RD Sharma solution class 8 chapter 16 Understanding Shapes II(Quadrilaterals) Exercise 16.1

Exercise 16.1

Page-16.15

Question 1:

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

Answer 1:

(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.

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 2:

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

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 3:

(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-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 4:

$$\begin{gathered} \left(\mathrm{i}\right) (\mathrm{AB}, \mathrm{BC}) \mathrm{or} (\mathrm{BC}, \mathrm{CD}) \mathrm{or} (\mathrm{CD}, \mathrm{DA}) \mathrm{or} (\mathrm{AD}, \mathrm{AB}) \\ \left(\mathrm{ii}\right) (\mathrm{AB}, \mathrm{CD}) \mathrm{or} (\mathrm{BC}, \mathrm{DA}) \\ \left(\mathrm{iii}\right) \mathrm{Four} \\ \left(\mathrm{iv}\right) \mathrm{Two} \\ \left(\mathrm{v}\right) (\angle \mathrm{A}, \angle \mathrm{B}) \mathrm{or} (\angle \mathrm{B}, \angle \mathrm{C}) \mathrm{or} (\angle \mathrm{C}, \angle \mathrm{D}) \mathrm{or} (\angle \mathrm{D}, \angle \mathrm{A}) \\ \left(\mathrm{vi}\right) (\angle \mathrm{A}, \angle \mathrm{C}) \mathrm{or} (\angle \mathrm{B}, \angle \mathrm{D}) \\ \left(\mathrm{vii}\right) \mathrm{Four} \\ \left(\mathrm{viii}\right) \mathrm{Two} \end{gathered}$$

Question 5:

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

Answer 5:

$$\begin{gathered} \mathrm{The} \mathrm{sum} \mathrm{of} \mathrm{anlges} \mathrm{of} \mathrm{a} \mathrm{quadilateral} \mathrm{is} 360°. \\ \mathrm{So}, \mathrm{we} \mathrm{get} 110°+72°+55°+x=360° \\ \Rightarrow 237°+x=360° \\ \Rightarrow x=360°-237° \\ \therefore x=123° \end{gathered}$$

Question 6:

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

Answer 6:

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

Question 7:

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

Answer 7:

$$\begin{gathered} \mathrm{Let} x \mathrm{be} \mathrm{the} \mathrm{fourth} \mathrm{angle}. \\ \mathrm{Since}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{quadrilateral} \mathrm{is} 360°, \mathrm{we} \mathrm{have}: \\ 80°+80°+80°+x=360° \\ \Rightarrow 240°+x=360° \\ \Rightarrow x=120° \\ \therefore \mathrm{The} \mathrm{fourth} \mathrm{angle} \mathrm{is} 120°. \end{gathered}$$

Question 8:

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

Answer 8:

$$\begin{gathered} \mathrm{Let} x \mathrm{be} \mathrm{the} \mathrm{measure} \mathrm{of} \mathrm{each} \mathrm{angle}. \\ \mathrm{Since}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{quadrilateral} \mathrm{is} 360°, \mathrm{we} \mathrm{have}: \\ x°+x°+x°+x°=360° \\ \Rightarrow 4x°=360° \\ \Rightarrow x°=90° \\ \therefore \mathrm{The} \mathrm{measure} \mathrm{of} \mathrm{each} \mathrm{angle} \mathrm{is} 90°. \end{gathered}$$

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 9:

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

Question 10:

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

Answer 10:

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

Question 11:

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

Answer 11:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{angles} \mathrm{be} \mathrm{in} \mathrm{the} \mathrm{ratio} 3x : 5x : 7x : 9x. \\ \mathrm{Since}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{quadrilateral} \mathrm{is} 360°, we have: \\ 3x+5x+7x+9x=360° \\ \Rightarrow 24x=360° \\ \Rightarrow x=15° \\ \mathrm{Thus}, \mathrm{the} \mathrm{angles} \mathrm{are}: \\ 3x=45° \\ 5x=75° \\ 7x=105° \\ 9x=135° \end{gathered}$$

Question 12:

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

Answer 12:

$$\begin{gathered} \mathrm{Let} \left(x+y\right) \mathrm{be} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{remaining} \mathrm{two} \mathrm{angles}. \\ \mathrm{Since}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{quadrilateral} \mathrm{is} 360°, \mathrm{we} \mathrm{have}: \\ 180°+\left(x+y\right)°=360° \\ \Rightarrow \left(x+y\right)°=180° \\ \therefore \mathrm{The} \mathrm{sum} \mathrm{of} \mathrm{the} \mathrm{remaining} \mathrm{two} \mathrm{angles} \mathrm{is} 180°. \end{gathered}$$

Question 13:

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

Answer 13:

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{quadrilateral} \mathrm{is} 360°, \mathrm{we} \mathrm{have}: \\ 45°+90°+90°+\angle \mathrm{MPN}=360° \\ \Rightarrow 225°+\angle \mathrm{MPN}=360° \\ \therefore \angle \mathrm{MPN}=135° \end{gathered}$$

Question 14:

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

Answer 14:

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

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 15:

$$\begin{gathered} \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360° \\ \Rightarrow \angle \mathrm{A}+\angle \mathrm{B}+100°+50°=360° \\ \Rightarrow \angle \mathrm{A}+\angle \mathrm{B}=210° ...\left(\mathrm{i}\right) \\ \mathrm{In} △\mathrm{APB}, \mathrm{we} \mathrm{have}: \\ \frac{1}{2}\angle \mathrm{A}+\frac{1}{2}\angle \mathrm{B}+\angle \mathrm{APB}=180° \\ \Rightarrow \angle \mathrm{APB}=180°-\frac{1}{2}\left(\angle \mathrm{A}+\angle \mathrm{B}\right) \\ \mathrm{From} \left(\mathrm{i}\right), \mathrm{we} \mathrm{get}: \\ \Rightarrow \angle \mathrm{APB}=180°-\left(\frac{1}{2}\times 210°\right) \\ \therefore \angle \mathrm{APB}=75° \end{gathered}$$

Page-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 16:

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

Question 17:

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

Answer 17:

$$\begin{gathered} \angle \mathrm{COD} =180°-\left(\angle \mathrm{OCD}+\angle \mathrm{ODC}\right) \\ =180°-\frac{1}{2}\left(\angle \mathrm{C}+\angle \mathrm{D}\right) \\ =180°-\frac{1}{2}\left[360°-\left(\angle \mathrm{A}+\angle \mathrm{B}\right)\right] \\ =180°-180°+\frac{1}{2}\left(\angle \mathrm{A}+\angle \mathrm{B}\right) \\ =\frac{1}{2}\left(\angle \mathrm{A}+\angle \mathrm{B}\right) \\ = \mathrm{RHS} \\ \mathrm{Hence} \mathrm{proved}. \end{gathered}$$

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 18:

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

Question 19:

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

Answer 19:

$$\begin{gathered} \mathrm{Each} \mathrm{exterior} \mathrm{angle}={\left(\frac{360}{\mathrm{n}}\right)}^{°} \\ \mathrm{For} \mathrm{a} \mathrm{regular} \mathrm{pentagon}, n=5. \\ \therefore \mathrm{Exterior} \mathrm{angle}={\left(\frac{360}{5}\right)}^{°}=72° \end{gathered}$$

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 20:

$$\begin{gathered} \mathrm{Since} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{hexagon} \mathrm{is} 720°, \mathrm{we} \mathrm{get}: \\ x°+\left(x-5\right)°+\left(x-5\right)°+\left(2x-5\right)°+\left(2x-5\right)°+\left(2x+20\right)°=720° \\ \Rightarrow x°+x°-5°+x°-5°+2x-5°+2x-5°+2x+20°=720° \\ \Rightarrow 9x-20°+20°=720° \\ \Rightarrow 9x=720° \\ \therefore x=80 \end{gathered}$$

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 21:

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

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 22:

$$\begin{gathered} \left\{\left(2\mathrm{n}-4\right)\times 90°\right\}=3\times \left(\frac{360°}{\mathrm{n}}\times \mathrm{n}\right) \\ \Rightarrow \left(\mathrm{n}-2\right)\times 180=3\times 360 \\ \Rightarrow \mathrm{n}-2=6 \\ \therefore \mathrm{n}=8 \end{gathered}$$

Question 23:

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

Answer 23:

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

Question 24:

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

Answer 24:

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