NCERT Exemplar · Class 9 Mathematics Areas of Parallelograms and Triangles

\[AB \parallel DF \quad \text{(F lies on DC produced)} \] \[\operatorname{ar}(\triangle ABC) = \operatorname{ar}(\triangle ABF) \quad \text{(same base } AB\text{, between } AB \parallel CF\text{)} \] \[\operatorname{ar}(ABFC) = \operatorname{ar}(\triangle ABF) + \operatorname{ar}(\triangle AFC) \] So \[\operatorname{ar}(ABFC) = \operatorname{ar}(\triangle ABC) + \operatorname{ar}(\triangle ACF) \] A diagonal of a parallelogram bisects its area, so \[\operatorname{ar}(\triangle ADC) = \operatorname{ar}(\triangle ABC) \] Since \(\displaystyle C\) lies on \(\displaystyle DF\), \[\operatorname{ar}(\triangle ADF) = \operatorname{ar}(\triangle ADC) + \operatorname{ar}(\triangle ACF) = \operatorname{ar}(\triangle ABC) + \operatorname{ar}(\triangle ACF) \] \[\therefore \operatorname{ar}(\triangle ADF) = \operatorname{ar}(ABFC) \] Answer: \(\displaystyle \operatorname{ar}(ADF) = \operatorname{ar}(ABFC)\)

Diagonals \(\displaystyle AC,BD\) of the parallelogram bisect each other at \(\displaystyle O\), and \(\displaystyle P,O,Q\) are collinear. \[\triangle AOB \cong \triangle COD \quad \text{(SAS: } OA=OC,\ OB=OD,\ \angle AOB=\angle COD\text{)} \] \[\triangle AOD \cong \triangle COB \quad \text{(SAS: } OA=OC,\ OD=OB,\ \angle AOD=\angle COB\text{)} \] Since \(\displaystyle AD \parallel BC\), alternate angles give \(\displaystyle \angle OAP=\angle OCQ\), so \[\triangle AOP \cong \triangle COQ \quad \text{(ASA: } OA=OC,\ \angle AOP=\angle COQ,\ \angle OAP=\angle OCQ\text{)} \] Subtracting \(\displaystyle \triangle AOP\) from \(\displaystyle \triangle AOD\), and \(\displaystyle \triangle COQ\) from \(\displaystyle \triangle COB\), \[\operatorname{ar}(\triangle DOP) = \operatorname{ar}(\triangle AOD)-\operatorname{ar}(\triangle AOP) = \operatorname{ar}(\triangle COB)-\operatorname{ar}(\triangle COQ) = \operatorname{ar}(\triangle BOQ) \] Adding the three equal pairs, \[\operatorname{ar}(\triangle AOB)+\operatorname{ar}(\triangle AOP)+\operatorname{ar}(\triangle BOQ) = \operatorname{ar}(\triangle COD)+\operatorname{ar}(\triangle COQ)+\operatorname{ar}(\triangle DOP) \] \[\operatorname{ar}(ABQP) = \operatorname{ar}(DPQC) \] Answer: \(\displaystyle \operatorname{ar}(ABQP)=\operatorname{ar}(DPQC)=\tfrac12\operatorname{ar}(ABCD)\)

\[\operatorname{ar}(\triangle ABE) = \operatorname{ar}(\triangle CBE) \quad \text{(}BE\text{ is a median)} \] \(\displaystyle F\) lies on \(\displaystyle AB\) and \(\displaystyle G\) on \(\displaystyle BE\), so \[\operatorname{ar}(AFGE) + \operatorname{ar}(\triangle BFG) = \operatorname{ar}(\triangle BGC) + \operatorname{ar}(\triangle GEC) \quad \text{...(i)} \] \[\operatorname{ar}(\triangle ACF) = \operatorname{ar}(\triangle BCF) \quad \text{(}CF\text{ is a median)} \] \(\displaystyle E\) lies on \(\displaystyle AC\) and \(\displaystyle G\) on \(\displaystyle CF\), so \[\operatorname{ar}(AFGE) + \operatorname{ar}(\triangle GEC) = \operatorname{ar}(\triangle BGC) + \operatorname{ar}(\triangle BFG) \quad \text{...(ii)} \] Adding (i) and (ii), \[2\operatorname{ar}(AFGE) = 2\operatorname{ar}(\triangle BGC) \] \[\therefore \operatorname{ar}(AFGE) = \operatorname{ar}(\triangle GBC) \] Answer: \(\displaystyle \operatorname{ar}(AFGE) = \operatorname{ar}(\triangle GBC)\)

\(\displaystyle X,Y\) lie on line \(\displaystyle BE\), with \(\displaystyle X\) on \(\displaystyle AC\) and \(\displaystyle Y\) on \(\displaystyle AD\). \[\operatorname{ar}(\triangle ABC) = \operatorname{ar}(\triangle ABY) \quad \text{(same base } AB\text{, between } AB \parallel CY\text{)} \] \[\operatorname{ar}(\triangle ABC) = \operatorname{ar}(\triangle ABX) + \operatorname{ar}(\triangle CBX) \quad (X \text{ on } AC) \] \[\operatorname{ar}(\triangle ABY) = \operatorname{ar}(\triangle ABX) + \operatorname{ar}(\triangle AXY) \quad (X \text{ on } BY) \] \[\therefore \operatorname{ar}(\triangle ABX) + \operatorname{ar}(\triangle CBX) = \operatorname{ar}(\triangle ABX) + \operatorname{ar}(\triangle AXY) \] \[\operatorname{ar}(CBX) = \operatorname{ar}(AXY) \] Answer: \(\displaystyle \operatorname{ar}(CBX) = \operatorname{ar}(AXY)\)

\[XY = \frac{AB+DC}{2} = \frac{50+30}{2} = 40 \text{ cm} \quad \text{(}X,Y\text{ midpoints of the legs)} \] \(\displaystyle XY\) lies midway between \(\displaystyle AB\) and \(\displaystyle DC\), so both parts have height \(\displaystyle h/2\). \[\operatorname{ar}(DCYX) = \frac12(DC+XY)\cdot\frac h2, \qquad \operatorname{ar}(XYBA) = \frac12(XY+AB)\cdot\frac h2 \] \[\frac{\operatorname{ar}(DCYX)}{\operatorname{ar}(XYBA)} = \frac{DC+XY}{XY+AB} = \frac{30+40}{40+50} = \frac{70}{90} = \frac79 \] \[\therefore \operatorname{ar}(DCYX) = \frac79\operatorname{ar}(XYBA) \] Answer: \(\displaystyle \operatorname{ar}(DCYX) = \tfrac79\operatorname{ar}(XYBA)\)

\(\displaystyle O\) is the intersection of \(\displaystyle CL\) and \(\displaystyle BM\). \[\operatorname{ar}(\triangle LBC) = \operatorname{ar}(\triangle MBC) \quad \text{(same base } BC\text{, between } LM \parallel BC\text{)} \] \[\operatorname{ar}(\triangle LBC) = \operatorname{ar}(\triangle LOB) + \operatorname{ar}(\triangle OBC) \quad (O \text{ on } CL) \] \[\operatorname{ar}(\triangle MBC) = \operatorname{ar}(\triangle MOC) + \operatorname{ar}(\triangle OBC) \quad (O \text{ on } BM) \] \[\therefore \operatorname{ar}(\triangle LOB) + \operatorname{ar}(\triangle OBC) = \operatorname{ar}(\triangle MOC) + \operatorname{ar}(\triangle OBC) \] \[\operatorname{ar}(\triangle LOB) = \operatorname{ar}(\triangle MOC) \] Answer: \(\displaystyle \operatorname{ar}(\triangle LOB) = \operatorname{ar}(\triangle MOC)\)

\[\operatorname{ar}(\mathrm{ABC}) = \operatorname{ar}(\mathrm{APC}) \quad \text{(base AC, } \mathrm{BP} \parallel \mathrm{AC}\text{)} \] \[\operatorname{ar}(\mathrm{AED}) = \operatorname{ar}(\mathrm{AQD}) \quad \text{(base AD, } \mathrm{EQ} \parallel \mathrm{AD}\text{)} \] \[\operatorname{ar}(\mathrm{ABCDE}) = \operatorname{ar}(\mathrm{ABC}) + \operatorname{ar}(\mathrm{ACD}) + \operatorname{ar}(\mathrm{AED}) \] \[= \operatorname{ar}(\mathrm{APC}) + \operatorname{ar}(\mathrm{ACD}) + \operatorname{ar}(\mathrm{AQD}) = \operatorname{ar}(\mathrm{APQ}) \]Answer: \(\displaystyle \operatorname{ar}(\mathrm{ABCDE}) = \operatorname{ar}(\mathrm{APQ})\)

\[\operatorname{ar}(\mathrm{ABD}) = \operatorname{ar}(\mathrm{ACD}) \quad \text{(AD median of } \triangle \mathrm{ABC}\text{)} \] \[\operatorname{ar}(\mathrm{GBD}) = \operatorname{ar}(\mathrm{GCD}) \quad \text{(GD median of } \triangle \mathrm{GBC}\text{)} \] \[\operatorname{ar}(\mathrm{AGB}) = \operatorname{ar}(\mathrm{AGC}) \quad \text{(subtracting the two lines above)} \] \[\operatorname{ar}(\mathrm{AGB}) = \operatorname{ar}(\mathrm{BGC}) \quad \text{(same argument, median BE)} \] \[\operatorname{ar}(\mathrm{AGB}) + \operatorname{ar}(\mathrm{AGC}) + \operatorname{ar}(\mathrm{BGC}) = \operatorname{ar}(\mathrm{ABC}) \] \[\Rightarrow\ 3\operatorname{ar}(\mathrm{AGB}) = \operatorname{ar}(\mathrm{ABC}) \ \Rightarrow\ \operatorname{ar}(\mathrm{AGB})=\operatorname{ar}(\mathrm{AGC})=\operatorname{ar}(\mathrm{BGC}) = \frac{1}{3}\operatorname{ar}(\mathrm{ABC}) \]Answer: \(\displaystyle \operatorname{ar}(\mathrm{AGB})=\operatorname{ar}(\mathrm{AGC})=\operatorname{ar}(\mathrm{BGC}) = \tfrac13\operatorname{ar}(\mathrm{ABC})\)

\[\operatorname{ar}(\mathrm{ABX}) = \operatorname{ar}(\mathrm{CBX}) = \tfrac{1}{2}\operatorname{ar}(\mathrm{ABC}) \quad \text{(BX median, X mid-point of AC)} \] \[\operatorname{ar}(\mathrm{ACY}) = \operatorname{ar}(\mathrm{BCY}) = \tfrac{1}{2}\operatorname{ar}(\mathrm{ABC}) \quad \text{(CY median, Y mid-point of AB)} \] \[\operatorname{ar}(\mathrm{QBC}) = \operatorname{ar}(\mathrm{PBC}) \quad \text{(same base BC, QP} \parallel \text{BC)} \] \[\operatorname{ar}(\mathrm{BYQ}) + \operatorname{ar}(\mathrm{BYC}) = \operatorname{ar}(\mathrm{CXP}) + \operatorname{ar}(\mathrm{CXB}) \quad \text{(Y on CQ, X on BP)} \] \[\operatorname{ar}(\mathrm{BYQ}) = \operatorname{ar}(\mathrm{CXP}) \quad \text{(} \operatorname{ar}(\mathrm{BYC}) = \operatorname{ar}(\mathrm{CXB}) \text{, from the first two lines)} \] \[\operatorname{ar}(\mathrm{AYQ}) = \operatorname{ar}(\mathrm{BYQ}), \quad \operatorname{ar}(\mathrm{AXP}) = \operatorname{ar}(\mathrm{CXP}) \quad \text{(Y, X mid-points; common vertex Q, P)} \] \[\operatorname{ar}(\mathrm{ABP}) = \operatorname{ar}(\mathrm{ABX}) + \operatorname{ar}(\mathrm{AXP}) = \operatorname{ar}(\mathrm{ACY}) + \operatorname{ar}(\mathrm{AYQ}) = \operatorname{ar}(\mathrm{ACQ}) \]Answer: \(\displaystyle \operatorname{ar}(\mathrm{ABP}) = \operatorname{ar}(\mathrm{ACQ})\)

\[\mathrm{EF} \parallel \mathrm{AD} \quad \text{(opposite sides, parallelogram AEFD)} \] \[\mathrm{AP} \parallel \mathrm{DQ} \quad \text{(P on AB, Q on DC, AB} \parallel \mathrm{DC}\text{)} \] \[\mathrm{APQD} \text{ is a parallelogram} \ \Rightarrow\ \operatorname{ar}(\mathrm{APD}) = \operatorname{ar}(\mathrm{DPQ}) \quad \text{(diagonal PD)} \] \[\operatorname{ar}(\mathrm{APD}) = \operatorname{ar}(\mathrm{ADF}) \quad \text{(base AD; P, F} \in \mathrm{EF} \parallel \mathrm{AD}\text{)} \] \[\operatorname{ar}(\mathrm{ADF}) = \operatorname{ar}(\mathrm{AEF}) \quad \text{(diagonal AF, parallelogram AEFD)} \] \[\operatorname{ar}(\mathrm{APF}) = \operatorname{ar}(\mathrm{DPF}) \quad \text{(base PF; A, D on line AD} \parallel \mathrm{EF}\text{)} \] \[\operatorname{ar}(\mathrm{PEA}) = \operatorname{ar}(\mathrm{AEF}) - \operatorname{ar}(\mathrm{APF}) = \operatorname{ar}(\mathrm{DPQ}) - \operatorname{ar}(\mathrm{DPF}) = \operatorname{ar}(\mathrm{QFD}) \]Answer: \(\displaystyle \operatorname{ar}(\mathrm{PEA}) = \operatorname{ar}(\mathrm{QFD})\)