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NCERT Exemplar · Class 9 Mathematics Areas of Parallelograms and Triangles

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EXERCISE 9.1 1–10 (part 1 of 4)

  1. Write the correct answer in each of the following :

    Exercise 1

    The median of a triangle divides it into two (A) triangles of equal area (B) congruent triangles (C) right triangles (D) isosceles triangles

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    (A)
    (A) triangles of equal areaNCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-1_Q1\[BD=DC \quad\text{(D is the mid-point of }BC\text{)} \]\[ar(\triangle ABD)=\tfrac12\,BD\cdot AH=\tfrac12\,DC\cdot AH=ar(\triangle ACD) \]Equal bases BD, DC and a common height AH from A give equal areas, not necessarily congruent triangles.

  2. Exercise 2

    In which of the following figures (Fig. 9.3\displaystyle 9.3), you find two polygons on the same base and between the same parallels? NCERT_Question_Class9_Maths_Exemplar_Ch9_Ex9-1_Q2

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    (D)
    (D)NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-1_Q2\[PS\parallel QR \] \[ar(\triangle AQR)=ar(\triangle BQR) \]A and B lie on PS; both triangles share base QR, so both sit between the same parallels.

  3. Exercise 3

    The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8\displaystyle 8 cm and 6\displaystyle 6 cm is : (A) a rectangle of area 24 cm2\displaystyle 24 \mathrm{~cm}^{2} (B) a square of area 25 cm2\displaystyle 25 \mathrm{~cm}^{2} (C) a trapezium of area 24 cm2\displaystyle 24 \mathrm{~cm}^{2} (D) a rhombus of area 24 cm2\displaystyle 24 \mathrm{~cm}^{2}

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    (D)
    (D) a rhombus of area $\displaystyle 24$ \(\displaystyle \text{cm}^2\)NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-1_Q3\[PQ=QR=RS=SP=\tfrac12\sqrt{8^{2}+6^{2}}=5\text{ cm} \] \[ar(PQRS)=\tfrac12\,ar(ABCD)=\tfrac12(8\times6)=24\ \text{cm}^{2} \]Equal sides but unequal diagonals rule out a square.

  4. Exercise 4

    In Fig. 9.4\displaystyle 9.4, the area of parallelogram ABCD is : (A) AB×BM\displaystyle \mathrm{AB} \times \mathrm{BM} (B) BC×BN\displaystyle \mathrm{BC} \times \mathrm{BN} (C) DC × DL (D) AD×DL\displaystyle \mathrm{AD} \times \mathrm{DL} NCERT_Question_Class9_Maths_Exemplar_Ch9_Ex9-1_Q4

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    NCERT’s answer
    (C)
    (C) \(\displaystyle DC\times DL\)NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-1_Q4\[ar(ABCD)=DC\times DL=AB\times DL \]DL and BN both drop onto DC (and AB); BM alone drops onto AD. So BN pairs with the wrong base in (B), leaving (C) correct.

  5. Exercise 5

    In Fig. 9.5\displaystyle 9.5, if parallelogram ABCD and rectangle ABEF are of equal area, then : (A) Perimeter of ABCD = Perimeter of ABEM (B) Perimeter of ABCD < Perimeter of ABEM (C) Perimeter of ABCD > Perimeter of ABEM (D) Perimeter of ABCD=12\displaystyle \mathrm{ABCD}=\frac{1}{2} (Perimeter of ABEM) NCERT_Question_Class9_Maths_Exemplar_Ch9_Ex9-1_Q5

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    NCERT’s answer
    (C)
    (C) Perimeter of ABCD > Perimeter of ABEMNCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-1_Q5\[ar(ABCD)=ar(ABEM)=AB\times AM \]\[AD=\sqrt{AM^{2}+MD^{2}}>AM \]\[\text{Perimeter}(ABCD)=2(AB+AD) > 2(AB+AM)=\text{Perimeter}(ABEM) \]Same base, equal area: same height AM.

  6. Exercise 6

    The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to (A) 12\displaystyle \frac{1}{2} ar (ABC) (B) 13\displaystyle \frac{1}{3} ar (ABC) (C) 14\displaystyle \frac{1}{4} ar (ABC) (D) ar (ABC)

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    (A)
    (A) \(\displaystyle \tfrac12\) ar(ABC)NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-1_Q6\[FD\parallel AE,\quad FD=AE=\tfrac12 AC \ \Rightarrow\ AFDE \text{ is a parallelogram} \]\[ar(\triangle AFE)=ar(\triangle FDE)=\tfrac14\,ar(ABC) \]\[ar(AFDE)=\tfrac12\,ar(ABC) \]F, D, E are the mid-points of AB, BC, CA; AFE and the medial triangle FDE are each a quarter of ABC.

  7. Exercise 7

    Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is (A) 1\displaystyle 1 : 2\displaystyle 2 (B) 1\displaystyle 1 : 1\displaystyle 1 (C) 2\displaystyle 2 : 1\displaystyle 1 (D) 3\displaystyle 3 : 1\displaystyle 1

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    NCERT’s answer
    (B)
    (B) $\displaystyle 1$ : $\displaystyle 1$NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-1_Q7\[ar(PQRS)=PQ\cdot h=WX\cdot h=ar(WXYZ) \]Equal bases and the same pair of parallels fix a common height h for both parallelograms.

  8. Exercise 8

    ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD (A) is a rectangle (B) is always a rhombus (C) is a parallelogram (D) need not be any of (A), (B) or (C)

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    NCERT’s answer
    (D)
    (D) need not be any of (A), (B) or (C)NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-1_Q8\[ar(\triangle ABC)=\tfrac12\,AC\times h_B,\quad ar(\triangle ACD)=\tfrac12\,AC\times h_D \] \[h_B=h_D\ \Rightarrow\ ar(\triangle ABC)=ar(\triangle ACD) \]Only the two heights from B and D to AC must match — nothing forces the sides to be parallel.

  9. Exercise 9

    If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is (A) 1\displaystyle 1 : 3\displaystyle 3 (B) 1\displaystyle 1 : 2\displaystyle 2 (C) 3\displaystyle 3 : 1\displaystyle 1 (D) 1\displaystyle 1 : 4\displaystyle 4

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    NCERT’s answer
    (B)
    (B) $\displaystyle 1$ : $\displaystyle 2$NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-1_Q9\[ar(\triangle ABC)=\tfrac12\,ar(BCEF) \]Same base BC and the same parallels fix the same height; a triangle is always half of a parallelogram sharing that base and height.

  10. Exercise 10

    ABCD is a trapezium with parallel sides AB=a cm\displaystyle \mathrm{AB}=a \mathrm{~cm} and DC=b cm\displaystyle \mathrm{DC}=b \mathrm{~cm} (Fig. 9.6\displaystyle 9.6). E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD) is (A) a:b\displaystyle a: b (B) (3a+b):(a+3b)\displaystyle (3 a+b):(a+3 b) (C) (a+3b):(3a+b)\displaystyle (a+3 b):(3 a+b) (D) (2a+b):(3a+b)\displaystyle (2 a+b):(3 a+b) NCERT_Question_Class9_Maths_Exemplar_Ch9_Ex9-1_Q10

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    (B)
    (B) \(\displaystyle (3a+b):(a+3b)\)NCERT_Solution_Class9_Maths_Exemplar_Ch9_Ex9-1_Q10\[EF=\tfrac12(AB+DC)=\tfrac{a+b}{2} \] \[ar(ABFE)=\tfrac12\cdot\tfrac{h}{2}\Big(a+\tfrac{a+b}{2}\Big)=\tfrac{h}{8}(3a+b) \] \[ar(EFCD)=\tfrac12\cdot\tfrac{h}{2}\Big(\tfrac{a+b}{2}+b\Big)=\tfrac{h}{8}(a+3b) \]E, F are mid-points of the legs, so EF is the mid-segment and each half-trapezium takes half the total height h.