Rd Sharma 2019 2020 for Class 8 Maths Chapter 17 - Understanding Shapes Iii Special Types Of Quadrilaterals

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{No}. This is because the \mathrm{oppos}ite \mathrm{angles} \mathrm{are} \mathrm{not} \mathrm{equal}. \\ \left(\mathrm{ii}\right) \\ \mathrm{Yes}. This is because the opposite sides are equal. \\ \left(\mathrm{ii}\right) \\ \mathrm{No}, This is because the \mathrm{diagonals} \mathrm{do} \mathrm{not} \mathrm{bisect} \mathrm{each} \mathrm{other}. \end{gathered}$$

Answer:

$$\begin{gathered} \angle \mathrm{HOP}+70°=180° (l\mathrm{inear} \mathrm{pair}) \\ \angle \mathrm{HOP}=180°-70°=110° \\ \mathrm{x}=\angle \mathrm{HOP}=110° (o\mathrm{pposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \\ \angle \mathrm{EHP}+\angle \mathrm{HEP}=180° (s\mathrm{um} \mathrm{of} \mathrm{adjacent} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{is} 180°) \\ 110°+40°+\mathrm{z}=180° \\ \mathrm{z}=180°-150°=30° \\ \mathrm{y}=40° \left(\mathrm{alternate} \mathrm{angle}s \right) \end{gathered}$$

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{Opposite} \mathrm{sides} \mathrm{are} \mathrm{equal} \mathrm{in} \mathrm{a} \mathrm{parallelogram}. \\ \therefore 3\mathrm{y}-1=26 \\ 3\mathrm{y}=27 \\ \mathrm{y}=9 \\ \mathrm{Similarly}, 3\mathrm{x}=18 \\ \mathrm{x}=6 \\ \left(\mathrm{ii}\right) \\ \mathrm{Diagonals} \mathrm{bisect} \mathrm{each} \mathrm{other} \mathrm{in} \mathrm{a} \mathrm{parallelogram}. \\ \therefore \mathrm{y}-7=20 \\ \mathrm{y}=27 \\ \mathrm{x}-\mathrm{y}=16 \\ \mathrm{x}-27=16 \\ \mathrm{x}=43 \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{In} \mathrm{the} \mathrm{parallelogram} \mathrm{RISK}: \\ \angle \mathrm{ISK}+\angle \mathrm{RKS}=180° (\mathrm{sum} \mathrm{of} \mathrm{adjacent} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{is} 180°) \\ \angle \mathrm{ISK}=180°-120°=60° \\ \mathrm{Similarly}, \mathrm{in} \mathrm{parallelogram} \mathrm{CLUE}: \\ \angle \mathrm{CEU}=\angle \mathrm{CLU}=70° (\mathrm{opposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \\ \mathrm{In} \mathrm{the} \mathrm{triangle}: \\ x+\angle \mathrm{ISK}+\angle \mathrm{CEU}=180° \\ x=180°-\left(70°+60°\right) \\ x=180°-\left(70°+60°\right)=50° \end{gathered}$$

Answer:

$$\begin{gathered} O\mathrm{ppostie} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{congurent}. \\ \therefore \left(3x-2\right)°=\left(50-x\right)° \\ 3x°-2°=50°-x° \\ 3x°+x°=50°+2° \\ 4x°=52° \\ x°=13° \\ Putting \mathrm{the} \mathrm{value} \mathrm{of} x \mathrm{in} \mathrm{one} \mathrm{angle}: \\ 3x°-2°=39°-2° \\ =37° \\ O\mathrm{pposite} \mathrm{angles} \mathrm{are} \mathrm{congurent}: \\ \therefore 50-x° \\ =37° \\ \mathrm{Let} \mathrm{the} \mathrm{remaining} \mathrm{two} \mathrm{angles} \mathrm{be} y and z. \\ \mathrm{Angles} y \mathrm{and} z \mathrm{are} \mathrm{congurent} \mathrm{because} \mathrm{they} \mathrm{are} \mathrm{also} \mathrm{opposite} \mathrm{angles}. \\ \therefore y=z \\ \mathrm{The} \mathrm{sum} \mathrm{of} \mathrm{adjacent} angles \mathrm{of} a paralle\log ram \mathrm{is} \mathrm{equal} \mathrm{to} 180°. \\ \therefore 37°+y=180° \\ y=180°-37° \\ y=143° \\ \mathrm{So}, \mathrm{the} \mathrm{anlges} \mathrm{measure} \mathrm{are}: \\ 37°, 37°, 143° and 143° \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Two} \mathrm{adjacent} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{add} \mathrm{up} \mathrm{to} 180°. \\ \mathrm{Let} x \mathrm{be} \mathrm{the} \mathrm{angle}. \\ \therefore x+\frac{2x}{3}=180° \\ \frac{5x}{3}==180° \\ x=72° \\ \frac{2x}{3}=\frac{2\times 72°}{3}=108° \\ \mathrm{Thus}, \mathrm{two} \mathrm{of} \mathrm{the} \mathrm{angles} \mathrm{in} \mathrm{the} \mathrm{parallelogram} \mathrm{are} 108° \mathrm{and} \mathrm{the} \mathrm{other} \mathrm{two} \mathrm{are} 72°. \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Given} \mathrm{that} \mathrm{one} \mathrm{angle} \mathrm{of} \mathrm{the} \mathrm{parallelogram} \mathrm{is} 70°. \\ \mathrm{Since} \mathrm{opposite} \mathrm{angles} \mathrm{have} \mathrm{same} \mathrm{value}, \mathrm{if} \mathrm{one} \mathrm{is} 70°, \mathrm{then} \mathrm{the} \mathrm{one} \mathrm{directly} \mathrm{opposite} \mathrm{will} \mathrm{also} \mathrm{be} 70°. \\ \mathrm{So}, \mathrm{let} \mathrm{one} \mathrm{angle} \mathrm{be} \mathrm{x}°. \\ \mathrm{x}°+70°=180° (\mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{adjacent} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{is} 180° ) \\ \mathrm{x}°=180°-70° \\ \mathrm{x}°=110° \\ \mathrm{Thus}, \mathrm{the} \mathrm{remaining} \mathrm{angles} \mathrm{are} 110°, 110° \mathrm{and} 70°. \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{angle} \mathrm{be} \mathrm{A} \mathrm{and} \mathrm{B}. \\ The a\mathrm{ngles} \mathrm{are} \mathrm{in} \mathrm{the} \mathrm{ratio} \mathrm{of} 1:2. \\ M\mathrm{easures} \mathrm{of} \angle \mathrm{A} \mathrm{and} \angle \mathrm{B} \mathrm{are} \mathrm{x}° \mathrm{and} 2\mathrm{x}°. \\ \mathrm{Then}, \angle \mathrm{C}=\angle \mathrm{A} \mathrm{and} \angle \mathrm{D}=\angle \mathrm{B} (o\mathrm{pposite} \mathrm{angles} \mathrm{of} a \mathrm{parallelogram} \mathrm{are} \mathrm{congruent}) \\ \mathrm{As} \mathrm{we} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{adjacent} \mathrm{angle}s \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{is} 180°. \\ \therefore \angle \mathrm{A}+\angle \mathrm{B}=180° \\ \Rightarrow \mathrm{x}°+2\mathrm{x}°=180° \\ \Rightarrow 3\mathrm{x}°=180° \\ \Rightarrow \mathrm{x}°=\frac{180°}{3}=60° \\ \mathrm{Thus}, \mathrm{measure} \mathrm{of} \angle \mathrm{A}=60°, \angle \mathrm{B}=120°, \angle \mathrm{C}=60° and \angle \mathrm{D}=120°. \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{In} a \mathrm{parallelogram}, \mathrm{opposite} \mathrm{angles} \mathrm{have} \mathrm{the} \mathrm{same} \mathrm{value}. \\ \therefore \angle \mathrm{D}=\angle \mathrm{B} \\ =135° \\ \mathrm{Also}, \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360° \\ \angle \mathrm{A}+\angle \mathrm{D}=180° \left(\mathrm{opposite} \mathrm{angles} \mathrm{have} \mathrm{the} \mathrm{same} \mathrm{value}\right) \\ \angle \mathrm{A}=180°-135°=45° \\ \angle \mathrm{A}=45° \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Opposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}. \\ \therefore \angle \mathrm{C}=70°=\angle \mathrm{A}. \\ \angle \mathrm{B}=\angle \mathrm{D} \\ \mathrm{Also}, \mathrm{the} \mathrm{sum} o\mathrm{f} \mathrm{the} \mathrm{adjacent} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{is} 180°. \\ \therefore \angle \mathrm{A}+\angle \mathrm{B}=180° \\ 70°+\angle \mathrm{B}=180° \\ \angle \mathrm{B}=110° \\ \therefore \angle \mathrm{B}=110°, \angle \mathrm{C}=70° \mathrm{and} \angle \mathrm{D}=110° \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{angles} \mathrm{be} \mathrm{A}, \mathrm{B}, \mathrm{C} \mathrm{and} \mathrm{D}. \\ It is g\mathrm{iven} \mathrm{that} the \mathrm{sum} \mathrm{of} \mathrm{two} \mathrm{opposite} \mathrm{angles} \mathrm{is} 130°. \\ \therefore \angle \mathrm{A}+\angle \mathrm{C}=130° \\ \angle \mathrm{A}+\angle \mathrm{A}=130° \left(\mathrm{opp}o\mathrm{site} \mathrm{angle}s \mathrm{of} a \mathrm{parallelogram} \mathrm{are} \mathrm{same}\right) \\ \angle \mathrm{A}=65° \\ \mathrm{and} \angle \mathrm{C}=65° \\ The s\mathrm{um} \mathrm{of} adjacent angles of a paralle\log ram is 180°. \\ \angle \mathrm{A}+\angle \mathrm{B}=180° \\ 65°+\angle \mathrm{B}=180° \\ \angle \mathrm{B}=180°-65° \\ \angle \mathrm{B}=115° \\ \angle \mathrm{D}=115° \\ \therefore \angle \mathrm{A}=65°, \mathrm{v}\angle \mathrm{B}=115°, \angle \mathrm{C}=65° \mathrm{and} \angle \mathrm{D}=115°. \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{angle} \mathrm{be} \mathrm{x}. \\ A\mathrm{ll} \mathrm{the} \mathrm{angles} \mathrm{are} \mathrm{equal}. \\ \therefore x+x+x+x=360° \\ 4x=360° \\ x=90° \\ \mathrm{So}, \mathrm{each} \mathrm{angle} \mathrm{is} 90° \mathrm{and} \mathrm{quadrilateral} \mathrm{is} \mathrm{a} \mathrm{parallelogram}. \mathrm{It} \mathrm{is} \mathrm{a} \mathrm{rectangle}. \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{opposite} \mathrm{sides} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}. \\ \mathrm{Two} \mathrm{sides} \mathrm{are} \mathrm{given}, \mathrm{i}.\mathrm{e}. 4 \mathrm{cm} \mathrm{and} 3 \mathrm{cm}. \\ \mathrm{Therefore}, \mathrm{the} \mathrm{rest} \mathrm{of} \mathrm{the} \mathrm{sies} \mathrm{will} \mathrm{also} \mathrm{be} 4 \mathrm{cm} \mathrm{and} 3 \mathrm{cm}. \\ \therefore P\mathrm{erimeter} = S\mathrm{um} \mathrm{of} \mathrm{all} the \mathrm{sides} \mathrm{of} a \mathrm{parallelogram} \\ =4+3+4+3 \\ =14 \mathrm{cm} \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Opposite} \mathrm{sides} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{same}. \\ \mathrm{Le}t \mathrm{two} \mathrm{sides} \mathrm{of} the \mathrm{parallelogram} \mathrm{be} x \mathrm{and} y. \\ \mathrm{Given}: \\ x=y+25 \\ \mathrm{Also}, x+y+x+y=150 (Perimeter= Sum of all the sides of a paralle\log ram) \\ y+25+y+y+25+y=150 \\ 4y=150-50 \\ 4y=100 \\ y=\frac{100}{4}=25 \\ \therefore x=y+25=25+25=50 \\ \mathrm{Thus}, \mathrm{the} \mathrm{length}s \mathrm{of} \mathrm{the} \mathrm{sides} \mathrm{of} \mathrm{the} \mathrm{parallelogram} are 50 \mathrm{cm} \mathrm{and} 25 \mathrm{cm}. \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Shorter} \mathrm{side}=4.8 \mathrm{cm} \\ \mathrm{Longer} \mathrm{side}=\frac{4.8}{2}+4.8=7.2 \mathrm{cm} \\ \mathrm{Perimeter}=\mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{the} \mathrm{sides} \\ =4.8+4.8+7.2+7.2 \\ =24 \mathrm{cm} \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{adjacent} \mathrm{angle}s \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{supplementry}. \\ \mathrm{Hence}, \left(3x+10\right)° and \left(3x-4\right)° \mathrm{are} \mathrm{supplementry}. \\ \left(3x+10\right)°+\left(3x-4\right)°=180° \\ 6x°+6°=180° \\ 6x°=174° \\ x=29° \\ \mathrm{First} \mathrm{angle} = \left(3\mathrm{x}+10\right)°=\left(3\times 29°+10°\right)=97° \\ \mathrm{Second} \mathrm{angle} = \left(3\mathrm{x}-4\right)°=83° \\ \mathrm{Thus}, \mathrm{the} \mathrm{angles} \mathrm{of} \mathrm{the} \mathrm{parallelogram} \mathrm{are} 97°, 83°, 97° \mathrm{and} 83°. \end{gathered}$$

Answer:

$$\begin{gathered} \angle ABC=30° \\ \therefore \angle ADC=30° \left(\mathrm{opposite} \mathrm{angle} \mathrm{of} the \mathrm{parallelogram}\right) \\ \mathrm{and} \angle BDA=\angle \mathrm{ADC}-\angle \mathrm{BDC}=30°-10°=20° \\ \angle BAC=\angle ACD=70° (alternate angle) \\ \mathrm{In} △ \mathrm{ABC}: \\ \angle \mathrm{CAB}+\angle \mathrm{ABC}+\angle \mathrm{BCA}=180° \\ 70°+30°+\angle \mathrm{BCA}=180° \\ \therefore \angle \mathrm{BCA}=80° \\ \angle \mathrm{DAB}=\angle \mathrm{DAC}+\angle \mathrm{CAB}=70°+80°=150° \\ \angle \mathrm{BCD}=150° \left(\mathrm{opposite} \mathrm{angle} \mathrm{of} the \mathrm{parallelogram}\right) \\ \angle DCA=\angle CAB=70° \\ \mathrm{In} △\mathrm{DOC}: \\ \angle \mathrm{ODC}+\angle \mathrm{DOC}+\angle \mathrm{OCD}=180 \\ 10°+70°+\angle \mathrm{DOC}=180° \\ \therefore \angle \mathrm{DOC}=100° \\ \angle \mathrm{DOC}+\angle \mathrm{BOC}=180° \\ \angle \mathrm{BOC}=180°-100° \\ \angle \mathrm{BOC}=80° \\ \angle AOD=\angle BOC=80° \left(\mathrm{ve}\mathrm{rtically} \mathrm{opposite} \mathrm{angles}\right) \\ \angle \mathrm{AOB}=\angle \mathrm{DOC}=100° \left(\mathrm{ve}\mathrm{rtically} \mathrm{opposite} \mathrm{angles}\right) \\ \angle \mathrm{CAB}=70° \left(\mathrm{given}\right) \\ \angle \mathrm{ADB}=20° \\ \angle DBA=\angle BDC=10° (alternate angle) \\ \angle ADB=\angle DBC=20° (alternate angle) \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{In} △\mathrm{CEB}: \\ \angle \mathrm{ECB}+\angle \mathrm{CBE}+\angle \mathrm{BEC}=180° (angle sum property of a triangle) \\ 40°+90°+\angle \mathrm{EBC}=180° \\ \therefore \angle \mathrm{EBC}=50° \\ \mathrm{Also}, \angle \mathrm{EBC}=\angle \mathrm{ADC}=50° \left(\mathrm{opposite} \mathrm{angle} \mathrm{of} \mathrm{a} \mathrm{parallelogram}\right) \\ \mathrm{In} △\mathrm{FDC}: \\ \angle \mathrm{FDC}+\angle \mathrm{DCF}+\angle \mathrm{CFD}=180° \\ 50°+90°+\angle \mathrm{DCF}=180° \\ \therefore \angle \mathrm{DCF}=40° \\ \mathrm{Now}, \angle \mathrm{BCE}+\angle \mathrm{ECF}+\angle \mathrm{FCD}+\angle FDC=180° (\mathrm{in} \mathrm{a} \mathrm{parallelogram}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{alternate} \mathrm{angles} \mathrm{is} 180° ) \\ 50°+40°+\angle \mathrm{ECF}+40°=180° \\ \angle \mathrm{ECF}=180°-50°+40°-40°=50° \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Draw} \mathrm{a} \mathrm{parallelogram} \mathrm{ABCD}. \\ \mathrm{Drop} \mathrm{a} \mathrm{perpendicular} \mathrm{from} \mathrm{B} \mathrm{to} \mathrm{the} \mathrm{side} \mathrm{AD}, \mathrm{at} \mathrm{the} \mathrm{point} \mathrm{E}. \\ \mathrm{Drop} \mathrm{perpendicular} \mathrm{from} \mathrm{B} \mathrm{to} \mathrm{the} \mathrm{side} \mathrm{CD}, \mathrm{at} \mathrm{the} \mathrm{point} \mathrm{F}. \\ \mathrm{In} \mathrm{the} \mathrm{quadrilateral} \mathrm{BEDF}: \\ \angle \mathrm{EBF}=60°,\angle \mathrm{BED}=90° \\ \angle \mathrm{BFD}=90° \\ \angle \mathrm{EDF}=360°-(60°+90°+90°)=120° \\ \mathrm{In} \mathrm{a} \mathrm{parallelogram}, \mathrm{opposite} \mathrm{angles} \mathrm{are} \mathrm{congruent} \mathrm{and} \mathrm{adjacent} \mathrm{angles} \mathrm{are} \mathrm{supplementary}. \\ \mathrm{In} \mathrm{the} \mathrm{parallelogram} \mathrm{ABCD}: \\ \angle \mathrm{B}=\angle \mathrm{D}=120° \\ \angle \mathrm{A}=\angle \mathrm{C}=180°-120°=60° \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Both} \mathrm{the} \mathrm{parallelograms} \mathrm{ABCD} \mathrm{and} \mathrm{AEFG} \mathrm{are} \mathrm{similar}. \\ \therefore \angle \mathrm{C}=\angle \mathrm{A}=55° (\mathrm{opposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \\ \therefore \angle \mathrm{A}=\angle F=55° (\mathrm{opposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{In} \mathrm{parallelogram} \mathrm{BDEF} \\ \therefore \mathrm{BD}=\mathrm{EF} ...\left(\mathrm{i}\right) (\mathrm{opposite} \mathrm{sides} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \\ \mathrm{In} \mathrm{parallelogram} \mathrm{DCEF} \\ \mathrm{CD}=\mathrm{EF} ...\left(\mathrm{ii}\right) (\mathrm{opposite} \mathrm{sides} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \\ \mathrm{From} \mathrm{equations} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) \\ \mathrm{BD}=\mathrm{CD} \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{In} ∆\mathrm{FDE}: \\ \mathrm{DE}=\mathrm{DF} \\ \therefore \angle \mathrm{FED}=\angle \mathrm{DFE}.............\left(\mathrm{i}\right) (\mathrm{angles} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{sides}) \\ \mathrm{In} \mathrm{the} {\mathrm{II}}^{\mathrm{gm} }\mathrm{BDEF}: \\ \angle \mathrm{FBD}= \angle \mathrm{FED}.......\left(\mathrm{ii}\right) (\mathrm{opposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \\ \mathrm{In} \mathrm{the} {\mathrm{II}}^{\mathrm{gm} } \mathrm{DCEF}: \\ \angle \mathrm{DCE}=\angle \mathrm{DFE}......\left(\mathrm{iii}\right) (\mathrm{opposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \\ \mathrm{From} \mathrm{equations} \left(\mathrm{i}\right), \left(\mathrm{ii}\right) \mathrm{and} \left(\mathrm{iii}\right): \\ \angle \mathrm{FBD}=\angle \mathrm{DCE} \\ \mathrm{In} △\mathrm{ABC}: \\ \mathrm{If} \angle \mathrm{FBD}=\angle \mathrm{DCE}, \mathrm{then} \mathrm{AB}=\mathrm{AC} (\mathrm{sides} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{angles}). \\ \mathrm{Hence}, △\mathrm{ABC} \mathrm{is} \mathrm{isosceles}. \end{gathered}$$

Answer:

(i) Diagonals of a parallelogram bisect each other.
(ii) Alternate angles
(iii) Vertically opposite angles
(iv)
 $$\begin{gathered} \mathrm{In} ∆\mathrm{BOY} \mathrm{and} ∆\mathrm{DOX}: \\ \mathrm{OB}=\mathrm{OD} (d\mathrm{iagonals} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{bisect} \mathrm{each} \mathrm{o}\mathrm{ther}) \\ \angle \mathrm{OBY}=\angle \mathrm{ODX} (a\mathrm{lternate} \mathrm{a}\mathrm{ngles}) \\ \angle \mathrm{BOY}=\angle \mathrm{DOX} (v\mathrm{ertically} \mathrm{o}\mathrm{pposite} \mathrm{a}\mathrm{ngles}) \end{gathered}$$


ASA congruence:
XO = YO (c.p.c.t)
So, XY is bisected at O.

Answer:

(i) True, since opposite angles of a parallelogram are equal.
(ii) True, as AF is the bisector of $\angle$A.
(iii) True, as CE is the bisector of $\angle$C.
(iv) True
               $\angle$CEB = $\angle$DCE........(i)  (alternate angles)
               $\angle$DCE= $\angle$ FAB.........(ii)    (opposite angles of a parallelogram are equal)
            
  From equations (i) and (ii):
              $\angle$CEB = $\angle$FAB
                
(v) True, as corresponding angles are equal ($\angle$CEB = $\angle$FAB).

Answer:

$$\begin{gathered} \mathrm{In} \mathrm{\Delta AOL} \mathrm{and} \mathrm{\Delta CMO}: \\ \angle \mathrm{AOL}=\angle \mathrm{COM}( v\mathrm{ertically} \mathrm{opposite} \mathrm{angle})....\left(i\right) \\ \angle \mathrm{ALO}=\angle \mathrm{CMO}=90° (e\mathrm{ach} \mathrm{right} \mathrm{angle}).....\left(ii\right) \\ \mathrm{Usin}g angle sum property: \\ \angle \mathrm{AOL}+\angle \mathrm{ALO}+\angle LAO=180°..........\left(iii\right) \\ \angle COM+\angle CMO+\angle OCM=180°......\left(iv\right) \\ From equations \left(iii\right) and \left(iv\right): \\ \angle \mathrm{AOL}+\angle \mathrm{ALO}+\angle LAO=\angle COM+\angle CMO+\angle OCM \\ \angle LAO=\angle OCM (from equations (i) and (ii) ) \\ \mathrm{In} \mathrm{\Delta AOL} \mathrm{and} \mathrm{\Delta CMO}: \\ \angle \mathrm{ALO}=\angle \mathrm{CMO} (e\mathrm{ach} \mathrm{right} \mathrm{angle}) \\ \mathrm{AO}=\mathrm{OC} (\mathrm{diagonals} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{bisect} \mathrm{each} \mathrm{other}) \\ \angle LAO=\angle OCM (proved above) \\ \mathrm{So}, \mathrm{\Delta AOL} \mathrm{is} \mathrm{congruent} \mathrm{to} \mathrm{\Delta CMO} \left(S\mathrm{AS}\right). \\ \Rightarrow \mathrm{AL}=\mathrm{CM} \left[\mathrm{cpct}\right] \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{In} \mathrm{the} {\mathrm{II}}^{\mathrm{gm}} \mathrm{ABCD}: \\ \mathrm{AO}=\mathrm{OC}......\left(\mathrm{i}\right) (\mathrm{diagonals} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{bisect} \mathrm{each} \mathrm{other}) \\ \mathrm{AE}=\mathrm{CF}.......\left(\mathrm{ii}\right) \left(\mathrm{given}\right) \\ \mathrm{Subtracting} \left(\mathrm{ii}\right) \mathrm{from} \left(\mathrm{i}\right): \\ \mathrm{AO}-\mathrm{AE}=\mathrm{OC}-\mathrm{CF} \\ \mathrm{EO}=\mathrm{OF}......... \left(\mathrm{iii}\right) \\ \mathrm{In} ∆\mathrm{DOE} \mathrm{and} ∆\mathrm{BOF}: \\ \mathrm{EO}=\mathrm{OF} (\mathrm{proved} \mathrm{above}) \\ \mathrm{DO}=\mathrm{OB} (\mathrm{diagonals} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{bisect} \mathrm{each} \mathrm{other}) \\ \angle \mathrm{DOE} =\angle \mathrm{BOF} (\mathrm{vertically} \mathrm{opposite} \mathrm{angles}) \\ \mathrm{By} \mathrm{SAS} \mathrm{congruence}: \\ ∆\mathrm{DOE}\cong ∆\mathrm{BOF} \\ \therefore \mathrm{DE}=\mathrm{BF} (\mathrm{c}.\mathrm{p}.\mathrm{c}.\mathrm{t}) \\ \mathrm{In} ∆\mathrm{BOE} \mathrm{and} ∆\mathrm{DOF}: \\ \mathrm{EO}=\mathrm{OF} (\mathrm{proved} \mathrm{above}) \\ \mathrm{DO}=\mathrm{OB} (\mathrm{diagonals} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{bisect} \mathrm{each} \mathrm{other}) \\ \angle \mathrm{DOF} =\angle \mathrm{BOE} (\mathrm{vertically} \mathrm{opposite} \mathrm{angles}) \\ \mathrm{By} \mathrm{SAS} \mathrm{congruence}: \\ ∆\mathrm{DOE}\cong ∆\mathrm{BOF} \\ \therefore \mathrm{DF}=\mathrm{BE} (\mathrm{c}.\mathrm{p}.\mathrm{c}.\mathrm{t}) \\ \mathrm{Hence}, \mathrm{the} \mathrm{pair} \mathrm{of} \mathrm{opposite} \mathrm{sides} \mathrm{are} \mathrm{equal}. \mathrm{Thus}, \mathrm{DEBF} \mathrm{is} \mathrm{a} \mathrm{parallelogram}. \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{AE} \mathrm{is} \mathrm{the} \mathrm{bisector} \mathrm{of} \angle \mathrm{A}. \\ \therefore \angle \mathrm{DAE}=\angle \mathrm{BAE} =x \\ \angle \mathrm{BAE}=\angle \mathrm{AED}=x (\mathrm{alternate} \mathrm{angles}) \\ \mathrm{Since} \mathrm{opposite} \mathrm{angles} \mathrm{in} ∆\mathrm{ADE} \mathrm{are} \mathrm{equal}, ∆\mathrm{ADE} \mathrm{is} \mathrm{an} \mathrm{isosceles} \mathrm{triangle}. \\ \therefore \mathrm{AD}=\mathrm{DE}= 6 \mathrm{cm} (\mathrm{sides} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{angles}) \\ \mathrm{AB}=\mathrm{CD}=10 \mathrm{cm} \\ \mathrm{CD}= \mathrm{DE}+ \mathrm{EC} \\ \Rightarrow \mathrm{EC}=\mathrm{CD}-\mathrm{DE} \\ \Rightarrow \mathrm{EC}=10-6=4 \mathrm{cm} \\ \angle \mathrm{DEA}=\angle \mathrm{CEF}=x (\mathrm{vertically} \mathrm{opposite} \mathrm{angle}) \\ \angle \mathrm{EAD}=\angle \mathrm{EFC}=x (\mathrm{alternate} \mathrm{angles}) \\ \mathrm{Since} \mathrm{opposite} \mathrm{angles} \mathrm{in} ∆\mathrm{EFC} \mathrm{are} \mathrm{equal}, ∆\mathrm{EFC} \mathrm{is} \mathrm{an} \mathrm{isosceles} \mathrm{triangle}. \\ \therefore \mathrm{CF}=\mathrm{CE}= 4 \mathrm{cm} (\mathrm{sides} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{angles}) \\ \therefore \mathrm{CF}= 4\mathrm{cm} \end{gathered}$$

Answer:

(i) True
(ii) True
(iii) True
(iv) False
(v) True
(vi) False

Diagonals of a rhombus are perpendicular, but not equal.

(vii) True
(viii) True

It is a parallelogram because it has two pairs of parallel sides.

(ix) True

It is a quadrilateral because it has four sides.

(x) True

It can be a square if each of the angle is a right angle.

(xi) False

It is not a square because each of the angle is a right angle in a square.

Answer:

(i) A rhombus is a parallelogram in which adjacent sides are equal.
(ii) A square is a rhombus in which all angles are right angled.
(iii) A rhombus has all its sides of equal length.
(iv) The diagonals of a rhombus bisect each other at right angles.
(v) If the diagonals of a parallelogram bisect each other at right angles, then it is a rhombus.

Answer:

No, it is not a rhombus. This is because diagonals of a rhombus must be perpendicular.

Answer:

No, it is not so.
Diagonals of a rhombus are perpendicular and bisect each other. Along with this, all of its sides are equal. In the figure given below, the diagonals are perpendicular to each other, but do not bisect each other.

Answer:

$$\begin{gathered} \mathrm{In} \mathrm{a} \mathrm{rhombus}, \mathrm{the} \mathrm{diagonals} \mathrm{are} \mathrm{perpendicular}. \\ \therefore \angle \mathrm{BPC}=90° \\ \mathrm{From} ∆ \mathrm{BPC}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{angles} \mathrm{is} 180°. \\ \therefore \angle \mathrm{CB}P+\angle \mathrm{BP}C+\angle PBC=180° \\ \angle \mathrm{CB}P=180°-\angle \mathrm{BP}C-\angle PBC \\ \angle \mathrm{CBP}=180°-40°-90°=50° \\ \angle ADB=\angle \mathrm{CBP}=50° (alternate angle) \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{All} \mathrm{sides} \mathrm{of} \mathrm{a} \mathrm{rhombus} \mathrm{are} \mathrm{equal} \mathrm{in} \mathrm{length}. \\ \mathrm{The} \mathrm{diagonals} \mathrm{intersect} \mathrm{at} 90° \mathrm{and} \mathrm{the} \mathrm{sides} \mathrm{of} \mathrm{the} \mathrm{rhombus} \mathrm{form} \mathrm{right} \mathrm{triangles}. \\ \mathrm{One} \mathrm{leg} \mathrm{of} \mathrm{these} \mathrm{right} \mathrm{triangles} \mathrm{is} \mathrm{equal} \mathrm{to} 8 \mathrm{cm} \mathrm{and} \mathrm{the} \mathrm{other} \mathrm{is} \mathrm{equal} \mathrm{to} 6 \mathrm{cm}. \\ \mathrm{The} \mathrm{sides} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{form} \mathrm{the} \mathrm{hypotenuse} \mathrm{of} \mathrm{these} \mathrm{right} \mathrm{triangles}. \\ \mathrm{So}, \mathrm{we} \mathrm{get}: \\ (8^2+6^2){\mathrm{cm}}^2 \\ =(64+36){\mathrm{cm}}^2 \\ =100 {\mathrm{cm}}^2 \\ \mathrm{The} \mathrm{hypotneuse} \mathrm{is} \mathrm{the} \mathrm{square} \mathrm{root} \mathrm{of} 100{\mathrm{cm}}^2. \mathrm{This} \mathrm{makes} \mathrm{the} \mathrm{hypotneuse} \mathrm{equal} \mathrm{to} 10. \\ \mathrm{Thus}, \mathrm{the} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{rhombus} \mathrm{is} \mathrm{equal} \mathrm{to} 10 \mathrm{cm}. \end{gathered}$$

Answer:

1. Draw AC equal to 10 cm.
2. Draw XY, the right bisector of AC, meeting it at O.
3. With O as centre and radius equal to half of the length of the other diagonal,
    i.e. 3 cm, cut OB = OD = 3 cm.
4. Join AB, AD and CB, CD.

Answer:

1. Draw a line segment AB of 3.5 cm.
2. Draw $\angle$BAX equal to 40$°$.
3. With A as centre and the radius equal to AB, cut AD at 3.5 cm.
4. With D as centre, cut an arc of radius 3.5 cm.
5. With B as centre, cut an arc of radius 3.5 cm. This arc cuts the arc of step 4 at C.
6. Join DC and BC.

Answer:

1. Draw a line segment AB of 4 cm.
2. Draw a perpendicular XY on AB, which intersects AB at P.
3. With P as centre, cut PE at 3.2 cm.
4. Draw a line WZ that passes through E. This line should be parallel to AB.
5. With A as centre, draw an arc of radius 4 cm that cuts WZ at D.
6. With D as centre and radius 4 cm, cut line DZ. Label it as point C.
4. Join AD and CB.

Answer:

1. Draw a line segment AC of 5 cm.
2. With A as centre, draw an arc of radius 6 cm on each side of AC.
3. With C as centre, draw an arc of radius 6 cm on each side of AC. These arcs intersect the arcs of step 2 at B and D.
4. Join AB, AD, CD and CB.

Answer:

(i) Yes

$$\begin{gathered} \mathrm{In} ∆\mathrm{BCO} \mathrm{and}∆\mathrm{DCO}: \\ \mathrm{OC}=\mathrm{OC} \left(\mathrm{common}\right) \\ \mathrm{BC}=\mathrm{DC} (\mathrm{all} \mathrm{sides} \mathrm{of} \mathrm{a} \mathrm{rhombus} \mathrm{are} \mathrm{equal}) \\ \mathrm{BO}=\mathrm{OD} (\mathrm{diagonal}s \mathrm{of} \mathrm{a} \mathrm{rhomus} \mathrm{bisect} \mathrm{each} \mathrm{other}) \\ \mathrm{By} \mathrm{SSS} \mathrm{congruence}: \\ ∆\mathrm{BCO} \cong ∆\mathrm{DCO} \end{gathered}$$

(ii) Yes
By c.p.c.t:
$\angle BCO=\angle DCO$

Answer:

$$\begin{gathered} \mathrm{In} ∆\mathrm{AED} \mathrm{and} ∆\mathrm{DEC}: \\ \mathrm{AE}=\mathrm{EC} (\mathrm{diagonals} \mathrm{bisect} \mathrm{each} \mathrm{other}) \\ \mathrm{AD}=\mathrm{DC} (\mathrm{sides} \mathrm{are} \mathrm{equal}) \\ \mathrm{DE}=\mathrm{DE} \left(\mathrm{common}\right) \\ \mathrm{By} \mathrm{SSS} \mathrm{congruence}: \\ ∆\mathrm{AED}\cong ∆\mathrm{CED} \\ \angle \mathrm{ADE}=\angle \mathrm{CDE} (\mathrm{c}.\mathrm{p}.\mathrm{c}.\mathrm{t}) \\ \mathrm{Similarly}, \mathrm{we} \mathrm{can} \mathrm{prove} ∆\mathrm{AEB} \mathrm{and} ∆\mathrm{BEC}, ∆\mathrm{BEC} \mathrm{and} ∆\mathrm{DEC}, ∆\mathrm{AED} \mathrm{and} ∆\mathrm{AEB} \mathrm{are} \mathrm{congruent} \mathrm{to} \mathrm{each} \mathrm{other}. \\ \mathrm{Hence}, \mathrm{diagonal} \mathrm{of} \mathrm{a} \mathrm{rhombus} \mathrm{bisects} \mathrm{the} \mathrm{angle} \mathrm{through} \mathrm{which} \mathrm{it} \mathrm{passes}. \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{diagonals} \mathrm{of} \mathrm{a} \mathrm{rhombus} \mathrm{bisect} \mathrm{each} \mathrm{other} \mathrm{at} \mathrm{right} \mathrm{angles}. \\ \therefore \mathrm{BO}=\frac{1}{2}\mathrm{BD}=(\frac{1}{2}\times 16) \mathrm{cm} \\ =8\mathrm{cm} \\ \mathrm{AB}=10 \mathrm{cm} \mathrm{and}\angle \mathrm{AOB}=90° \\ \mathrm{From} \mathrm{right} ∆\mathrm{OAB}: \\ {\mathrm{AB}}^2={\mathrm{AO}}^2+{\mathrm{BO}}^2 \\ \Rightarrow {\mathrm{AO}}^2=({\mathrm{AB}}^2–{\mathrm{BO}}^2) \\ \Rightarrow {\mathrm{AO}}^2=(10{)}^2-(8{)}^2{\mathrm{cm}}^2 \\ \Rightarrow {\mathrm{AO}}^2=(100-64) {\mathrm{cm}}^2=36{\mathrm{cm}}^2 \\ \Rightarrow \mathrm{AO}=\sqrt{36} \mathrm{cm}=6\mathrm{cm} \\ \therefore \mathrm{AC}=2\times \mathrm{AO}=(2\times 6) \mathrm{cm}=12 \mathrm{cm} \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{given} \mathrm{quadrilateral} \mathrm{be} \mathrm{ABCD} \mathrm{in} \mathrm{which} \mathrm{diagonals} \mathrm{AC} \mathrm{is} \mathrm{equal} \mathrm{to} 6 \mathrm{cm} \mathrm{and} \mathrm{BD} \mathrm{is} \mathrm{equal} \mathrm{to} 8 \mathrm{cm}. \\ \mathrm{Also}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that} \mathrm{the} \mathrm{diagonals} \mathrm{bisect} \mathrm{each} \mathrm{other} \mathrm{at} \mathrm{right} \mathrm{angle}, \mathrm{at} \mathrm{point} \mathrm{O}. \\ \therefore \mathrm{AO}=\mathrm{OC}=\frac{1}{2}\mathrm{AC}=3 \mathrm{cm} \\ \mathrm{Also}, \mathrm{OB}=\mathrm{OD}=\frac{1}{2}\mathrm{BD}=4 \mathrm{cm} \\ \mathrm{In} \mathrm{right} △\mathrm{AOB}: \\ {\mathrm{AB}}^2={\mathrm{OA}}^2+{\mathrm{OB}}^2 \\ \Rightarrow {\mathrm{AB}}^2=(9+16) {\mathrm{cm}}^2 \\ \Rightarrow {\mathrm{AB}}^2=25 {\mathrm{cm}}^2 \\ \Rightarrow \mathrm{AB}=5 \mathrm{cm} \\ \mathrm{Thus}, \mathrm{the} \mathrm{length} \mathrm{of} \mathrm{each} \mathrm{side} \mathrm{of} \mathrm{the} \mathrm{quadrilateral} \mathrm{is} 5 \mathrm{cm}. \end{gathered}$$

Answer:

(i) True
(ii) False
(iii) True
(iv) True
(v) False
(vi) False
The diagonals are not perpendicular to each other.
(vii) True
(viii) False
The diagonals are not perpendicular to each other.
(ix) False
All sides are not equal.
(x) True
(xi) True
(xii) False
All squares are parallelogram.

Answer:

(i) True
(ii) True
(iii) True
(iv) False
This is because the hypotenuse in any right angle triangle is always greater than its two sides.

Answer:

(i) A rectangle is a parallelogram in which each angle is a right angle.

(ii) A square is a rhombus in which each angle is a right angle.

(iii) A square is a rectangle in which the adjacent sides are equal.

Answer:

No, since diagonals of a rectangle are equal.

Answer:

In ∆ACB and ∆CAD:
AB = CD (rectangle property)

AD = BC (rectangle property)

AC ( common side )

Hence, by SSS criterion, it is proved that $∆ACB\cong ∆CAD$.

Answer:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{side} \mathrm{be} x \mathrm{cm} \mathrm{and} y cm. \\ \mathrm{So}, \mathrm{we} \mathrm{have}: \\ 2\left(x+y\right)=20 \\ \mathrm{Sides} \mathrm{are} \mathrm{in} \mathrm{the} \mathrm{ratio} 2:3. \\ \therefore y=\frac{3x}{2} \\ \mathrm{Putting} \mathrm{the} \mathrm{v}alue\mathit{ }of y: \\ 2\left(x+\frac{3x}{2}\right)=20 \\ \frac{2x+3x}{2}=10 \\ 5x=20 \\ x=4 \\ \therefore y=\frac{3\times 4}{2}=6 \\ \mathrm{Thus}, \mathrm{sides} \mathrm{of} \mathrm{the} \mathrm{rectangle} \mathrm{will} \mathrm{be} 4 \mathrm{cm} \mathrm{and} 6 \mathrm{cm}. \\ \mathrm{ABCD} \mathrm{is} \mathrm{the} \mathrm{rectangle} \mathrm{having} \mathrm{side}s 4 \mathrm{cm} \mathrm{and} 6 \mathrm{cm}. \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{side} \mathrm{be} x \mathrm{cm} \mathrm{and} y cm. \\ \mathrm{So}, \mathrm{we} \mathrm{have}: \\ 2\left(x+y\right)=90 \\ \mathrm{Sides} \mathrm{are} \mathrm{in} \mathrm{the} \mathrm{ratio} 4:5. \\ \therefore y=\frac{5x}{4} \\ \mathrm{Putting} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{y}: \\ 2\left(x+\frac{5x}{4}\right)=90 \\ \frac{4x+5x}{4}=45 \\ 9x=180 \\ x=20 \\ \therefore y=\frac{5\times 20}{4}=25 \\ \mathrm{Thus}, \mathrm{the} \mathrm{sides} \mathrm{of} \mathrm{the} \mathrm{rectangle} \mathrm{will} \mathrm{be} 20 \mathrm{cm} \mathrm{and} 25 \mathrm{cm}. \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Using} \mathrm{Pythagoras} \mathrm{theorem}: \\ A{D}^2+D{C}^2=A{C}^2 \\ {5}^2+{12}^2=A{C}^2 \\ 25+14=A{C}^2 \\ 169=A{C}^2 \\ AC=\sqrt{169} \\ =13 \mathrm{cm} \\ \mathrm{Thus}, \mathrm{length} \mathrm{of} \mathrm{the} \mathrm{diagonal} \mathrm{is} 13 \mathrm{cm}. \end{gathered}$$

Answer:

(i) Draw a side AB, equal to 8 cm.
(ii) With A as the centre, draw an arc of length 10 cm.
(iii) Draw $\angle$ABX = 90$°$, which intersects the arc at C.
(iv)Draw $\angle$BAY = 90$°$.
(v) With C as the centre, draw an arc of length 8 cm.
(vi) Join CD.
Thus, ABCD is the required rectangle.

Answer:

 $$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Draw} \mathrm{side} \mathrm{AB}=4.8 \mathrm{cm}. \\ \left(\mathrm{ii}\right) \mathrm{From} \mathrm{A}, \mathrm{make} \mathrm{an} \mathrm{angle} \mathrm{of} 90° \mathrm{and} \mathrm{cut} \mathrm{it} \mathrm{at} 4.8 \mathrm{cm} \mathrm{and} \mathrm{mark} \mathrm{it} \mathrm{point} \mathrm{D}. \\ \left(\mathrm{iii}\right) \mathrm{From} \mathrm{B}, \mathrm{make} \mathrm{an} \mathrm{angle} \mathrm{of} 90° \mathrm{and} \mathrm{cut} \mathrm{it} \mathrm{at} 4.8 \mathrm{cm} \mathrm{and} \mathrm{mark} \mathrm{it} \mathrm{point} \mathrm{C}. \\ \left(\mathrm{iv}\right) \mathrm{Join} \mathrm{C} \mathrm{and} \mathrm{D}. \\ \mathrm{Thus}, \mathrm{ABCD} \mathrm{is} \mathrm{the} \mathrm{required} \mathrm{square}. \end{gathered}$$

Answer:

(i) If all four sides are equal, then it can be either a square or a rhombus.

(ii) All four right angles, make it either a rectangle or a square.

Answer:

(i) Since a square has four sides, it is a quadrilateral.

(ii) Since the opposite sides are parallel and equal, it is a parallelogram.

(iii) Since the diagonals bisect each other and all the sides are equal, it is a rhombus.

(iv) Since the opposite sides are equal and all the angles are right angles, it is a rectangle.

Answer:

(i) Rhombus, parallelogram, rectangle and square
(ii) Rhombus and square
(iii) Rectangle and square

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Construct} \mathrm{a} \mathrm{triangle} \mathrm{ABC} \mathrm{right} \mathrm{angle} \mathrm{at} \mathrm{B}. \\ \left(\mathrm{ii}\right) \mathrm{Suppose} \mathrm{O} \mathrm{is} \mathrm{the} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{AC}. \\ \left(\mathrm{iii}\right)\mathrm{Complete} \mathrm{the} \mathrm{rectangle} \mathrm{ABCD} \mathrm{having} \mathrm{AC} \mathrm{as} \mathrm{its} \mathrm{diagonal}. \\ \mathrm{Since} \mathrm{diagonals} \mathrm{of} \mathrm{a} \mathrm{rectangle} \mathrm{are} \mathrm{equal} \mathrm{and} \mathrm{they} \mathrm{bisect} \mathrm{each} \mathrm{other}, \mathrm{O} \mathrm{is} \mathrm{the} \mathrm{midpoint} \mathrm{of} \mathrm{both} \mathrm{AC} \mathrm{and} \mathrm{BD}. \\ \therefore \mathrm{OA}=\mathrm{OB}=\mathrm{OC} \end{gathered}$$

Answer:

(i) By measuring each angle - Each angle of a rectangle is 90$°$.

(ii) By measuring the length of the diagonals - Diagonals of a rectangle are equal.

(iii) By measuring the sides of rectangle - Each pair of opposite sides are equal.

Answer:

The correct figure is

$$\begin{gathered} \left(\mathrm{i}\right) \\ \mathrm{AD}=\mathrm{BC} (\mathrm{opposite} \mathrm{sides} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{equal}) \\ \left(\mathrm{ii}\right) \\ \angle \mathrm{DCB}=\angle \mathrm{BAD} \left(\mathrm{opposite} \mathrm{angles} \mathrm{are} \mathrm{equal}\right) \\ \left(\mathrm{iii}\right) \\ \mathrm{OC}=\mathrm{OA} \left(\mathrm{diagonals} \mathrm{of} \mathrm{a} \mathrm{prallelogram} \mathrm{bisect} \mathrm{each} \mathrm{other}\right) \\ \left(\mathrm{iv}\right) \\ \angle \mathrm{DAB}+\angle \mathrm{CDA}=180° \left(\mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{two} \mathrm{adjacent} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{is} {180}^0\right) \end{gathered}$$

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \\ O\mathrm{pposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{same}. \\ \therefore \mathrm{x}=\mathrm{z} \mathrm{and} \mathrm{y}=100° \\ \mathrm{Also}, \mathrm{y}+\mathrm{z}=180° (s\mathrm{um} \mathrm{of} \mathrm{adjacent} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{quadrilateral} \mathrm{is} 180°) \\ \mathrm{z}+100°=180° \\ \mathrm{x}=180°-100° \\ \mathrm{x}=80° \\ \therefore \mathrm{x}=80°, \mathrm{y}=100° \mathrm{and} \mathrm{z}=80° \\ \left(\mathrm{ii}\right) \\ O\mathrm{pposite} \mathrm{angles} \mathrm{of} \mathrm{a} \mathrm{parallelogram} \mathrm{are} \mathrm{same}. \\ \therefore \mathrm{x}=\mathrm{y} \mathrm{and} \angle \mathrm{RQP}=100° \\ \angle \mathrm{PSR}+\angle \mathrm{SRQ}=180° \\ \mathrm{y}+50°=180° \\ \mathrm{x}=180°-50° \\ \mathrm{x}=130° \\ \therefore \mathrm{x}=130°, \mathrm{y}=130° \\ Since \mathrm{y} \mathrm{and} \mathrm{z} \mathrm{are} \mathrm{alternate} \mathrm{angles}, \mathrm{z}=130°. \\ \left(\mathrm{iii}\right) \\ S\mathrm{um} \mathrm{of} \mathrm{all} \mathrm{angles} \mathrm{in} \mathrm{a} \mathrm{triangle} \mathrm{is} {180}^{°}. \\ \therefore 30°+90°+\mathrm{z}=180° \\ \mathrm{z}=60° \\ O\mathrm{pposite} \mathrm{angles} \mathrm{are} \mathrm{equal} \mathrm{in} \mathrm{parallelogram}. \\ \therefore \mathrm{y}=\mathrm{z}=60° \\ \mathrm{and} \mathrm{x}=30° (\mathrm{alternate} \mathrm{angles}) \\ \left(\mathrm{iv}\right)x=90° (\mathrm{vertically} \mathrm{opposite} \mathrm{angle}) \\ S\mathrm{um} \mathrm{of} \mathrm{all} \mathrm{angles} \mathrm{in} \mathrm{a} \mathrm{triangle} \mathrm{is} {180}^{°}. \\ \therefore \mathrm{y}+90°+30°=180° \\ \mathrm{y}=180°-(90°+30°)=60° \\ \mathrm{y}=\mathrm{z}=60° (a\mathrm{lternate} \mathrm{angles}) \\ \left(\mathrm{v}\right) \\ O\mathrm{pposite} \mathrm{angles} \mathrm{are} \mathrm{equal} \mathrm{in} a \mathrm{parallelogram}. \\ \therefore \mathrm{y}=80° \\ \mathrm{y}+\mathrm{x}=180° \\ \mathrm{x}=180°-100°=80° \\ \mathrm{z}=\mathrm{y}=80° (\mathrm{alternate} \mathrm{angles}) \\ \left(\mathrm{vi}\right) \\ \mathrm{y}=112° (\mathrm{opposite} \mathrm{angles} \mathrm{are} \mathrm{equal} \mathrm{in} \mathrm{a} \mathrm{parallelogram}) \\ \mathrm{In}∆ \mathrm{UTW} : \\ \mathrm{x}+\mathrm{y}+40°=180° (a\mathrm{ngle} \mathrm{sum} \mathrm{property} \mathrm{of} a \mathrm{triangle}) \\ \mathrm{x}=180°-(112°-40°)=28° \\ B\mathrm{ottom} \mathrm{left} \mathrm{vertex}=180°-112°=68° \\ \therefore \mathrm{z}=\mathrm{x}=28° (\mathrm{alternate} \mathrm{angles}) \end{gathered}$$