Rd Sharma Solutions Class 7 Chapter 15 Properties of Triangles
Welcome to NCTB Solution. Here with this post we are going to help 7th class students for the Solutions of Rd Sharma Class 7 Mathematics, Chapter 15, Properties of Triangles. Here students can easily find Exercise wise solution for chapter 15, Properties of Triangles. Students will find proper solutions for Exercise 15.1, 15.2, 15.3, 15.4 and 15.5. Our teachers solved every problem with easily understandable methods so that every students can understand easily. Here all solutions are based on the CBSE latest curriculum.
Properties of Triangles Exercise 15.1 Solution
Question no – (1)
Solution :
Triangle △ ABC
(i) the side opposite to <B
= AC
(ii) the angle opposite to side AB
= ∠ACB
(iii) the vertex opposite to side BC
= A
(iv) the side opposite to vertex B.
= AC
Question no – (4)
Solution :
All the tingles in the given figure are, △ACD, △ADB, and △ABC.
Question no – (5)
Solution :
In the given figure all the triangles are,
△AOB, △AOD, △DOC, △BOC,
△ABD, △BCD, △ADC and △ABC.
Question no – (7)
Solution :
(i) Triangle : A plane figure formed by three non-parallel line segments is called a Triangle.
(ii) Parts or elements of a Triangle : The three sides AB, BC, CA and three angles ∠A, ∠B, ∠C of a △ABC are together called six parts or elements of the △ABC.
(iii) Scalene Triangle : A triangle whose no two sides are equal is called a scalene Triangle.
(iv) Isosceles triangle : A triangle whose two sides are equal is called Isosceles triangle.
(v) Equilateral triangle : A triangle whose all sides are equal to one another is called an Equilateral triangle.
(vi) Acute Triangle : A triangle whose all the angles are acute is called an Acute Triangle or Acute-Angled-Triangle.
(vii) Right triangle : A triangle whose one angle is a right angle, is called a Right triangle or Right-Angled-Triangle.
(viii) Obtuse Triangle : A triangle whose one angle is obtuse is called a Obtuse Triangle or Obtuse-Angled-Triangle.
(ix) Interior of a triangle : The part made up by all such points P which are enclosed by △ABC is called the interior of △ABC.
(x) Exterior of a triangle : The part made up by all such points Q which are not enclosed by △ABC is called the Exterior of △ABC.
Question no – (8)
Solution :
Figure – (i)
= Scalene triangle
Figure – (ii)
= Isosceles triangle
Figure – (iii)
= Equilateral triangle
Figure – (iv)
= Scalene triangle
Figure – (v)
= Isosceles triangle.
Question no – (9)
Solution :
Figure – (i)
= Right angle triangle.
Figure – (ii)
= Obtuse triangle.
Figure – (iii)
= Acute triangle.
Figure – (iv)
= Right angle triangle.
Figure – (v)
= Obtuse triangle.
Question no – (10)
Solution :
(i) A triangle has Three sides.
(ii) A triangle has Three vertices.
(iii) A triangle has Three angles.
(iv) A triangle has 6 parts. (3 sides and 3 angles.)
(v) A triangle whose no two sides are equal is known as Scalene.
(vi) A triangle whose two sides are equal is known as Isosceles.
(vii) A triangle whose all the sides are equal is known as Equilateral.
(viii) A triangle whose one angle is a right angle is known as Right angle triangle.
(ix) A triangle whose all the angles are of measure less than 90° is known as Acute triangle.
(x) A triangle whose one angle is more than 90° is known as Obtuse triangle.
Question no – (11)
Solution :
(i) A triangle has three sides – True.
(ii) A triangle may have four vertices – False.
(iii) Any three line segments make up a triangle – False.
(iv) The interior of a triangle includes its vertices – False.
(v) The triangular region includes the vertices of the corresponding triangle – True.
(vi) The vertices of a triangle are three collinear points – False.
(vii) An equilateral triangle is isosceles also – True.
(viii) Every right triangle is scalene – False.
(ix) Each acute triangle is equilateral – False.
(x) No isosceles triangle is obtuse – False.
Properties of Triangles Exercise 15.2 Solution
Question no – (1)
Solution :
As per the question,
Two angles of a triangle are of measures 105° and 30°
∴ The third angle is,
= 180° – (105° + 30°)
= 180° – 135°
= 45°
Thus, the measure of the third angle is 45°
Question no – (2)
Solution :
Let the equal angles one x
∴ 130° + x + x = 180°
= 2x = 180° – 130°
= x = 50/2
= x = 25°
Hence, the measure of each of these equal angles will be 25°
Question no – (3)
Solution :
Let the three equal angles are y
∴ y + y + y = 180°
= 3y = 180°
= y = 180/3
= y = 60°
Thus, the measure of each of the angles is 60°.
Question no – (4)
Solution :
In the given question,
Angles of a triangle are in the ratio 1 : 2 : 3
∴ Let, the angles are x, 2x and 3x,
= x + 2x + 3x = 180
= 6x = 180
= x = 30
∴ All the Angle are,
x = 30°
2 × 30 = 60°
3 × 30 = 90°
Therefore, the three angles will be 30°, 60° and 90°.
Question no – (5)
Solution :
According to the question,
Angles of a triangle are,
(x – 40)°
(x – 20)°
(1/2x – 10)°
∴ x – 40° + x – 20° + 1/2x – 10° = 180°
= 2x + 1/2x = 180 + 70
= 4x + x/2 = 250
= x = 250 × 2/100
∴ x = 100°
Thus, the value of x will be 100°.
Question no – (6)
Solution :
Let, the angles x, x + 10, x + 10 + 10
∴ x + x + 10 + x + 10 + 10 = 180°
= 3x = 180° – 30°
= x = 150/3
= x = 50°
∴ The angles are,
= x = 50°
= 50 + 10 = 60°
= 50 + 10 + 10 = 70°
Therefore, the required three angles are 50°, 60° and 70°.
Question no – (7)
Solution :
Let an angle = x
∴ x + x + x + 30 = 180°
= 3x = 180 – 30
= x = 150/3
= x = 50°
Thus, the all the angles of the triangle will be 50°, 50°, 80°
Question no – (8)
Solution :
Let, the angles will be x, yz
∴ x + y + z = 180°
= z + z = 180°
= z = 90°
∴ x + y = z
Hence, the triangle is right angle triangle.
Question no – (10)
Solution :
(i) 63°, 37°, 80°
= 63 + 37 + 80 = 180
= 180 = 180
∴ Triangle.
(ii) 45°, 61°, 73°
= 45 + 61 + 73 = 189
= 189 ≠ 180
∴ not triangle.
(iii) 59°, 72°, 61°
= 59 + 72 + 61 = 182
= 182 ≠ 180
∴ Not triangle
(iv) 45°, 45°, 90°
= 45 + 45 + 90 = 180
= 180 = 180
∴ Triangle.
(v) 30°, 20°, 125°
= 30 + 20 + 125 = 175
= 175 ≠ 180
∴ Not triangle.
Question no – (11)
Solution :
Let, Angle are 3x, 4x and 5x
∴ 3x – 4x + 5x = 180°
= 12x = 180°
= x = 180/12
= x = 15
∴ Smallest angle,
= 3 × 15
= 45°
Therefore, the smallest angle will be 45°
Question no – (12)
Solution :
Let acute angle is = x
∴ x + x + 90° = 180°
= 2x = 180° – 90°
= x = 90/2
= x = 45°
Hence, the two angles will be 45° and 45°
Question no – (14)
Solution :
∠FAE + ∠AEF + ∠AFE = 180° … (i)
∠AED + ∠ADE + ∠DAE = 180° …. (ii)
∠ADE + ∠ACD + ∠DAC = 180° …. (iii)
∠ABC + ∠ACB + ∠BAC = 180° …. (iv)
Adding we get,
∠FAE + ∠AEF + ∠AFE + ∠AED + ∠ADE + ∠DAE + ∠ADE + ∠ACD + ∠DAC + ∠ABC + ∠ACB + ∠BAC = 720°
⇒ ∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA = 720°
Question no – (15)
Solution :
Figure – (i)
BC || DE and AB divider
∴ ∠ABC = ∠ADE = 40
And ∠ACB = ∠AED = 30
And Z = 180 – 930 + 40)
= 180 – 70
= 110°
Figure – (ii)
In △ ACD
y = 180 – (90 + 45)
= 180 – 90 – 45
= 45°
△ ABC,
x = 180 – 90 – 45
= 50°
Figure – (iii)
x = 180 – 50 – 50
= 80°
z = 180 – [50 + (50 + 30°)]
= 180 – 130
= z = 50°
∴ y = 180 – (30 + 50)
= 100°
Question no – (16)
Solution :
Let, the angles be x, 2x
∴ 60 + x + 2x = 180°
= 3x = 180 – 60
= x = 120/3
= x = 40°
Thus, the other two angles will be 40° and 80°.
Question no – (17)
Solution :
Let, the two angles be 2x, 3x
∴ 100 + 2x + 3x = 180
= 5x = 180 – 100
= x = 80/5
= x = 16°
∴ The angle are,
x = 16°
16 × 2 = 32°
16 × 3 = 48°
Therefore, the required angles will be 32° and 48°.
Question no – (18)
Solution :
As per the given question,
3∠A = 4∠B = 6∠C
∴ 3∠A = 4∠B = 6 ∠C = 180
= 3∠A + 3∠A + 3∠A = 180
= ∠A = 180/9
= ∠A = 20°
∴ Angles are,
= 3 × 20 = 60°
= ∠B = 60/4 = 15°
= ∠C = 60/10 = 10°
Hence, the required angles will be 60°, 15° and 10°.
Question no – (20)
Solution :
In the given question,
∠A = 100°
We know, ∠BAD = ∠DAC
∴ ∠BAD = 50° and ∠BDA = 90°
= ∠B = 180 – 90 – 50
= 180 – 140
= 40°
∴ ∠B = 40°
Therefore, ∠B ill be 40°.
Question no – (21)
Solution :
According to the given question,
△ABC, ∠A = 50°, ∠B = 70°
Bisector of ∠C meets AB in D
∴ ∠C = 180° – 50° – 70°
= 180 – 120
= 60°
∠ACD = 30 and ∠DCA = 30°
∴ ∠ADC,
= 180 – 50 – 30
= 100°
∴ ∠CDB,
= 180 – 30 – 70
= 80°
Therefore, the ∠ADC will be 100° and ∠CDB will be 80°.
Question no – (22)
Solution :
(i) ∠C = 180° – 60 – 80
= 40°
So, ∠C = 40°
(ii) ∠BOC
We know,
∠BOC = 2 × ∠BAC
= 2 × 60°
= 120°
Therefore, ∠BOC = 120°
Question no – (23)
Solution :
∠B = 90°
∴ Other angles ,
∠A + ∠C = 180° – 90
= 90°
And ∠OAC + ∠OCA
= 90/2
= 45° ….(i)
∴ In △ABC
∠AOC + ∠OAC + ∠OCA = 180°
= ∠AOC = 180° – 45°
∴ ∠AOC = 135°
Question no – (26)
Solution :
Given in the question,
= AC || DB
∴ ∠CAB = ∠ABD = 35° …alternative angle.
∴ ∠BOD = 180° – 53 – 35
= 180° – 90
= 90°
Hence, ∠BOD = 90°
Question no – (27)
Solution :
∠B + ∠C = 180°– 90= 90°
∠ABC = ∠QRP
And ∠CQP = ∠ACQ
∠B + ∠C = 90°
= ∠R + ∠C = 90°
∠P + ∠Q + ∠R = 180
= ∠P = 180° – 90
= 90°
Therefore, ∠P = 90°
Properties of Triangles Exercise 15.3 Solution
Question no – (1)
Solution :
(i) The interior adjacent angle.
= ∠ABC
(ii) The interior opposite angles to exterior ∠CBX.
= ∠BAC, ∠ACB ; ∠ABC; ∠ACB
Question no – (2)
Solution :
We know,
∠ACX = ∠CAB + ∠ABC
= 50 + 55
= 105°
∴ ∠ACB = 180° – 105°
= 75°
Therefore, the measures of ∠ACX will be 105° and ∠ACB will be 75°.
Question no – (3)
Solution :
∠C = 95°
Let, opposite angle ∠B = 55°
∴ ∠ACD = ∠B + ∠A
= ∠A = 95 – 55
= 40°
∴ ∠C = 180 – 95
= 85°
Question no – (4)
Solution :
As per the given question,
∠A = ∠B
∴ ∠ACX = ∠A + ∠B = ∠A + ∠A
= 2∠A = 80°
∠A = 40
∠A + ∠B + ∠C = 180
= 40 + 40 + ∠C = 180
= ∠C = 180 – 80
∴ (∠C) = 100°
Hence, the measures of these angles will be ∠A = 40 and ∠C = 100°.
Question no – (5)
Solution :
Let, ∠ACD = 105° and ∠EAF = 45°
∴ ∠ACB = 180° – 105
= 75°
∠CAB = ∠EAF = 45°
∴ ∠ABC = 180° – 75° – 45°
= 180° – 120°
= 60°
Question no – (7)
Solution :
Let the angles 3x, 2x, x
∴ x + 2x + 3x = 180°
= x = 30°
∠A = 3 × 30°
= 90
∴ ∠B = 2 × 30°
= 60°
∠C = 30°
∴ ∠ECD = 180° – 90° + 30°
= 180 °– 120°
= 60°
Thus, the value of ∠ECD will be 60°
Question no – (8)
Solution :
Sum of A and B,
= 103° + 74°
= 183°
∴ 183° ≠ 180°
Hence, this is not possible, because the sum of the interior angles A and B is 183° > 180°.
Question no – (10)
Solution :
According to the question,
∠AED = 180° – 120 = 60°
∠ADE = 180° – 60 – 30 = 90°
∴ ∠FDC = 180 – 90 = 90°
∠CDF + ∠DCF + ∠DFC = 180°
= ∠DCF
= 180° – 90° – 60°
= 30°
Also, ∠DCF + X = 180°
= X = 180° – 30°
= 150°
Therefore, the value ox will be 150°
Question no – (12)
Solution :
∠DAX = ∠DAC
∠DAC = 70°
∠DAX + ∠DAC + ∠DAC = 180°
= ∠BAC = 180 – 70°
= 40°
Let, ∠B = ∠C = x
= 40° + x + x = 180°
= 2x = 180° – 40°
= x = 140/2
= 70°
Therefore, the angle ∠ACB will be 70°
Question no – (13)
Solution :
∠ACD = 115°
∠ABC = 30°
∠ACB = 180° – 115° = 65°
30° + 65° + ∠BAC = 180°
⇒ ∠BAC = 180° – 95°
∠BAC = 75°
∠BAL = ∠CAL = 75/2
= 42.5°
∴ ∠ALC
= 180° – 42.5° – 65°
= 72.5°
Hence, The angle ∠ALC will be 72.5°
Question no – (15)
Solution :
Figure – (i)
∠ACD = 180° – 75 = 105°
∠y = 180 – 75 – 40 = 65°
Figure – (ii)
x = 180° – 80 = 100°
and y = 180° – 100 – 30°
= 50°
Figure – (iii)
y = 180 – 100 – 30 = 50°
and 45 + x + 30 + 50 = 180°
= x = 180° – 125°
= 55°
Figure – (iv)
We know ∠DBA = ∠BDC + ∠BCD
= 30 + 50 = 80°
∴ X = 80°
∠AEB = 180° – (30 + 80)
= 180 – 110°
= 70°
∴ Y = 180° – 70°
= 110°
Question no – (16)
Solution :
Figure – (i)
∠BAC = 180° – 120° = 60°
∠BCA = 180° – 112° = 68°
x = 180° – 60° – 68°
= 52°
Figure – (ii)
∠ABC = 180° – 120°
= 60°
∠BCA = 180° – 60° – 70°
= 180° – 130°
= 50°
Figure – (iii)
Sine BA || DC and AD divider
∴ ∠BAD = ∠ADC = 52°
∠ADC = 52°
In △ADC,
x = 180° – 52° – 40°
= 180° – 92°
= 88°
Figure – (iv)
∠ADC = 360° – 45° – 35° – 50°
= 360° – 80° – 50°
= 360° – 130°
= 230°
∴ Extended ∠ADC = 360° – 230°
= 130°
Properties of Triangles Exercise 15.4 Solution
Question no – (1)
Solution :
(i) Given, 5, 7, 9
= 5 + 7 = 12 >9,
= 5 + 9 = 14 > 7
and 9 + 7 = 16 > 5
So, triangle possible
(ii) Given, 2, 10, 15
2 + 10 = 12 > 15
Triangle not possible
(iii) Given, 3, 4, 5
3 + 4 = 7 > 5
4 + 5 = 9> 3
3 + 5 = 8 > 5
Hence, Triangle possible.
(iv) Given, 2, 5, 7
= 5 + 8 = 13 ∠ 20
∴ Triangle not possible.
Question no – (2)
Solution :
(i) AP < AB + BP
(ii) AP < AC + PC
(iii) AP < 1/2 (AB + AC + BC)
Question no – (3)
Solution :
(i) AP + PB < AB – This statement is False.
(ii) AP + PC > AC – This statement is True.
(iii) BP + PC = BC – This statement is False.
Question no – (5)
Solution :
∴ BC is the largest side.
And AC is the smallest side.
Properties of Triangles Exercise 15.5 Solution
Question no – (1)
Solution :
Pythagoras Theorem :
In a right triangle the square of the Hypotenuse equal the sum of square of its remaining two sides.
Converse Pythagoras theorem :
If the square of one side of a triangle is equal to the sum of the square of the other two sides then the triangle is a right triangle with the angle opposite the first side as right angle.
Question no – (2)
Solution :
(i) a2 + b2 = c2
= c = √a2 + b2
= √62 + 82
= √36 + 64
= √100
= 10 cm.
Therefore, the length will be 10 cm.
(ii) C = √82 + 152
= √64 + 225
= √289
= 17 cm.
Thus, the length will be 17 cm.
(iii) c = √32 + 42
= √9 + 16
= √25
= 5 cm.
Hence, the length will be 5 cm.
(iv) C = √22 + 1.52
= √4 + 2.25
= 2.5 cm.
So, the length will be 2.5 cm.
Question no – (3)
Solution :
In the given question,
The hypotenuse of a triangle = 2.5 cm
One of the sides = 1.5 cm
length of the other side = ?
∴ Length of other Side,
= √2.52 – 1.52
= √6.25 – 1.25
= √4
= 2
Therefore, the length of the other side will be 2 cm.
Question no – (4)
Solution :
As per the given question,
Length of the ladder = 3.7 m
Distance from wall = 1.2 m
∴ Height= √3.72 + 1.22
= √13.69 – 1.44
= √12.25
= 3.5
Hence, the height of the wall to which the ladder reaches will be 3.5 cm.
Question no – (5)
Solution :
= 3² + 4²
= 9 + 16
= 25 ≠ 62
Therefore, it is not a right-angled triangle.
Question no – (6)
Solution :
(i) a = 7 cm, b = 24 cm and c = 25 cm
= 72 + 242
= 49 + 516
= 625
= 252
Thus, the triangle is right triangle.
(ii) a = 9 cm, b = 16 cm and = 18 cm
= 92 + 162
= 81 + 256
= 337 ≠ 182
Therefore, the side is not right triangle.
Question no – (7)
Solution :
Height = 11 – 6 = 5
∴ The hypotenuse,
= √52 + 122
= √25 + 144
= √169
= 13.
Therefore, the distance between their tops is 13 m.
Question no – (8)
Solution :
As per the given question,
A man goes 15 m due west and then 8 m due north
∴ Hypotenuse = √152 + 82
= √225 + 64
= √289
= 17
Hence, he is 17 m away from the starting point.
Question no – (9)
Solution :
For height = 8 m
Hypotenuse = 6² + 8²
= 36 + 64
= √100
= 10
and so on,
= 10² = 8² + height²
Height = √100 – 64
= √36
= 6 m
Thus, its top will reach 6 m high.
Question no – (10)
Solution :
Base2 + height2 = ladder space
= Base2 = √502 – 482
= Base = √2500 – 2304
= √196
= 14
Thus, the base of the wall will be 14 dm.
Question no – (11)
Solution :
Since two legs are equal height
So, Let height be x.
∴ x2 + x2 = 502
= 2x2 = 502 … (Square of hypotenuse = 50)
= x2 = 25
= x = 5
Hence, the length of each leg is 5 units.
Question no – (12)
Solution :
(i) Given, 12, 35, 37
= 122 + 352
= 144 + 1235 = 1369
= 372
So, it is triplet.
(ii) Given,7, 24, 25
= 72 + 242
= 49 + 576 = 625
= 252
Thus, it is triplet.
(iii) Given, 27, 36, 45
= 272 + 362
= 729 + 1296 = 2025
= 452
Hence, it is triplet.
(iv) Given, 15, 36, 39
= 152 + 362
= 225 + 1296 = 1521
= 392
Therefore, it is triplet.
Question no – (13)
Solution :
According to the question,
∠ABC = 100°, ∠BAC = 35°
∴ 100° + 35° + ∠C = 180°
= ∠ = 180 – 135
= 45°
Since, ∠BDC = 90° and ∠DCB = 90°
Obviously, ∠DBC = 45°
If BD = 2 cm
Therefore, then DC = 2 cm.
Question no – (14)
Solution :
So, BA2 = 92 + 122
= BA2 = 81 + 144
∴ BA = √225
∴ AB = 15 cm
And 122 + 162 = AC2
= AC2 = 144 + 256
= AC = √400
∴ AC = 20
∴ AC = 20 cm
Therefore, Yes, △ABC is right angled at A.
Next chapter solution :
👉 Chapter 16 👈
