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RD Sharma Class 10 Solutions Chapter 12 Heights and Distances MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 12 Heights and Distances MCQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 12 Heights and Distances Ex 12.1
• RD Sharma Class 10 Solutions Chapter 12 Heights and Distances VSAQS
• RD Sharma Class 10 Solutions Chapter 12 Heights and Distances MCQS

Mark the correct alternative in each of the following :
Question 1.
The ratio of the length of a rod and its shadow is 1 : \(\sqrt { 3 } \) The angle of elevation of the sum is
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
Let AB be rod and BC be its shadow
So that AB : BC = 1 : \(\sqrt { 3 } \)
Let θ be the angle of elevation

Question 2.
If the angle of elevation of a tower from a distance of 100 metres from its foot is 60?, the height of the tower is

Solution:
Let AB be the tower and a point P at a distance of 100 m from its foot, angle of elevation of the top of the tower is 60°

Let height of the tower = h

Question 3.
If the altitude of the sun is at 60?, then the height of the vertical tower that will cast a shadow of length 30 m is

Solution:
Let AB be tower and a point P distance of 30 m from its foot of the tower which form an angle of elevation pf the sun of 60°

Question 4.
If the angles of elevation of a tower from two points distance a and b (a > b) from its foot and in the same straight line from it are 30? and 60?, then the height of the tower is

Solution:
Let AB be the tower and P and Q are such points that PB = a, QB = b and angles of elevation at P and Q are 30° and 60° respectively

Question 5.
If the anglr of elevation of the top of a tower from two points distant a and b from the base and in the same straight line with It are complementary, then the height of the tower is

Solution:
Let AB be the tower and P and Q are two points such that PB = a and QB = b and angles of elevation are θ and (90° – θ)
Let height of tower = h

Question 6.
From a light house the angles of depression of two ships on opposite sides of the light house are observed to be 30° and 45°. If the height of the light house is h metres, the distance between the ships is

Solution:
Let AB be light house and P and Q are two ships on its opposite sides which form angle of elevation of A as 45° and 30° respectively AB = h
Let PB = x and QB = y

Question 7.
The angle of elevation of the top of a tower stahding on a horizontal plane from a point A is a. After walking a distance d towards the foot of the tower the angle of elevation is found to be p. The height of the tower is

Solution:
Let AB be the tower and C is a point such that the angle of elevation of A is a. After walking towards the foot B of the tower, at D the angle of elevation is p Let h be the height of the tower and DB = x Now in right AACB,

Question 8.
The tops of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of 30° with horizontal, then the length of the wire is
(a) 12 m
(b) 10 in
(c) 8 m
(d) 6 m
Solution:
Let AB and CD be two poles
AB = 20 m, CD = 14 m
A and C are joined by a wire
CE || DB and angle of elevation of A is 30° Class 10 Solutions Chapter 12 Heights and Distances MCQS – 9.png
Let CE = DB = x and AC = l

Question 9.
From the top of a cliff 25 m high the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. The height of the tower is
(a) 25 m
(b) 50 in
(c) 75 m
(d) 100 m
Solution:
Let AB be the tower and CD be cliff Angle of elevation of A is equal to the angle of depression of B at C
Let angle be Q and CD = 25 m

Question 10.
The angles of depression of two ships from the top of a light house are 45° and 30° towards east. If the ships are 100 m apart, the height of the light house is

Solution:
Let AB be the light house C and D are two ships whose angles of depression on A are 30° and 45° respectively

Question 11.
if the angle of elevation of a cloud from a point 200 m above a lake is 30° and the angle of depression of its reflection in the lake is 60°, then the height of the cloud above the lake, is
(a) 200 m
(b) 500 m
(c) 30 m
(d) 400 m
Solution:
Let C be the cloud and R is its reflection in the lake
L is a point 200 m above the lake. Such that
LM = 200 m
Angle of elevation of C with L is 30° and angle of depression of R is 60°

Let height of cloud CB = h
∴ BR = h and NB = LM = 200 m
∴ CN = (h – 200) m and NR = (h + 200) m

Question 12.
The height of a tower is 100 m. When the angle of elevation of the sun changes from 30° to 45°, the shadow of the tower becomes x metres less. The value of x is

Solution:
Let AB be tower and AB = 100 m and angles of elevation of A at C and D are 30° and 45° respectively and CD = x
Let BD =y

Question 13.
Two persons are a metres apart and the height of one is double that of the other. If from the middle point of the line joining their feet, an observer finds the angular elevation of their tops to be complementary, then the height of the shorter post is

Solution:
Let AB and CD are two persons standing ‘a’ metres apart
M is the mid-point of BD and from M, the angles of elevation of A and C are complementary

Question 14.
The angle of elevation of a cloud from a point h metre above a lake is θ. The angle of depression of its reflection in the lake is 45°. The height of the cloud is
(a) It tan (45° + θ)
(b) h cot (45° – θ)
(c) h tan (45° – θ)
(d) h cot (45° + θ)
Solution:
Let C is the cloud and R is its reflection in the lake
From the lake, ‘7’ m aboves it, E is point
where angle of elevation of C is θ
and angle of depression of reflection is 45°

Question 15.
A lower subtends an angle of 30° at a point on the same level as its foot. At a second point h metres above the first, the depression of the foot of the tower is 60°. The height of the tower is

Solution:
Let CD is the tower and A is a point such that the angle of elevation of C is 30°
B is and their point h m high of A and angle of depressio of D is 60°

Question 16.
It is found that on walking x metres towards a chimney in a horizontal line through its base, the elevation of its top changes from 30° to 60° . The height of the chimney is

Solution:
In the figure, AB is chimney and CB and DB are its shadow

Question 17.
The length of the shadow of a tower standing on level ground is found to be 2.v metres longer when the sun’s elevation is 30° than when it was 45° . The height of the tower in metres is

Solution:
AB is a tower
BD and BC are its shadows and CD = 2x

Question 18.
Two poles are ‘a’ metres apart and the height of one is double of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the smaller is


Solution:
Let height of pole CD = h
and AB = 2h, BD = a
M is mid-point of BD

Question 19.
The tops of two poles of height 16 m and 10 m are connected by a wire of length l metres. If the wire makes an angle of 30° with the horizontal, then l =
(a) 26
(b) 16
(c) 12
(d) 10
Solution:
Let AB and CD are two poles AB = 10 m and CD = 16 m

Question 20.
If a 1.5 m tall girl stands at a distance of 3 m from a lamp-post and casts a shadow of length 4.5 m on the ground, then the height of the lamp-post is
(a) 1.5 m
(b) 2 m
(c) 2.5 m
(d) 2.8 m
Solution:
Let AB is girls and CD is lamp-post AB = 1.5
which casts her shadow EB

Question 21.
The length of shadow of a tower on the plane ground is \(\sqrt { 3 } \) times the height of the tower. The angle of elevation of sun is
(a) 45°
(b) 30°
(c) 60°
(d) 90° [CBSE 2012]
Solution:
Let AB be tower and BC be its shadow
∴Let AB = x

Question 22.
The angle of depression of a car, standing on the ground, from the top of a 75 m tower, is 30°. The distance of the car from the base of the tower (in metres) is
(a) 25 \(\sqrt { 3 } \)
(b) 50 \(\sqrt { 3 } \)
(c) 75 \(\sqrt { 3 } \)
(d) 150 [CBSE 2013]
Solution:
AB is a tower and AB = 75 m
From A, the angle of depression of a car C
on the ground is 30°

Question 23.
A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, then the height of the wall is

Solution:
Let AB is a wall and AC is the ladder 15 m long which makes an angle of 60° with the ground
∴In ∆ABC, ∠B = 90°
Let height of wall AB = h

Question 24.
The angle of depression of a car parked on the road from the top of a 150 m high tower is 30°. The distance of the car from the tower (in metres) is
(a) 50\(\sqrt { 3 } \)
(b) 150 \(\sqrt { 3 } \)
(c) 150 \(\sqrt { 3 } \)
(d) 75
Solution:
Let AB be the tower of height 150 m
C is car and angle of depression is 30°

Question 25.
If the hei8ht of a vertical pole is \(\sqrt { 3 } \) times the length of its shadow on the ground, then the angle of elevation of the sun at that time is
(a) 30°
(b) 60°
(c) 45°
(d) 75° [CBSE 2014]
Solution:
Let AB be a vertical pole and let its shadow be BC

Question 26.
The angle of elevation of the top of a tower at a point on the ground 50 m away from the foot of the tower is 45°. Then the height of the tower (in metres) is

Solution:
Let AB be tower and C is a point on the ground 50 m away

Question 27.
A ladder makes an angle of 60° with the ground when placed against a wall. If the foot of the ladder is 2 m away from the wall, then the length of the ladder (in metres) is

Solution:
Suppose AB is the ladder of length x m
∴ OA = 2m, ∠OAB = 60°

Hope given RD Sharma Class 10 Solutions Chapter 12 Heights and Distances MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.7

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.7

Other Exercises

• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.2
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.3
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.7
• RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise
• RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS
• RD Sharma Class 10 Solutions Chapter 7 Triangles MCQS

Question 1.
If the sides of a triangle are 3 cm, 4 cm and 6 cm long, determine whether the triangle is a right-angled triangle. (C.B.S.E. 1992)
Solution:
We know that if the square of the hypotenuse (longest side) is equal to the sum of squares of other two sides then it is right triangle
Now the sides of a triangle are 3 cm, 4 cm and 6 cm
(Longest side)² = (6)² = 36
and sum of two smaller sides = (3)² + (4)² = 9 + 16 = 25
36 ≠ 25
It is not a right-angled triangle

Question 2.
The sides of certain triangles are given below. Determine which of them are right triangles :
(i) a = 1 cm, b = 24 cm and c = 25 cm
(ii) a = 9 cm, b = 16 cm and c = 18 cm
(iii) a = 1.6 cm, b = 3.8 cm and c = 4 cm
(iv) a = 8 cm, b = 10 cm and c = 6 cm (C.B.S.E. 1992)
Solution:
We know that if the square of hypotenuse is equal to the sum of squares of other two sides, then it is a right triangle
(i) Sides of a triangle are a = 7 cm, b = 5.24 cm and c = 25 cm
(Longest side)² = (25)² = 625
Sum of square of shorter sides = (7)² + (24)² = 49 + 576 = 625
625 = 625
This is right triangle
(ii) Sides of the triangle are a = 9 cm, b = 16 cm, c = 18 cm
(Longest side)² = (18)² = 324
and sum of squares of shorter sides = (9)² + (16)² = 81 + 256 = 337
324 ≠ 337
It is not a right-angled triangle
(iii) Sides of the triangle are a = 1.6 cm, 6 = 3.8 cm, c = 4 cm
(Longest side)² = (4)² =16
Sum of squares of shorter two sides + (1.6)² + (3.8)² = 2.56 + 14.44 = 17.00
16 ≠ 17
It is not a right triangle
(iv) Sides of the triangle are a = 8 cm, b = 10 cm, c = 6 cm
(Longest side)² = (10)² = 100
Sum of squares of shorter sides = (8)² + (6)² = 64 + 36 = 100
100 = 100
It is a right triangle

Question 3.
A man goes 15 metres due west and then 8 metres due north. How far is he from the starting point ?
Solution:
Let a man starts from O, the starting point to west 15 m at A and then from A, 8 m due north at B
Join OB

Now in right ∆OAB
OB² = OA² + AB² (Pythagoras Theorem)
OB² = (15)² + (8)² = 225 + 64 = 289 = (17)²
OB = 17
The man is 17 m away from the starting point

Question 4.
A ladder 17 m long reaches a window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.
Solution:
Length of ladder = 17 m
Height of window = 15 m

Let the distance of the foot of ladder from the building = x
Using Pythagoras Theorem
AC² = AB² + BC²
=> (17)² = (15)² + x²
=> 289 = 225 + x²
=> x² = 289 – 225
=> x² = 64 = (8)²
x = 8
Distance of the foot of the ladder from the building = 8m

Question 5.
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops. (C.B.S.E. 1996C, 2002C)
Solution:
Two poles AB and CD which are 6 m and 11 m long respectively are standing oh the ground 12 m apart

Draw AE || BD so that AE = BD = 12 m and ED = AB = 6 m
Then CE = CD – ED = 11 – 6 = 5 m
Now in right ∆ACE
Using Pythagoras Theorem,
AC² = AE² + EC² = (12)² + (5)² = 144 + 25 = 169 = (13)²
AC = 13
Distance between their tops = 13 m

Question 6.
In an isosceles triangle ABC, AB = AC = 25 cm, BC = 14 cm. Calculate the altitude from A on BC. (C.B.S.E. 1994)
Solution:
∆ABC is an isosceles triangle in which AB = AC = 25 cm .
AD ⊥ BC BC = 14 cm

Perpendicular AD bisects the base i.e . BD = DC = 7 cm
Let perpendicular AD = x
In right ∆ABD,
AB² = AD² + BD² (Pythagoras Theorem)
=> (25)² = AD² + (7)²
=> 625 = AD² + 49
=> AD² = 625 – 49
=> AD² = 576 = (24)²
=> AD = 24
Perpendicular AD = 24 cm

Question 7.
The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach ?
Solution:
In first case,
The foot of the ladder are 6 m away from the wall and its top reaches window 8 m high
Let AC be ladder and BC = 6 m, AB = 8 m

Now in right ∆ABC,
Using Pythagoras Theorem
AC² = BC² + AB² = (6)² + (8)² = 36 + 64 = 100 = (10)²
AC = 10 m
In second case,
ED = AC = 10 m
BD = 8 m, let ED = x
ED² = BD² + EB²
=> (10)² = (8)² + x²
=> 100 = 64 + x²
=> x² = 100 – 64 = 36 = (6)²
x = 6
Height of the ladder on the wall = 6 m

Question 8.
Two poles of height 9 m and 14 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
Solution:
Let CD and AB be two poles which are 12 m apart
AB = 14 m, CD = 9 m and BD = 12 m
From C, draw CE || DB
CB = DB = 12 m
EB = CD = 9 m
and AE = 14 – 9 = 5 m

Now in right ∆ACE,
AC² = AE² + CE² (Pythagoras Theorem)
= (5)² + (12)²
= 25 + 144 = 169 = (13)²
AC = 13
Distance between their tops = 13 m

Question 9.
Using Pythagoras theorem, determine the length of AD in terms of b and c shown in the figure. (C.B.S.E. 1997C)

Solution:
In right ∆ABC, ∠A = 90°
AB = c, AC = b
AD ⊥ BC

Question 10.
A triangle has sides 5 cm, 12 cm and 13 cm. Find the length to one decimal place, of the perpendicular from the opposite vertex to the side whose length is 13 cm. (C.B.S.E. 1992C)
Solution:
A triangle has sides 5 cm, 12 cm and 13 cm
(Longest side)² = (13)² = 169
Sum of squares of shorter sides = (5)² + (12)² = 25 + 144= 169
169 = 169
It is a right triangle whose hypotenuse is 13 cm

BD = 4.6 cm

Question 11.
ABCD is a square, F is the mid point of AB. BE is one third of BC. If the area of ∆FBE = 108 cm² find the length of AC. (C.B.S.E. 1995)
Solution:
In square ABCD, F is mid point of AB i.e.,

Question 12.
In an isosceles triangle ABC, if AB = AC = 13 cm and the altitude from A on BC is 5 cm, find BC. (C.B.S.E. 2000)
Solution:
In ∆ABC, AB = AC
AD ⊥ BC
AB = AC = 13 cm, AD = 5 cm
AD ⊥ BC
AD bisects BC at D
BD = \(\frac { 1 }{ 2 }\) BC
=> BC = 2BD

Question 13.
In a ∆ABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that
(i) AD = a √3
(ii) area (∆ABC) = √3 a² (C.B.S.E. 1991)
Solution:
In ∆ABC, AB = BC = AC = 2a
AD ⊥ BC
AD bisects BC at D
BD = DC = \(\frac { 1 }{ 2 }\) BC = a

Question 14.
The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus. (C.B.S.E. 1993C)
Solution:
ABCD is a rhombus whose diagonals AC = 24 cm and BD = 10 cm
The diagonals of a rhombus bisect eachother at right angles

AO = OC = \(\frac { 24 }{ 2 }\) = 12 cm
and BO = OD = \(\frac { 10 }{ 2 }\) = 5 cm
Now in right ∆AOB,
AB² = AO² + BO² (Pythagoras Theorem)
(12)² + (5)² = 144 + 25 = 169 = (13)²
AB = 13
Each side of rhombus = 13 cm

Question 15.
Each side of a rhombus is 10 cm. If one of its diagonals is 16 cm, find the length of the other diagonal.
Solution:
In rhombus ABCD, diagonals AC and BD bisect eachother at O at right angles
Each side = 10 cm and one diagonal AC = 16 cm

AO = OC = \(\frac { 16 }{ 2 }\) = 8 cm
Now in right angled triangle AOB,
AB² = AO² + OB² (Pythagoras Theorem)
(10)² = (8)² + (BO)²
=> 100 = 64 + BO²
=> BO² = 100 – 64 = 36 = (6)²
BO = 6
BD = 2BO = 2 x 6 = 12 cm

Question 16.
Calculate the height of an equilateral triangle each of whose sides measures 12 cm.
Solution:
Each side of the equilateral ∆ABC = 12 cm
AD ⊥ BC which bisects BC at D


BD = DC = \(\frac { 12 }{ 2 }\) = 6 cm

Question 17.
In the figure, ∠B < 90° and segment AD ⊥ BC. Show that:
(i) b² = h² + a² + x² – 2ax
(ii) b² = a² + c² – 2ax

Solution:
Given : In ∆ABC, ∠B < 90°
AD ⊥ BC
AD = c, BC = a, CA = b AD = h, BD = x, DC = a – x
To prove: (i) b² = h² + a² + x² – 2ax
(ii) b² = a² + c² – 2ax
Proof: (i) In right ∆ADC, AC² = AD² + DC² (Pythagoras Theorem)
=> b² = h² + (a – x)² = h² + a² + x² – 2ax
(ii) Similarly in right ∆ADB
AB² = AD² + BD²
c² = h² + x² ….(i)
b² = h² + a² + x² – 2ax = h² + x² + a² – 2ax
= c² + a² – 2ax {From (i)}
= a² + c² – 2ax
Hence proved.

Question 18.
In an equilateral ∆ABC, AD ⊥ BC, prove that AD² = 3 BD². (C.B.S.E. 2002C)
Solution:
Given : ∆ABC is an equilateral in which AB = BC = CA .
AD ⊥ BC

Question 19.
∆ABD is a right triangle right-angled at A and AC ⊥ BD. Show that
(i) AB² = BC.BD
(ii) AC² = BC.DC
(iii) AD² = BD.CD
(iv) \(\frac { { AB }^{ 2 } }{ { AC }^{ 2 } }\) = \(\frac { BD }{ DC }\) [NCERT]
Solution:
In ∆ABD, ∠A = 90°
AC ⊥ BD

Question 20.
A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut ?
Solution:
Let AB be the vertical pole whose height = 18 m

Question 21.
Determine whether the triangle having sides (a – 1) cm, 2 √a cm and (a + 1) cm is a right angled triangle. [CBSE 2010]
Solution:
Sides of a triangle are (a – 1) cm, 2 √a cm and (a + 1) cm
Let AB = (a – 1) cm BC = (a + 1) cm
and AC = 2 √a

Question 22.
In an acute-angled triangle, express a median in terms of its sides.
Solution:
In acute angled ∆ABC,
AD is median and AL ⊥ BC
Using result of theorem, sum of the squares of any two sides is equal to the twice the square of half of the third side together with twice the square of the median which bisects the third side

Question 23.
In right-angled triangle ABC in which ∠C = 90°, if D is the mid point of BC, prove that AB² = 4AD² – 3AC².
Solution:
Given : In right ∆ABC, ∠C = 90°
D is mid point of BC

Question 24.
In the figure, D is the mid-point of side BC and AE ⊥ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that :


Solution:

Question 25.
In ∆ABC, ∠A is obtuse, PB x AC and QC x AB. Prove that:
(i) AB x AQ = AC x AP
(ii) BC² = (AC x CP + AB x BQ)
Solution:
Given : In ∆ABC, ∠A is an obtuse angle PB x AC and QC x AB on producing them

Question 26.
In a right ∆ABC right-angled at C, if D is the mid-point of BC, prove that BC² = 4 (AD² – AC²).
Solution:
Given : In ∆ABC, ∠C = 90°
D is mid-point of BC
AD is joined

=> 4AD² = 4AC² + BC²
=> BC² = 4AD² – 4AC²
=> BC² = 4 (AD² – AC²)
Hence proved.

Question 27.
In a quadrilateral ABCD, ∠B = 90°, AD² = AB² + BC² + CD², prove that ∠ACD = 90°.
Solution:
Given : ABCD is a quadrilateral in which ∠B = 90° and AD² = AB² + BC² + CD²
To prove : ∠ACD = 90°
Construction : Join AC

proof : In right ∆ABC, ∠B = 90°
AC² = AB² + BC² ….(i) (Pythagoras Theorem)
But AD² = AB² + BC² + CD² (given)
=> AD² = AC² + CD² {From (i)}
∆ACD is a right angle with right angle ACD
Hence proved.

Question 28.
An aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/ hr. How far apart will be the two planes after 1\(\frac { 1 }{ 2 }\) hours?
Solution:

Speed of the first plane = 1000 km/hr
Distance travelled in 1\(\frac { 1 }{ 2 }\) hour due north = 1000 x \(\frac { 3 }{ 2 }\) = 1500 km
Speed of the second plane = 1200 km/hr
Distance travelled in 1\(\frac { 1 }{ 2 }\) hours due west

Hope given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.7 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 12 Heights and Distances VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 12 Heights and Distances VSAQS

Other Exercises

• RD Sharma Class 10 Solutions Chapter 12 Heights and Distances Ex 12.1
• RD Sharma Class 10 Solutions Chapter 12 Heights and Distances VSAQS
• RD Sharma Class 10 Solutions Chapter 12 Heights and Distances MCQS

Answer each of the following questions either in one word or one sentence or per requirement of the questions :
Question 1.
The height of a tower is 10 m. What is the length of its shadow when Sun’s altitude is 45° ?
Solution:
Let AB be the tower and BC is its shadow
Height of AB = 10 m
Let length of BC = x
Angle of elevation of A at C is 45°

Question 2.
If the ratio of the height of a angle tower and the length of 7 its shadow is \(\sqrt { 3 } \) : 1, what is the of elevation of the Sun?
Solution:
Ratio in the height of a tower and its shadow = \(\sqrt { 3 } \) : 1

Question 3.
What is the angle of elevation of the Sun when the length of the shadow of a vertical pole is equal to its height ?
Solution:
Let height of a vertical pole = h
Then its shadow = x

Question 4.
From a point on the ground, 20 m away from the foot of a vertical tower, the angle of elevation of the top of the tower is 60°, what is the height of the tower ?
Solution:
Let AB be the tower and the angles of elevation of its top from a point C on the ground is 60° and BC = 20 m
Let height of the tower AB = h

Question 5.
If the angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line, with it are complementary, find the height of the tower.
Solution:
Let CD be the tower and let the angles of elevation of the top C of the tower from two points A and B are θ and (90° – θ) as the angles are complementary

Let height of the tower CD = h
and AD = 9 m, BD = 4 m

Question 6.
In the figure, what are the angles of depression from the observing positions O₁ and O₂ of the object at A ?
Solution:
In the given figure, draw XY || ABC from O₁,O₂ (by joinging them)
∵∠AO₁C = 60°
∠AO₁O₂ = 90° – 60° = 30°
Similarly,
∵∠O₂ AB = 45°
∠XO₂A = ∠O₂AB = 45° (alternate angles)
Hence angles of depression are 30° and 45°

Question 7.
The tops of two towers of height x and y, standing on level ground, subtend angles of 30° and 60° respectively at the centre of the line joining their feet, then find x : y. [CBSE2015]
Solution:
Let AB and CD be two towers which are apart from each other at a distance of BD and M is mid point of BD.
Angles of elevation are 30° and 60°

Question 8.
The angle of elevation of the top of a tower at a point on the ground is 30°. What will be the angle of elevation, if the height of the tower is tripled?     [CBSE 2015]
Solution:
Let AB be the tower and apgle of elevation of A at C is 30°

Question 9.
AB is a pole of height 6 m standing at a point B and CD is a ladder inclined at angle of 60° to the horizontal and reaches upto a point D of pole. If AD = 2.54 m, find the length of the ladder. (Use \(\sqrt { 3 } \)  = 1.73)       [CBSE 2016]
Solution:
BD = AB – AD = 6 – 2.54 = 3.46 m

Question 10.
An observer, 1.7 m tall, is 20\(\sqrt { 3 } \) m away from a tower. The angle of elevation from the eye of an observer to the top of tower is 30°. Find the height of the tower. [CBSE 2016]
Solution:

Question 11.
An observer, 1.5 m tall, is 28.5 m away from a 30 m high tower. Determine the angle of elevation of the top-of the tower from the eye of the observer. [CBSE 2017]
Solution:

Here, AE = 28.5 m
ED = BC = 28.5 m
In ΔADE,
tan ∠ADE = \(\frac { 28.5 }{ 28.5 }\)
= ∠ADE = tan^-1(1) – 45°
∴ The angle of elevation at the top of the tower from the eye of the observer is 45°.

Hope given RD Sharma Class 10 Solutions Chapter 12 Heights and Distances VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6

Other Exercises

• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.2
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.3
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6
• RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.7
• RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise
• RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS
• RD Sharma Class 10 Solutions Chapter 7 Triangles MCQS

Question 1.
Triangles ABC and DEF are similar.
(i) If area (∆ABC) = 16 cm², area (∆DEF) = 25 cm² and BC = 2.3 cm, find EF. (C.B.S.E. 1992)
(ii) If area (∆ABC) = 9 cm², area (∆DEF) = 64 cm² and DE = 5.1 cm, find AB.
(iii) If AC = 19 cm and DF = 8 cm, find the ratio of the area of two triangles. (C.B.S.E. 1992C)
(iv) If area (∆ABC) = 36 cm², area (∆DEF) = 64 cm² and DE = 6.2 cm, find AB. (C.B.S.E. 1992)
(v) If AB = 1.2 cm and DE = 1.4 cm, find the ratio of the areas of ∆ABC and ∆DEF. (C.B.S.E. 1991C)
Solution:

Question 2.
In the figure, ∆ACB ~ ∆APQ. If BC = 10 cm, PQ = 5 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ. Also, find the area (∆ACB) : area (∆APQ).

Solution:

Question 3.
The areas of two similar triangles are 81 cm² and 49 cm² respectively. Find the ratio of their corresponding heights, what is the ratio of their corresponding medians ?
Solution:
Areas of two similar triangles are 81 cm² and 49 cm²
The ratio of the areas of two similar triangles are proportion to the square of their corresponding altitudes and also squares of their corresponding medians
Ratio in their altitudes = √81 : √49 = 9 : 7
Similarly, the ratio in their medians = √81 : √49 = 9 : 7

Question 4.
The areas of two similar triangles are 169 cm² and 121 cm² respectively. If the longest side of the larger triangle is 26 cm, find the longest side of the smaller triangle.
Solution:
Triangles are similar Area of larger triangle = 169 cm²
and area of the smaller triangle =121 cm²
Length of longest sides of the larger triangles = 26 cm
Let the length of longest side of the smaller triangle = x

Question 5.
The areas of two similar triangles are 25 cm² and 36 cm² respectively. If the altitude of the first triangle is 2.4 cm, find the corresponding altitude of the other.
Solution:
Area of first triangle = 25 cm²
Area of second = 36 cm²
Altitude of the first triangle = 2.4 cm
Let altitude of the second triangle = x
The triangles are similar

Question 6.
The corresponding altitudes of two similar triangles are 6 cm and 9 cm respectively. Find the ratio of their areas.
Solution:
Length of the corresponding altitude of two triangles are 6 cm and 9 cm
triangles are similar

Question 7.
ABC is a triangle in which ∠A = 90°, AN ⊥ BC, BC = 12 cm and AC = 5 cm. Find the ratio of the areas of the ∆ANC and ∆ABC.
Solution:
In ∆ABC, ∠A = 90°
AN ⊥ BC
BC = 12 cm, AC = 5 cm

Question 8.
In the figure, DE || BC
(i) If DE = 4 cm, BC = 6 cm and area (∆ADE) = 16 cm², find the area of ∆ABC.
(ii) If DE = 4 cm, BC = 8 cm and area of (∆ADE) = 25 cm², find the area of ∆ABC. (C.B.S.E. 1991)
(iii) If DE : BC = 3 : 5, calculate the ratio of the areas of ∆ADE and the trapezium BCED

Solution:

Question 9.
In ∆ABC, D and E are the mid-points of AB and AC respectively. Find the ratio of the areas of ∆ADE and ∆ABC.
Solution:
In ∆ABC, D and E are the mid points of AB and AC respectively

Question 10.
The areas of two similar triangles are 100 cm² and 49 cm² respectively. If the altitude of the bigger triangle is 5 cm, find the corresponding altitude of the other. (C.B.S.E. 2002)
Solution:
∆ABC ~ ∆DEF
area ∆ABC = 100 cm²
and area ∆DEF = 49 cm²

Question 11.
The areas of two similar triangles are 121 cm² and 64 cm² respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other. (C.B.S.E. 2001)
Solution:
∆ABC ~ ∆DEF
area of ∆ABC = 121 cm² area of ∆DEF = 64 cm²
AL and DM are the medians of ∆ABC and ∆DEF respectively
AL = 12.1 cm

Question 12.
In ∆ABC ~ ∆DEF such that AB = 5 cm and (∆ABC) = 20 cm² and area (∆DEF) = 45 cm², determine DE.
Solution:
∆ABC ~ ∆DEF
area (∆ABC) = 20 cm²
area (∆DEF) = 45 cm²

Question 13.
In ∆ABC, PQ is a line segment intersecting AB at P and AC at Q such that PQ || BC and PQ divides ∆ABC into two parts equal in area. Find \(\frac { BP }{ AB }\).
Solution:
In ∆ABC, PQ || BC and PQ divides ∆ABC in two parts ∆APQ and trap. BPQC of equal in area
i.e., area ∆APQ = area BPQC

Question 14.
The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR. (C.B.S.E. 2004)
Solution:
∆ABC ~ ∆PQR
area (∆ABC) : area (∆PQR) = 9 : 16
and BC = 4.5 cm

Question 15.
ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of ∆APQ is one sixteenth of the area of ∆ABC. (C.B.S.E. 2005)
Solution:
In ∆ABC, P and Q are two points on AB and AC respectively such that
AP = 1 cm, PB = 3 cm, AQ = 1.5 cm and QC = 4.5 cm

Question 16.
If D is a point on the side AB of ∆ABC such that AD : DB = 3 : 2 and E is a point on BC such that DE || AC. Find the ratio of areas of ∆ABC and ∆BDE. (C.B.S.E. 2006C)
Solution:
In ∆ABC, D is a point on AB such that AD : DB = 3 : 2

Question 17.
If ∆ABC and ∆BDE are equilateral triangles, where D is the mid point of BC, find the ratio of areas of ∆ABC and ∆BDE. [CBSE 2010]
Solution:
∆ABC and ∆DBE are equilateral triangles Where D is mid point of BC

Question 18.
Two isosceles triangles have equal vertical angles and their areas are in the ratio 36 : 25. Find the ratio of their corresponding heights.
Solution:
Two isosceles triangles have equal vertical angles
So their base angles will also be the equal to each other
Triangles will be similar Now, ratio in their areas = 36 : 25

Question 19.
In the figure, ∆ABC and ∆DBC are on the same base BC. If AD and BC intersect

Solution:
Given : Two ∆ABC and ∆DBC are on the same base BC as shown in the figure
AC and BD intersect eachother at O

Question 20.
ABCD is a trapezium in which AB || CD. The diagonals AC and BD intersect at O. Prove that
(i) ∆AOB ~ ∆COD
(ii) If OA = 6 cm, OC = 8 cm, find

Solution:
Given : ABCD is a trapezium in which AB || CD
Diagonals AC and BD intersect each other at O

Question 21.
In ∆ABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ || BC. Find the ratio of the areas of ∆APQ and trapezium BPQC.
Solution:
In ∆ABC, P is a point on AB such that AP : PQ = 1 : 2
PQ || BC
Now we have to find the ratio between area ∆APQ and area trap BPQC

Question 22.
AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed. Prove that Area (∆ADE) : Area (∆ABC) = 3 : 4. [CBSE 2010]
Solution:
Given: In equilateral ∆ABC, AD ⊥ BC and with base AD, another equilateral ∆ADE is constructed

Hope given RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.2

Other Exercises

• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.1
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.2
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.3
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios VSAQS
• RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios MCQS

Evaluate each of the following (1-19) : 1.
Question 1.
sin45° sin30° + cos45° cos30°.
Solution:
sin 45° sin 30° + cos 45° cos 30°

Question 2.
sin60° cos30° + cos60° sin30°.
Solution:

Question 3.
cos60° cos45° – sin60° sin45°.
Solution:

Question 4.
sin²30° + sin²45° + sin²60° + sin²290°.
Solution:

Question 5.
cos²30° + cos²45° + cos²60° + cos²90°.
Solution:

Question 6.
tan²30° + tan²60° + tan²45°.
Solution:

Question 7.
2sin²30° – 3cos²45° + tan²60°.
Solution:

Question 8.
sin²30°cos²4S° + 4tan²30° + \(\frac { 1 }{ 2 }\) sin²90° -2cos²90° + \(\frac { 1 }{ 24 }\) cos20°.
Solution:

Question 9.
4 (sin⁴ 60° + cos⁴ 30°) – 3 (tan² 60° – tan² 45°) + 5cos² 45°
Solution:

Question 10.
(cosecc² 45° sec² 30°) (sin² 30° + 4cot² 45° – sec² 60°).
Solution:

Question 11.
cosec³ 30° cos 60° tan³ 45° sin2 90° sec² 45° cot 30°.
Solution:

Question 12.
cot² 30° – 2cocs² 60° – \(\frac { 3 }{ 4 }\)sec2 45° – 4sec² 30°.
Solution:

Question 13.
(cos0° + sin45° + sin30°) (sin90° – cos45° + cos60°)
Solution:

Question 14.

Solution:

Question 15.

Solution:

Question 16.
4 (sin⁴ 30° + cos² 60°) – 3 (cos² 45° – sin² 90°) – sin² 60°
Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Find the value of x in each of the following : (20-25)

Question 20.
2sin 3x = √3
Solution:

Question 21.
2sin \(\frac { x }{ 2 }\) = 1
Solution:

Question 22.
√3 sin x=cos x
Solution:

Question 23.
tan x = sin 45° cos 45° + sin 30°
Solution:

Question 24.
√3 tan 2x = cos 60° +sin 45° cos 45°
Solution:

Question 25.
cos 2x = cos 60° cos 30° + sin 60° sin 30°
Solution:

Question 26.
If θ = 30°, verify that :

Solution:

Question 27.
If A = B = 60°, verify that:
(i) cos (A – B) = cos A cos B + sin A sin B
(ii) sin (A – B) = sin A cos B – cos A sin B tan A – tan B
(iii) tan (A – B) = \(\frac { tanA-tanB }{ 1+tanA-tanB }\)
Solution:

Question 28.
If A = 30° and B = 60°, verify that :
(i) sin (A + B) = sin A cos B + cos A sin B
(ii) cos (A + B) = cos A cos B – sin A sin B
Solution:

Question 29.
If sin (A + B) = 1 and cos (A,-B) = 1,0° < A + B < 90°, A > B find A and B.
Solution:

Question 30.

Solution:

Question 31.

Solution:

Question 32.
In a ∆ABC right angle at B, ∠A = ∠C. Find the values of
(i) sin A cos C + cos A sin C
(ii) sin A sin B + cos A cos B
Solution:

Question 33.
Find acute angles A and B, if sin (A + 2B)=\(\frac { \sqrt { 3 } }{ 2 }\) and cos (A + 4B) = 0, A > B.
Solution:

Question 34.
In ΔPQR, right-angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠P and ∠R.
Solution:

Question 35.
If sin (A – B) = sin A cos B – cos A sin B and cos (A – B) = cos A cos B + sin A sin B, find the values of sin 15° and cos 15°.
Solution:

Question 36.
In a right triangle ABC, right angled at ∠C if ∠B = 60° and AB – 15 units. Find the remaining angles and sides.
Solution:

Question 37.
If ΔABC is a right triangle such that ∠C = 90°, ∠A = 45° and BC = 7 units. Find ∠B, AB and AC.
Solution:

Question 38.
In a rectangle ABCD, AB = 20 cm, ∠BAC = 60°, calculate side BC and diagonals AC and BD.
Solution:

Question 39.
If A and B are acute angles such that

Solution:

Question 40.
Prove that (√3 + 1) (3 – cot 30°) = tan³ 60° – 2sin 60°. [NCERT Exemplar]
Solution:

Hope given RD Sharma Class 10 Solutions Chapter 10 Trigonometric Ratios Ex 10.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.