# NCERT Solutions for Class 7 Maths

Find 100% accurate solutions for NCERT Class VII Science. All questions have been explained and solved step-by-step, to help you understand thoroughly. Free Download option available!

Sign Up to Download as PDF

1.    Given: QD = DR

PD is median.

No, QM ≠ MR as D is the mid-point of QR.

2.    (a) Herem 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 Δ XYZ. 3.   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. 4.   Since, Exterior angle = Sum of interior opposite angles, therefore

(i) x = 50° + 70° = 120°

(ii) x = 65° + 45° = 110°

(iii) x = 30°+ 40° = 70°

(iv) x = 60° + 60° = 120°

(v) x = 50° + 50° = 100°

(vi) x = 60° +30° = 90°

5.    Since, Exterior angle = Sum of interior opposite angles, therefore

(i) x + 50° = 115°   ⇒     x = 115° – 50° = 65°

(ii) 70° + x = 100°  ⇒   x = 100° – 70° = 30°

(iii) x + 90° = 125°    ⇒    x = 120° – 90° = 35°

(iv) 60° + x = 120°    ⇒   x = 120° – 60° = 60°

(v) 30° + x = 80°    ⇒   x = 80° -30° = 50°

(vi) x + 35° = 75°   ⇒   x = 75° – 35° = 40°

6.    (i) In ΔABC,

∠ BAC + ∠ ACB + ∠ ABC = 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 Δ XYZ,

∠ 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° (v) In the given equilateral triangle,

x + x + x = 180°                          [By angle sum property of a triangle]

⇒ 3x = 180° (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° 7.    (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° + y = 180°                                          [From eq. (i)]

⇒ 130° + y = 180°                            ⇒           y = 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 ]

⇒ 50° + 60°+ y = 180°                     [From eq. (i)]

⇒ 90° + y = 180°           ⇒           y = 180° – 90° = 90°

(v) y = 90°                ….(i)                  [Vertically opposite angles]

Now, y + x + x = 180°                       [Angle sum property of a ]

⇒ 90°+ 2x = 180°                                   [From eq. (i)]

⇒ 2x = 180° – 90°           ⇒            2x = 90° (vi) x = y             ….. (i)                        [Vertically opposite angle]

Now, x + x + y = 180°                         [Angle sum property of a ]

⇒ 2x + x = 180°                                    [From eq.(i)]

8.   Since, a triangle is possible whose sum of the lengths of any two sides would 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.

9.    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.

10.  Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

Therefore, In Δ ABM, AB + BM > AM       ….(i)

In Δ AMC, 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.

11.  Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side.

Therefore,

In Δ ABC, AB + BC > AC   ….(i)

In Δ ADC, AD + DC > AC  ….(ii)

In Δ DCB, DC + CB > DB  ….(iii)

In Δ ADB, 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.

12.  Since, the sum of lengths of any two sides in a triangle should be greater than the length of third side. Therefore,

In Δ AOB,             AB < OA + OB    …(i)

In Δ BOC,             BC < OB + OC     …(ii)

In Δ COD,             CD < OC + OD    …(iii)

In Δ AOD,            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 < 20A + 20B + 20C + 20D

⇒ AB + BC + CD + DA < 2[(A0 + OC) + (DO + OB)]

⇒ AB + BC + CD + DA < 2(AC + BD)

Hence, it is proved.

13.  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.

14.   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 = (13Q)2 (PR)2

⇒ x2 = (10)2 + (24)2

⇒ x2 = 100 + 576 = 676

⇒ x = √676 = 26 cm Thus, the length of QR is 26 cm.

15.  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.

16.  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 = (C13)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.

17.  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)

L.H.S. = (5)2 = 25

R.H.S. = (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) 5 cm, 2 cm, 2.5 cm In Δ PQR,     (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.

18.  Let NCB 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 AABC

(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.

19.  In Δ PQR,

∠ PQR + ∠ QRP + ∠ RPQ = 180°                    [By Angle sum property of a Δ]

⇒  25° + 65° + ∠ RPQ = 180°            ⇒                 90° + ∠ RPQ = 180°

⇒  ∠ RPQ = 180° – 90° = 90°

Thus, Δ PQR 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.

20.  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 + 49)

= 2 × 49 = 98 cm

Hence the perimeter of the rectangle is 98 cm.

21.  Given: Diagonals AC = 30 cm and DB = 16 cm. Since the diagonals of the rhombus bisect at right angle to each other.

Now,       In right angle triangle DOC,

(DC)2 = (OD)2 + (OC)2

⇒ (DC)2 = (8)+ (15)2

⇒ (DC)2 = 64 + 225 = 289

⇒ DC = √289 = 17 cm

Perimeter of rhombus = 4 × side

= 4 × 17 = 68 cm

Thus, the perimeter of rhombus is 68 cm.

MySchoolPage connects you with exceptional, certified math tutors who help you stay focused, understand concepts better and score well in exams!