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NCERT Exemplar · Class 9 Mathematics Areas of Parallelograms and Triangles

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EXERCISE 9.2 1–5 (part 2 of 4)

  1. Write True or False and justify your answer :

    Exercise 1

    ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD)=24 cm2\displaystyle (\mathrm{AXCD})=24 \mathrm{~cm}^{2}, then ar⁡(ABC)=24 cm2\displaystyle \operatorname{ar}(\mathrm{ABC})=24 \mathrm{~cm}^{2}.

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    NCERT’s answer
    False, since $\displaystyle \operatorname{ar}(\mathrm{AXCD})=\operatorname{ar}(\mathrm{ABCD})-\operatorname{ar}(\mathrm{BCX})=48-12=36 \mathrm{~cm}^{2}$
    False.NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-2_Q1\[\operatorname{ar}(\mathrm{BXC}) = \frac{1}{2}\cdot BX\cdot h, \quad BX=\frac{1}{2}AB \quad \text{(same height \(\displaystyle h\) as } \mathrm{ABCD}\text{)} \] \[\operatorname{ar}(\mathrm{BXC}) = \frac{1}{4}\operatorname{ar}(\mathrm{ABCD}) \] \[\operatorname{ar}(\mathrm{AXCD}) = \operatorname{ar}(\mathrm{ABCD}) - \operatorname{ar}(\mathrm{BXC}) = \frac{3}{4}\operatorname{ar}(\mathrm{ABCD}) \] \[24 = \frac{3}{4}\operatorname{ar}(\mathrm{ABCD}) \implies \operatorname{ar}(\mathrm{ABCD}) = 32\text{ cm}^2 \] \[\operatorname{ar}(\mathrm{ABC}) = \frac{1}{2}\operatorname{ar}(\mathrm{ABCD}) \quad \text{(diagonal \(\displaystyle AC\) bisects the parallelogram)} \] \[\operatorname{ar}(\mathrm{ABC}) = 16\text{ cm}^2 \]

  2. Exercise 2

    PQRS is a rectangle inscribed in a quadrant of a circle of radius 13\displaystyle 13 cm. A is any point on PQ . If PS=5 cm\displaystyle \mathrm{PS}=5 \mathrm{~cm}, then ar⁡(PAS)=30 cm2\displaystyle \operatorname{ar}(\mathrm{PAS})=30 \mathrm{~cm}^{2}.

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    False.NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-2_Q2\[SQ^2 = SP^2+PQ^2 \quad \text{(diagonal of rectangle } \mathrm{PQRS}\text{, } SQ=13) \] \[13^2 = 5^2+PQ^2 \implies PQ=12\text{ cm} \] \[\operatorname{ar}(\mathrm{PAS}) = \frac{1}{2}\cdot PA\cdot PS \quad (\angle SPQ=90^\circ) \] \[\operatorname{ar}(\mathrm{PAS}) = \frac{5}{2}PA, \quad 0\le PA\le 12 \] Area grows with \(\displaystyle PA\); it equals \(\displaystyle 30\text{ cm}^2\) only when \(\displaystyle A=Q\), not for every point on \(\displaystyle PQ\).Answer: False — \(\displaystyle \operatorname{ar}(\mathrm{PAS})=30\text{ cm}^2\) only when \(\displaystyle A=Q\).NCERT prints: True — but the stem as printed asks about ar(PAS), and the key's own working (half the rectangle, same base and parallels) is about ar(ASR).

  3. Exercise 3

    PQRS is a parallelogram whose area is 180 cm2\displaystyle 180 \mathrm{~cm}^{2} and A is any point on the diagonal QS . The area of ΔASR=90 cm2\displaystyle \Delta \mathrm{ASR}=90 \mathrm{~cm}^{2}.

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    False, because area of $\displaystyle \Delta \mathrm{QSR}=90 \mathrm{~cm}^{2}$ and area of $\displaystyle \Delta \mathrm{ASR}<$ area of $\displaystyle \Delta \mathrm{QRS}$.
    False.NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-2_Q3\[\operatorname{ar}(\mathrm{QSR}) = \frac{1}{2}\operatorname{ar}(\mathrm{PQRS}) = 90\text{ cm}^2 \quad \text{(diagonal \(\displaystyle QS\) bisects the parallelogram)} \] \[\operatorname{ar}(\mathrm{ASR}) = \frac{SA}{SQ}\cdot\operatorname{ar}(\mathrm{QSR}) \quad \text{(same base \(\displaystyle SR\), heights} \propto SA:SQ\text{)} \] Since \(\displaystyle SA<SQ\) for \(\displaystyle A\) strictly between \(\displaystyle Q\) and \(\displaystyle S\), \(\displaystyle \operatorname{ar}(\mathrm{ASR})<90\text{ cm}^2\); equality holds only at \(\displaystyle A=Q\).

  4. Exercise 4

    ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then ar⁡(BDE)=14ar⁡(ABC)\displaystyle \operatorname{ar}(\mathrm{BDE})=\frac{1}{4} \operatorname{ar}(\mathrm{ABC}).

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    True, $\displaystyle \frac{\operatorname{ar} \mathrm{BDE}}{\operatorname{ar} \mathrm{ABC}}=\frac{\sqrt{3}(\mathrm{BD})^{2}}{\dfrac{\sqrt{3}(\mathrm{BC})^{2}}{4}}=\frac{(\mathrm{BC})^{2}}{(\mathrm{BC})^{2}}=\frac{1}{4}$
    True.NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-2_Q4\[BD=\frac{1}{2}BC \quad \text{(\(\displaystyle D\) is the mid-point of \(\displaystyle BC\))} \] \[\operatorname{ar}(\mathrm{BDE})=\frac{\sqrt3}{4}BD^2, \quad \operatorname{ar}(\mathrm{ABC})=\frac{\sqrt3}{4}BC^2 \quad \text{(area of an equilateral triangle)} \] \[\frac{\operatorname{ar}(\mathrm{BDE})}{\operatorname{ar}(\mathrm{ABC})}=\left(\frac{BD}{BC}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4} \] \[\operatorname{ar}(\mathrm{BDE})=\frac{1}{4}\operatorname{ar}(\mathrm{ABC}) \]

  5. Exercise 5

    In Fig. 9.8\displaystyle 9.8, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then ar⁡(DPC)=12ar⁡(EFGD).\operatorname{ar}(\mathrm{DPC})=\frac{1}{2} \operatorname{ar}(\mathrm{EFGD}) . (D) Short Answer Questions Sample Question 1\displaystyle 1 : PQRS is a square. T and U are respectively, the mid-points of PS and QR (Fig. 9.9\displaystyle 9.9). Find the area of ΔOTS\displaystyle \Delta \mathrm{OTS}, if PQ=8 cm\displaystyle \mathrm{PQ}=8 \mathrm{~cm}, where O is the point of intersection of TU and QS. Solution : PS=PQ=8 cm\displaystyle \mathrm{PS}=\mathrm{PQ}=8 \mathrm{~cm} and TU∥PQ\displaystyle \mathrm{TU} \| \mathrm{PQ} ST=12PS=12×8=4 cmPQ=TU=8 cm\begin{aligned} \mathrm{ST} & =\frac{1}{2} \mathrm{PS}=\frac{1}{2} \times 8=4 \mathrm{~cm} \\ \mathrm{PQ} & =\mathrm{TU}=8 \mathrm{~cm} \end{aligned} OT=12TU=12×8=4 cm\mathrm{OT}=\frac{1}{2} \mathrm{TU}=\frac{1}{2} \times 8=4 \mathrm{~cm} Area of triangle OTS =12×ST×OT [Since OTS is a right angled triangle] =12×4×4 cm2=8 cm2\begin{aligned} & =\frac{1}{2} \times \mathrm{ST} \times \mathrm{OT} \text { [Since OTS is a right angled triangle] } \\ & =\frac{1}{2} \times 4 \times 4 \mathrm{~cm}^{2}=8 \mathrm{~cm}^{2} \end{aligned} Sample Question 2\displaystyle 2 : ABCD is a parallelogram and BC is produced to a point Q such that AD=CQ\displaystyle \mathrm{AD}=\mathrm{CQ} (Fig. 9.10\displaystyle 9.10). If AQ intersects DC at P , show that ar⁡(BPC)=ar⁡(DPQ)\displaystyle \operatorname{ar}(\mathrm{BPC})=\operatorname{ar}(\mathrm{DPQ}) Solution: ar⁡(ACP)=ar⁡(BCP)\displaystyle \operatorname{ar}(\mathrm{ACP})=\operatorname{ar}(\mathrm{BCP}) [Triangles on the same base and between same parallels] ar⁡(ADQ)=ar⁡(ADC)\displaystyle \operatorname{ar}(\mathrm{ADQ})=\operatorname{ar}(\mathrm{ADC}) ar⁡(ADC)−ar⁡(ADP)=ar⁡(ADQ)−ar⁡(ADP)\displaystyle \operatorname{ar}(\mathrm{ADC})-\operatorname{ar}(\mathrm{ADP})=\operatorname{ar}(\mathrm{ADQ})-\operatorname{ar}(\mathrm{ADP}) ar⁡(APC)=ar⁡(DPQ)\displaystyle \operatorname{ar}(\mathrm{APC})=\operatorname{ar}(\mathrm{DPQ}) From (1\displaystyle 1) and (3\displaystyle 3), we get ar⁡(BCP)=ar⁡(DPQ)\displaystyle \operatorname{ar}(\mathrm{BCP})=\operatorname{ar}(\mathrm{DPQ}) NCERT_Question_Class9_Maths_Exemplar_Ch9_Ex9-2_Q5 NCERT_Question_Class9_Maths_Exemplar_Ch9_Ex9-2_Q5 NCERT_Question_Class9_Maths_Exemplar_Ch9_Ex9-2_Q5

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    False, because $\displaystyle \operatorname{ar}(\mathrm{DPC})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD})=\operatorname{ar}(\mathrm{EFGD})$
    False.NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-2_Q5\[GD=\frac{1}{2}DC \quad \text{(\(\displaystyle G\) is the mid-point of \(\displaystyle DC\))} \] \[\operatorname{ar}(\mathrm{EFGD})=GD\cdot h \quad \text{(parallelogram, height \(\displaystyle h\) between \(\displaystyle AB\) and \(\displaystyle DC\))} \] \[\operatorname{ar}(\mathrm{DPC})=\frac{1}{2}\cdot DC\cdot h \quad \text{(\(\displaystyle P\) on \(\displaystyle AB\), the same height \(\displaystyle h\) above base \(\displaystyle DC\))} \] \[\operatorname{ar}(\mathrm{DPC})=\frac{1}{2}DC\cdot h = GD\cdot h = \operatorname{ar}(\mathrm{EFGD}) \]