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NCERT Solutions · Class 10 Mathematics Triangles

29 questions · 29 still being checked

EXERCISE 6.2 1–10 (part 2 of 4)

  1. Exercise 1

    In Fig. 6.17\displaystyle 6.17, (i) and (ii), DE∥BC\displaystyle \mathrm{DE} \| \mathrm{BC}. Find EC in (i) and AD in (ii).NCERT_Question_Class10_Maths_Ch6_Ex6-2_Q1
    NCERT’s answer
    (i)
    $\displaystyle 2$ cm (ii) $\displaystyle 2.4$ cm

    Working being prepared

  2. Exercise 2

    E and F are points on the sides PQ and PR respectively of a △PQR\displaystyle \triangle \mathrm{PQR}. For each of the following cases, state whether EF∥QR\displaystyle \mathrm{EF} \| \mathrm{QR} :
    (i)
    PE=3.9 cm,EQ=3 cm,PF=3.6 cm\displaystyle \mathrm{PE}=3.9 \mathrm{~cm}, \mathrm{EQ}=3 \mathrm{~cm}, \mathrm{PF}=3.6 \mathrm{~cm} and FR=2.4 cm\displaystyle \mathrm{FR}=2.4 \mathrm{~cm}
    (ii)
    PE=4 cm,QE=4.5 cm,PF=8 cm\displaystyle \mathrm{PE}=4 \mathrm{~cm}, \mathrm{QE}=4.5 \mathrm{~cm}, \mathrm{PF}=8 \mathrm{~cm} and RF=9 cm\displaystyle \mathrm{RF}=9 \mathrm{~cm}
    (iii)
    PQ=1.28 cm,PR=2.56 cm,PE=0.18 cm\displaystyle \mathrm{PQ}=1.28 \mathrm{~cm}, \mathrm{PR}=2.56 \mathrm{~cm}, \mathrm{PE}=0.18 \mathrm{~cm} and PF=0.36 cm\displaystyle \mathrm{PF}=0.36 \mathrm{~cm}
    NCERT’s answer
    (i)
    No (ii) Yes (iiii) Yes

    Working being prepared

  3. Exercise 3

    In Fig. 6.18\displaystyle 6.18, if LM∥CB\displaystyle \mathrm{LM} \| \mathrm{CB} and LN∥CD\displaystyle \mathrm{LN} \| \mathrm{CD}, prove that AMAB=ANAD.\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}} . NCERT_Question_Class10_Maths_Ch6_Ex6-2_Q3

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  4. Exercise 4

    In Fig. 6.19\displaystyle 6.19, DE∥AC\displaystyle \mathrm{DE} \| \mathrm{AC} and DF∥AE\displaystyle \mathrm{DF} \| \mathrm{AE}. Prove that BFFE=BEEC.\frac{B F}{F E}=\frac{B E}{E C} . Fig. 6.19\displaystyle 6.19NCERT_Question_Class10_Maths_Ch6_Ex6-2_Q4

    Solution being prepared

  5. Exercise 5

    In Fig. 6.20\displaystyle 6.20, DE∥OQ\displaystyle \mathrm{DE} \| \mathrm{OQ} and DF∥OR\displaystyle \mathrm{DF} \| \mathrm{OR}. Show that EF∥QR\displaystyle \mathrm{EF} \| \mathrm{QR}.NCERT_Question_Class10_Maths_Ch6_Ex6-2_Q5

    Solution being prepared

  6. Exercise 6

    In Fig. 6.21\displaystyle 6.21, A,B\displaystyle \mathrm{A}, \mathrm{B} and C are points on OP,OQ\displaystyle \mathrm{OP}, \mathrm{OQ} and OR respectively such that AB∥PQ\displaystyle \mathrm{AB} \| \mathrm{PQ} and AC ∥PR\displaystyle \| \mathrm{PR}. Show that BC∥QR\displaystyle \mathrm{BC} \| \mathrm{QR}.NCERT_Question_Class10_Maths_Ch6_Ex6-2_Q6

    Solution being prepared

  7. Exercise 7

    Using Theorem 6.1\displaystyle 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

    Solution being prepared

  8. Exercise 8

    Using Theorem 6.2\displaystyle 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

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  9. Exercise 9

    ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O . Show that AOBO=CODO\displaystyle \frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}.
    NCERT’s answer
    Through O , draw a line parallel to DC , intersecting AD and BC at E and F respectively.

    Working being prepared

  10. Exercise 10

    The diagonals of a quadrilateral ABCD intersect each other at the point O such that AOBO=CODO\displaystyle \frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}. Show that ABCD is a trapezium.

    Solution being prepared