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NCERT Exemplar · Class 10 Mathematics Triangles

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EXERCISE 6.3 1–10 (part 3 of 6)

  1. Exercise 1

    In a ΔPQR,PR2−PQ2=QR2\displaystyle \Delta \mathrm{PQR}, \mathrm{PR}^2-\mathrm{PQ}^2=\mathrm{QR}^2 and M is a point on side PR such that QM⊥PR\displaystyle \mathrm{QM} \perp \mathrm{PR}. Prove that QM2=PM×MR.\mathrm{QM}^2=\mathrm{PM} \times \mathrm{MR} .

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    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-3_Q1 \[PR^2 = PQ^2+QR^2 \quad \text{(given, rearranged)} \] \[\angle PQR = 90^\circ \quad \text{(converse of Pythagoras theorem)} \] In \(\displaystyle \Delta PMQ\) and \(\displaystyle \Delta QMR\): \[\angle PMQ = \angle QMR = 90^\circ \quad (QM \perp PR) \] \[\angle MPQ = 90^\circ-\angle PQM \quad \text{(angle sum in } \Delta PMQ\text{)} \] \[\angle MPQ = \angle MQR \quad \text{(} \angle PQM+\angle MQR = \angle PQR = 90^\circ\text{)} \] \[\Delta PMQ \sim \Delta QMR \quad \text{(AA)} \] \[\frac{PM}{QM} = \frac{QM}{MR} \] \[QM^2 = PM \times MR \] Answer: \(\displaystyle QM^2 = PM \times MR \)

  2. Exercise 2

    Find the value of x\displaystyle x for which DE∥AB\displaystyle \mathrm{DE} \| \mathrm{AB} in Fig. 6.8. NCERT_Question_Class10_Maths_Exemplar_Ch6_Ex6-3_Q2

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    NCERT’s answer
    \(\displaystyle x=2\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-3_Q2 \[\frac{CD}{DA} = \frac{CE}{EB} \quad \text{(converse of the basic proportionality theorem, } DE \parallel AB\text{)} \] \[\frac{x+3}{3x+19} = \frac{x}{3x+4} \] \[(x+3)(3x+4) = x(3x+19) \] \[3x^2+13x+12 = 3x^2+19x \] \[6x = 12 \] \[x = 2 \] Answer: \(\displaystyle x = 2 \)

  3. Exercise 3

    In Fig. 6.9\displaystyle 6.9, if ∠1=∠2\displaystyle \angle 1=\angle 2 and ΔNSQ≅ΔMTR\displaystyle \Delta \mathrm{NSQ} \cong \Delta \mathrm{MTR}, then prove that ΔPTS∼ΔPRQ\displaystyle \Delta \mathrm{PTS} \sim \Delta \mathrm{PRQ}. NCERT_Question_Class10_Maths_Exemplar_Ch6_Ex6-3_Q3

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    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-3_Q3 \[\angle PST = \angle 1 = \angle 2 = \angle PTS \quad \text{(given)} \] \[PS = PT \quad \text{(sides opposite equal angles of } \Delta PST\text{)} \] \[SQ = TR \quad \text{(CPCT, } \Delta NSQ \cong \Delta MTR\text{)} \] \[PQ = PS+SQ = PT+TR = PR \] In \(\displaystyle \Delta PTS\) and \(\displaystyle \Delta PRQ\): \[\angle TPS = \angle RPQ \quad \text{(common, since } S \in PQ,\ T \in PR\text{)} \] \[\frac{PT}{PR} = \frac{PS}{PQ} \quad (PT=PS,\ PR=PQ) \] \[\Delta PTS \sim \Delta PRQ \quad \text{(SAS)} \] Answer: \(\displaystyle \Delta PTS \sim \Delta PRQ \)

  4. Exercise 4

    Diagonals of a trapezium PQRS intersect each other at the point O, PQ ∥RS\displaystyle \| \mathrm{RS} and PQ=3RS\displaystyle \mathrm{PQ}=3 \mathrm{RS}. Find the ratio of the areas of triangles POQ and ROS .

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    NCERT’s answer
    $\displaystyle 9$:$\displaystyle 1$
    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-3_Q4 In \(\displaystyle \Delta POQ\) and \(\displaystyle \Delta ROS\): \[\angle POQ = \angle ROS \quad \text{(vertically opposite angles)} \] \[\angle OPQ = \angle ORS \quad \text{(alternate angles, } PQ \parallel RS\text{)} \] \[\Delta POQ \sim \Delta ROS \quad \text{(AA)} \] \[\frac{ar(\Delta POQ)}{ar(\Delta ROS)} = \left(\frac{PQ}{RS}\right)^2 = \left(\frac{3RS}{RS}\right)^2 = 9 \] Answer: \(\displaystyle ar(\Delta POQ) : ar(\Delta ROS) = 9:1 \)

  5. Exercise 5

    In Fig. 6.10\displaystyle 6.10, if AB∥DC\displaystyle \mathrm{AB} \| \mathrm{DC} and AC and PQ intersect each other at the point O, prove that OA.CQ=OC.AP\displaystyle \mathrm{OA} . \mathrm{CQ}=\mathrm{OC} . \mathrm{AP}. NCERT_Question_Class10_Maths_Exemplar_Ch6_Ex6-3_Q5

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-3_Q5 In \(\displaystyle \Delta OAP\) and \(\displaystyle \Delta OCQ\): \[\angle AOP = \angle COQ \quad \text{(vertically opposite angles)} \] \[\angle OAP = \angle OCQ \quad \text{(alternate angles, } AB \parallel DC,\ AC \text{ transversal)} \] \[\Delta OAP \sim \Delta OCQ \quad \text{(AA)} \] \[\frac{OA}{OC} = \frac{AP}{CQ} \] \[OA \times CQ = OC \times AP \] Answer: \(\displaystyle OA \times CQ = OC \times AP \)

  6. Exercise 6

    Find the altitude of an equilateral triangle of side 8\displaystyle 8 cm.

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    NCERT’s answer
    \(\displaystyle 4 \sqrt{3} \mathrm{~cm}\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-3_Q6 \[BD = DC = \frac{8}{2} = 4 \text{ cm} \quad (AD \perp BC \text{ bisects } BC \text{ in an equilateral } \Delta) \] \[AD^2 = AB^2-BD^2 = 8^2-4^2 = 48 \] \[AD = 4\sqrt3 \text{ cm} \] Answer: \(\displaystyle 4\sqrt3 \) cm

  7. Exercise 7

    If ΔABC∼ΔDEF,AB=4 cm,DE=6 cm,EF=9 cm\displaystyle \Delta \mathrm{ABC} \sim \Delta \mathrm{DEF}, \mathrm{AB}=4 \mathrm{~cm}, \mathrm{DE}=6 \mathrm{~cm}, \mathrm{EF}=9 \mathrm{~cm} and FD=12 cm\displaystyle \mathrm{FD}=12 \mathrm{~cm}, find the perimeter of ΔABC\displaystyle \Delta \mathrm{ABC}.

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    NCERT’s answer
    $\displaystyle 18$ cm
    \[\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} \quad (\Delta ABC \sim \Delta DEF) \] \[\frac{4}{6} = \frac{BC}{9} = \frac{CA}{12} \] \[BC = 6 \text{ cm}, \quad CA = 8 \text{ cm} \] \[\text{Perimeter} = AB+BC+CA = 4+6+8 = 18 \text{ cm} \] Answer: \(\displaystyle 18 \) cm

  8. Exercise 8

    In Fig. 6.11\displaystyle 6.11, if DE∥BC\displaystyle \mathrm{DE} \| \mathrm{BC}, find the ratio of ar (ADE) and ar (DECB). NCERT_Question_Class10_Maths_Exemplar_Ch6_Ex6-3_Q8

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    NCERT’s answer
    $\displaystyle 1$:$\displaystyle 3$
    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-3_Q8 \[\Delta ADE \sim \Delta ABC \quad (DE \parallel BC,\ \angle A \text{ common}) \] \[\frac{ar(\Delta ADE)}{ar(\Delta ABC)} = \left(\frac{DE}{BC}\right)^2 = \left(\frac{6}{12}\right)^2 = \frac14 \] \[ar(DECB) = ar(\Delta ABC)-ar(\Delta ADE) = 4k-k = 3k \quad (ar(\Delta ADE)=k) \] Answer: \(\displaystyle ar(\Delta ADE) : ar(DECB) = 1:3 \)

  9. Exercise 9

    ABCD is a trapezium in which AB∥DC\displaystyle \mathrm{AB} \| \mathrm{DC} and P and Q are points on AD and BC, respectively such that PQ∥DC\displaystyle \mathrm{PQ} \| \mathrm{DC}. If PD=18 cm,BQ=35 cm\displaystyle \mathrm{PD}=18 \mathrm{~cm}, \mathrm{BQ}=35 \mathrm{~cm} and QC=15 cm\displaystyle \mathrm{QC}=15 \mathrm{~cm}, find AD.

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    NCERT’s answer
    $\displaystyle 60$ cm
    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-3_Q9 Join AC meeting PQ at R. \[\frac{AP}{PD} = \frac{AR}{RC} \quad \text{(BPT in } \triangle ADC,\ PR \parallel DC\text{)} \] \[\frac{AR}{RC} = \frac{BQ}{QC} \quad \text{(BPT in } \triangle CAB,\ RQ \parallel AB\text{)} \] \[\frac{AP}{PD} = \frac{BQ}{QC} = \frac{35}{15} = \frac{7}{3} \] \[AP = \frac{7}{3} \times 18 = 42 \text{ cm} \] \[AD = AP + PD = 42 + 18 = 60 \text{ cm} \]Answer: \(\displaystyle AD = 60\) cm.

  10. Exercise 10

    Corresponding sides of two similar triangles are in the ratio of 2\displaystyle 2 : 3. If the area of the smaller triangle is 48 cm2\displaystyle 48 \mathrm{~cm}^2, find the area of the larger triangle.

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    NCERT’s answer
    \(\displaystyle 108 \mathrm{~cm}^2\)
    \[\frac{\text{side}_1}{\text{side}_2} = \frac{2}{3} \quad \text{(given)} \] \[\frac{\text{ar}(\triangle_1)}{\text{ar}(\triangle_2)} = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \quad \text{(areas of similar triangles)} \] \[\frac{48}{\text{ar}(\triangle_2)} = \frac{4}{9} \] \[\text{ar}(\triangle_2) = 48 \times \frac{9}{4} = 108 \text{ cm}^2 \]Answer: \(\displaystyle 108\) cm\(\displaystyle ^2\).