Rd Sharma 2019 2020 for Class 8 Maths Chapter 16 - Understanding Shapes Ii Quadrilaterals

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Page No 16.15:

Question 1:

Define the following terms:
(i) Quadrilateral
(ii) Convex Quadrilateral

Answer:

(i) A quadrilateral is a polygon that has four sides (or edges) and four vertices (or corners).
It can be any four-sided closed shape.

(ii) A convex quadrilateral is a mathematical figure whose every internal angle is less than or equal to 180 degrees.

Page No 16.15:

Question 2:

In a quadrilateral, define each of the following:
(i) Sides
(ii) Vertices
(iii) Angles
(iv) Diagonals
(v) Adjacent angles
(vi) Adjacent sides
(vii) Opposite sides
(viii) Opposite angles
(ix) Interior
(x) Exterior

Answer:

$(i) In a quadrilateral ABCD, the four line segments AB, BC, CD and DA are called its sides . (ii) The vertices of a quadrilateral are the corner s of the quadrilateral . A quadrilateral has four vertices . (iii) The meeting point of two sides of a quadrilateral is called an angle . A quadrilteral has four angles . (iv) In a quadrilateral ABCD, the line segments AC and BD are called its diagonals . (v) The angles which have a common side as their arm are called adjacent angles . (vi) Two sides are adjacent if they have a common end point . (vii) Two sides are opposite if they do not have a common end point . (viii) The two angles of a quadrilateral which are not adjacent are called opposite angles . (ix) The part of the plane made up by all such points that are enclosed by quadrilateral is called the interior . (x) The part of the plane made up by all the points that are not enclosed by quadrilateral is called the exterior .$

Page No 16.15:

Question 3:

Complete each of the following, so as to make a true statement:
(i) A quadrilateral has ....... sides.
(ii) A quadrilateral has ...... angles.
(iii) A quadrilateral has ..... vertices, no three of which are .....
(iv) A quadrilateral has .... diagonals.
(v) The number of pairs of adjacent angles of a quadrilateral is .......
(vi) The number of pairs of opposite angles of a quadrilateral is .......
(vii) The sum of the angles of a quadrilateral is ......
(viii) A diagonal of a quadrilateral is a line segment that joins two ...... vertices of the quadrilateral.
(ix) The sum of the angles of a quiadrilateral is .... right angles.
(x) The measure of each angle of a convex quadrilateral is ..... 180°.
(xi) In a quadrilateral the point of intersection of the diagonals lies in .... of the quadrilateral.
(xii) A point is in the interior of a convex quadrilateral, if it is in the ..... of its two opposite angles.
(xiii) A quadrilateral is convex if for each side, the remaining .... lie on the same side of the line containing the side.

Answer:

(i) four
(ii) four
(iii) four, collinear
(iv) two
(v) four
(vi) two
(vii) 360°
(viii) opposite
(ix) four
(x) less than
(xi) the interior
(xii) interiors
(xiii) vertices

Page No 16.16:

Question 4:

In Fig. 16.19, ABCD is a quadrilateral.
(i) Name a pair of adjacent sides.
(ii) Name a pair of opposite sides.
(iii) How many pairs of adjacent sides are there?
(iv) How many pairs of opposite sides are there?
(v) Name a pair of adjacent angles.
(vi) Name a pair of opposite angles.
(vii) How many pairs of adjacent angles are there?
(viii) How many pairs of opposite angles are there?

Answer:

$(i) (AB, BC) or (BC, CD) or (CD, DA) or (AD, AB) (ii) (AB, CD) or (BC, DA) (iii) Four (iv) Two (v) (∠ A, ∠ B) or (∠ B, ∠ C) or (∠ C, ∠ D) or (∠ D, ∠ A) (vi) (∠ A, ∠ C) or (∠ B, ∠ D) (vii) Four (viii) Two$

Page No 16.16:

Question 5:

The angles of a quadrilateral are 110°, 72°, 55° and x°. Find the value of x.

Answer:

$The sum of anlges of a quadilateral is 360 ° . So, we get 110 ° + 72 ° + 55 ° + x = 360 ° \Rightarrow 237 ° + x = 360 ° \Rightarrow x = 360 ° - 237 ° ∴ x = 123 °$

Page No 16.16:

Question 6:

The three angles of a quadrilateral are respectively equal to 110°, 50° and 40°. Find its fourth angle.

Answer:

$Let x be the the fourth angle . Since, the sum of all the angles of a quadrilateral is 360 °, we have : 110 ° + 50 ° + 40 ° + x = 360 ° \Rightarrow 200 ° + x = 360 ° \Rightarrow x = 160 ° ∴ The fourth angle is 160 ° .$

Page No 16.16:

Question 7:

A quadrilateral has three acute angles each measures 80°. What is the measure of the fourth angle?

Answer:

$Let x be the fourth angle . Since, the sum of all angles of a quadrilateral is 360 °, we have : 80 ° + 80 ° + 80 ° + x = 360 ° \Rightarrow 240 ° + x = 360 ° \Rightarrow x = 120 ° ∴ The fourth angle is 120 ° .$

Page No 16.16:

Question 8:

A quadrilateral has all its four angles of the same measure. What is the measure of each?

Answer:

$Let x be the measure of each angle . Since, the sum of all the angles of a quadrilateral is 360 °, we have : x ° + x ° + x ° + x ° = 360 ° \Rightarrow 4 x ° = 360 ° \Rightarrow x ° = 90 ° ∴ The measure of each angle is 90 ° .$

Page No 16.16:

Question 9:

Two angles of a quadrilateral are of measure 65° and the other two angles are equal. What is the measure of each of these two angles?

Answer:

$Let x be the measure of each angle . Since, the sum of all the angles of a quadrilateral is 360 °, we have : 65 ° + 65 ° + x ° + x ° = 360 ° \Rightarrow 2 x ° + 130 ° = 360 ° \Rightarrow 2 x ° = 230 ° \Rightarrow x ° = 115 ° ∴ The measure of each angle is 115 ° .$

Page No 16.16:

Question 10:

Three angles of a quadrilateral are equal. Fourth angle is of measure 150°. What is the measure of equal angles.

Answer:

$Let x be the measure of the equal angle s of the quadrilateral . Since, the sum of all the angles of a quadrilateral is 360 °, we have : x ° + x ° + x ° + 150 ° = 360 ° \Rightarrow 3 x ° = 360 ° - 150 ° \Rightarrow x ° = 210 ° ∴ The measure of each angle is 70 ° .$

Page No 16.16:

Question 11:

The four angles of a quadrilateral are as 3 : 5 : 7 : 9. Find the angles.

Answer:

$Let the angles be in the ratio 3 x : 5 x : 7 x : 9 x . Since, the sum of all the angles of a quadrilateral is 360 °, w e h a v e : 3 x + 5 x + 7 x + 9 x = 360 ° \Rightarrow 24 x = 360 ° \Rightarrow x = 15 ° Thus, the angles are : 3 x = 45 ° 5 x = 75 ° 7 x = 105 ° 9 x = 135 °$

Page No 16.16:

Question 12:

If the sum of the two angles of a quadrilateral is 180°. What is the sum of the remaining two angles?

Answer:

$Let (x + y) be the sum of the remaining two angles . Since, the sum of all the angles of a quadrilateral is 360 °, we have : 180 ° + (x + y) ° = 360 ° \Rightarrow (x + y) ° = 180 ° ∴ The sum of the remaining two angles is 180 ° .$

Page No 16.16:

Question 13:

In Fig. 16.20, find the measure of ∠MPN.

Answer:

$Since the sum of all the angles of a quadrilateral is 360 °, we have : 45 ° + 90 ° + 90 ° + ∠ MPN = 360 ° \Rightarrow 225 ° + ∠ MPN = 360 ° ∴ ∠ MPN = 135 °$

Page No 16.16:

Question 14:

The sides of a quadrilateral are produced in order. What is the sum of the four exterior angles?

Answer:

$The sides of the quadrilateral ABCD are produced in order (according to figure) . Now, we need to find the sum of the exterior angles . Since the angles made on the same side of straight line are 180 °, i . e ., linear pair, we have : a + x + b + y + c + z + w + d = 180 ° + 180 ° + 180 ° + 180 ° = 720 ° OR Sum of the interior angles + sum of exterior the angles = 180 ° \times 4 = 720 ° Since the sum of the interior angles of a quadrilateral is 360 °, w e h a v e : w + x + y + z = 360 ° Substituing the value, we get : a + b + c + d = 360 ° ∴ Sum of the exterior angles = 360 °$

Page No 16.16:

Question 15:

In Fig. 16.21, the bisectors of ∠A and ∠B meet at a point P. If ∠C = 100° and ∠D = 50°, find the measure of ∠APB.

Answer:

$∠ A + ∠ B + ∠ C + ∠ D = 360 ° \Rightarrow ∠ A + ∠ B + 100 ° + 50 ° = 360 ° \Rightarrow ∠ A + ∠ B = 210 ° . . . (i) In △ APB, we have : \frac{1}{2} ∠ A + \frac{1}{2} ∠ B + ∠ APB = 180 ° \Rightarrow ∠ APB = 180 ° - \frac{1}{2} (∠ A + ∠ B) From (i), we get : \Rightarrow ∠ APB = 180 ° - (\frac{1}{2} \times 210 °) ∴ ∠ APB = 75 °$

Page No 16.17:

Question 16:

In a quadrilateral ABCD, the angles A, B, C and D are in the ratio 1 : 2 : 4 : 5. Find the measure of each angle of the quadrilateral.

Answer:

$Let x be the measure of each angle . Then the ratio becomes x : 2 x : 4 x : 5 x . Since, the sum of all angles in a quadrilateral is 360 °, we have : x + 2 x + 4 x + 5 x = 360 ° \Rightarrow 12 x = 360 ° \Rightarrow x = \frac{360 °}{12} \Rightarrow x = 30 ° Thus, the angles are : x = 30 ° 2 x = 60 ° 4 x = 120 ° 5 x = 150 °$

Page No 16.17:

Question 17:

In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively. Prove that $∠ C O D = \frac{1}{2} (∠ A + ∠ B) .$

Answer:

$∠ COD = 180 ° - (∠ OCD + ∠ ODC) = 180 ° - \frac{1}{2} (∠ C + ∠ D) = 180 ° - \frac{1}{2} [360 ° - (∠ A + ∠ B)] = 180 ° - 180 ° + \frac{1}{2} (∠ A + ∠ B) = \frac{1}{2} (∠ A + ∠ B) = RHS Hence proved .$

Page No 16.17:

Question 18:

Find the number of sides of a regular polygon, when each of its angles has a measure of
(i) 160°
(ii) 135°
(iii) 175°
(iv) 162°
(v) 150°

Answer:

$(i) Each interior angle = {(\frac{2 n - 4}{n} \times 90)}^{°} S o, {(\frac{2 n - 4}{n} \times 90)}^{°} = 160 ° \Rightarrow \frac{2 n - 4}{n} = \frac{160 °}{90 °} \Rightarrow \frac{2 n - 4}{n} = \frac{16}{9} \Rightarrow 18 n - 36 = 16 n \Rightarrow 2 n = 36 ∴ n = 18 (ii) Each interior angle = {(\frac{2 n - 4}{n} \times 90)}^{°} So, {(\frac{2 n - 4}{n} \times 90)}^{°} = 135 ° \Rightarrow \frac{2 n - 4}{n} = \frac{135 °}{90 °} \Rightarrow \frac{2 n - 4}{n} = \frac{3}{2} \Rightarrow 4 n - 8 = 3 n ∴ n = 8 (iii) Each interior angle = {(\frac{2 n - 4}{n} \times 90)}^{°} So, {(\frac{2 n - 4}{n} \times 90)}^{°} = 175 ° \Rightarrow \frac{2 n - 4}{n} = \frac{175 °}{90 °} \Rightarrow \frac{2 n - 4}{n} = \frac{35}{18} \Rightarrow 36 n - 72 = 35 n ∴ n = 72 (iv) Each interior angle = {(\frac{2 n - 4}{n} \times 90)}^{°} So, {(\frac{2 n - 4}{n} \times 90)}^{°} = 162 ° \Rightarrow \frac{2 n - 4}{n} = \frac{162 °}{90 °} \Rightarrow \frac{2 n - 4}{n} = \frac{9}{5} \Rightarrow 10 n - 20 = 9 n ∴ n = 20 (v) Each interior angle = {(\frac{2 n - 4}{n} \times 90)}^{°} So, {(\frac{2 n - 4}{n} \times 90)}^{°} = 150 ° \Rightarrow \frac{2 n - 4}{n} = \frac{150 °}{90 °} \Rightarrow \frac{2 n - 4}{n} = \frac{5}{3} \Rightarrow 6 n - 12 = 5 n ∴ n = 12$

Page No 16.17:

Question 19:

Find the number of degrees in each exterior exterior angle of a regular pentagon.

Answer:

$Each exterior angle = {(\frac{360}{n})}^{°} For a regular pentagon, n = 5 . ∴ Exterior angle = {(\frac{360}{5})}^{°} = 72 °$

Page No 16.17:

Question 20:

The measure of angles of a hexagon are x°, (x − 5)°, (x − 5)°, (2x − 5)°, (2x − 5)°, (2x + 20)°. Find the value of x.

Answer:

$Since the sum of all the angles of a hexagon is 720 °, we get : x ° + (x - 5) ° + (x - 5) ° + (2 x - 5) ° + (2 x - 5) ° + (2 x + 20) ° = 720 ° \Rightarrow x ° + x ° - 5 ° + x ° - 5 ° + 2 x - 5 ° + 2 x - 5 ° + 2 x + 20 ° = 720 ° \Rightarrow 9 x - 20 ° + 20 ° = 720 ° \Rightarrow 9 x = 720 ° ∴ x = 80$

Page No 16.17:

Question 21:

In a convex hexagon, prove that the sum of all interior angle is equal to twice the sum of its exterior angles formed by producing the sides in the same order.

Answer:

$For a convex hexagon, interior angle = (\frac{2 n - 4}{n} \times 90 °) For a hexagon, n = 6 ∴ Interior angle = (\frac{12 - 4}{6} \times 90 °) = (\frac{8}{6} \times 90 °) = 120 ° So, the sum of all the interior angles = 120 ° + 120 ° + 120 ° + 120 ° + 120 ° + 120 ° = 720 ° ∴ Exterior angle = {(\frac{360}{n})}^{°} = {(\frac{360}{6})}^{°} = 60^{°} So, sum of all the exterior angles = 60^{°} + 60^{°} + 60^{°} + 60^{°} + 60^{°} + 60^{°} = 360^{°} Now, sum of all interior angles = 720 ° = 2 (360 °) = twice the exterior angles Hence proved .$

Page No 16.17:

Question 22:

The sum of the interior angles of a polygon is three times the sum of its exterior angles. Determine the number of sided of the polygon.

Answer:

$\{(2 n - 4) \times 90 °\} = 3 \times (\frac{360 °}{n} \times n) \Rightarrow (n - 2) \times 180 = 3 \times 360 \Rightarrow n - 2 = 6 ∴ n = 8$

Page No 16.17:

Question 23:

Determine the number of sides of a polygon whose exterior and interior angles are in the ratio 1 : 5.

Answer:

$Let n be the number of sides of a polygon . Let x and 5 x b e the exterior and interior angles . Since the sum of an interior and the corresponding exterior angle is 180 °, we have : x + 5 x = 180 ° \Rightarrow 6 x = 180 ° \Rightarrow x = 30 ° The polygon has n sides . So, sum of all the exterior angles = (30 n) ° We know that the sum of all the exterior angles of a polygon is 360 ° . i . e ., 30 n = 360 ∴ n = 12$

Page No 16.17:

Question 24:

PQRSTU is a regular hexagon. Determine each angle of ΔPQT.

Answer:

$A regular hexagon is made up of 6 equilateral triangles . So, ∠ PQT = 60 ° and ∠ QTP = 30 ° Since the sum of the angles of ∆ PQT is 180 °, we have : ∠ P + ∠ Q + ∠ T = 180 ° \Rightarrow ∠ P + 60 ° + 30 ° = 180 ° \Rightarrow ∠ P = 180 ° - 90 ° \Rightarrow ∠ QPT = 90 ° ∴ The angles of the triangle are 90 °, 60 ° and 30 ° .$

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