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NCERT Solutions · Class 10 Mathematics Triangles

29 questions · 29 still being checked

EXERCISE 6.3 1–10 (part 3 of 4)

  1. Exercise 1

    State which pairs of triangles in Fig. 6.34\displaystyle 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:NCERT_Question_Class10_Maths_Ch6_Ex6-3_Q1
    NCERT’s answer
    (i)
    Yes. AAA, $\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ (ii) Yes. SSS, $\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{QRP}$ (iii) No (iv) Yes. SAS, $\displaystyle \Delta \mathrm{MNL} \sim \Delta \mathrm{QPR}$ (v) No (vi) Yes. AA, $\displaystyle \triangle \mathrm{DEF} \sim \triangle \mathrm{PQR}$

    Working being prepared

  2. Exercise 2

    In Fig. 6.35\displaystyle 6.35, ΔODC∼ΔOBA,∠BOC=125∘\displaystyle \Delta \mathrm{ODC} \sim \Delta \mathrm{OBA}, \angle \mathrm{BOC}=125^{\circ} and ∠CDO=70∘\displaystyle \angle \mathrm{CDO}=70^{\circ}. Find ∠DOC,∠DCO\displaystyle \angle \mathrm{DOC}, \angle \mathrm{DCO} and ∠OAB\displaystyle \angle \mathrm{OAB}.NCERT_Question_Class10_Maths_Ch6_Ex6-3_Q2
    NCERT’s answer
    $\displaystyle 55^{\circ}, 55^{\circ}, 55^{\circ}$

    Working being prepared

  3. Exercise 3

    Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OAOC=OBOD\displaystyle \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}.

    Solution being prepared

  4. Exercise 4

    In Fig. 6.36\displaystyle 6.36, QRQS=QTPR\displaystyle \frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}} and ∠1=∠2\displaystyle \angle 1=\angle 2. Show that ΔPQS∼ΔTQR\displaystyle \Delta \mathrm{PQS} \sim \Delta \mathrm{TQR}.NCERT_Question_Class10_Maths_Ch6_Ex6-3_Q4

    Solution being prepared

  5. Exercise 5

    S and T are points on sides PR and QR of ΔPQR\displaystyle \Delta \mathrm{PQR} such that ∠P=∠RTS\displaystyle \angle \mathrm{P}=\angle \mathrm{RTS}. Show that ΔRPQ∼ΔRTS\displaystyle \Delta \mathrm{RPQ} \sim \Delta \mathrm{RTS}.

    Solution being prepared

  6. Exercise 6

    In Fig. 6.37\displaystyle 6.37, if ΔABE≅ΔACD\displaystyle \Delta \mathrm{ABE} \cong \Delta \mathrm{ACD}, show that △ADE∼△ABC\displaystyle \triangle \mathrm{ADE} \sim \triangle \mathrm{ABC}.NCERT_Question_Class10_Maths_Ch6_Ex6-3_Q6

    Solution being prepared

  7. Exercise 7

    In Fig. 6.38\displaystyle 6.38, altitudes AD and CE of △ABC\displaystyle \triangle \mathrm{ABC} intersect each other at the point P. Show that:
    (i)
    △AEP∼△CDP\displaystyle \triangle \mathrm{AEP} \sim \triangle \mathrm{CDP}
    (ii)
    △ABD∼△CBE\displaystyle \triangle \mathrm{ABD} \sim \triangle \mathrm{CBE}
    (iii)
    △AEP∼△ADB\displaystyle \triangle \mathrm{AEP} \sim \triangle \mathrm{ADB}
    (iv)
    △PDC∼△BEC\displaystyle \triangle \mathrm{PDC} \sim \triangle \mathrm{BEC}
    NCERT_Question_Class10_Maths_Ch6_Ex6-3_Q7

    Solution being prepared

  8. Exercise 8

    E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE∼ΔCFB\displaystyle \Delta \mathrm{ABE} \sim \Delta \mathrm{CFB}.

    Solution being prepared

  9. Exercise 9

    In Fig. 6.39\displaystyle 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
    (i)
    ΔABC∼ΔAMP\displaystyle \Delta \mathrm{ABC} \sim \Delta \mathrm{AMP}
    (ii)
    CAPA=BCMP\displaystyle \frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}
    NCERT_Question_Class10_Maths_Ch6_Ex6-3_Q9

    Solution being prepared

  10. Exercise 10

    CD and GH are respectively the bisectors of ∠ACB and ∠ EGF such that D and H lie on sides AB and FE of ΔABC\displaystyle \Delta \mathrm{ABC} and ΔEFG\displaystyle \Delta \mathrm{EFG} respectively. If ΔABC∼ΔFEG\displaystyle \Delta \mathrm{ABC} \sim \Delta \mathrm{FEG}, show that:
    (i)
    CDGH=ACFG\displaystyle \frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}
    (ii)
    △DCB∼△HGE\displaystyle \triangle \mathrm{DCB} \sim \triangle \mathrm{HGE}
    (iii)
    △DCA∼△HGF\displaystyle \triangle \mathrm{DCA} \sim \triangle \mathrm{HGF}

    Solution being prepared