\displaystyle \textbf{Exercise - 16}

\displaystyle \textbf{Question 1: }\text{In the adjoining figure, }BD\text{ is a diagonal of quadrilateral }ABCD.
\displaystyle \text{Show that }ABCD\text{ is a parallelogram and calculate its area.} \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }BD\text{ is the diagonal of quadrilateral }ABCD,
\displaystyle AB=6\text{ cm},\ CD=6\text{ cm},\ BD=8\text{ cm}
\displaystyle \text{and }\angle ABD=\angle BDC=90^\circ.
\displaystyle \textbf{To prove: }\text{(i) }ABCD\text{ is a parallelogram.}
\displaystyle \text{(ii) Find the area of parallelogram }ABCD.
\displaystyle \textbf{Proof: }\angle ABD=\angle BDC=90^\circ.
\displaystyle \text{These are alternate angles.}
\displaystyle \therefore AB\parallel DC.
\displaystyle \text{Also, }AB=DC=6\text{ cm.}
\displaystyle \therefore ABCD\text{ is a parallelogram.}
\displaystyle \text{Area of parallelogram }ABCD=\text{Base}\times\text{Altitude}.
\displaystyle =6\times8=48\text{ cm}^2.
\displaystyle {\therefore \text{The area of parallelogram }ABCD\text{ is }48\text{ cm}^2.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In parallelogram }ABCD,\text{ it is given that }AB=16\text{ cm and the}
\displaystyle \text{altitudes corresponding to the sides }AB\text{ and }AD\text{ are }6\text{ cm and }8\text{ cm} \displaystyle \text{respectively. Find the length of }AD.
\displaystyle \textbf{Answer:}
\displaystyle \text{In parallelogram }ABCD,\ AB=16\text{ cm.}
\displaystyle \text{The altitudes }DE\text{ and }BF\text{ are drawn on }AB\text{ and }AD\text{ respectively.}
\displaystyle DE=6\text{ cm and }BF=8\text{ cm.}
\displaystyle \text{Area of parallelogram }ABCD=AB\times DE.
\displaystyle =16\times6=96\text{ cm}^2. \qquad \cdots(i)
\displaystyle \text{Again, area of parallelogram }ABCD=AD\times BF.
\displaystyle =AD\times8\text{ cm}^2. \qquad \cdots(ii)
\displaystyle \text{From }(i)\text{ and }(ii),
\displaystyle 8AD=96.
\displaystyle \therefore AD=\frac{96}{8}=12\text{ cm}.
\displaystyle {\therefore \text{The length of }AD\text{ is }12\text{ cm}.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the area of a rhombus, the lengths of whose diagonals are}
\displaystyle 18\text{ cm and }24\text{ cm respectively.} \displaystyle \textbf{Answer:}
\displaystyle \text{Let the first diagonal of the rhombus be }d_1=18\text{ cm}
\displaystyle \text{and the second diagonal be }d_2=24\text{ cm}.
\displaystyle \text{Area of a rhombus}=\frac{d_1\times d_2}{2}.
\displaystyle =\frac{18\times24}{2}=216\text{ cm}^2.
\displaystyle {\therefore \text{The area of the rhombus is }216\text{ cm}^2.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the area of a trapezium whose parallel sides measure }10\text{ cm}
\displaystyle \text{and }8\text{ cm respectively and the distance between these sides is }6\text{ cm}. \displaystyle \textbf{Answer:}
\displaystyle \text{In trapezium }ABCD,\ AB\parallel DC\text{ and }DL\perp AB.
\displaystyle AB=10\text{ cm},\ DC=8\text{ cm and }DL=6\text{ cm}.
\displaystyle \text{Area of trapezium }ABCD
\displaystyle =\frac{\text{Sum of parallel sides}}{2}\times\text{Height}.
\displaystyle =\frac{10+8}{2}\times6.
\displaystyle =\frac{18}{2}\times6=54\text{ cm}^2.
\displaystyle {\therefore \text{The area of the trapezium is }54\text{ cm}^2.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Show that the line segment joining the mid-points of a pair of}
\displaystyle \text{opposite sides of a parallelogram divides it into two equal parallelograms.} \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In parallelogram }ABCD,\ P\text{ and }Q\text{ are the mid-points of sides}
\displaystyle AB\text{ and }DC\text{ respectively. }PQ\text{ is joined.}
\displaystyle \textbf{To prove: }APQD\text{ and }PBCQ\text{ are parallelograms of equal areas.}
\displaystyle \textbf{Proof: }\text{Since }P\text{ and }Q\text{ are the mid-points of }AB\text{ and }DC,
\displaystyle AP=PB\text{ and }DQ=QC.
\displaystyle \text{Also, }AB\parallel DC.
\displaystyle \therefore AP\parallel DQ.
\displaystyle \text{In parallelogram }ABCD,\ AB=DC.
\displaystyle \therefore \frac{1}{2}AB=\frac{1}{2}DC.
\displaystyle \Rightarrow AP=DQ.
\displaystyle \therefore APQD\text{ is a parallelogram.}
\displaystyle \text{Similarly, }PBCQ\text{ is a parallelogram.}
\displaystyle \text{The parallelograms }APQD\text{ and }PBCQ\text{ are on equal bases }AP\text{ and }PB
\displaystyle \text{and between the same parallels }AB\text{ and }DC.
\displaystyle \therefore \text{ar}(APQD)=\text{ar}(PBCQ).
\displaystyle \text{Hence, }PQ\text{ divides parallelogram }ABCD\text{ into two equal parallelograms.}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In the given figure, the area of parallelogram }ABCD\text{ is }90\text{ cm}^2.
\displaystyle \text{State giving reasons: (i) Area of parallelogram }ABEF\text{ (ii) Area of }\triangle ABD
\displaystyle \text{(iii) Area of }ABEF. \displaystyle \textbf{Answer:}
\displaystyle \text{Area of parallelogram }ABCD=90\text{ cm}^2.
\displaystyle AF\parallel BE\text{ are drawn and }BF\text{ is joined.}
\displaystyle \therefore ABEF\text{ is a parallelogram.}
\displaystyle \text{(i) Parallelogram }ABCD\text{ and parallelogram }ABEF\text{ are on the same}
\displaystyle \text{base and between the same parallels.}
\displaystyle \therefore \text{ar}(ABCD)=\text{ar}(ABEF).
\displaystyle \therefore \text{ar}(ABEF)=90\text{ cm}^2.
\displaystyle \text{(ii) }BD\text{ is the diagonal of parallelogram }ABCD.
\displaystyle \therefore \text{ar}(\triangle ABD)=\frac{1}{2}\text{ar}(ABCD).
\displaystyle =\frac{1}{2}\times90=45\text{ cm}^2.
\displaystyle \text{(iii) }BF\text{ is the diagonal of parallelogram }ABEF.
\displaystyle \therefore\text{ar}(\triangle ABF)=\frac{1}{2}\text{ar}(ABEF).
\displaystyle =\frac{1}{2}\times90=45\text{ cm}^2.
\displaystyle {\therefore \text{ar}(ABEF)=90\text{ cm}^2,\ \text{ar}(\triangle ABD)=45\text{ cm}^2,}
\displaystyle {\text{ar}(\triangle ABF)=45\text{ cm}^2.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{In the given figure, the area of }\triangle ABC\text{ is }64\text{ cm}^2.
\displaystyle \text{Find: (i) Area of parallelogram }ABCD\text{ (ii) Area of rectangle }ABEF. \displaystyle \textbf{Answer:}
\displaystyle \text{Area of }\triangle ABC=64\text{ cm}^2.
\displaystyle \text{Parallelogram }ABCD\text{ and rectangle }ABEF\text{ are drawn on the same}
\displaystyle \text{base }AB\text{ of }\triangle ABC.
\displaystyle \text{(i) In parallelogram }ABCD,\ AC\text{ is its diagonal.}
\displaystyle \therefore \text{ar}(\triangle ABC)=\frac{1}{2}\text{ar}(ABCD).
\displaystyle \therefore \text{ar}(ABCD)=2\times64=128\text{ cm}^2.
\displaystyle \text{(ii) Parallelogram }ABCD\text{ and rectangle }ABEF\text{ are on the same}
\displaystyle \text{base }AB\text{ and between the same parallels.}
\displaystyle \therefore \text{ar}(ABCD)=\text{ar}(ABEF).
\displaystyle \therefore \text{ar}(ABEF)=128\text{ cm}^2.
\displaystyle {\therefore \text{ar}(ABCD)=128\text{ cm}^2\text{ and }}
\displaystyle {\text{ar}(ABEF)=128\text{ cm}^2.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In the given figure, }ABCD\text{ is a quadrilateral. A line through }D
\displaystyle \text{is drawn parallel to }AC,\text{ meeting }BC\text{ produced at }P.\text{ Prove that}
\displaystyle \text{Area }(\triangle ABP)=\text{Area (quadrilateral }ABCD). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In quadrilateral }ABCD,\text{ a line through }D\text{ is drawn}
\displaystyle \text{parallel to }AC\text{ and meets }BC\text{ produced at }P.
\displaystyle \textbf{To prove: }\text{Area }(\triangle ABP)=\text{Area (quadrilateral }ABCD).
\displaystyle \textbf{Proof: }AC\parallel PD.
\displaystyle \text{Hence, }\triangle ACD\text{ and }\triangle ACP\text{ are on the same base }AC
\displaystyle \text{and between the same parallel lines.}
\displaystyle \therefore \text{Area }(\triangle ACD)=\text{Area }(\triangle ACP).
\displaystyle \text{Adding Area }(\triangle ABC)\text{ to both sides,}
\displaystyle \text{Area }(\triangle ACD)+\text{Area }(\triangle ABC)
\displaystyle =\text{Area }(\triangle ACP)+\text{Area }(\triangle ABC).
\displaystyle \therefore \text{Area (quadrilateral }ABCD)=\text{Area }(\triangle ABP).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In quadrilateral }ABCD,\text{ if }AL\perp BD\text{ and }CM\perp BD,\text{ prove that}
\displaystyle \text{Area (quadrilateral }ABCD)=\frac{1}{2}\times BD\times(AL+CM). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In quadrilateral }ABCD,\ AL\perp BD\text{ and }CM\perp BD.
\displaystyle \textbf{To prove: }\text{Area (quadrilateral }ABCD)=\frac{1}{2}\times BD\times(AL+CM).
\displaystyle \textbf{Proof: }\text{Area }(\triangle ABD)=\frac{1}{2}\times\text{Base}\times\text{Height}
\displaystyle =\frac{1}{2}\times BD\times AL.\qquad\cdots(i)
\displaystyle \text{Again, Area }(\triangle BCD)=\frac{1}{2}\times BD\times CM.\qquad\cdots(ii)
\displaystyle \text{Adding }(i)\text{ and }(ii),
\displaystyle \text{Area }(\triangle ABD)+\text{Area }(\triangle BCD)
\displaystyle =\frac{1}{2}\times BD\times AL+\frac{1}{2}\times BD\times CM.
\displaystyle \therefore \text{Area (quadrilateral }ABCD)=\frac{1}{2}\times BD\times(AL+CM).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{In the given figure, }D\text{ is the mid-point of }BC\text{ and }E\text{ is any}
\displaystyle \text{point on }AD.\text{ Prove that: (i) Area }(\triangle EBD)=\text{Area }(\triangle EDC)
\displaystyle \text{(ii) Area }(\triangle ABE)=\text{Area }(\triangle ACE). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In }\triangle ABC,\ D\text{ is the mid-point of }BC
\displaystyle \text{and }E\text{ is any point on }AD.
\displaystyle \textbf{To prove: }\text{(i) Area }(\triangle EBD)=\text{Area }(\triangle EDC)
\displaystyle \text{(ii) Area }(\triangle ABE)=\text{Area }(\triangle ACE).
\displaystyle \textbf{Proof: }\text{In }\triangle ABC,\ AD\text{ is the median.}
\displaystyle \therefore \text{Area }(\triangle ABD)=\text{Area }(\triangle ACD).\qquad\cdots(i)
\displaystyle \text{Again, in }\triangle EBC,\ ED\text{ is the median of }\triangle EBC.
\displaystyle \therefore \text{Area }(\triangle EBD)=\text{Area }(\triangle EDC).\qquad\cdots(ii)
\displaystyle \text{Subtracting }(ii)\text{ from }(i),
\displaystyle \text{Area }(\triangle ABD)-\text{Area }(\triangle EBD)
\displaystyle =\text{Area }(\triangle ACD)-\text{Area }(\triangle EDC).
\displaystyle \therefore \text{Area }(\triangle ABE)=\text{Area }(\triangle ACE).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In the given figure, }D\text{ is the mid-point of }BC\text{ and }E\text{ is the}
\displaystyle \text{mid-point of }AD.\text{ Prove that Area }(\triangle ABE)=\frac{1}{4}\text{Area }(\triangle ABC). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In }\triangle ABC,\ D\text{ is the mid-point of }BC
\displaystyle \text{and }E\text{ is the mid-point of }AD.\ CE\text{ and }BE\text{ are joined.}
\displaystyle \textbf{To prove: }\text{Area }(\triangle ABE)=\frac{1}{4}\text{Area }(\triangle ABC).
\displaystyle \textbf{Proof: }\text{In }\triangle ABC,\ AD\text{ is the median.}
\displaystyle \therefore \text{Area }(\triangle ABD)=\text{Area }(\triangle ACD)
\displaystyle =\frac{1}{2}\text{Area }(\triangle ABC).\qquad\cdots(i)
\displaystyle \text{Again, in }\triangle ABD,\ BE\text{ is the median.}
\displaystyle \therefore \text{Area }(\triangle ABE)=\text{Area }(\triangle EBD)
\displaystyle =\frac{1}{2}\text{Area }(\triangle ABD).
\displaystyle =\frac{1}{2}\times\frac{1}{2}\text{Area }(\triangle ABC)\qquad\text{[from (i)]}
\displaystyle =\frac{1}{4}\text{Area }(\triangle ABC).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In the given figure, a point }D\text{ is taken on side }BC\text{ of }\triangle ABC
\displaystyle \text{and }AD\text{ is produced to }E,\text{ making }DE=AD.\text{ Show that:}
\displaystyle \text{Area }(\triangle BEC)=\text{Area }(\triangle ABC). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In }\triangle ABC,\ D\text{ is any point on }BC.
\displaystyle AD\text{ is joined and produced to }E\text{ such that }DE=AD.
\displaystyle BE\text{ and }CE\text{ are joined.}
\displaystyle \textbf{To prove: }\text{Area }(\triangle BEC)=\text{Area }(\triangle ABC).
\displaystyle \textbf{Proof: }AD=DE.
\displaystyle \text{(Given)}
\displaystyle \therefore D\text{ is the mid-point of }AE.
\displaystyle \text{In }\triangle ABE,\ BD\text{ is the median.}
\displaystyle \therefore \text{Area }(\triangle BDE)=\text{Area }(\triangle ABD).\qquad\cdots(i)
\displaystyle \text{Similarly, in }\triangle ACE,\ CD\text{ is the median.}
\displaystyle \therefore \text{Area }(\triangle CDE)=\text{Area }(\triangle ACD).\qquad\cdots(ii)
\displaystyle \text{Adding }(i)\text{ and }(ii),
\displaystyle \text{Area }(\triangle BDE)+\text{Area }(\triangle CDE)
\displaystyle =\text{Area }(\triangle ABD)+\text{Area }(\triangle ACD).
\displaystyle \therefore \text{Area }(\triangle BEC)=\text{Area }(\triangle ABC).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If the medians of }\triangle ABC\text{ intersect at }G,\text{ show that:}
\displaystyle \text{Area }(\triangle AGB)=\text{Area }(\triangle AGC)=\text{Area }(\triangle BGC)
\displaystyle =\frac{1}{3}\text{Area }(\triangle ABC). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In }\triangle ABC,\ AD,\ BE\text{ and }CF\text{ are the medians of sides}
\displaystyle BC,\ CA\text{ and }AB\text{ respectively, intersecting at }G.
\displaystyle \textbf{To prove: }\text{Area }(\triangle AGB)=\text{Area }(\triangle AGC)
\displaystyle =\text{Area }(\triangle BGC)=\frac{1}{3}\text{Area }(\triangle ABC).
\displaystyle \textbf{Proof: }\text{In }\triangle ABC,\ AD\text{ is the median.}
\displaystyle \therefore \text{Area }(\triangle ABD)=\text{Area }(\triangle ACD).\qquad\cdots(i)
\displaystyle \text{Again, in }\triangle GBC,\ GD\text{ is the median.}
\displaystyle \therefore \text{Area }(\triangle GBD)=\text{Area }(\triangle GCD).\qquad\cdots(ii)
\displaystyle \text{Subtracting }(ii)\text{ from }(i),
\displaystyle \text{Area }(\triangle ABD)-\text{Area }(\triangle GBD)
\displaystyle =\text{Area }(\triangle ACD)-\text{Area }(\triangle GCD).
\displaystyle \therefore \text{Area }(\triangle AGB)=\text{Area }(\triangle AGC).\qquad\cdots(iii)
\displaystyle \text{Similarly, }\text{Area }(\triangle AGC)=\text{Area }(\triangle BGC).\qquad\cdots(iv)
\displaystyle \text{From }(iii)\text{ and }(iv),
\displaystyle \text{Area }(\triangle AGB)=\text{Area }(\triangle AGC)=\text{Area }(\triangle BGC).
\displaystyle \text{Also, }\text{Area }(\triangle AGB)+\text{Area }(\triangle AGC)
\displaystyle +\text{Area }(\triangle BGC)=\text{Area }(\triangle ABC).
\displaystyle \therefore 3\text{Area }(\triangle AGB)=\text{Area }(\triangle ABC).
\displaystyle \therefore \text{Area }(\triangle AGB)=\frac{1}{3}\text{Area }(\triangle ABC).
\displaystyle \text{Hence, }\text{Area }(\triangle AGB)=\text{Area }(\triangle AGC)
\displaystyle =\text{Area }(\triangle BGC)=\frac{1}{3}\text{Area }(\triangle ABC).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{D is a point on base }BC\text{ of }\triangle ABC\text{ such that }2BD=DC.
\displaystyle \text{Prove that Area }(\triangle ABD)=\frac{1}{3}\text{Area }(\triangle ABC). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In }\triangle ABC,\ D\text{ is a point on }BC\text{ such that }2BD=DC.
\displaystyle \textbf{To prove: }\text{Area }(\triangle ABD)=\frac{1}{3}\text{Area }(\triangle ABC).
\displaystyle \textbf{Proof: }\text{In }\triangle ABC,
\displaystyle 2BD=DC.
\displaystyle \therefore \frac{BD}{DC}=\frac{1}{2}.
\displaystyle \Rightarrow BD:DC=1:2.
\displaystyle \therefore \text{Area }(\triangle ABD):\text{Area }(\triangle ADC)=1:2.
\displaystyle \text{But Area }(\triangle ABD)+\text{Area }(\triangle ADC)=\text{Area }(\triangle ABC).
\displaystyle \therefore \text{Area }(\triangle ABD)+2\text{Area }(\triangle ABD)=\text{Area }(\triangle ABC).
\displaystyle \Rightarrow 3\text{Area }(\triangle ABD)=\text{Area }(\triangle ABC).
\displaystyle \therefore \text{Area }(\triangle ABD)=\frac{1}{3}\text{Area }(\triangle ABC).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In the given figure, }AD\text{ is a median of }\triangle ABC\text{ and }P\text{ is a point}
\displaystyle \text{on }AC\text{ such that Area }(\triangle ADP):\text{Area }(\triangle ABD)=2:3.
\displaystyle \text{Find: (i) }AP:PC\text{ (ii) Area }(\triangle PDC):\text{Area }(\triangle ABC). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In }\triangle ABC,\ AD\text{ is a median and }P\text{ is a point on }AC
\displaystyle \text{such that Area }(\triangle ADP):\text{Area }(\triangle ABD)=2:3.
\displaystyle \textbf{To find: }\text{(i) }AP:PC\text{ and (ii) Area }(\triangle PDC):\text{Area }(\triangle ABC).
\displaystyle \textbf{(i) Proof: }\text{Since }AD\text{ is the median of }\triangle ABC,
\displaystyle \text{Area }(\triangle ABD)=\text{Area }(\triangle ADC).\qquad\cdots(i)
\displaystyle \text{Given, Area }(\triangle ADP):\text{Area }(\triangle ABD)=2:3.
\displaystyle \therefore \text{Area }(\triangle ADP):\text{Area }(\triangle ADC)=2:3.
\displaystyle \text{[From }(i)\text{]}
\displaystyle \therefore \frac{\text{Area }(\triangle ADC)}{\text{Area }(\triangle ADP)}=\frac{3}{2}.
\displaystyle \Rightarrow \frac{\text{Area }(\triangle ADC)}{\text{Area }(\triangle ADP)}-1=\frac{3}{2}-1.
\displaystyle \therefore \frac{\text{Area }(\triangle ADC)-\text{Area }(\triangle ADP)} {\text{Area }(\triangle ADP)}=\frac{1}{2}.
\displaystyle \Rightarrow \frac{\text{Area }(\triangle PDC)}{\text{Area }(\triangle ADP)}=\frac{1}{2}.
\displaystyle \therefore \text{Area }(\triangle ADP):\text{Area }(\triangle PDC)=2:1.
\displaystyle \therefore AP:PC=2:1.
\displaystyle \textbf{(ii) }\frac{\text{Area }(\triangle ADP)}{\text{Area }(\triangle PDC)}=\frac{2}{1}.
\displaystyle \text{Adding }1\text{ to both sides,}
\displaystyle \frac{\text{Area }(\triangle ADP)}{\text{Area }(\triangle PDC)}+1=\frac{2}{1}+1.
\displaystyle \therefore \frac{\text{Area }(\triangle ADP)+\text{Area }(\triangle PDC)} {\text{Area }(\triangle PDC)}=\frac{3}{1}.
\displaystyle \Rightarrow \frac{\text{Area }(\triangle ADC)}{\text{Area }(\triangle PDC)}=\frac{3}{1}.
\displaystyle \text{But Area }(\triangle ADC)=\text{Area }(\triangle ABD).
\displaystyle \therefore \frac{\text{Area }(\triangle ABD)}{\text{Area }(\triangle PDC)}=\frac{3}{1}.
\displaystyle \Rightarrow \frac{\text{Area }(\triangle PDC)}{\text{Area }(\triangle ABD)}=\frac{1}{3}.
\displaystyle \text{Also, Area }(\triangle ABD)=\frac{1}{2}\text{Area }(\triangle ABC).
\displaystyle \therefore \frac{\text{Area }(\triangle PDC)} {\frac{1}{2}\text{Area }(\triangle ABC)}=\frac{1}{3}.
\displaystyle \Rightarrow \frac{2\text{Area }(\triangle PDC)}{\text{Area }(\triangle ABC)}=\frac{1}{3}.
\displaystyle \therefore \frac{\text{Area }(\triangle PDC)}{\text{Area }(\triangle ABC)}=\frac{1}{6}.
\displaystyle {\therefore \text{(i) }AP:PC=2:1\text{ and (ii) Area }(\triangle PDC):\text{Area }(\triangle ABC)=1:6.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In the given figure, }P\text{ is a point on side }BC\text{ of }\triangle ABC
\displaystyle \text{such that }BP:PC=1:2\text{ and }Q\text{ is a point on }AP\text{ such that }PQ:QA=2:3.
\displaystyle \text{Show that Area }(\triangle AQC):\text{Area }(\triangle ABC)=2:5. \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In }\triangle ABC,\ P\text{ is a point on }BC\text{ such that }BP:PC=1:2.
\displaystyle Q\text{ is a point on }AP\text{ such that }PQ:QA=2:3.
\displaystyle \textbf{To prove: }\text{Area }(\triangle AQC):\text{Area }(\triangle ABC)=2:5.
\displaystyle \textbf{Proof: }\text{In }\triangle ABC,\ P\text{ is a point on }BC\text{ such that}
\displaystyle BP:PC=1:2.
\displaystyle \therefore \text{Area }(\triangle APB):\text{Area }(\triangle APC)=1:2.
\displaystyle \therefore \text{Area }(\triangle APC)=\frac{2}{3}\text{Area }(\triangle ABC).
\displaystyle \text{Again, in }\triangle APC,\ Q\text{ is a point on }AP\text{ such that }PQ:QA=2:3.
\displaystyle \therefore \text{Area }(\triangle AQC):\text{Area }(\triangle PQC)=3:2.
\displaystyle \therefore \text{Area }(\triangle AQC)=\frac{3}{5}\text{Area }(\triangle APC).
\displaystyle =\frac{3}{5}\times\frac{2}{3}\text{Area }(\triangle ABC).
\displaystyle =\frac{2}{5}\text{Area }(\triangle ABC).
\displaystyle \therefore \frac{\text{Area }(\triangle AQC)}{\text{Area }(\triangle ABC)}=\frac{2}{5}.
\displaystyle \therefore \text{Area }(\triangle AQC):\text{Area }(\triangle ABC)=2:5.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In the adjoining figure, }ABCD\text{ is a parallelogram. }P\text{ and }Q
\displaystyle \text{are any two points on the sides }AB\text{ and }BC\text{ respectively. Prove that}
\displaystyle \text{Area }(\triangle CPD)=\text{Area }(\triangle AQD). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In parallelogram }ABCD,\ P\text{ and }Q\text{ are points on }AB\text{ and }BC
\displaystyle \text{respectively. }AQ,\ DQ,\ CP\text{ and }DP\text{ are joined.}
\displaystyle \textbf{To prove: }\text{Area }(\triangle CPD)=\text{Area }(\triangle AQD).
\displaystyle \textbf{Proof: }\triangle CPD\text{ and parallelogram }ABCD\text{ are on the same base }CD
\displaystyle \text{and between the same parallel lines.}
\displaystyle \therefore \text{Area }(\triangle CPD)=\frac{1}{2}\text{Area }(\text{parallelogram }ABCD).\qquad\cdots(i)
\displaystyle \text{Similarly, }\triangle AQD\text{ and parallelogram }ABCD\text{ are on the same base }AD
\displaystyle \text{and between the same parallel lines.}
\displaystyle \therefore \text{Area }(\triangle AQD)=\frac{1}{2}\text{Area }(\text{parallelogram }ABCD).\qquad\cdots(ii)
\displaystyle \text{From }(i)\text{ and }(ii),
\displaystyle \therefore \text{Area }(\triangle CPD)=\text{Area }(\triangle AQD).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{In the adjoining figure, }DE\parallel BC.\text{ Prove that:}
\displaystyle \text{(i) Area }(\triangle ABE)=\text{Area }(\triangle ACD)
\displaystyle \text{(ii) Area }(\triangle OBD)=\text{Area }(\triangle OCE). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In }\triangle ABC,\ DE\parallel BC.
\displaystyle \textbf{To prove: }\text{(i) Area }(\triangle ABE)=\text{Area }(\triangle ACD)
\displaystyle \text{(ii) Area }(\triangle OBD)=\text{Area }(\triangle OCE).
\displaystyle \textbf{Proof:}
\displaystyle \textbf{(i)}\text{ In }\triangle ABC,\ DE\parallel BC.
\displaystyle \triangle BDE\text{ and }\triangle CDE\text{ are on the same base }DE
\displaystyle \text{and between the same parallel lines.}
\displaystyle \therefore \text{Area }(\triangle BDE)=\text{Area }(\triangle CDE).\qquad\cdots(i)
\displaystyle \text{Adding Area }(\triangle ADE)\text{ to both sides of }(i),
\displaystyle \text{Area }(\triangle BDE)+\text{Area }(\triangle ADE)
\displaystyle =\text{Area }(\triangle CDE)+\text{Area }(\triangle ADE).
\displaystyle \therefore \text{Area }(\triangle ABE)=\text{Area }(\triangle ACD).
\displaystyle \textbf{(ii)}\text{ Subtracting Area }(\triangle DOE)\text{ from both sides of }(i),
\displaystyle \text{Area }(\triangle BDE)-\text{Area }(\triangle DOE)
\displaystyle =\text{Area }(\triangle CDE)-\text{Area }(\triangle DOE).
\displaystyle \therefore \text{Area }(\triangle OBD)=\text{Area }(\triangle OCE).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{In the given figure, }ABCD\text{ is a parallelogram and }P\text{ is a point on }BC.
\displaystyle \text{Prove that Area }(\triangle ABP)+\text{Area }(\triangle DPC)=\text{Area }(\triangle APD). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In parallelogram }ABCD,\ P\text{ is a point on }BC.
\displaystyle \textbf{To prove: }\text{Area }(\triangle ABP)+\text{Area }(\triangle DPC)=\text{Area }(\triangle APD).
\displaystyle \textbf{Construction: }\text{From }P,\text{ draw }PQ\parallel AB\text{ or }DC.\displaystyle \textbf{Proof: }\text{Quadrilateral }QPCD\text{ is a parallelogram and }PD\text{ is its diagonal.}
\displaystyle \therefore \text{Area }(\triangle DPC)=\text{Area }(\triangle QPD).\qquad\cdots(i)
\displaystyle \text{Similarly, }ABPQ\text{ is a parallelogram and }AP\text{ is its diagonal.}
\displaystyle \therefore \text{Area }(\triangle ABP)=\text{Area }(\triangle APQ).\qquad\cdots(ii)
\displaystyle \text{Adding }(i)\text{ and }(ii),
\displaystyle \text{Area }(\triangle DPC)+\text{Area }(\triangle ABP)
\displaystyle =\text{Area }(\triangle QPD)+\text{Area }(\triangle APQ).
\displaystyle \therefore \text{Area }(\triangle ABP)+\text{Area }(\triangle DPC)=\text{Area }(\triangle APD).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{In the adjoining figure, }ABCDE\text{ is a pentagon. }BP\parallel AC
\displaystyle \text{meets }DC\text{ produced at }P\text{ and }EQ\parallel AD\text{ meets }CD\text{ produced at }Q.
\displaystyle \text{Prove that Area (pentagon }ABCDE)=\text{Area }(\triangle APQ). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In pentagon }ABCDE,\ AC\text{ and }AD\text{ are joined.}
\displaystyle \text{From }B,\ BP\parallel AC\text{ and from }E,\ EQ\parallel AD\text{ are drawn to meet }CD
\displaystyle \text{produced on both sides at }P\text{ and }Q\text{ respectively.}
\displaystyle \textbf{To prove: }\text{Area (pentagon }ABCDE)=\text{Area }(\triangle APQ).
\displaystyle \textbf{Construction: }\text{Join }AP\text{ and }AQ.\displaystyle \textbf{Proof: }\text{Since }BP\parallel AC,\ \triangle ABC\text{ and }\triangle APC\text{ are on the}
\displaystyle \text{same base }AC\text{ and between the same parallel lines.}
\displaystyle \therefore \text{Area }(\triangle ABC)=\text{Area }(\triangle APC).\qquad\cdots(i)
\displaystyle \text{Similarly, since }EQ\parallel AD,
\displaystyle \text{Area }(\triangle ADE)=\text{Area }(\triangle ADQ).\qquad\cdots(ii)
\displaystyle \text{Also, }\text{Area }(\triangle ACD)=\text{Area }(\triangle ACD).\qquad\cdots(iii)
\displaystyle \text{Adding }(i),\ (ii)\text{ and }(iii),
\displaystyle \text{Area }(\triangle ABC)+\text{Area }(\triangle ADE)+\text{Area }(\triangle ACD)
\displaystyle =\text{Area }(\triangle APC)+\text{Area }(\triangle ADQ)+\text{Area }(\triangle ACD).
\displaystyle \therefore \text{Area (pentagon }ABCDE)=\text{Area }(\triangle APQ).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{In the adjoining figure, two parallelograms }ABCD\text{ and }AEFB
\displaystyle \text{are drawn on opposite sides of }AB.\text{ Prove that:}
\displaystyle \text{Area (parallelogram }ABCD)+\text{Area (parallelogram }AEFB)
\displaystyle =\text{Area (parallelogram }EFCD). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{Parallelograms }ABCD\text{ and }AEFB\text{ are drawn on opposite sides of }AB.
\displaystyle DE\text{ and }FC\text{ are joined.}
\displaystyle \textbf{To prove: }\text{Area (parallelogram }ABCD)+\text{Area (parallelogram }AEFB)
\displaystyle =\text{Area (parallelogram }EFCD).
\displaystyle \textbf{Proof: }\text{In }\triangle ADE\text{ and }\triangle BFC,
\displaystyle AD=BC.
\displaystyle \text{(Opposite sides of a parallelogram are equal)}
\displaystyle AE=BF.
\displaystyle \text{(Opposite sides of a parallelogram are equal)}
\displaystyle DE=CF.
\displaystyle \therefore \triangle ADE\cong\triangle BFC.
\displaystyle \therefore \text{Area }(\triangle ADE)=\text{Area }(\triangle BFC).\qquad\cdots(i)
\displaystyle \text{(Congruent triangles are equal in area)}
\displaystyle \text{Now, Area (parallelogram }ABCD)+\text{Area (parallelogram }AEFB)
\displaystyle =\text{Area (parallelogram }EFCD)-\text{Area }(\triangle ADE)
\displaystyle +\text{Area }(\triangle BFC).
\displaystyle =\text{Area (parallelogram }EFCD)-\text{Area }(\triangle ADE)
\displaystyle +\text{Area }(\triangle ADE).\qquad\text{[From }(i)\text{]}
\displaystyle =\text{Area (parallelogram }EFCD).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{In the adjoining figure, }ABCD\text{ is a parallelogram and }O\text{ is any}
\displaystyle \text{point on its diagonal }AC.\text{ Show that Area }(\triangle AOB)=\text{Area }(\triangle AOD). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In parallelogram }ABCD,\ O\text{ is any point on its diagonal }AC.
\displaystyle OB\text{ and }OD\text{ are joined.}
\displaystyle \textbf{To prove: }\text{Area }(\triangle AOB)=\text{Area }(\triangle AOD).
\displaystyle \textbf{Construction: }\text{Join }BD,\text{ which intersects }AC\text{ at }P.
\displaystyle \textbf{Proof: }\text{The diagonals of a parallelogram bisect each other.}
\displaystyle \therefore AP=PC\text{ and }BP=PD.
\displaystyle \text{In }\triangle ABD,\ AP\text{ is the median.}
\displaystyle \therefore \text{Area }(\triangle APB)=\text{Area }(\triangle APD).\qquad\cdots(i)
\displaystyle \text{Similarly, in }\triangle OBD,\ OP\text{ is the median.}
\displaystyle \therefore \text{Area }(\triangle OBP)=\text{Area }(\triangle ODP).\qquad\cdots(ii)
\displaystyle \text{Adding }(i)\text{ and }(ii),
\displaystyle \text{Area }(\triangle APB)+\text{Area }(\triangle OBP)
\displaystyle =\text{Area }(\triangle APD)+\text{Area }(\triangle ODP).
\displaystyle \therefore \text{Area }(\triangle AOB)=\text{Area }(\triangle AOD).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{In the given figure, }XY\parallel BC,\ BE\parallel CA\text{ and }FC\parallel AB.
\displaystyle \text{Prove that Area }(\triangle ABE)=\text{Area }(\triangle ACF). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In the figure, }XY\parallel BC,\ BE\parallel CA\text{ and }FC\parallel AB.
\displaystyle \textbf{To prove: }\text{Area }(\triangle ABE)=\text{Area }(\triangle ACF).
\displaystyle \textbf{Proof: }\triangle ABE\text{ and parallelogram }BCYE\text{ are on the same base }BE
\displaystyle \text{and between the same parallel lines.}
\displaystyle \therefore \text{Area }(\triangle ABE)=\frac{1}{2}\text{Area (parallelogram }BCYE).\qquad\cdots(i)
\displaystyle \text{Similarly, }\triangle ACF\text{ and parallelogram }BCFX\text{ are on the same base }CF
\displaystyle \text{and between the same parallel lines.}
\displaystyle \therefore \text{Area }(\triangle ACF)=\frac{1}{2}\text{Area (parallelogram }BCFX).\qquad\cdots(ii)
\displaystyle \text{But parallelograms }BCFX\text{ and }BCYE\text{ are on the same base }BC
\displaystyle \text{and between the same parallel lines.}
\displaystyle \therefore \text{Area (parallelogram }BCFX)=\text{Area (parallelogram }BCYE).\qquad\cdots(iii)
\displaystyle \text{From }(i),\ (ii)\text{ and }(iii),
\displaystyle \therefore \text{Area }(\triangle ABE)=\text{Area }(\triangle ACF).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{In the given figure, the side }AB\text{ of parallelogram }ABCD\text{ is}
\displaystyle \text{produced to a point }P.\text{ A line through }A\text{ drawn parallel to }CP\text{ meets }CB
\displaystyle \text{produced at }Q\text{ and the parallelogram }PBQR\text{ is completed. Prove that}
\displaystyle \text{Area (parallelogram }ABCD)=\text{Area (parallelogram }BPRQ). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{Side }AB\text{ of parallelogram }ABCD\text{ is produced to }P.
\displaystyle CP\text{ is joined. Through }A,\text{ a line parallel to }CP\text{ is drawn, meeting }CB
\displaystyle \text{produced at }Q.\text{ Parallelogram }PBQR\text{ is completed.}
\displaystyle \textbf{To prove: }\text{Area (parallelogram }ABCD)=\text{Area (parallelogram }BPRQ).
\displaystyle \textbf{Construction: }\text{Join }AC\text{ and }PQ.
\displaystyle \textbf{Proof: }\triangle AQC\text{ and }\triangle AQP\text{ are on the same base }AQ
\displaystyle \text{and between the same parallel lines }AQ\parallel CP.
\displaystyle \therefore \text{Area }(\triangle AQC)=\text{Area }(\triangle AQP).
\displaystyle \text{Subtracting Area }(\triangle AQB)\text{ from both sides,}
\displaystyle \text{Area }(\triangle AQC)-\text{Area }(\triangle AQB)
\displaystyle =\text{Area }(\triangle AQP)-\text{Area }(\triangle AQB).
\displaystyle \therefore \text{Area }(\triangle ABC)=\text{Area }(\triangle BPQ).\qquad\cdots(i)
\displaystyle \text{But Area }(\triangle ABC)=\frac{1}{2}\text{Area (parallelogram }ABCD).\qquad\cdots(ii)
\displaystyle \text{Also, Area }(\triangle BPQ)=\frac{1}{2}\text{Area (parallelogram }BPRQ).\qquad\cdots(iii)
\displaystyle \text{From }(i),\ (ii)\text{ and }(iii),
\displaystyle \frac{1}{2}\text{Area (parallelogram }ABCD)
\displaystyle =\frac{1}{2}\text{Area (parallelogram }BPRQ).
\displaystyle \therefore \text{Area (parallelogram }ABCD)=\text{Area (parallelogram }BPRQ).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{In the adjoining figure, }CE\text{ is drawn parallel to }DB\text{ to meet}
\displaystyle AB\text{ produced at }E.\text{ Prove that Area (quadrilateral }ABCD)
\displaystyle =\text{Area }(\triangle DAE). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In the adjoining figure, }CE\parallel DB\text{ and }CE\text{ meets }AB
\displaystyle \text{produced at }E.\ DE\text{ is joined.}
\displaystyle \textbf{To prove: }\text{Area (quadrilateral }ABCD)=\text{Area }(\triangle DAE).
\displaystyle \textbf{Proof: }\triangle DBE\text{ and }\triangle DBC\text{ are on the same base }BD
\displaystyle \text{and between the same parallel lines }BD\text{ and }CE.
\displaystyle \therefore \text{Area }(\triangle DBE)=\text{Area }(\triangle DBC).
\displaystyle \text{Adding Area }(\triangle ABD)\text{ to both sides,}
\displaystyle \text{Area }(\triangle DBE)+\text{Area }(\triangle ABD)
\displaystyle =\text{Area }(\triangle DBC)+\text{Area }(\triangle ABD).
\displaystyle \therefore \text{Area }(\triangle DAE)=\text{Area (quadrilateral }ABCD).
\displaystyle \therefore \text{Area (quadrilateral }ABCD)=\text{Area }(\triangle DAE).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{In the adjoining figure, }ABCD\text{ is a parallelogram. }AB\text{ is}
\displaystyle \text{produced to a point }P\text{ and }DP\text{ intersects }BC\text{ at }Q.\text{ Prove that}
\displaystyle \text{Area }(\triangle APD)=\text{Area (quadrilateral }BPCD). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In parallelogram }ABCD,\ AB\text{ is produced to }P
\displaystyle \text{and }DP\text{ intersects }BC\text{ at }Q.
\displaystyle \textbf{To prove: }\text{Area }(\triangle APD)=\text{Area (quadrilateral }BPCD).
\displaystyle \textbf{Construction: }\text{Join }BD.
\displaystyle \textbf{Proof: }\triangle BPD\text{ and }\triangle BPC\text{ are on the same base }BP
\displaystyle \text{and between the same parallel lines }BP\text{ and }DC.
\displaystyle \therefore \text{Area }(\triangle BPD)=\text{Area }(\triangle BPC).\qquad\cdots(i)
\displaystyle \text{In parallelogram }ABCD,\ BD\text{ is its diagonal.}
\displaystyle \therefore \text{Area }(\triangle ABD)=\text{Area }(\triangle DBC).\qquad\cdots(ii)
\displaystyle \text{Adding }(i)\text{ and }(ii),
\displaystyle \text{Area }(\triangle BPD)+\text{Area }(\triangle ABD)
\displaystyle =\text{Area }(\triangle BPC)+\text{Area }(\triangle DBC).
\displaystyle \therefore \text{Area }(\triangle APD)=\text{Area (quadrilateral }BPCD).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{In the adjoining figure, }ABCD\text{ is a parallelogram. Any line through }A
\displaystyle \text{cuts }DC\text{ at }P\text{ and }BC\text{ produced at }Q.\text{ Prove that}
\displaystyle \text{Area }(\triangle BPC)=\text{Area }(\triangle DPQ). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }ABCD\text{ is a parallelogram. A line through }A\text{ intersects }DC
\displaystyle \text{at }P\text{ and }BC\text{ produced at }Q.
\displaystyle \textbf{To prove: }\text{Area }(\triangle BPC)=\text{Area }(\triangle DPQ).
\displaystyle \textbf{Construction: }\text{Join }AC\text{ and }BP.
\displaystyle \textbf{Proof: }\triangle BPC\text{ and }\triangle APC\text{ are on the same base }PC
\displaystyle \text{and between the same parallel lines }AB\text{ and }DC.
\displaystyle \therefore \text{Area }(\triangle BPC)=\text{Area }(\triangle APC).\qquad\cdots(i)
\displaystyle \text{Again, }\triangle AQC\text{ and }\triangle DQC\text{ are on the same base }QC
\displaystyle \text{and between the same parallel lines }AD\text{ and }BC.
\displaystyle \therefore \text{Area }(\triangle AQC)=\text{Area }(\triangle DQC).\qquad\cdots(ii)
\displaystyle \text{Now, Area }(\triangle BPC)=\text{Area }(\triangle APC).
\displaystyle =\text{Area }(\triangle AQC)-\text{Area }(\triangle PQC).
\displaystyle =\text{Area }(\triangle DQC)-\text{Area }(\triangle PQC).
\displaystyle =\text{Area }(\triangle DPQ).
\displaystyle \therefore \text{Area }(\triangle BPC)=\text{Area }(\triangle DPQ).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{In the adjoining figure, }ABCD\text{ is a parallelogram. }P\text{ is a point}
\displaystyle \text{on }BC\text{ such that }BP:PC=1:2.\ DP\text{ produced meets }AB\text{ produced at }Q.
\displaystyle \text{Given Area }(\triangle CPQ)=20\text{ cm}^2.\text{ Calculate:}
\displaystyle \text{(i) Area }(\triangle CDP)\text{ (ii) Area (parallelogram }ABCD). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }ABCD\text{ is a parallelogram. }P\text{ is a point on }BC\text{ such that}
\displaystyle BP:PC=1:2.\ DP\text{ produced meets }AB\text{ produced at }Q.
\displaystyle \text{Area }(\triangle CPQ)=20\text{ cm}^2.\ CQ\text{ is joined.}
\displaystyle \textbf{(i)}\text{ Since }BP:PC=1:2,
\displaystyle \text{Area }(\triangle BPQ):\text{Area }(\triangle CPQ)=1:2.
\displaystyle \therefore \text{Area }(\triangle BPQ)=\frac{1}{2}\text{Area }(\triangle CPQ).
\displaystyle =\frac{1}{2}\times20=10\text{ cm}^2.
\displaystyle \text{In }\triangle BPQ\text{ and }\triangle CPD,
\displaystyle \angle BPQ=\angle DPC.
\displaystyle \text{(Vertically opposite angles)}
\displaystyle \angle BQP=\angle PDC.
\displaystyle \text{(Alternate angles, since }AB\parallel DC\text{)}
\displaystyle \therefore \triangle BPQ\sim\triangle CPD.
\displaystyle \therefore \frac{\text{Area }(\triangle CPD)}{\text{Area }(\triangle BPQ)} =\left(\frac{PC}{BP}\right)^2.
\displaystyle =\left(\frac{2}{1}\right)^2=\frac{4}{1}.
\displaystyle \therefore \text{Area }(\triangle CPD)=4\text{Area }(\triangle BPQ).
\displaystyle =4\times10=40\text{ cm}^2.
\displaystyle \textbf{(ii)}\ \text{Area }(\triangle CQD)=\text{Area }(\triangle CPD)+\text{Area }(\triangle CPQ).
\displaystyle =40+20=60\text{ cm}^2.
\displaystyle \triangle CQD\text{ and parallelogram }ABCD\text{ are on the same base }DC
\displaystyle \text{and between the same parallel lines }DC\text{ and }AB.
\displaystyle \therefore \text{Area }(\triangle CQD)=\frac{1}{2}\text{Area (parallelogram }ABCD).
\displaystyle \therefore \text{Area (parallelogram }ABCD)=2\text{Area }(\triangle CQD).
\displaystyle =2\times60=120\text{ cm}^2.
\displaystyle {\therefore \text{(i) Area }(\triangle CDP)=40\text{ cm}^2\text{ and}}
\displaystyle {\text{(ii) Area (parallelogram }ABCD)=120\text{ cm}^2.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{In the adjoining figure, }ABCD\text{ is a parallelogram. }P\text{ is a point}
\displaystyle \text{on }DC\text{ such that Area }(\triangle APD)=25\text{ cm}^2\text{ and Area }(\triangle BPC)=15\text{ cm}^2.
\displaystyle \text{Calculate: (i) Area (parallelogram }ABCD)\text{ (ii) }DP:PC. \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }ABCD\text{ is a parallelogram. }P\text{ is a point on }DC\text{ such that}
\displaystyle \text{Area }(\triangle APD)=25\text{ cm}^2\text{ and Area }(\triangle BPC)=15\text{ cm}^2.
\displaystyle \text{Through }P,\text{ draw }PQ\parallel AD\text{ or }BC.\displaystyle \textbf{(i)}\text{ In parallelogram }AQPD,\ AP\text{ is its diagonal.}
\displaystyle \therefore \text{Area }(\triangle APD)=\text{Area }(\triangle APQ).\qquad\cdots(i)
\displaystyle \text{Similarly, in parallelogram }QBCP,\ PB\text{ is its diagonal.}
\displaystyle \therefore \text{Area }(\triangle BPC)=\text{Area }(\triangle PQB).\qquad\cdots(ii)
\displaystyle \text{Adding }(i)\text{ and }(ii),
\displaystyle \text{Area }(\triangle APQ)+\text{Area }(\triangle PQB)
\displaystyle =\text{Area }(\triangle APD)+\text{Area }(\triangle BPC).
\displaystyle \therefore \text{Area }(\triangle APB)=25+15=40\text{ cm}^2.
\displaystyle \triangle APB\text{ and parallelogram }ABCD\text{ are on the same base }AB
\displaystyle \text{and between the same parallel lines }AB\text{ and }DC.
\displaystyle \therefore \text{Area }(\triangle APB)=\frac{1}{2}\text{Area (parallelogram }ABCD).
\displaystyle \therefore \text{Area (parallelogram }ABCD)=2\text{Area }(\triangle APB).
\displaystyle =2\times40=80\text{ cm}^2.
\displaystyle \textbf{(ii)}\ \text{Area }(\triangle APQ):\text{Area }(\triangle PQB)=AQ:QB.
\displaystyle \therefore \text{Area }(\triangle APD):\text{Area }(\triangle BPC)=DP:PC.
\displaystyle \therefore 25:15=DP:PC.
\displaystyle \therefore DP:PC=5:3.
\displaystyle {\therefore \text{(i) Area (parallelogram }ABCD)=80\text{ cm}^2\text{ and (ii) }DP:PC=5:3.}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{In the given figure, }AB\parallel DC\parallel EF,\ AD\parallel BE
\displaystyle \text{and }DE\parallel AF.\text{ Prove that Area (parallelogram }DEFH)
\displaystyle =\text{Area (parallelogram }ABCD). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{In the figure, }AB\parallel DC\parallel EF,\ AD\parallel BE
\displaystyle \text{and }DE\parallel AF.
\displaystyle \textbf{To prove: }\text{Area (parallelogram }DEFH)
\displaystyle =\text{Area (parallelogram }ABCD).
\displaystyle \textbf{Proof: }\text{Parallelograms }ABCD\text{ and }ADGE\text{ are on the same base }AD
\displaystyle \text{and between the same parallel lines.}
\displaystyle \therefore \text{Area (parallelogram }ABCD)=\text{Area (parallelogram }ADGE).\qquad\cdots(i)
\displaystyle \text{Similarly, parallelograms }DEFH\text{ and }ADEG\text{ are on the same base }DE
\displaystyle \text{and between the same parallel lines.}
\displaystyle \therefore \text{Area (parallelogram }DEFH)=\text{Area (parallelogram }ADGE).\qquad\cdots(ii)
\displaystyle \text{From }(i)\text{ and }(ii),
\displaystyle \therefore \text{Area (parallelogram }ABCD)=\text{Area (parallelogram }DEFH).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{In the given figure, squares }ABDE\text{ and }AFGC\text{ are drawn}
\displaystyle \text{on side }AB\text{ and hypotenuse }AC\text{ of right triangle }ABC,\text{ and }BH\perp FG.
\displaystyle \text{Prove that: (i) }\triangle EAC\cong\triangle BAF
\displaystyle \text{(ii) Area (square }ABDE)=\text{Area (rectangle }ARHF). \displaystyle \textbf{Answer:}
\displaystyle \textbf{Given: }\text{Squares }ABDE\text{ and }AFGC\text{ are drawn on side }AB
\displaystyle \text{and hypotenuse }AC\text{ of right triangle }ABC.
\displaystyle BH\perp FG\text{ and intersects }AC\text{ at }R.
\displaystyle \textbf{To prove: }\text{(i) }\triangle EAC\cong\triangle BAF
\displaystyle \text{(ii) Area (square }ABDE)=\text{Area (rectangle }ARHF).
\displaystyle \textbf{Construction: }\text{Join }BF\text{ and }CE.
\displaystyle \textbf{Proof: }\angle CAE=\angle CAB+\angle BAE.
\displaystyle =\angle CAB+90^\circ.\qquad\cdots(i)
\displaystyle \text{Similarly, }\angle BAF=\angle CAB+\angle CAF.
\displaystyle =\angle CAB+90^\circ.\qquad\cdots(ii)
\displaystyle \text{From }(i)\text{ and }(ii),
\displaystyle \angle CAE=\angle BAF.
\displaystyle \textbf{(i)}\text{ In }\triangle EAC\text{ and }\triangle BAF,
\displaystyle \angle CAE=\angle BAF.
\displaystyle AE=AB.
\displaystyle \text{(Sides of a square)}
\displaystyle AC=AF.
\displaystyle \text{(Sides of a square)}
\displaystyle \therefore \triangle EAC\cong\triangle BAF.
\displaystyle \text{(S.A.S. criterion of congruency)}
\displaystyle \therefore \text{Area }(\triangle EAC)=\text{Area }(\triangle BAF).
\displaystyle \textbf{(ii)}\text{ Square }ABDE\text{ and }\triangle EAC\text{ are on the same base }AE
\displaystyle \text{and between the same parallel lines.}
\displaystyle \therefore \text{Area }(\triangle AEC)=\frac{1}{2}\text{Area (square }ABDE).\qquad\cdots(iii)
\displaystyle \text{Similarly, }\text{Area }(\triangle BAF)=\frac{1}{2}\text{Area (rectangle }ARHF).\qquad\cdots(iv)
\displaystyle \text{But Area }(\triangle AEC)=\text{Area }(\triangle BAF).
\displaystyle \text{From }(iii)\text{ and }(iv),
\displaystyle \frac{1}{2}\text{Area (square }ABDE)=\frac{1}{2}\text{Area (rectangle }ARHF).
\displaystyle \therefore \text{Area (square }ABDE)=\text{Area (rectangle }ARHF).
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{Construct a quadrilateral }ABCD\text{ in which }AB=3.2\text{ cm},
\displaystyle BC=2.8\text{ cm},\ CD=4\text{ cm},\ DA=4.5\text{ cm and }BD=5.2\text{ cm}.
\displaystyle \text{Also, construct a triangle equal in area to this quadrilateral.}
\displaystyle \textbf{Answer:}
\displaystyle \textbf{Steps of Construction:}\displaystyle \text{(i) Draw a line segment }AB=3.2\text{ cm.}
\displaystyle \text{(ii) With centre }A\text{ and radius }4.5\text{ cm, and with centre }B
\displaystyle \text{and radius }5.2\text{ cm, draw arcs intersecting each other at }D.
\displaystyle \text{(iii) Join }AD\text{ and }BD.
\displaystyle \text{(iv) Again, with centre }B\text{ and radius }2.8\text{ cm, and with centre }D
\displaystyle \text{and radius }4\text{ cm, draw two arcs intersecting each other at }C.
\displaystyle \text{(v) Join }BC\text{ and }CD.
\displaystyle \therefore ABCD\text{ is the required quadrilateral.}
\displaystyle \text{(vi) Produce }AB.
\displaystyle \text{(vii) From }C,\text{ draw a line parallel to }BD,\text{ meeting }AB\text{ produced at }E.
\displaystyle \text{(viii) Join }DE.
\displaystyle \therefore \triangle ADE\text{ is the required triangle whose area is equal to the area}
\displaystyle \text{of quadrilateral }ABCD.
\displaystyle \\

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