NCERT Grade 8-Understanding Quadrilaterals-Answers – MySchoolPage

NCERT Grade 8-Understanding Quadrilaterals-Answers

NCERT Solutions for Class 8 Maths

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1.    (a) Sample curve

(b) Simple closed curve

(c) Polygons

(d) Convex polygons

(e) Concave polygon

2.   (a) A convex quadrilateral has two diagonal

Here, AC and BD are two diagonal

(b) A regular hexagon has 9 diagonals.

Here, diagonals are AD, AE BD, FC, FB, AC, EC and FD.

(c) A triangle has no diagonal.

3.    Let ABCD is convex quadrilateral, then we draw a diagonal AC which divides the quadrilateral in two triangles.

∠A + ∠B + ∠C + ∠D = ∠1 + ∠6 + ∠5 + ∠4 + ∠3 + ∠2

= (∠1 + ∠2 + 3) + (∠4 + ∠5 + ∠6)

= 180° ± 180°                              [By Angle sum property of triangle]

= 360°

Hence, the sum of measures of the triangles of a convex quadrilateral is 360°.

Yes, if quadrilateral is not convex then, this property will also be applied.

Let ABCD is a non-convex quadrilateral and join BD, which also divides the quadrilateral in two triangles

Using angle sum property of triangle,

In Δ ABD,                 ∠1 + ∠2 + ∠3 = 180° ………….. (i)

In Δ BDC,                 ∠4 + ∠5 + ∠6 = 180°  …………. (i)

Adding eq. (i) and (ii),

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 = 360°

⇒ ∠1 + ∠2 + (∠3 + ∠4) + ∠5 + ∠6 = 360°

⇒ ∠A + ∠B + ∠C + ∠D = 360°

Hence proved.

4.   (a) When n = 7, then

Angle sum of a polygon = (n – 2) × 180° = (7 – 2) × 180° = 5 × 180° = 900°

(b) When n = 8, then

Angle sum of a polygon = (n – 2) × 180° = [8 – 2) × 180° = 6 × 180° = 1080°

(c) When n = 10, then

Angle sum of a polygon = (n – 2) × 180° = (10 – 2) × 180° = 8 × 180° = 1440°

(d) When n = n, then

Angle sum of a polygon = (n – 2) × 180°

5.   A regular polygon: A polygon having all sides of equal length and the interior angles of equal size is known as regular polygon.

(i) 3 sides

      Polygon having three sides is called a triangle.

(ii) 4 sides

       Polygon having four sides is called a quadrilateral.

(ii) 6 sides

       Polygon having six sides is called a hexagon.

6.   (a) Using angle sum property of a quadrilateral,

50° + 130° + 120° + x = 360°

⇒ 300° + x360°

⇒ x = 360° – 300°

⇒ x = 60°

(b) Using angle sum property of a quadrilateral,

90° + 60° + 70° + x = 360°

⇒ 220° + x =360°

⇒ x = 360° – 220°

⇒ x = 140°

(c) First base interior angle = 180° – 70° = 110°

Second base interior angle = 180° – 60° = 120°

There are 5 sides, n = 5

∴ Angle sum of a polygon = (n – 2) × 180°

= (5 – 2) × 180° = 3 × 180° = 540°

∴ 30° + x + 110° + 120° + x = 540°

⇒ 260° + 2x = 540°

⇒ 2x = 540° – 260°

⇒ 2x = 280°

⇒ x = 140°

(d) Angle sum of a polygon = (n – 2) × 180°

= [5 – 2) × 180° = 3 × 180° = 540°

∴ x + x + x + x + x = 540°

⇒ 5x = 540°                     x = 108°

Hence each interior angle is 108°.

7.   (a) Since sum of linear pair angles is 180°.

∴ 90° + x = 180°

⇒ x = 180° – 90° = 90°

And z + 30° = 180°

⇒ z = 180° – 30° = 150°

Also y = 90° + 30° = 120°                  [Exterior angle property]

∴ x + y + x = 90° ± 120° ± 150° = 360°

(b) Using angle sum property of a quadrilateral,

60° + 80° + 120° + n = 360°

⇒ 260° + n = 360°

⇒ n = 360° – 260°

⇒ n = 100°.

Since sum of linear pair angles is 180°.

∴ w + 100 = 180° ……………………….. (i)

   x  + 120° = 180° ………………………. (ii)

   y +  80° = 180° ………………………… (iii)

   z +  60° = 180° ………………………… (iv)

Adding eq. (i), (ii), (iii) and (iv),

⇒ x + y + z + w + 100° + 120° + 80° + 60° = 180° + 180° + 180°±180°

⇒ x + y + z + w + 360° = 720°

⇒ x + y + z + w = 720° – 360°

⇒ x + y + z + w = 360°.

8.   (a) Here, 125°+m =180° [Linear pair]

⇒ m = 180° – 125° = 55°

And 125°+ n = 180°   [Linear pair]

⇒ n = 180° – 125° = 55°

∵ Exterior angle x° = Sum of opposite interior angles

∴ x° = 55° + 55° = 110°

(b) Sum of angles of a pentagon (n – 2) × 180°

= (5 – 2) × 180°

= 3 × 180° = 540°

By linear pairs of angles,

∠1 + 90° = 180° …………………………… (i)

∠2 + 60° = 180° …………………………… (ii)

∠3 + 90° = 180° ……………………………. (iii)

∠4 + 70° = 180° ……………………………. (iv)

∠5 + x = 180° ………………………………… (v)

Adding eq. (i), (ii), (iii), (iv) and (v), x + (∠1 + ∠2 + ∠3 + ∠4 + ∠5) + 310° = 900

⇒ x + 540° + 310° = 900°                          ⇒            x + 850° = 900°

⇒ x = 900° – 850° = 50°

9.    (i) Sum of angles of a regular polygon = (n2) × 180°

= (9 – 2) × 180°= 7 × 180° = 1260°

Each interior angle = 

Each exterior angle = 180° – 140° = 40°

(ii) Sum of exterior angles of a regular polygon = 360°

Each interior angle = 

10.  Let number of sides be n. Sum of exterior angles of a regular polygon = 360°

Number of sides 

Hence, the regular polygon has 15 sides.

11.  Let number of sides be

Exterior angle = 180° – 165° = 15°

Sum of exterior angles of a regular polygon = 360°

Number of sides 

Hence, the regular polygon has 24 sides.

12.  (a) No. (Since 22 is not a divisor of 360°)

(b) No, (Because each exterior angle is 180° – 22° = 158°, which is not a divisor of 360°)

13.  (a) The equilateral triangle being a regular polygon of 3 sides has the least measure of an interior angle of 60°.

∵ Sum of all the angles of a triangle = 180°

∴ x + x + x = 180°

⇒ 3x = 180°

⇒ x = 60°

(b) By (a), we can observe that the greatest exterior angle is 180° – 60° = 120°.

14.  (i) AD = BC        [Since opposite sides of a parallelogram are equal]

(ii) ∠ DCB = ∠ DAB [Since opposite angles of a parallelogram are equal]

(iii) OC = OA [Since diagonals of a parallelogram bisect each other]

(iv)∠ DAB + m ∠ CDA = 180° [Adjacent angles in a parallelogram are supplementary]

15.  (i) ∠ B + ∠ C = 180°           [Adjacent angles in a parallelogram are supplementary]

⇒ 100° + x = 180°

⇒ x = 180° – 100° = 80°

And z = x = 80°

[Since opposite angles of a parallelogram are equal]

Also y = 100°         [Since opposite angles of a parallelogram are equal]

(ii) x + 50° = 180°

[Adjacent angles in a gm are supplementary]

⇒ x = 180° – 50° = 130°

⇒ z = x = 130° [Corresponding angles]

(iii) x = 90°            [Vertically opposite angles]

⇒ y + x + 30° = 180°           [Angle sum property of a triangle]

⇒ y + 90° + 30° = 180°

⇒ y + 120° = 180°

⇒ y = 180° – 120° = 60°

z = y = 60°           [Alternate angles]

(iv) z = 80°    [Corresponding angles]

 x + 80° = 180°

[Adjacent angles in a || gm are supplementary]

⇒ x = 180° – 80° = 100°

And y = 80° [Opposite angles are equal in a || gm]

(v) y = 112°          [Opposite angles are equal in a || gm]

⇒ 40° + y + x = 180°          [Angle sum property of a triangle]

⇒ 40° + 112° + x = 180°       ⇒           152° + x = 180°

⇒ x = 180° – 152° = 28°

And z = x = 28°      [Alternate angles]

16.  (i) ∠D + ∠B = 180°

It can be, but here, it needs not to be.

(ii) No, in this case because one pair of opposite sides are equal and another pair of opposite sides are unequal. So, it is not a parallelogram.

(iii) No.  ∠A ≠ ∠C

Since opposite angles are equal in parallelogram and here opposite angles are not equal in quadrilateral ABCD.

Therefore it is not a parallelogram.

17.  ABCD is a quadrilateral in which angles ∠A = ∠C = 110°.

Therefore, it could be a kite.

18.  Let two adjacent angles be 3x and 2x.

Since the adjacent angles in a parallelogram are supplementary.

∴ 3x + 2x = 180°        ⇒          5x =180°

∴ One angle = 3x = 3 × 36°= 108°

And Another angle = 2x = 2 × 36° = 72°.

19.  Let each adjacent angle be x.

Since the adjacent angles in a parallelogram are supplementary.

∴ x + x = 180°              ⇒         2x = 180°

Hence, each adjacent angle is 90°.

20. Here      ∠ HOP = 180° – 70° = 110°                   [Angles of linear pair]

And        ∠ E = ∠ HOP               [Opposite angles of a || gm are equal]

⇒ x = 110°

∠ PHE = ∠ HPO                                   [Alternate angles]

∴ y = 40°

Now        ∠ EHO = ∠ O = 70°                             [Corresponding angles]

⇒ 40° + z = 70°                ⇒           z = 70° – 40° = 30°

Hence, x =110°, y = 40° and z = 30°

21.  (i) In parallelogram GUNS,

GS = UN                                [Opposite sides of parallelogram are equal]

⇒ 3x = 18                               ⇒        x =  = 6 cm

Also GU = SN                           [Opposite sides of parallelogram are equal]

⇒ 3y – 1 = 26                          ⇒           3y = 26 + 1

⇒ 3y = 27                               ⇒            y = = 9 cm

Hence, x = 6 cm and y = 9 cm.

(ii) In parallelogram RUNS,

y + 7 = 20                                            [Diagonals of || gm bisects each other]

⇒ y = 20 – 7 = 13 cm

And x + y = 16                     ⇒          x + 13 = 16          ⇒             x = 16 – 13

⇒ x = 3 cm

Hence, x = 3 cm and y = 13 cm.

22.  In parallelogram RISK

∠ RIS = ∠ K = 120°

[Opposite angles of a || gm are equal]

m + 120° = 180°          [Linear pair]

⇒ ∠= 180° – 120° = 60°

And          ∠ ECI = ∠L =  70°

[Corresponding angles]

⇒ m + n +  ECI = 180°                      [Angle sum property of a triangle]

⇒ 60° + n + 70° = 180°            ⇒        130° + n = 180°

⇒ n = 180° – 130° = 50°

Also        x = n = 50°                                [Vertically opposite angles]

23.  Here,         ∠M + ∠L = 100° + 80° = 180°

[Sum of interior opposite angles is 180°]

∴ NM and KL are parallel.

Hence, KLMN is a trapezium.

24.  Here,     ∠B + ∠C = 180°       

∴ 120° +  if m∠C = 180°

m∠C = 180° – 120° = 60°

25.  Here,      ∠ P + Q = 180°               [Sum of co-interior angles is 1800]

∠ P + 130° = 180°

∠ P = 180° – 130°

∠ P =  50° L

∵ R = 90°                   [Given]

∴  S + 90° = 180°

⇒  S = 180° – 90°

⇒ ∠ S =  90°

Yes, one more method is there to find ∠ P.

∠ S + ∠ R + ∠ Q + ∠ P = 360°              [Angle sum property of quadrilateral]

90° + 90° + 130° + ∠ P = 360°

310° + ∠ P = 360°

∠ P = 360° – 310°

∠ P =  50°

26.  (a) False.    Since, squares have all sides are equal.

(b) True.     Since, in rhombus, opposite angles are equal and diagonals intersect at mid-point.

(c) True.     Since, squares have the same property of rhombus but not a rectangle.

(d) False.    Since, all squares have the same property of parallelogram.

(e) False.    Since, all kites do not have equal sides.

(f) True.     Since, all rhombuses have equal sides and diagonals bisect each other.

(g) True.    Since, trapezium has only two parallel sides.

(h) True.    Since, all squares have also two parallel lines.

27.  (a) Rhombus and square have sides of equal length.

(b) Square and rectangle have four right angles.

28.  (i) A square is a quadrilateral, if it has four unequal lengths of sides.

(ii) A square is a parallelogram, since it contains both pairs of opposite sides equal.

(iii) A square is already a rhombus. Since, it has four equal sides and diagonals bisect at 90° to each other.

(iv) A square is a parallelogram, since having each adjacent angle a right angle and
opposite sides are equal.

29.  (i) If diagonals of a quadrilateral bisect each other then it is a rhombus, parallelogram, rectangle or square.

(ii) If diagonals of a quadrilateral are perpendicular bisector of each other, then it is a rhombus or square.

(iii) If diagonals are equal, then it is a square or rectangle.

30.  A rectangle is a convex quadrilateral since its vertex are raised and both of its diagonals lie in its interior.

31.  Since, two right triangles make a rectangle where 0 is equidistant point from A, B, C and D because 0 is the mid-point of the two diagonals of a rectangle.

Since AC and BD are equal diagonals and intersect at mid-point.

So, 0 is the equidistant from A, B, C and D.

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