MCQ Test: Height & Distance - 1 - SSC CGL MCQ

We can solve it through ratio proportion rule,
let the tower is x m long, then
12 cm stick casts → 8 cm shadow
x m tower casts → 40 m shadow
On cross multiplying, we get

Hence, option B is correct.

AB = Height of balloon = h km
BD = x km, CD = 1 km
From Δ ABD,

From Δ ABC,

Hence, option A is correct.

AP = CP = BP
It is possible only when OA = OB = OC i.e. radii of circum circle.
Hence, option B is correct.

Given that ther are two poles
AE = 11 m and, CD = 6 m
∴ BE = 6 m
[∵CD = BE]
∴ AB = AE – BE = 11 – 6 = 5m
distance between their feet
ED = 12 m
∴ BC = 12 m [∵ED = BC] Now, AC = ?
In ΔABC,
From Pythagorus theorum,
AC² = AB² + BC²
AC² = 5² + 12²
AC² = 25 + 144 = 169
AC = √169
AC = 13
Hence, option C is correct.

Let AB = x
Then, BC = √ x
and θ = ?


tan θ = tan 30°
[∵ tan 30° = 1/√3]
∴ θ = 30°
Hence, option A is correct.

Given, ∠ADB = 30° and ∠ ACB = 60°
When the sun's elevation is 30°, the shadow of tower is "BD = 15 m" and when the sun's elevation is 60°, the shadow of tower is "BC = ?"
Let, BC = x m In ΔABD, tan 30° = AB/BD

In ΔABC, tan 60° = AB/BC
√3 = AB/x
∴ AB = x √3 ...(ii)
From Eqs. (i) and (ii), we get
x √3 = 15/√3
x = 5 m
Hence, option C is correct.

Hence, option B is correct.

Given, AB = 90 m
∠ADE = 30°
And ∠ACB = 60°
Then, DC = ?
Ratio of angles,

AE = 30 m
Now, DC = EB
= AB – AE
= 90 – 30 = 60 m
Hence, option C is correct.

Given, BC = 150 m
∠ACB = 30°
and, ∠DCB = 45°
Then, AD = ?

DB = 150
AD + AB = 150
[∵ DB = AD + AB]
∴ AD = 150 – AB
= 150 – 86.6 = 63.4m
Hence, option D is correct.

Given, ∠ ACB = 45°
∠ ADB = 30°
and distance between two ships, i.e.,
CD = 200 m
Then, AB = ?
Let BC = x m
In ΔABC, tan 45° = AB/BC
(∵ tan 45° = 1)
1 = AB/x
∴ AB = x m ....(i)
In ΔABD, tan30° = AB/BD

(∵ tan 30° = 1/√3 )
x = √ AB – 200 ...(ii)
From Eqs. (i) and (ii),
AB = √ AB – 200
√ AB – AB = 200
0.732 AB = 200
(∵√ = 1.732)

= 273 m
Hence, option D is correct.

Given, BC = 8 √ m
In ΔABC,

AB = 8 m
Again, sin 30° = AB/AC

AC = 16 m
∴ The height of the post = AC + AB = 16 + 8 = 24 m.
Hence, option C is correct.

Given, BC = 100 m,
Let, the height of each pole = h metre
And, BE = x metre
∴ CE = (100 – x) m
In ΔCDE,

h = x √3
h = (100 – h√ 3) √3 [From eq. (i)]
h = 100 √3 – 3h
4h = 100 √3
h = 25√3
∴ The height of each pole is 25 √3 meter.
Hence, option A is correct.

Given, AB = 20 m
Let, the height of the tower = h metre
And, BC = x metre
∴ AC = AB + BC = (20 + x) m
In ΔACD,

h = (h √3 – 20) √3 [From eq. (i)]
h = 3h – 20 √3
2h = 20 √3
h = 10 √3
∴ The height of the tower is 10 √3 meter.
Hence, option C is correct.

Given, the height of the tree, CD = h metre
Let, the height of the building, AB = H metre
And, BC = a metre
∴ AE = AB – EB = (H – h) metre [∵ CD = BE]
In ΔABC,

a = H cot y ...(i)
Now, in ΔADE,

a = (H – h) cot x ...(ii)
From equations (i) and (ii),
H cot y = (H – h) cot x = H cot x – h cot x
H(cot x – cot y) = h cot x

∴ The height of the building

Hence, option C is correct.

Given, AB = 16 m, CD = 9 m and BC = x metre
And, ∠ACB and ∠CBD are complementary.
∴ Let, ∠ACB = Θ and ∠CBD = (90° – Θ)
In ΔABC,

Now, In ΔBCD,

[∵ tan (90° – Θ) = cot Θ]
By multiplying eq. (i) & (ii),

x = 12 m
Hence, option C is correct.