
then, the shadow of the tower, BC = h√3 metre
In ΔABC,

Θ = 30°
Hence, option B is correct.

Total height of post = AB + AC = 15 ft
Let, the post is broken at point A, ∴ AB = h
And, AC = 15 – AB = (15 – h) ft
In ΔABC,

15 – h = 2h
3h = 15
h = 5 ft
Hence, option B is correct.

Let, the height of the tower, AB = h metre.
When the sun's angle of elevation was 45°, then the length of shadow of the tower is BD = x (let).
When the sun changes from 45° to 30°, then the length of shadow of the tower increases CD = 60 m (given)
And, when the sun's angle of elevation is 30°, then the length of shadow of the tower is BC = CD + BD = (60 + x) metre.
In ΔABD,

h = x
Now, in ΔABC,

h√3 – x = 60
h√3 – h = 60 [∵ h = x]
h(√3 – 1) = 60

= 30(√3 + 1) m
Hence, option C is correct.

Given, the height of an aeroplane from the ground, BD = 3125 m
Let, the distance between the aeroplanes, AD = x metre
And, BC = y metre
∴ AB = AD + BD = (3125 + x) m
In ΔBCD,

y√3 = x + 3125
3125√3 × √3 = x + 3125
x = 9375 – 3125 = 6250 m
∴ The distance between the aeroplanes is 6250 metres.
Hence, option D is correct.

Given, BC = 60 metre,
Let E is the mid-point of BC
∴ BE = EC = 30 metre
Let, the height of the pole AB = h metre
∴ The height of the pole CD = 2h metre
And, ∠AEB and ∠DEC are complementary.
∴ ∠AEB = (90° – Θ) and ∠DEC = Θ
In ΔABE,

Now, in ΔCDE,

By multiplying both equation (i) and (ii),

∴ The height of the pole AB = h = 15 √2
And, the height of the pole CD = 2h = 2 × 15 √2 = 30 √2
Hence, option D is correct.

Given, the height of the tower, AB = 30 m
Let, the man moves the distance, CP = y metre
And, BC = x metre
∴ BP = BC + CP = (x + y) m
In ΔABC,


Hence, option A is correct.

Given, the height of an aeroplane from the ground, AB = 5000 m,
Let, the distance between the aeroplanes, AD = x metre
And, BC = y metre
∴ BD = AB – AD = (5000 – x) m
In ΔABC,

Hence, option C is correct.

Given, BC = x metre,
Let E is the mid-point of BC
∴ BE = EC = x/2 metre
Let, the height of the pole AB = h metre
∴ The height of the pole CD = 2h metre
And, ∠AEB and ∠DEC are complementary.
∴ ∠AEB = (90° – Θ) and ∠DEC = Θ
In ΔABE,


[∵ tan (90° – Θ) = cot Θ]
Now, in ΔCDE,

By multiplying both equation (i) and (ii),

Hence, option A is correct.

Given, the height of the building, BD = h metre
Let, the height of the chimney, AD = x metre
and, BC = a metre
∴ AB = BD + AD = (h + x) m
In ΔABC,


a = h cot x ...(ii)
From eq. (i) and (ii),
x + h = h cot x
x = h cot x – h
Hence, option B is correct.