## RD Sharma Solutions Class 8 Chapter 16 Exercise 16.1

**Exercise 16.1**

**Q 1. Define the following terms:**

**(i) Quadrilateral (ii) Convex Quadrilateral**

**SOLUTION:**

**(i)** A quadrilateral is a polygon having 4 sides (or edges) and 4 vertices (or corners). So, we can say that any closed shape having four sides is a quadrilateral.

**(ii)**Â If a mathematical figure has all internal angles equal to or less than 180 degrees, then it is known as a convex quadrilateral.

**Q 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**

**SOLUTION:**

**(i)** In a quadrilateral PQRS, the 4 line segments PQ, QR, RS and SP are called its sides.

**(ii)**Â The four corners of a quadrilateral are called the four vertices.

**(iii)**An angleÂ is the meeting point of 2 sides of a quadrilateral and so a quadrilateral has 4 angles.

**(iv)** In a quadrilateral PQRS, the line segments PR and QS will be its diagonals.

**(v)** The angles having their arm as a common side is called adjacent angles.

**(vi)**Â if 2 sides have a common endpoint then they are adjacent.

**(vii)**Â ifÂ 2 sides areÂ opposite, they do not have a common endpoint.

**(viii)** The two non-adjacent angles of a quadrilateral are called opposite angles.

**(ix) **Interior is that part of the plane of a quadrilateral which encloses all the points in it.

**(x)** ExteriorÂ is that part of the plane of a quadrilateral which does not encloses all the points in it.

**Q 3. Complete each of the following, so as to make a true statement:**

**(i) A quadrilateral has â€¦.. sides. **

**(ii) A quadrilateral has â€¦â€¦. angles. **

**(iii) A quadrilateral has â€¦â€¦.. vertices, no three of which are â€¦â€¦.. **

**(iv) A quadrilateral has …. diagonals. **

**(v) The number of pairs of adjacent angles of a quadrilateral is â€¦â€¦â€¦ **

**(vi) The number of pairs of opposite angles of a quadrilateral is â€¦â€¦. **

**(vii) The sum of the angles of a quadrilateral is â€¦â€¦â€¦ **

**(viii) A diagonal of a quadrilateral is a line segment that joins two â€¦â€¦â€¦ vertices of the quadrilateral. **

**(ix) The sum of the angles of a quadrilateral is …. right angles.**

**(x) The measure of each angle of a convex quadrilateral is ….. 180Â°. **

**(xi) In a quadrilateral, the point of intersection of the diagonals lies in â€¦â€¦ 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.**

**SOLUTION:**

**(i)** four

**(ii)** four

**(iii)** four, collinear

**(iv)** two

**(v)** four

**(vi)** two

**(vii)** 360Â°

**(viii)** opposite

**(ix)** four

**(x)** less than

**(xi)** the interior

**(xii)** interiors

**(xiii)** vertices

**Q 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?**

**SOLUTION:**

**(i)** (AB, BC) or (BC, CD) or (CD, DA) or (AD, AB)

**(ii)** (AB, CD) or (BC, DA)

**(iii)** Four

**(iv)** Two

**(v)** (\(\angle\)A, \(\angle\)B) or (\(\angle\)B, \(\angle\)C) or (\(\angle\)C, \(\angle\)D) or (\(\angle\)D, \(\angle\)A)

**(vi)** (\(\angle\)A, \(\angle\)C) or (\(\angle\)B, \(\angle\)D)

**(vii)** Four

**(viii)** Two

**Q 5. The angles of a quadrilateral are 110Â°, 72Â°, 55Â° and xÂ°. Find the value of x.**

**SOLUTION:**

**The sum of the four angles of a quadrilateral is 360Â°.**

Hence, we get 110Â° + 72Â° + 55Â° + x = 360Â°

237Â° + x = 360Â°

x = 360Â° – 237Â°

Thus, x = 123Â°

**Q 6. The three angles of a quadrilateral are respectively equal to 110Â°, 50Â° and 40Â°. Find its fourth angle.**

**SOLUTION:**

Let us assume y be the 4th angle.

As we know, **the sum of all the 4 angles of a quadrilateral is 360Â°**, we will have :

110Â° + 50Â° + 40Â° + y = 360Â°

200Â° + y = 360 Â°

y = 160Â°

Thus, The 4th angle is 160Â°.

**Q 7. A quadrilateral has three acute angles each measure 80Â°. What is the measure of the fourth angle?**

**SOLUTION:**

Let us assume y be the 4th angle.

As the sum of all 4 angles of a quadrilateral is 360Â°, we will have :

80Â° + 80Â° + 80Â° + y = 360Â°

240Â° + y = 360Â°

y = 120Â°

Thus, The 4th angle is 120Â°.

**Q 8. A quadrilateral has all its four angles of the same measure. What is the measure of each?**

**SOLUTION:**

Let us assume y be the measure of each angle.

As **the sum of all the 4 angles of a quadrilateral is 360Â°**, we willÂ have :

yÂ° + yÂ° + yÂ° + yÂ° = 360Â°

4yÂ° = 360Â°

yÂ° = 90Â°

Thus, the measure of each angle of the quadrilateral is 90Â°.

**Q 9. Two angles of a quadrilateral are of measure 65Â° and the other two angles are equal. What is the measure of each of these two angles?**

**SOLUTION:**

Let us assume y be the measure of each angle.

As **the sum of all the 4Â angles of a quadrilateral is 360Â°**, we will have :

65Â° + 65Â° + yÂ° + yÂ° = 360Â°

2yÂ° + 130Â° = 360Â°

2yÂ° = 230Â°

yÂ° = 115Â°

Thus, the measure of each angle of the quadrilateral is 115Â°.

**Q 10. Three angles of a quadrilateral are equal. The fourth angle is of measure 150Â°. What is the measure of equal angles?**

**SOLUTION:**

Let us assume yÂ be the measure of all the equal angles of the quadrilateral.

As, **the sum of all the 4 angles of a quadrilateral is 360Â°**, we will have :

yÂ° + yÂ° + yÂ° + 150Â° = 360Â°

3yÂ° = 360Â° – 150Â°

yÂ° = 210Â°

Thus, the measure of each angle of the quadrilateral is 70Â°.

**Q 11. The four angles of a quadrilateral are as 3 : 5 : 7 : 9. Find the angles.**

**SOLUTION:**

Let all the angles of the quadrilateral be in the ratio 3y : 5y: 7y : 9y.

As **the sum of all the 4 angles of a quadrilateral is 360Â°**, we will have :

3y + 5y + 7y + 9y = 360Â°

24y = 360Â°

y = 15Â°

Therefore, the angles are :

3y = 45Â°

5y = 75Â°

7y = 105Â°

9y = 135Â°

**Q 12. If the sum of the two angles of a quadrilateral is 180Â°. What is the sum of the remaining two angles?**

**SOLUTION:**

Let us assume (y + z) be the sum of the remaining 2 angles.

As **the sum of all the 4 angles of a quadrilateral is 360Â°**, we willÂ have :

180Â° + (y + z) Â° = 360Â°

(y + z) Â° = 180Â°

Thus, the sum of the remaining 2 angles is 180Â°.

**Q 13. In Fig. 16.20, find the measure of \(\angle\)MPN.**

**SOLUTION:**

As **the sum of all the 4 angles of a quadrilateral is 360Â°**, we will have :

45Â° + 90Â° + 90Â° + \(\angle\)MPN = 360Â°

225Â° + \(\angle\)MPN = 360Â°

Thus, \(\angle\)MPN = 135Â°

**Q 14. The sides of a quadrilateral are produced in order. What is the sum of the four exterior angles?**

**SOLUTION:**

**The sum of the interior angles of quadrilateral + sum of exterior angles of the quadrilateral = 180Â°** \(\times\) 4 = 720Â°

As** the sum of the 4 interior angles of a quadrilateral is 360Â°**, we will have :

w + x + y + z = 360Â°

Putting the value, we will get :

a + b + c + d = 360Â°

Sum of the 4 exterior angles = 360 Â°

**Q 15. In Fig.16.21, the bisectors of \(\angle\)A and \(\angle\)B meet at a point P. If \(\angle\)C =100Â° and \(\angle\)D = 50Â°, find the measure of \(\angle\)APB.**

**SOLUTION:**

\(\angle\)A + \(\angle\)B + \(\angle\)C + \(\angle\)D = 360Â°

\(\angle\)A + \(\angle\)B + 100Â° + 50Â° = 360Â°

\(\angle\)A + \(\angle\)B = 210Â° … (a)

In the triangle APB, we have :

\(\frac{1}{2}\) \(\angle\)A + \(\frac{1}{2}\) \(\angle\)B + \(\angle\)APB = 180Â°

\(\angle\)APB = 180Â° – \(\frac{1}{2}\) (\(\angle\)A + \(\angle\)B)

From (a), we will get :

\(\angle\)APB = 180Â° – (\(\frac{1}{2}\) \(\times\)210Â°)

Thus, \(\angle\)APB = 75Â°

**Q 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.**

**SOLUTION:**

Let us assume yÂ be the measure of each angle.

Then the ratio will become y : 2y : 4y : 5y.

Thus, **the sum of all 4 angles in a quadrilateral is 360Â°**, we will have :

y + 2y + 4y + 5y = 360Â°

12y = 360Â°

y = \(\frac{360^{\circ}}{12^{\circ}}\)

y = 30Â°

Therefore, the angles are :

y = 30Â°

2y = 60Â°

4y = 120Â°

5y = 150Â°

**Q 17. In a quadrilateral ABCD, CO and DO are the bisectors of \(\angle\)C and \(\angle\)D respectively. Prove that \(\angle\)COD = â€”1 (\(\angle\)A + \(\angle\)B).**

**SOLUTION:**

\(\angle\)COD = 180Â° – (\(\angle\)OCD + \(\angle\)ODC)

= 180Â° – \(\frac{1}{2}\) (\(\angle\)C + \(\angle\)D)

= 180Â° – \(\frac{1}{2}\) [360Â° – (\(\angle\)A + \(\angle\)B)]

= 180Â° – 180Â° + \(\frac{1}{2}\) (\(\angle\)A + \(\angle\)B)

= \(\frac{1}{2}\) (\(\angle\)A + \(\angle\)B)

= RHS

Hence it is proved.

**Q 18. Find the number of sides of a regular polygon, when each of its angles has a measure of**

**(i) 160Â° (ii) 135Â° (iii) 175Â° (iv) 162Â° (v) 150Â°**

**SOLUTION:**

**(i)** Each interior angle of polygon = \(\left ( \frac{2n-4}{n}\times 90 \right )^{\circ}\)

So, \(\left ( \frac{2n-4}{n}\times 90 \right )^{\circ}\) = 160Â°

\(\Rightarrow \frac{2n-4}{n}=\frac{160^{\circ}}{90^{\circ}}\)

\(\Rightarrow \frac{2n-4}{n}=\frac{16}{9}\)

\(\Rightarrow 18n-36=16n\)

\(\Rightarrow 2n=36\)

Thus, n = 18

**(ii)** Each interior angleÂ of polygon = \(\left ( \frac{2n-4}{n}\times 90 \right )^{\circ}\)

So, \(\left ( \frac{2n-4}{n}\times 90 \right )^{\circ}\) = 135Â°

\(\Rightarrow \frac{2n-4}{n}=\frac{135^{\circ}}{90^{\circ}}\)

\(\Rightarrow \frac{2n-4}{n}=\frac{3}{2}\)

\(\Rightarrow 4n-8=3n\)

\(\Rightarrow n=8\)

Thus, n = 8

**(iii)** Each interior angleÂ of polygon = \(\left ( \frac{2n-4}{n}\times 90 \right )^{\circ}\)

Thus, \(\left ( \frac{2n-4}{n}\times 90 \right )^{\circ}\) = 175Â°

\(\Rightarrow \frac{2n-4}{n}=\frac{175^{\circ}}{90^{\circ}}\)

\(\Rightarrow \frac{2n-4}{n}=\frac{35}{18}\)

\(\Rightarrow 36n-72=35n\)

\(\Rightarrow n=72\)

therefore, n = 72

**(iv)** Each interior angle of polygon = \(\left ( \frac{2n-4}{n}\times 90 \right )^{\circ}\)

Therefore, \(\left ( \frac{2n-4}{n}\times 90 \right )^{\circ}\) = 162Â°

\(\Rightarrow \frac{2n-4}{n}=\frac{162^{\circ}}{90^{\circ}}\)

\(\Rightarrow \frac{2n-4}{n}=\frac{9}{5}\)

\(\Rightarrow 10n-20=9n\)

\(\Rightarrow n=20\)

Thus, n = 20

**(v)** Each interior angleÂ of polygon = \(\left ( \frac{2n-4}{n}\times 90 \right )^{\circ}\)

Therefore, \(\left ( \frac{2n-4}{n}\times 90 \right )^{\circ}\) = 150Â°

\(\Rightarrow \frac{2n-4}{n}=\frac{150^{\circ}}{90^{\circ}}\)

\(\Rightarrow \frac{2n-4}{n}=\frac{5}{3}\)

\(\Rightarrow 6n-12=5n\)

\(\Rightarrow n=12\)

Thus, n = 12.

**Q 19. Find the number of degrees in each exterior angle of a regular pentagon. **

**SOLUTION:**

Each exterior angle of pentagon = \(\left (\frac{360}{n} \right )^{\circ}\)

For a regular pentagon, n will be = 5.

Thus, exterior angle = \(\left (\frac{360}{5} \right )^{\circ}\) = 72 \(^{\circ}\)

**Q 20. The measure of angles of a hexagon are xÂ°, (x – 5)Â°, (x – 5)Â°, (2x – 5)Â°, (2x – 5)Â°, (2x + 20)Â°. Find the value of x.**

**SOLUTION:**

Therefore **the sum of all the 6 angles of a hexagon is 720Â°**, we will get :

xÂ° + (x – 5)Â° + (x – 5)Â° + (2x – 5)Â° + (2x – 5)Â° + (2x + 20) Â° = 720Â°

xÂ° + xÂ° – 5Â° + xÂ° – 5Â° + 2x – 5Â° + 2x – 5Â° + 2x + 20Â° = 720Â°

9x – 20Â° + 20Â° = 720Â°

9x = 720Â°

Thus, x = 80

**Q 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.**

**SOLUTION:**

For a convex hexagon, interior angle = \(\left ( \frac{2n-4}{n}\times 90 \right )^{\circ}\)

For a hexagon, n = 6

Thus, Interior angle = \(\left ( \frac{12-4}{6}\times 90 \right )^{\circ}\)

= \(\left ( \frac{8}{6}\times 90 \right )^{\circ}\)

= 120Â°

Therefore, the sum of all the 6 interior angles = 120Â° + 120Â° + 120Â° + 120Â° + 120Â° + 120Â° = 720Â°

Thus, Exterior angle = \(\left (\frac{360}{n} \right )^{\circ}\) = \(\left (\frac{360}{6} \right )^{\circ}\) = 60Â°

Hence, sum of all the exterior angles = 60Â° + 60Â° + 60Â° + 60Â° + 60Â° + 60Â° = 360Â°

Now, sum of all interior angles = 720Â°

= 2(360Â°)

= two times the exterior angles

Hence it is proved.

**Q 22. The sum of the interior angles of a polygon is three times the sum of its exterior angles. Determine the number of sides of the polygon.**

**SOLUTION:**

{(2n – 4) \(\times\) 90Â°} = 3 \(\times\) (\(\left (\frac{360}{n} \right )^{\circ}\) \(\times\) n)

=> (n – 2) \(\times\)180 = 3 \(\times\)360

=> n – 2 = 6

Thus, n = 8.

**Q 23. Determine the number of sides of a polygon whose exterior and interior angles are in the ratio 1 : 5.**

**SOLUTION:**

Let us assume p be the number of sides of the polygon.

Let us assume a and 5b be the exterior and interior angles respectively.

Therefore **the sum of an interior & the corresponding exterior angle is 180**Â°, we will have :

a + 5b = 180Â°

6b = 180Â°

b = 30Â°

As the polygon has p sides.

Hence, the sum of all the exterior angles = (30p)Â°

The sum of all the exterior angles of a polygon is 360Â°.

i.e., 30p = 360

Thus, p = 12.

**Q 24. PQRSTU is a regular hexagon. Determine each angle of triangle PQT.**

**SOLUTION:**

As we know that regular hexagon is made up of six equilateral triangles.

Therefore, \(\angle\)PQT = 60Â° and \(\angle\)QTP = 30Â°

As** the sum of the angles of trianglePQT is 180**Â°, we willÂ have :

\(\angle\)P + \(\angle\)Q + \(\angle\)T = 180Â°

\(\angle\)P + 60Â° + 30Â° = 180Â°

\(\angle\)P = 180Â° – 90Â°

\(\angle\)QQPT = 90Â°

So, the angles of the triangle are 90Â°, 60Â° and 30Â°.