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NCERT Exemplar · Class 10 Mathematics Triangles

57 questions · 57 still being checked

EXERCISE 6.3 11–15 (part 4 of 6)

  1. Exercise 11

    In a triangle PQR,N\displaystyle \mathrm{PQR}, \mathrm{N} is a point on PR such that QN⊥PR\displaystyle \mathrm{QN} \perp \mathrm{PR}. If PN.NR=QN2\displaystyle \mathrm{PN} . \mathrm{NR}=\mathrm{QN}^2, prove that ∠PQR=90∘\displaystyle \angle \mathrm{PQR}=90^{\circ}.

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    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-3_Q11 \[\angle PNQ = \angle QNR = 90^\circ \quad \text{(} QN \perp PR\text{, given)} \] \[PN \cdot NR = QN^2 \Rightarrow \frac{PN}{QN} = \frac{QN}{NR} \quad \text{(given)} \] \[\triangle PNQ \sim \triangle QNR \quad \text{(SAS similarity)} \] \[\angle NPQ = \angle NQR, \ \ \angle PQN = \angle QRN \quad \text{(corr. angles)} \] \[\angle PQR = \angle PQN + \angle NQR = \angle R + \angle P \] \[\angle P + \angle Q + \angle R = 180^\circ \quad \text{(angle sum, } \triangle PQR\text{)} \] \[\angle Q = \angle P + \angle R = 180^\circ - \angle Q \Rightarrow 2\angle Q = 180^\circ \] \[\angle Q = 90^\circ \]Answer: \(\displaystyle \angle PQR = 90^\circ\).

  2. Exercise 12

    Areas of two similar triangles are 36 cm2\displaystyle 36 \mathrm{~cm}^2 and 100 cm2\displaystyle 100 \mathrm{~cm}^2. If the length of a side of the larger triangle is 20\displaystyle 20 cm, find the length of the corresponding side of the smaller triangle.

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    NCERT’s answer
    $\displaystyle 12$ cm
    \[\frac{\text{ar}(\triangle_1)}{\text{ar}(\triangle_2)} = \frac{36}{100} \quad \text{(given)} \] \[\frac{\text{ar}(\triangle_1)}{\text{ar}(\triangle_2)} = \left(\frac{s_1}{s_2}\right)^2 \quad \text{(areas of similar triangles)} \] \[\left(\frac{s_1}{20}\right)^2 = \frac{36}{100} \Rightarrow \frac{s_1}{20} = \frac{6}{10} \] \[s_1 = 20 \times \frac{6}{10} = 12 \text{ cm} \]Answer: \(\displaystyle 12\) cm.

  3. Exercise 13

    In Fig. 6.12\displaystyle 6.12, if ∠ACB=∠CDA,AC=8 cm\displaystyle \angle \mathrm{ACB}=\angle \mathrm{CDA}, \mathrm{AC}=8 \mathrm{~cm} and AD=3 cm\displaystyle \mathrm{AD}=3 \mathrm{~cm}, find BD. NCERT_Question_Class10_Maths_Exemplar_Ch6_Ex6-3_Q13

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    NCERT’s answer
    \(\displaystyle \frac{55}{3} \mathrm{~cm}\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-3_Q13 \[\angle DAC = \angle CAB \quad \text{(common angle)} \] \[\angle ADC = \angle ACB \quad \text{(given)} \] \[\triangle ADC \sim \triangle ACB \quad \text{(AA similarity)} \] \[\frac{AD}{AC} = \frac{AC}{AB} \Rightarrow AC^2 = AD \cdot AB \] \[64 = 3(3 + BD) \Rightarrow 3 + BD = \frac{64}{3} \] \[BD = \frac{64}{3} - 3 = \frac{55}{3} \text{ cm} \]Answer: \(\displaystyle BD = \dfrac{55}{3}\) cm.

  4. Exercise 14

    A 15\displaystyle 15 metres high tower casts a shadow 24\displaystyle 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16\displaystyle 16 metres long. Find the height of the telephone pole.

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    NCERT’s answer
    $\displaystyle 10$ m
    \[\frac{\text{height of tower}}{\text{its shadow}} = \frac{\text{height of pole}}{\text{its shadow}} \quad \text{(equal angle of elevation, same instant)} \] \[\frac{15}{24} = \frac{h}{16} \] \[h = \frac{15 \times 16}{24} = 10 \text{ m} \]Answer: \(\displaystyle 10\) m.

  5. Exercise 15

    Foot of a 10\displaystyle 10 m long ladder leaning against a vertical wall is 6\displaystyle 6 m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 8$ m
    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-3_Q15 \[h^2 + 6^2 = 10^2 \quad \text{(Pythagoras theorem)} \] \[h^2 = 100 - 36 = 64 \] \[h = 8 \text{ m} \]Answer: \(\displaystyle 8\) m.