NCERT Solutions for Chapter 6 The Triangle and its Properties Class 7 Maths
Book Solutions1
In ΔPQR, D is the mid-point of QR.
PM is _______________
PD is ________________
Is QM = MR ?
Answer
Given: QD = DR
∴ PM is altitude.
∴ PM is altitude.
PD is median.
No, QM ≠ MR as D is the mid-point of QR.
Exercise 6.1
Page Number 116
2
Draw rough sketches for the following:
1. In ΔABC, BE is a median.
2. In ΔPQR, PQ and PR are altitudes of the triangle.
3. In ΔXYZ, YL is an altitude in the exterior of the triangle.
Answer
(a) Here, BE is a median in ΔABC and AE = EC.
(b) Here, PQ and PR are the altitudes of the ΔPQR and RP ⊥ QP.
(c) YL is an altitude in the exterior of ΔXYZ.
Exercise 6.1
Page Number 116
3
Verify by drawing a diagram if the median and altitude of a isosceles triangle can be same.
Answer
Isosceles triangle means any two sides are same.
Take ΔABC and draw the median when AB = AC.
AL is the median and altitude of the given triangle.
Exercise 6.1
Page Number 116
1
Find the value of the unknown exterior angle xx in the following diagrams:
Answer
Since, Exterior angle = Sum of opposite interior angles, therefore
1. x=50 ̊+70 ̊=120 ̊
2. x=65 ̊+45 ̊=110 ̊
3. x=30 ̊+40 ̊=70 ̊
4. x=60 ̊+60 ̊=120 ̊
5. x=50∘+50∘=100 ̊
6. x=60 ̊+30 ̊=90 ̊
Exercise 6.2
Page Number 118
2
Find the value of the unknown interior angle xx in the following figures:
Answer
Since, Exterior angle = Sum of opposite interior angles, therefore
1. x+50 ̊=115 ̊ ⇒ x=115 ̊−50 ̊=65 ̊
2. 70 ̊+x=100 ̊⇒ x=100 ̊−70 ̊=30 ̊
3. x+90 ̊=125 ̊⇒ x=125 ̊−90 ̊=35 ̊
4. 60 ̊+x=120 ̊⇒ x=120 ̊−60 ̊=60 ̊
5. 30 ̊+x=80 ̊⇒ x=80 ̊−30 ̊=50 ̊
6. x+35 ̊=75 ̊⇒ x=75 ̊−35 ̊=40 ̊
Exercise 6.2
Page Number 119
1
Find the value of unknown x in the following diagrams:
Answer
(i) In ΔABC,
∠BAC + ∠ABC + ∠ACB = 180∘ [By angle sum property of a triangle]
⇒ x+50∘+60∘=180∘
⇒ x+110∘=180∘⇒ x=180∘−110∘=70∘
(ii) In ΔPQR,
∠RPQ + ∠PQR + ∠RPQ = 180∘ [By angle sum property of a triangle]
⇒ 90∘+30∘+x=180∘
⇒ x+120∘=180∘⇒ x=180∘−120∘=60∘
(iii) In ΔXYZ,
∠ZXY + ∠XYZ + ∠YZX = 180∘ [By angle sum property of a triangle]
⇒ 30∘+110∘+x=180∘
⇒ x+140∘=180∘⇒ x=180∘−140∘=40∘
(iv) In the given isosceles triangle,
x+x+50∘=180∘ [By angle sum property of a triangle]
⇒ 2x+50∘=180∘
⇒ 2x=180∘−50∘⇒ 2x=130∘
⇒ x=130∘/2=65∘
(v) In the given equilateral triangle,
x+x+x=180∘ [By angle sum property of a triangle]
⇒ 3x=180∘
⇒ x=180∘/3=60∘
(vi) In the given right angled triangle,
x+2x+90∘=180∘ [By angle sum property of a triangle]
⇒ 3x+90∘=180∘
⇒ 3x=180∘−90∘⇒ 3x=90∘
⇒ x=90∘/3=30∘
Exercise 6.3
Page Number 121
2
Find the values of the unknowns xx and yy in the following diagrams:
Answer
(i) 50∘+x=120∘ [Exterior angle property of a Δ ]
⇒ x=120∘−50∘=70∘
Now, 50∘+x+y=180∘ [Angle sum property of a Δ ]
⇒ 50∘+70∘+y=180∘
⇒ 120∘+y=180∘⇒ y=180∘−120∘=60∘
(ii) y=80∘ ……….(i) [Vertically opposite angle]
Now, 50∘+x+y=180∘ [Angle sum property of a Δ ]
⇒ 50∘+80∘+x=180∘
[From eq. (i)
⇒ 130∘+x=180∘⇒ x=180∘−130∘=50∘
(iii) 50∘+60∘=x [Exterior angle property of a Δ ]
⇒ x=110∘
Now 50∘+60∘+y=180∘ [Angle sum property of a Δ ]
⇒ 110∘+y=180∘
⇒ y=180∘−110∘⇒ y=70∘
(iv) x=60∘……….(i) [Vertically opposite angle]
Now, 30∘+x+y=180∘ [Angle sum property of a Δ ]
⇒ 30∘+60∘+y=180∘[From eq. (i)]
⇒ 90∘+y=180∘⇒ y=180∘−90∘=90∘
(v) y=90∘ ……….(i) [Vertically opposite angle]
Now, y+x+x=180∘ [Angle sum property of aΔ ]
⇒ 90∘+2x=180∘ [From eq. (i)]
⇒ 2x=180∘−90∘⇒ 2x=90
⇒ x=90∘/2=45∘
(vi) x=y ……….(i) [Vertically opposite angle]
Now, x+x+y=180∘ [Angle sum property of a Δ ]
⇒ 2x+x=180∘ [From eq. (i)]
⇒ 3x=180∘⇒ x=180∘/3=60∘
Exercise 6.3
Page Number 121
1
Is it possible to have a triangle with the following sides?
(i) 2 cm, 3 cm, 5 cm
(ii) 3 cm, 6 cm, 7 cm
(iii) 6 cm, 3 cm, 2 cm
Answer
Since, a triangle is possible whose sum of the lengths of any two sides should be greater than the length of third side.
(i) 2 cm, 3 cm, 5 cm
2 + 3 = 5 No
2 + 5 > 3 Yes
3 + 5 > 2 Yes
This triangle is not possible.
(ii) 3 cm, 6 cm, 7 cm
3 + 6 > 7 Yes
6 + 7 > 3 Yes
3 + 7 > 6 Yes
This triangle is possible.
(iii) 6 cm, 3 cm, 2 cm
6 + 3 > 2 Yes
6 + 2 > 3 Yes
2 + 3 < 6 No
This triangle is not possible.
Exercise 6.4
Page Number 126
2
Take any point O in the interior of a triangle PQR. Is:
(i) OP + OQ > PQ ?
(ii) OQ + OR > QR ?
(iii) OR + OP > RP ?
(ii) OQ + OR > QR ?
(iii) OR + OP > RP ?
Since sum of two sides is greater than third side.
Answer
Join OR, OQ and OP.
(i) Is OP + OQ > PQ ?
Yes, POQ form a triangle.
(ii) Is OQ + OR > QR ?
Yes, RQO form a triangle.
(iii) Is OR + OP > RP ?
Yes, ROP form a triangle.
Exercise 6.4
Page Number 126
3
AM is a median of a triangle ABC. Is AB + BC + CA > 2AM ?
(Consider the sides of triangles ABM and AMC.)
Answer
Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.
Therefore, In ΔABM, AB + BM > AM ……….(i)
In ΔAMC, AC + MC > AM ……….(ii)
Adding eq. (i) and (ii),
AB + BM + AC + MC > AM + AM
⇒ AB + AC + (BM + MC) > 2AM
⇒ AB + AC + BC > 2AM
Hence, it is true.
Exercise 6.4
Page Number 126
4
ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD ?
Answer
Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.
Therefore, In ΔABC, AB + BC > AC ……….(i)
In ΔADC, AD + DC > AC ……….(ii)
In ΔDCB, DC + CB > DB ……….(iii)
In ΔADB, AD + AB > DB ……….(iv)
Adding eq. (i), (ii), (iii) and (iv),
AB + BC + AD + DC + DC + CB + AD + AB > AC + AC + DB + DB
⇒ (AB + AB) + (BC + BC) + (AD + AD) + (DC + DC) > 2AC + 2DB
⇒ 2AB + 2BC + 2AD + 2DC > 2(AC + DB)
⇒ 2(AB + BC + AD + DC) > 2(AC + DB)
⇒ AB + BC + AD + DC > AC + DB
⇒ AB + BC + CD + DA > AC + DB
Hence, it is true.
Exercise 6.4
Page Number 126
5
ABCD is quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD) ?
Answer
Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.
Therefore, In ΔAOB, AB < OA + OB ……….(i)
In ΔBOC, BC < OB + OC ……….(ii)
In ΔCOD, CD < OC + OD ……….(iii)
In ΔAOD, DA < OD + OA ……….(iv)
Adding eq. (i), (ii), (iii) and (iv),
AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OD + OA
⇒ AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD
⇒ AB + BC + CD + DA < 2[(AO + OC) + (DO + OB)]
⇒ AB + BC + CD + DA < 2(AC + BD)
Hence, it is proved.
Exercise 6.4
Page Number 126
6
The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?
Answer
Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.
It is given that two sides of triangle are 12 cm and 15 cm.
Therefore, the third side should be less than 12 + 15 = 27 cm.
And also the third side cannot be less than the difference of the two sides.
Therefore, the third side has to be more than 15 – 12 = 3 cm.
Therefore, the third side could be the length more than 3 cm and less than 27 cm.
Exercise 6.4
Page Number 127
1
PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
Answer
Given: PQ = 10 cm, PR = 24 cm
Let QR be x cm.
In right angled triangle QPR,
(Hypotenuse)2= (Base)2+ (Perpendicular)2
[By Pythagoras theorem]
⇒ (QR)2= (PQ)2+ (PR)2
⇒ x2=(10)2+(24)2
⇒ x2 = 100 + 576 = 676
⇒ x=√676 = 26 cm
Thus, the length of QR is 26 cm.
Exercise 6.5
Page Number 130
2
ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.
Answer
Given: AB = 25 cm, AC = 7 cm
Let BC be x cm.
In right angled triangle ACB,
(Hypotenuse)2= (Base)2+ (Perpendicular)2
[By Pythagoras theorem]
⇒ (AB)2= (AC)2+ (BC)2
⇒ (25)2=(7)2+x2
⇒ 625 = 49 + x2
⇒ x2= 625 – 49 = 576
⇒ x=√576 = 24 cm
Thus, the length of BC is 24 cm.
Exercise 6.5
Page Number 130
3
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.
Answer
Let AC be the ladder and A be the window.
Given: AC = 15 m, AB = 12 m, CB = a m
In right angled triangle ACB,
(Hypotenuse)2= (Base)2+ (Perpendicular)2
[By Pythagoras theorem]
⇒ (AC)2= (CB)2+ (AB)2
⇒ (15)2=(a)2+(12)2
⇒ 225 = a2+ 144
⇒ a2 = 225 – 144 = 81
⇒ a=√81 = 9 cm
Thus, the distance of the foot of the ladder from the wall is 9 m.
Exercise 6.5
Page Number 130
4
Which of the following can be the sides of a right triangle?
(i) 2.5 cm, 6.5 cm, 6 cm
(ii) 2 cm, 2 cm, 5 cm
(iii) 1.5 cm, 2 cm, 2.5 cm
In the case of right angled triangles, identify the right angles.
Answer
Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem,
(Hypotenuse)2= (Base)2+ (Perpendicular)2
(i) 2.5 cm, 6.5 cm, 6 cm
In ΔABC, (AC)2=(AB)2+(BC)2
L.H.S. = (6.5)2 = 42.25 cm
R.H.S. = (6)2+(2.5)2= 36 + 6.25 = 42.25 cm
Since, L.H.S. = R.H.S.
Therefore, the given sides are of the right angled triangle.
Right angle lies on the opposite to the greater side 6.5 cm, i.e., at B.
(ii) 2 cm, 2 cm, 5 cm
(5)2=(2)2+(2)2
L.H.S. = (5)2 = 25
R.H.S. = (2)2+(2)2 = 4 + 4 = 8
Since, L.H.S. ≠ R.H.S.
Therefore, the given sides are not of the right angled triangle.
(iii) 1.5 cm, 2 cm, 2.5 cm
In ΔPQR, (PR)2=(PQ)2+(RQ)2
L.H.S. = (2.5)2 = 6.25 cm
R.H.S. = (1.5)2+(2)2 = 2.25 + 4 = 6.25 cm
Since, L.H.S. = R.H.S.
Therefore, the given sides are of the right angled triangle.
Right angle lies on the opposite to the greater side 2.5 cm, i.e., at Q.
Exercise 6.5
Page Number 130
5
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.
Answer
Let ACB represents the tree before it broken at the point C and let the top A’ touches the ground at A after it broke. Then ΔABC is a right angled triangle, right angled at B.
AB = 12 m and BC = 5 m
Using Pythagoras theorem, In ΔABC
(AC)2=(AB)2+(BC)2
⇒ (AC)2=(12)2+(5)2
⇒ (AC)2=144+25
⇒ (AC)2=169
⇒ AC = 13 m
Hence, the total height of the tree = AC + CB = 13 + 5 = 18 m.
Exercise 6.5
Page Number 130
6
Angles Q and R of a ΔPQR are 25∘ and 65∘.
Write which of the following is true:
(i) PQ2+ QR2= RP2
(ii) PQ2+ RP2= QR2
(iii) RP2+ QR2= PQ2
Answer
In ΔPQR,
∠PQR + ∠QRP + ∠RPQ = 180∘
[By Angle sum property of a Δ ]
⇒ 25∘+65∘+∠RPQ = 180∘
⇒ 90∘+∠RPQ = 180∘
⇒ ∠RPQ = 180∘−90∘=90∘
Thus, ΔPQR is a right angled triangle, right angled at P.
∴ (Hypotenuse)2 = (Base)2 + (Perpendicular)2 [By Pythagoras theorem]
⇒ (QR)2=(PR)2+(QP)2
Hence, Option (ii) is correct.
Exercise 6.5
Page Number 130
7
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Answer
Given diagonal (PR) = 41 cm, length (PQ) = 40 cm
Let breadth (QR) be x cm.
Now, in right angled triangle PQR,
(PR)2=(RQ)2+(PQ)2
[By Pythagoras theorem]
⇒ (41)2=x2+(40)2
⇒ 1681 = x2 + 1600 ⇒ x2 = 1681 – 1600
⇒ x2 = 81 ⇒ x=√81 =9 cm
Therefore the breadth of the rectangle is 9 cm.
Perimeter of rectangle = 2(length + breadth)
= 2 (9 + 40)
= 2 x 49 = 98 cm
Hence, the perimeter of the rectangle is 98 cm.
Exercise 6.5
Page Number 130
8
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Answer
Given: Diagonals AC = 30 cm and DB = 16 cm.
Since the diagonals of the rhombus bisect at right angle to each other.
Therefore, OD = DB/2=16/2 = 8 cm
And OC = AC/2=30/2= 15 cm
Now, In right angle triangle DOC,
(DC)2=(OD)2+(OC)2 [By Pythagoras theorem]
⇒ (DC)2=(8)2+(15)2
⇒ (DC)2 = 64 + 225 = 289
⇒ DC = √289 = 17 cm
Perimeter of rhombus = 4 x side
= 4 x 17 = 68 cm
Thus, the perimeter of rhombus is 68 cm.
Exercise 6.5
Page Number 130