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NCERT Exemplar · Class 10 Mathematics Triangles

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EXERCISE 6.4 1–10 (part 5 of 6)

  1. Exercise 1

    In Fig. 6.16\displaystyle 6.16, if ∠A=∠C,AB=6 cm,BP=15 cm\displaystyle \angle \mathrm{A}=\angle \mathrm{C}, \mathrm{AB}=6 \mathrm{~cm}, \mathrm{BP}=15 \mathrm{~cm}, AP=12 cm\displaystyle \mathrm{AP}=12 \mathrm{~cm} and CP=4 cm\displaystyle \mathrm{CP}=4 \mathrm{~cm}, then find the lengths of PD and CD. NCERT_Question_Class10_Maths_Exemplar_Ch6_Ex6-4_Q1

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    NCERT’s answer
    $\displaystyle 5$ cm, $\displaystyle 2$ cm
    \[\angle APB = \angle CPD \quad \text{(vertically opposite angles)} \] \[\angle A = \angle C \quad \text{(given)} \] \[\triangle APB \sim \triangle CPD \quad \text{(AA similarity)} \] NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-4_Q1 \[\frac{AP}{CP} = \frac{PB}{PD} = \frac{AB}{CD} \] \[\frac{12}{4} = \frac{15}{PD} = \frac{6}{CD} \] \[PD = \frac{15 \times 4}{12} = 5 \text{ cm} \] \[CD = \frac{6 \times 4}{12} = 2 \text{ cm} \] Answer: \(\displaystyle PD = 5 \) cm, \(\displaystyle CD = 2 \) cm.

  2. Exercise 2

    It is given that ΔABC∼ΔEDF\displaystyle \Delta \mathrm{ABC} \sim \Delta \mathrm{EDF} such that AB=5 cm\displaystyle \mathrm{AB}=5 \mathrm{~cm}, AC=7 cm,DF=15 cm\displaystyle \mathrm{AC}=7 \mathrm{~cm}, \mathrm{DF}=15 \mathrm{~cm} and DE=12 cm\displaystyle \mathrm{DE}=12 \mathrm{~cm}. Find the lengths of the remaining sides of the triangles.

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    NCERT’s answer
    BC = $\displaystyle 6.25$ cm, EF = $\displaystyle 16.8$ cm.
    \[\triangle ABC \sim \triangle EDF \] \[\frac{AB}{ED} = \frac{BC}{DF} = \frac{AC}{EF} \] \[\frac{5}{12} = \frac{BC}{15} = \frac{7}{EF} \] \[BC = \frac{5 \times 15}{12} = 6.25 \text{ cm} \] \[EF = \frac{7 \times 12}{5} = 16.8 \text{ cm} \] Answer: \(\displaystyle BC = 6.25 \) cm, \(\displaystyle EF = 16.8 \) cm.

  3. Exercise 3

    Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.

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    \[DE \parallel BC, \quad D \text{ on } AB, \quad E \text{ on } AC \quad \text{(given)} \] Construction: \(\displaystyle BE\), \(\displaystyle CD\) joined. NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-4_Q3 \[\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DBE)} = \frac{AD}{DB} \quad \text{(equal height from } E\text{, bases on } AB\text{)} \] \[\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DEC)} = \frac{AE}{EC} \quad \text{(equal height from } D\text{, bases on } AC\text{)} \] \[\text{ar}(\triangle DBE) = \text{ar}(\triangle DEC) \quad \text{(same base } DE\text{, } DE \parallel BC\text{)} \] \[\frac{AD}{DB} = \frac{AE}{EC} \] Answer: \(\displaystyle \dfrac{AD}{DB} = \dfrac{AE}{EC} \)

  4. Exercise 4

    In Fig 6.17\displaystyle 6.17, if PQRS is a parallelogram and AB∥PS\displaystyle \mathrm{AB} \| \mathrm{PS}, then prove that OC∥SR\displaystyle \mathrm{OC} \| \mathrm{SR}. NCERT_Question_Class10_Maths_Exemplar_Ch6_Ex6-4_Q4

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    \[PS \parallel AB \implies \triangle OPS \sim \triangle OAB \quad \text{(AA)} \] \[\frac{OP}{OA} = \frac{OS}{OB} = \frac{PS}{AB} = k \] \[QR \parallel PS \parallel AB \quad \text{(opp. sides of the } \|\text{gm)} \] \[\triangle CQR \sim \triangle CAB \quad \text{(AA)} \implies \frac{CQ}{CA} = \frac{CR}{CB} = \frac{QR}{AB} = m \] \[QR = PS \quad \text{(} \|\text{gm sides)} \implies m = k \] NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-4_Q4 \[\frac{AQ}{QC} = \frac{CA-CQ}{CQ} = \frac{1-k}{k} = \frac{OA-OP}{OP} = \frac{AP}{PO} \] \[\text{In } \triangle AOC: \frac{AP}{PO} = \frac{AQ}{QC} \implies PQ \parallel OC \quad \text{(converse BPT)} \] \[PQ \parallel SR \quad \text{(opp. sides of the } \|\text{gm)} \implies OC \parallel SR \] Answer: \(\displaystyle OC \parallel SR \)

  5. Exercise 5

    A 5\displaystyle 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4\displaystyle 4 m high. If the foot of the ladder is moved 1.6\displaystyle 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.

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    NCERT’s answer
    0.$\displaystyle 8$ m
    \[\text{Ground distance}_1 = \sqrt{5^2-4^2} = \sqrt{9} = 3 \text{ m} \] \[\text{New ground distance} = 3 - 1.6 = 1.4 \text{ m} \] \[\text{New height} = \sqrt{5^2-1.4^2} = \sqrt{23.04} = 4.8 \text{ m} \] NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-4_Q5 \[\text{Rise} = 4.8 - 4 = 0.8 \text{ m} \] Answer: The top slides up by \(\displaystyle 0.8 \) m.

  6. Exercise 6

    For going to a city B from city A, there is a route via city C such that AC⊥CB\displaystyle \mathrm{AC} \perp \mathrm{CB}, AC=2x km\displaystyle \mathrm{AC}=2 x \mathrm{~km} and CB=2(x+7)km\displaystyle \mathrm{CB}=2(x+7) \mathrm{km}. It is proposed to construct a 26\displaystyle 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway.

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    NCERT’s answer
    $\displaystyle 8$ km
    \[AC^2 + CB^2 = AB^2 \quad \text{(Pythagoras, } \angle C = 90^\circ\text{)} \] \[(2x)^2 + [2(x+7)]^2 = 26^2 \] \[4x^2 + 4(x+7)^2 = 676 \] \[x^2 + (x+7)^2 = 169 \] \[2x^2 + 14x - 120 = 0 \implies x^2 + 7x - 60 = 0 \] \[x = \frac{-7+\sqrt{49+240}}{2} = \frac{-7+17}{2} = 5 \quad (x>0) \] \[AC = 2x = 10 \text{ km}, \quad CB = 2(x+7) = 24 \text{ km} \] NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-4_Q6 \[\text{Distance saved} = (AC+CB) - AB = 34 - 26 = 8 \text{ km} \] Answer: \(\displaystyle 8 \) km.

  7. Exercise 7

    A flag pole 18\displaystyle 18 m high casts a shadow 9.6\displaystyle 9.6 m long. Find the distance of the top of the pole from the far end of the shadow.

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    NCERT’s answer
    20.$\displaystyle 4$ m
    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-4_Q7 \[BT = 18\text{ m (pole)}, \quad BS = 9.6\text{ m (shadow)}, \quad \angle TBS = 90^\circ \quad \text{(given)} \] \[TS^2 = BT^2+BS^2 \quad \text{(Pythagoras theorem)} \] \[TS^2 = 18^2+9.6^2 = 324+92.16 = 416.16 \] \[TS = \sqrt{416.16} = 20.4\text{ m} \]Answer: \(\displaystyle 20.4\) m

  8. Exercise 8

    A street light bulb is fixed on a pole 6\displaystyle 6 m above the level of the street. If a woman of height 1.5\displaystyle 1.5 m casts a shadow of 3m, find how far she is away from the base of the pole.

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    NCERT’s answer
    $\displaystyle 9$ m
    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-4_Q8 \[AB = 6\text{ m (pole)}, \quad DC = 1.5\text{ m (woman)}, \quad CE = 3\text{ m (shadow)} \quad \text{(given)} \] \[\triangle ABE \sim \triangle DCE \quad \text{(} \angle E \text{ common, } \angle ABE=\angle DCE=90^\circ\text{)} \] \[\frac{AB}{DC}=\frac{BE}{CE} \] \[\frac{6}{1.5}=\frac{BE}{3} \implies BE=12\text{ m} \] \[BC = BE-CE = 12-3 = 9\text{ m} \]Answer: \(\displaystyle 9\) m

  9. Exercise 9

    In Fig. 6.18\displaystyle 6.18, ABC is a triangle right angled at B and BD⊥AC\displaystyle \mathrm{BD} \perp \mathrm{AC}. If AD=4 cm\displaystyle \mathrm{AD}=4 \mathrm{~cm}, and CD=5 cm\displaystyle \mathrm{CD}=5 \mathrm{~cm}, find BD and AB. NCERT_Question_Class10_Maths_Exemplar_Ch6_Ex6-4_Q9

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    NCERT’s answer
    \(\displaystyle 2 \sqrt{5} \mathrm{~cm}, 6 \mathrm{~cm}\)
    \[\triangle ADB \sim \triangle BDC \quad \text{(altitude-on-hypotenuse: } \angle ADB=\angle BDC=90^\circ\text{)} \] \[\frac{AD}{BD}=\frac{BD}{DC} \implies BD^2 = AD \cdot DC = 4\times5 = 20 \] \[BD = 2\sqrt5 \text{ cm} \] \[AC = AD + DC = 4+5 = 9 \text{ cm} \] \[\triangle ADB \sim \triangle ABC \quad \text{(} \angle A \text{ common, } \angle ADB=\angle ABC=90^\circ\text{)} \] \[\frac{AD}{AB}=\frac{AB}{AC} \implies AB^2 = AD \cdot AC = 4\times9 = 36 \] \[AB = 6 \text{ cm} \]Answer: \(\displaystyle BD=2\sqrt5\) cm, \(\displaystyle AB=6\) cm

  10. Exercise 10

    In Fig. 6.19\displaystyle 6.19, PQR is a right triangle right angled at Q and QS⊥PR\displaystyle \mathrm{QS} \perp \mathrm{PR}. If PQ=6 cm\displaystyle \mathrm{PQ}=6 \mathrm{~cm} and PS=4 cm\displaystyle \mathrm{PS}=4 \mathrm{~cm}, find QS,RS\displaystyle \mathrm{QS}, \mathrm{RS} and QR. NCERT_Question_Class10_Maths_Exemplar_Ch6_Ex6-4_Q10

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    NCERT’s answer
    \(\displaystyle 2 \sqrt{5} \mathrm{~cm}, 5 \mathrm{~cm}, 3 \sqrt{5} \mathrm{~cm}\)
    \[\triangle PSQ \sim \triangle PQR \quad \text{(} \angle P \text{ common, } \angle PSQ=\angle PQR=90^\circ\text{)} \] \[\frac{PS}{PQ}=\frac{PQ}{PR} \implies PQ^2 = PS \cdot PR \] \[36 = 4\times PR \implies PR = 9 \text{ cm} \] \[RS = PR-PS = 9-4 = 5 \text{ cm} \] \[\triangle PSQ \sim \triangle QSR \quad \text{(altitude-on-hypotenuse: } \angle PSQ=\angle QSR=90^\circ\text{)} \] \[QS^2 = PS \cdot RS = 4\times5 = 20 \implies QS = 2\sqrt5 \text{ cm} \] \[QR^2 = QS^2+RS^2 = 20+25 = 45 \implies QR = 3\sqrt5 \text{ cm} \]Answer: \(\displaystyle QS=2\sqrt5\) cm, \(\displaystyle RS=5\) cm, \(\displaystyle QR=3\sqrt5\) cm