SolveItClass 10 · NCERT

NCERT Solutions · Class 10 Mathematics Triangles

29 questions · 29 still being checked

EXERCISE 6.3 11–16 (part 4 of 4)

  1. Exercise 11

    In Fig. 6.40\displaystyle 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB=AC\displaystyle \mathrm{AB}=\mathrm{AC}. If AD⊥BC\displaystyle \mathrm{AD} \perp \mathrm{BC} and EF⊥AC\displaystyle \mathrm{EF} \perp \mathrm{AC}, prove that △ABD∼△ECF\displaystyle \triangle \mathrm{ABD} \sim \triangle \mathrm{ECF}.NCERT_Question_Class10_Maths_Ch6_Ex6-3_Q11

    Solution being prepared

  2. Exercise 12

    Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR\displaystyle \Delta \mathrm{PQR} (see Fig. 6.41\displaystyle 6.41). Show that △ABC∼△PQR\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}.NCERT_Question_Class10_Maths_Ch6_Ex6-3_Q12

    Solution being prepared

  3. Exercise 13

    D is a point on the side BC of a triangle ABC such that ∠ADC=∠BAC\displaystyle \angle \mathrm{ADC}=\angle \mathrm{BAC}. Show that CA2=CB.CD\displaystyle \mathrm{CA}^{2}=\mathrm{CB} . \mathrm{CD}.

    Solution being prepared

  4. Exercise 14

    Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that △ABC∼△PQR\displaystyle \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}.
    NCERT’s answer
    Produce AD to a point E such that $\displaystyle \mathrm{AD}=\mathrm{DE}$ and produce PM to a point N such that $\displaystyle \mathrm{PM}=\mathrm{MN}$. Join EC and NR.

    Working being prepared

  5. Exercise 15

    A vertical pole of length 6\displaystyle 6 m casts a shadow 4\displaystyle 4 m long on the ground and at the same time a tower casts a shadow 28\displaystyle 28 m long. Find the height of the tower.
    NCERT’s answer
    42m

    Working being prepared

  6. Exercise 16

    If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC∼ΔPQR\displaystyle \Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}, prove that ABPQ=ADPM\displaystyle \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}.

    Solution being prepared