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NCERT Exemplar · Class 9 Mathematics Introduction to Euclid's Geometry

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EXERCISE 5.3 1–12 (part 4 of 5)

  1. Solve each of the following question using appropriate Euclid's axiom :

    Exercise 1

    Two salesmen make equal sales during the month of August. In September, each salesman doubles his sale of the month of August. Compare their sales in September.

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    Let the August sales of the two salesmen be \(\displaystyle x\) and \(\displaystyle y\). \[x=y \quad \text{(equal sales, given)} \] September sales are \(\displaystyle 2x\) and \(\displaystyle 2y\). \[2x=2y \quad \text{(doubles of equals are equal --- Euclid's Axiom 6)} \] Answer: their sales in September are equal.

  2. Exercise 2

    It is known that x+y=10\displaystyle x+y=10 and that x=z\displaystyle x=z. Show that z+y=10\displaystyle z+y=10 ?

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    \[x=z \quad \text{(given)} \] \[x+y=z+y \quad \text{(equals added to equals --- Euclid's Axiom 2)} \] \[x+y=10 \quad \text{(given)} \] \[\therefore\ z+y=10 \quad \text{(equal to the same thing --- Euclid's Axiom 1)} \] Answer: \(\displaystyle z+y=10\).

  3. Exercise 3

    Look at the Fig. 5.3. Show that length AH>\displaystyle \mathrm{AH}> sum of lengths of AB+BC+CD\displaystyle \mathrm{AB}+\mathrm{BC}+\mathrm{CD}. NCERT_Question_Class9_Maths_Exemplar_Ch5_Ex5-3_Q3

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    NCERT_Solution_Class9_Maths_Exemplar_Ch5_Ex5-3_Q3 \[AH = AB+BC+CD+DE+EF+FG+GH \quad \text{(the whole is the sum of its parts)} \] \[DE+EF+FG+GH > 0 \] \[\Rightarrow\ AH > AB+BC+CD \quad \text{(the whole is greater than the part --- Euclid's Axiom 5)} \] Answer: \(\displaystyle AH>AB+BC+CD\).

  4. Exercise 4

    In the Fig.5.4, we have AB=BC,BX=BY\displaystyle \mathrm{AB}=\mathrm{BC}, \mathrm{BX}=\mathrm{BY}. Show that AX=CY\displaystyle \mathrm{AX}=\mathrm{CY}. NCERT_Question_Class9_Maths_Exemplar_Ch5_Ex5-3_Q4

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    NCERT_Solution_Class9_Maths_Exemplar_Ch5_Ex5-3_Q4 \[AB = AX+XB, \quad BC = BY+YC \quad \text{(whole = sum of parts)} \] \[AB=BC, \quad BX=BY \quad \text{(given)} \] \[\Rightarrow\ AX = YC \quad \text{(equals subtracted from equals --- Euclid's Axiom 3)} \] Answer: \(\displaystyle AX=CY\).

  5. Exercise 5

    In the Fig.5.5, we have X and Y are the mid-points of AC and BC and AX=CY\displaystyle \mathrm{AX}=\mathrm{CY}. Show that AC=BC\displaystyle \mathrm{AC}=\mathrm{BC}. NCERT_Question_Class9_Maths_Exemplar_Ch5_Ex5-3_Q5

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    NCERT_Solution_Class9_Maths_Exemplar_Ch5_Ex5-3_Q5 \[AX=XC, \quad CY=YB \quad \text{(X, Y bisect AC, BC)} \] \[AX=CY \quad \text{(given)} \ \Rightarrow\ XC=YB \] \[AC=AX+XC=2AX, \quad BC=CY+YB=2CY \quad \text{(whole = sum of parts)} \] \[2AX=2CY \quad \text{(doubles of equals are equal --- Euclid's Axiom 6)} \] Answer: \(\displaystyle AC=BC\).

  6. Exercise 6

    In the Fig.5.6, we have BX=12ABBY=12BC and AB=BC. Show that BX=BY.\begin{aligned} & \mathrm{BX}=\frac{1}{2} \mathrm{AB} \\ & \mathrm{BY}=\frac{1}{2} \mathrm{BC} \text { and } \mathrm{AB}=\mathrm{BC} . \text { Show that } \\ & \mathrm{BX}=\mathrm{BY} . \end{aligned} NCERT_Question_Class9_Maths_Exemplar_Ch5_Ex5-3_Q6

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    NCERT_Solution_Class9_Maths_Exemplar_Ch5_Ex5-3_Q6 \[AB=BC \quad \text{(given)} \] \[BX=\tfrac{1}{2}AB, \quad BY=\tfrac{1}{2}BC \quad \text{(given)} \] \[\Rightarrow\ BX=BY \quad \text{(halves of equals are equal --- Euclid's Axiom 7)} \] Answer: \(\displaystyle BX=BY\).

  7. Exercise 7

    In the Fig.5.7, we have ∠1=∠2,∠2=∠3\displaystyle \angle 1=\angle 2, \angle 2=\angle 3. Show that ∠1=∠3\displaystyle \angle 1=\angle 3. NCERT_Question_Class9_Maths_Exemplar_Ch5_Ex5-3_Q7

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    NCERT_Solution_Class9_Maths_Exemplar_Ch5_Ex5-3_Q7 \[\angle 1=\angle 2, \quad \angle 2=\angle 3 \quad \text{(given)} \] \[\Rightarrow\ \angle 1=\angle 3 \quad \text{(things equal to the same thing are equal to one another --- Euclid's Axiom 1)} \] Answer: \(\displaystyle \angle 1=\angle 3\).

  8. Exercise 8

    In the Fig. 5.8\displaystyle 5.8, we have ∠1=∠3\displaystyle \angle 1=\angle 3 and ∠2=∠4\displaystyle \angle 2=\angle 4. Show that ∠A=∠C\displaystyle \angle \mathrm{A}=\angle \mathrm{C}. NCERT_Question_Class9_Maths_Exemplar_Ch5_Ex5-3_Q8

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    NCERT_Solution_Class9_Maths_Exemplar_Ch5_Ex5-3_Q8 \[\angle A=\angle 1+\angle 2, \quad \angle C=\angle 3+\angle 4 \quad \text{(whole = sum of parts)} \] \[\angle 1=\angle 3, \quad \angle 2=\angle 4 \quad \text{(given)} \] \[\Rightarrow\ \angle 1+\angle 2=\angle 3+\angle 4 \quad \text{(equals added to equals --- Euclid's Axiom 2)} \] Answer: \(\displaystyle \angle A=\angle C\).

  9. Exercise 9

    In the Fig. 5.9\displaystyle 5.9, we have ∠ABC=∠ACB,∠3=∠4\displaystyle \angle \mathrm{ABC}=\angle \mathrm{ACB}, \angle 3=\angle 4. Show that ∠1=∠2\displaystyle \angle 1=\angle 2. NCERT_Question_Class9_Maths_Exemplar_Ch5_Ex5-3_Q9

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    NCERT_Solution_Class9_Maths_Exemplar_Ch5_Ex5-3_Q9 \[\angle 4=\angle ABD,\ \angle 1=\angle DBC,\ \angle 3=\angle ACD,\ \angle 2=\angle DCB \quad \text{(as marked)} \] \[\angle ABC = \angle 4+\angle 1 \quad \text{(angle addition at }B\text{)} \] \[\angle ACB = \angle 3+\angle 2 \quad \text{(angle addition at }C\text{)} \] \[\angle ABC=\angle ACB \quad \text{(given)} \] \[\angle 4+\angle 1=\angle 3+\angle 2 \] \[\angle 3=\angle 4 \quad \text{(given)} \] \[\angle 1=\angle 2 \quad \text{(Euclid's Axiom 3: equals subtracted from equals)} \]Answer: \(\displaystyle \angle 1=\angle 2\).

  10. Exercise 10

    In the Fig. 5.10\displaystyle 5.10, we have AC=DC,CB=CE\displaystyle \mathrm{AC}=\mathrm{DC}, \mathrm{CB}=\mathrm{CE}. Show that AB=DE\displaystyle \mathrm{AB}=\mathrm{DE}. NCERT_Question_Class9_Maths_Exemplar_Ch5_Ex5-3_Q10

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    NCERT_Solution_Class9_Maths_Exemplar_Ch5_Ex5-3_Q10 \[AB = AC + CB \quad \text{(}C\text{ lies on }AB\text{)} \] \[DE = DC + CE \quad \text{(}C\text{ lies on }DE\text{)} \] \[AC = DC, \quad CB = CE \quad \text{(given)} \] \[AB = DE \quad \text{(Euclid's Axiom 2: equals added to equals)} \]Answer: \(\displaystyle AB = DE\).

  11. Exercise 11

    In the Fig. 5.11\displaystyle 5.11, if OX=12XY,PX=12XZ\displaystyle \mathrm{OX}=\frac{1}{2} \mathrm{XY}, \mathrm{PX}=\frac{1}{2} \mathrm{XZ} and OX=PX\displaystyle \mathrm{OX}=\mathrm{PX}, show that XY=XZ\displaystyle \mathrm{XY}=\mathrm{XZ}. NCERT_Question_Class9_Maths_Exemplar_Ch5_Ex5-3_Q11

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    NCERT_Solution_Class9_Maths_Exemplar_Ch5_Ex5-3_Q11 \[XY = 2\,OX \quad \text{(}OX = \tfrac{1}{2}XY\text{, given)} \] \[XZ = 2\,PX \quad \text{(}PX = \tfrac{1}{2}XZ\text{, given)} \] \[OX = PX \quad \text{(given)} \] \[2\,OX = 2\,PX \] \[XY = XZ \quad \text{(Euclid's Axiom 6: doubles of equal things are equal)} \]Answer: \(\displaystyle XY = XZ\).

  12. Exercise 12

    In the Fig. 5.12\displaystyle 5.12 : (i) AB=BC,M\displaystyle \mathrm{AB}=\mathrm{BC}, \mathrm{M} is the mid-point of AB and N is the mid- point of BC. Show that AM=NC\displaystyle \mathrm{AM}=\mathrm{NC}. (ii) BM=BN,M\displaystyle \mathrm{BM}=\mathrm{BN}, \mathrm{M} is the mid-point of AB and N is the mid-point of BC. Show that AB=BC\displaystyle \mathrm{AB}=\mathrm{BC}. NCERT_Question_Class9_Maths_Exemplar_Ch5_Ex5-3_Q12

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    NCERT_Solution_Class9_Maths_Exemplar_Ch5_Ex5-3_Q12
    (i)
    \[AM = \tfrac{1}{2}AB \quad \text{(}M\text{ is the mid-point of }AB\text{)} \]
    \[NC = \tfrac{1}{2}BC \quad \text{(}N\text{ is the mid-point of }BC\text{)} \]
    \[AB = BC \quad \text{(given)} \]
    \[AM = NC \quad \text{(Euclid's Axiom 7: halves of equal things are equal)} \]
    (ii)
    \[BM = \tfrac{1}{2}AB \quad \text{(}M\text{ is the mid-point of }AB\text{)} \]
    \[BN = \tfrac{1}{2}BC \quad \text{(}N\text{ is the mid-point of }BC\text{)} \]
    \[BM = BN \quad \text{(given)} \]
    \[\tfrac{1}{2}AB = \tfrac{1}{2}BC \]
    \[AB = BC \quad \text{(Euclid's Axiom 6: doubles of equal things are equal)} \]
    Answer: (i) \(\displaystyle AM = NC\) (ii) \(\displaystyle AB = BC\).