SolveItNCERT · CBSE Boards

NCERT Exemplar · Class 9 Mathematics Areas of Parallelograms and Triangles

34 questions · 34 still being checked

EXERCISE 9.3 1–9 (part 3 of 4)

  1. Exercise 1

    In Fig.9.11, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ=QR=RS\displaystyle \mathrm{PQ}=\mathrm{QR}=\mathrm{RS} and PA∥QB∥RC\displaystyle \mathrm{PA}\|\mathrm{QB}\| \mathrm{RC}. Prove that ar⁡(PQE)=ar⁡(CFD)\displaystyle \operatorname{ar}(\mathrm{PQE})=\operatorname{ar}(\mathrm{CFD}). NCERT_Question_Class9_Maths_Exemplar_Ch9_Ex9-3_Q1

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-3_Q1 \[PS \parallel AD,\ SD \parallel PA \quad \text{(opposite sides of } \square PSDA\text{)} \] \[PA \parallel QB \parallel RC \parallel SD \] \[RS \parallel DC \quad \text{(parts of } PS,\ AD\text{)}, \quad SD \parallel RC\ \Rightarrow\ \square RSDC \text{ is a parallelogram} \] \[RS = DC \quad \text{(opposite sides of } \square RSDC\text{)} \] \[PQ = RS = DC \quad (PQ = RS \text{ given}) \] \[\angle QPE = \angle CDF \quad \text{(alt. angles, } PS\parallel AD,\ PD \text{ transversal)} \] \[\angle PQE = \angle QBD \quad \text{(alt. angles, } PS\parallel AD,\ QB \text{ transversal)} \] \[\angle QBD = \angle RCD \quad \text{(corr. angles, } QB\parallel RC,\ AD \text{ transversal)} \] \[\angle PQE = \angle RCD = \angle DCF \] \[\triangle PQE \cong \triangle DCF \quad (ASA:\ PQ=DC,\ \angle QPE=\angle CDF,\ \angle PQE=\angle DCF) \] \[\operatorname{ar}(PQE) = \operatorname{ar}(CFD) \]Answer: \(\displaystyle \operatorname{ar}(PQE)=\operatorname{ar}(CFD)\).

  2. Exercise 2

    X and Y are points on the side LN of the triangle LMN such that LX=XY=YN\displaystyle \mathrm{LX}=\mathrm{XY}=\mathrm{YN}. Through X, a line is drawn parallel to LM to meet MN at Z (See Fig. 9.12\displaystyle 9.12). Prove that ar⁡(LZY)=ar⁡(MZYX)\displaystyle \operatorname{ar}(\mathrm{LZY})=\operatorname{ar}(\mathrm{MZYX}) NCERT_Question_Class9_Maths_Exemplar_Ch9_Ex9-3_Q2

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-3_Q2 \[LX = XY = YN \quad \text{(given)} \] \[XZ \parallel LM \quad \text{(given)} \] \[\operatorname{ar}(LXZ) = \operatorname{ar}(MXZ) \quad \text{(same base } XZ,\text{ between } LM\parallel XZ\text{)} \] \[\operatorname{ar}(LZY) = \operatorname{ar}(LXZ) + \operatorname{ar}(XZY) \] \[\operatorname{ar}(MZYX) = \operatorname{ar}(MXZ) + \operatorname{ar}(XZY) \] \[\operatorname{ar}(LZY) = \operatorname{ar}(MZYX) \]Answer: \(\displaystyle \operatorname{ar}(LZY)=\operatorname{ar}(MZYX)\)

  3. Exercise 3

    The area of the parallelogram ABCD is 90 cm2\displaystyle 90 \mathrm{~cm}^{2} (see Fig.9.13). Find (i) ar (ABEF) (ii) ar (ABD) (iii) ar (BEF) NCERT_Question_Class9_Maths_Exemplar_Ch9_Ex9-3_Q3

    Matches the book, not yet reviewed

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

    NCERT’s answer
    (i)
    $\displaystyle 90 \mathrm{~cm}^{2}$ (ii) $\displaystyle 45 \mathrm{~cm}^{2}$ (iii) $\displaystyle 45 \mathrm{~cm}^{2}$
    NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-3_Q3 \[\text{(i)}\quad \square ABEF,\ \square ABCD \text{ share base } AB,\text{ between } AB \parallel FC \] \[\operatorname{ar}(ABEF) = \operatorname{ar}(ABCD) = 90 \text{ cm}^2 \] \[\text{(ii)}\quad \operatorname{ar}(ABD) = \tfrac12 \operatorname{ar}(ABCD) \quad \text{(diagonal } BD\text{)} \] \[\operatorname{ar}(ABD) = 45 \text{ cm}^2 \] \[\text{(iii)}\quad \operatorname{ar}(BEF) = \tfrac12 \operatorname{ar}(ABEF) \quad \text{(diagonal } BF\text{)} \] \[\operatorname{ar}(BEF) = 45 \text{ cm}^2 \]Answer: (i) \(\displaystyle \operatorname{ar}(\mathrm{ABEF}) = 90\text{ cm}^2\) (ii) \(\displaystyle \operatorname{ar}(\mathrm{ABD}) = 45\text{ cm}^2\) (iii) \(\displaystyle \operatorname{ar}(\mathrm{BEF}) = 45\text{ cm}^2\)

  4. Exercise 4

    In ΔABC,D\displaystyle \Delta \mathrm{ABC}, \mathrm{D} is the mid-point of AB\displaystyle A B and P\displaystyle P is any point on BC\displaystyle B C. If CQ∥PD\displaystyle CQ \parallel PD meets AB in Q (Fig. 9.14\displaystyle 9.14), then prove that ar⁡(BPQ)=12ar⁡(ABC)\displaystyle \operatorname{ar}(\mathrm{BPQ})=\frac{1}{2} \operatorname{ar}(\mathrm{ABC}). NCERT_Question_Class9_Maths_Exemplar_Ch9_Ex9-3_Q4

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-3_Q4 \[\operatorname{ar}(ADC) = \operatorname{ar}(DBC) = \tfrac12\operatorname{ar}(ABC) \quad (CD \text{ is a median}) \] \[\operatorname{ar}(DPQ) = \operatorname{ar}(DPC) \quad \text{(same base } DP,\ DP\parallel QC\text{)} \] \[\operatorname{ar}(BPQ) = \operatorname{ar}(BPD) + \operatorname{ar}(DPQ) \quad (D \text{ on } QB) \] \[\operatorname{ar}(BPQ) = \operatorname{ar}(BPD) + \operatorname{ar}(DPC) = \operatorname{ar}(DBC) \] \[\operatorname{ar}(BPQ) = \tfrac12 \operatorname{ar}(ABC) \]Answer: \(\displaystyle \operatorname{ar}(BPQ)=\tfrac12\operatorname{ar}(ABC)\)

  5. Exercise 5

    ABCD is a square. E and F are respectively the mid-points of BC and CD. If R is the mid-point of EF (Fig. 9.15\displaystyle 9.15), prove that ar⁡(AER)=ar⁡(AFR)\displaystyle \operatorname{ar}(\mathrm{AER})=\operatorname{ar}(\mathrm{AFR}). NCERT_Question_Class9_Maths_Exemplar_Ch9_Ex9-3_Q5

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-3_Q5 \[ER = FR \quad (R \text{ is the midpoint of } EF) \] \[\operatorname{ar}(AER) = \tfrac12 \cdot ER \cdot h,\quad \operatorname{ar}(AFR) = \tfrac12 \cdot FR \cdot h \quad (h:\text{ height from } A \text{ to } EF) \] \[\operatorname{ar}(AER) = \operatorname{ar}(AFR) \]Answer: \(\displaystyle \operatorname{ar}(AER)=\operatorname{ar}(AFR)\)

  6. Exercise 6

    O is any point on the diagonal PR of a parallelogram PQRS (Fig. 9.16\displaystyle 9.16). Prove that ar⁡(PSO)=ar⁡(PQO)\displaystyle \operatorname{ar}(\mathrm{PSO})=\operatorname{ar}(\mathrm{PQO}). NCERT_Question_Class9_Maths_Exemplar_Ch9_Ex9-3_Q6

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-3_Q6 \[\operatorname{ar}(PQR) = \operatorname{ar}(PSR) = \tfrac12 \operatorname{ar}(PQRS) \quad \text{(diagonal } PR\text{)} \] \[\Rightarrow\ d(Q, PR) = d(S, PR) \quad \text{(equal areas, common base } PR\text{)} \] \[\operatorname{ar}(PQO) = \tfrac12 \cdot PO \cdot d(Q,PR) \] \[\operatorname{ar}(PSO) = \tfrac12 \cdot PO \cdot d(S,PR) \] \[\operatorname{ar}(PSO) = \operatorname{ar}(PQO) \]Answer: \(\displaystyle \operatorname{ar}(PSO)=\operatorname{ar}(PQO)\)

  7. Exercise 7

    ABCD is a parallelogram in which BC is produced to E such that CE=BC\displaystyle \mathrm{CE}=\mathrm{BC} (Fig. 9.17\displaystyle 9.17). AE intersects CD at F. If ar⁡(DFB)=3 cm2\displaystyle \operatorname{ar}(\mathrm{DFB})=3 \mathrm{~cm}^{2}, find the area of the parallelogram ABCD. NCERT_Question_Class9_Maths_Exemplar_Ch9_Ex9-3_Q7

    Matches the book, not yet reviewed

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

    NCERT’s answer
    $\displaystyle 12 \mathrm{~cm}^{2}$
    NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-3_Q7 \[CE = BC \ \Rightarrow\ C \text{ is the midpoint of } BE \] \[CF \parallel AB,\ BC = CE \ \Rightarrow\ CF = \tfrac12 AB = \tfrac12 DC \quad \text{(converse of mid-point theorem, } \triangle ABE\text{)} \] \[DF = DC - CF = \tfrac12 DC \quad \Rightarrow\ DF = FC = \tfrac12 DC \] \[\operatorname{ar}(DFB) = \tfrac12 \operatorname{ar}(DCB) \quad (F \text{ midpoint of } DC) \] \[\operatorname{ar}(DCB) = \tfrac12 \operatorname{ar}(ABCD) \quad \text{(diagonal } BD\text{)} \] \[\operatorname{ar}(DFB) = \tfrac14 \operatorname{ar}(ABCD) \] \[\operatorname{ar}(ABCD) = 4 \times 3 = 12 \text{ cm}^2 \]Answer: ar(ABCD) = $\displaystyle 12$ cm².

  8. Exercise 8

    In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ∥AD\displaystyle PQ \parallel AD has been drawn which meets AB in P and DC produced in Q (Fig. 9.18\displaystyle 9.18). Prove that ar⁡(ABCD)=ar⁡(APQD)\displaystyle \operatorname{ar}(\mathrm{ABCD})=\operatorname{ar}(\mathrm{APQD}) NCERT_Question_Class9_Maths_Exemplar_Ch9_Ex9-3_Q8

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-3_Q8 \[BL = CL \quad (L \text{ is the midpoint of } BC) \] \[\angle PBL = \angle QCL \quad \text{(alternate angles, } AB \parallel DQ\text{)} \] \[\angle PLB = \angle QLC \quad \text{(vertically opposite)} \] \[\triangle PBL \cong \triangle QCL \quad (ASA) \] \[\operatorname{ar}(PBL) = \operatorname{ar}(QCL) \] \[\operatorname{ar}(ABCD) = \operatorname{ar}(APLCD) + \operatorname{ar}(PBL),\quad \operatorname{ar}(APQD) = \operatorname{ar}(APLCD) + \operatorname{ar}(QCL) \] \[\operatorname{ar}(ABCD) = \operatorname{ar}(APQD) \]Answer: \(\displaystyle \operatorname{ar}(ABCD)=\operatorname{ar}(APQD)\)

  9. Exercise 9

    If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram so formed will be half of the area of the given quadrilateral (Fig. 9.19\displaystyle 9.19). [Hint: Join BD and draw perpendicular from A on BD.] (E) Long Answer Questions Sample Question 1\displaystyle 1 : In Fig. 9.20\displaystyle 9.20, ABCD is a parallelogram. Points P and Q on BC trisects BC in three equal parts. Prove that ar⁡(APQ)=ar⁡(DPQ)=16ar⁡(ABCD)\operatorname{ar}(\mathrm{APQ})=\operatorname{ar}(\mathrm{DPQ})=\frac{1}{6} \operatorname{ar}(\mathrm{ABCD}) NCERT_Question_Class9_Maths_Exemplar_Ch9_Ex9-3_Q9 NCERT_Question_Class9_Maths_Exemplar_Ch9_Ex9-3_Q9

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-3_Q9\[S=\text{mid}(AB),\ P=\text{mid}(BC),\ F=\text{mid}(CD),\ R=\text{mid}(DA) \]Join \(\displaystyle BD\) and \(\displaystyle AC\).\[\operatorname{ar}(ADS)=\operatorname{ar}(DSB)=\tfrac12\operatorname{ar}(ABD) \quad \text{(}DS\text{ median of }\triangle ABD\text{)} \] \[\operatorname{ar}(ASR)=\operatorname{ar}(DSR)=\tfrac14\operatorname{ar}(ABD) \quad \text{(}SR\text{ median of }\triangle ADS\text{)} \] \[\operatorname{ar}(CPF)=\tfrac14\operatorname{ar}(CBD),\quad \operatorname{ar}(BSP)=\tfrac14\operatorname{ar}(ABC),\quad \operatorname{ar}(DRF)=\tfrac14\operatorname{ar}(ACD) \quad \text{(same double-median step at each corner)} \] \[\operatorname{ar}(ASR)+\operatorname{ar}(CPF)+\operatorname{ar}(BSP)+\operatorname{ar}(DRF)=\tfrac14\big[\operatorname{ar}(ABD)+\operatorname{ar}(CBD)\big]+\tfrac14\big[\operatorname{ar}(ABC)+\operatorname{ar}(ACD)\big]=\tfrac12\operatorname{ar}(ABCD) \] \[\operatorname{ar}(SPFR)=\operatorname{ar}(ABCD)-\tfrac12\operatorname{ar}(ABCD)=\tfrac12\operatorname{ar}(ABCD) \]Answer: \(\displaystyle \operatorname{ar}(SPFR)=\dfrac12\,\operatorname{ar}(ABCD)\)