MathsImportant Questions for Class 10 Maths Chapter 6 Triangles

Important Questions for Class 10 Maths Chapter 6 Triangles

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Important Questions for Class 10 Maths Chapter 6 Triangles

Triangles Class 10 Important Questions Very Short Answer (1 Mark)

Question 1.
If ∆ABC ~ ∆PQR, perimeter of ∆ABC = 32 cm, perimeter of ∆PQR = 48 cm and PR = 6 cm, then find the length of AC. (2012)
Solution:
∆ABC ~ ∆PQR …[Given
Important Questions for Class 10 Maths Chapter 6 Triangles 1

Question 2.
∆ABC ~ ∆DEF. If AB = 4 cm, BC = 3.5 cm, CA = 2.5 cm and DF = 7.5 cm, find the perimeter of ∆DEF. (2012, 2017D)
Solution:
∆ABC – ∆DEF …[Given
Important Questions for Class 10 Maths Chapter 6 Triangles 2
Important Questions for Class 10 Maths Chapter 6 Triangles 3

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    Question 3.
    If ∆ABC ~ ∆RPQ, AB = 3 cm, BC = 5 cm, AC = 6 cm, RP = 6 cm and PQ = 10, then find QR. (2014)
    Solution:
    ∆ABC ~ ∆RPQ …[Given
    Important Questions for Class 10 Maths Chapter 6 Triangles 4
    ∴ QR = 12 cm

    Question 4.
    In ∆DEW, AB || EW. If AD = 4 cm, DE = 12 cm and DW = 24 cm, then find the value of DB. (2015)
    Solution:
    Let BD = x cm
    then BW = (24 – x) cm, AE = 12 – 4 = 8 cm
    In ∆DEW, AB || EW
    Important Questions for Class 10 Maths Chapter 6 Triangles 5

    Question 5.
    In ∆ABC, DE || BC, find the value of x. (2015)
    Important Questions for Class 10 Maths Chapter 6 Triangles 6
    Solution:
    In ∆ABC, DE || BC …[Given
    Important Questions for Class 10 Maths Chapter 6 Triangles 7
    x(x + 5) = (x + 3)(x + 1)
    x2 + 5x = x2 + 3x + x + 3
    x2 + 5x – x2 – 3x – x = 3
    ∴ x = 3 cm

    Question 6.
    In the given figure, if DE || BC, AE = 8 cm, EC = 2 cm and BC = 6 cm, then find DE. (2014)
    Important Questions for Class 10 Maths Chapter 6 Triangles 8
    Solution:
    In ∆ADE and ∆ABC,
    ∠DAE = ∠BAC …Common
    ∠ADE – ∠ABC … [Corresponding angles
    ∆ADE – ∆ΑΒC …[AA corollary
    Important Questions for Class 10 Maths Chapter 6 Triangles 9

    Question 7.
    In the given figure, XY || QR, \(\frac{P Q}{X Q}=\frac{7}{3}\) and PR = 6.3 cm, find YR. (2017OD)
    Important Questions for Class 10 Maths Chapter 6 Triangles 10
    Solution:
    Let YR = x
    \(\frac{\mathrm{PQ}}{\mathrm{XQ}}=\frac{\mathrm{PR}}{\mathrm{YR}}\) … [Thales’ theorem
    Important Questions for Class 10 Maths Chapter 6 Triangles 11

    Question 8.
    The lengths of the diagonals of a rhombus are 24 cm and 32 cm. Calculate the length of the altitude of the rhombus. (2013)
    Solution:
    Diagonals of a rhombus are ⊥ bisectors of each other.
    ∴ AC ⊥ BD,
    OA = OC = \(\frac{A C}{2} \Rightarrow \frac{24}{2}\) = 12 cm
    OB = OD = \(\frac{B D}{2} \Rightarrow \frac{32}{2}\) = 16 cm
    In rt. ∆BOC,
    Important Questions for Class 10 Maths Chapter 6 Triangles 12

    Question 9.
    If PQR is an equilateral triangle and PX ⊥ QR, find the value of PX2. (2013)
    Solution:
    Altitude of an equilateral ∆,
    Important Questions for Class 10 Maths Chapter 6 Triangles 13

    Triangles Class 10 Important Questions Short Answer-I (2 Marks)

    Question 10.
    The sides AB and AC and the perimeter P, of ∆ABC are respectively three times the corresponding sides DE and DF and the perimeter P, of ∆DEF. Are the two triangles similar? If yes, find \(\frac { ar\left( \triangle ABC \right) }{ ar\left( \triangle DEF \right) } \) (2012)
    Solution:
    Given: AB = 3DE and AC = 3DF
    Important Questions for Class 10 Maths Chapter 6 Triangles 14
    …[∵ The ratio of the areas of two similar ∆s is equal to the ratio of the squares of their corresponding sides

    Question 11.
    In the figure, EF || AC, BC = 10 cm, AB = 13 cm and EC = 2 cm, find AF. (2014)
    Important Questions for Class 10 Maths Chapter 6 Triangles 15
    Solution:
    BE = BC – EC = 10 – 2 = 8 cm
    Let AF = x cm, then BF = (13 – x) cm
    In ∆ABC, EF || AC … [Given
    Important Questions for Class 10 Maths Chapter 6 Triangles 16

    Question 12.
    X and Y are points on the sides AB and AC respectively of a triangle ABC such that \(\frac{\mathbf{A X}}{\mathbf{A B}}=\frac{1}{4}\), AY = 2 cm and YC = 6 cm. Find whether XY || BC or not. (2015)
    Solution:
    Given: \(\frac{A X}{A B}=\frac{1}{4}\)
    Important Questions for Class 10 Maths Chapter 6 Triangles 17
    AX = 1K, AB = 4K
    ∴ BX = AB – AX
    = 4K – 1K = 3K
    Important Questions for Class 10 Maths Chapter 6 Triangles 18
    ∴ XY || BC … [By converse of Thales’ theorem

    Question 13.
    In the given figure, ∠A = 90°, AD ⊥ BC. If BD = 2 cm and CD = 8 cm, find AD. (2012; 2017D)
    Important Questions for Class 10 Maths Chapter 6 Triangles 19
    Solution:
    ∆ADB ~ ∆CDA …[If a perpendicular is drawn from the vertex of the right angle of a rt. ∆ to the hypotenuse then As on both sides of the ⊥ are similar to the whole D and to each other
    ∴ \(\frac{B D}{A D}=\frac{A D}{C D}\) …[∵ Sides are proportional
    AD2 = BD , DC
    AD2 = (2) (8) = 16 ⇒ AD = 4 cm

    Question 14.
    In ∆ABC, ∠BAC = 90° and AD ⊥ BC. Prove that AD\frac{B D}{A D}=\frac{A D}{C D} = BD × DC. (2013)
    Solution:
    In 1t. ∆BDA, ∠1 + ∠5 = 90°
    In rt. ∆BAC, ∠1 + ∠4 = 90° …(ii)
    ∠1 + ∠5 = ∠1 + ∠4 …[From (i) & (ii)
    .. ∠5 = ∠4 …(iii)
    In ∆BDA and ∆ADC,
    Important Questions for Class 10 Maths Chapter 6 Triangles 20
    ∠5 = 24 … [From (iii)
    ∠2 = ∠3 …[Each 90°
    ∴ ∆BDA ~ ∆ADC…[AA similarity
    \(\frac{B D}{A D}=\frac{A D}{C D}\)
    … [In ~ As corresponding BA sides are proportional
    ∴ AD2 = BD × DC

    Question 15.
    A 6.5 m long ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall. Find the height of the wall where the top of the ladder touches it. (2015)
    Solution:
    Let AC be the ladder and AB be the wall.
    Important Questions for Class 10 Maths Chapter 6 Triangles 21
    ∴Required height, AB = 6 m

    Question 16.
    In the figure ABC and DBC are two right triangles. Prove that AP × PC = BP × PD. (2013)
    Important Questions for Class 10 Maths Chapter 6 Triangles 22
    Solution:
    Important Questions for Class 10 Maths Chapter 6 Triangles 23
    In ∆APB and ∆DPC,
    ∠1 = ∠4 … [Each = 90°
    ∠2 = ∠3 …[Vertically opp. ∠s
    ∴ ∆APB ~ ∆DPC …[AA corollary
    ⇒ \(\frac{\mathrm{BP}}{\mathrm{PC}}=\frac{\mathrm{AP}}{\mathrm{PD}}\) … [Sides are proportional
    ∴ AP × PC = BP × PD

    Question 17.
    In the given figure, QA ⊥ AB and PB ⊥ AB. If AO = 20 cm, BO = 12 cm, PB = 18 cm, find AQuestion (2017OD)
    Important Questions for Class 10 Maths Chapter 6 Triangles 24
    Solution:
    Important Questions for Class 10 Maths Chapter 6 Triangles 25
    In ∆OAQ and ∆OBP,
    ∠OAQ = ∠OBP … [Each 90°
    ∠AOQ = ∠BOP … [vertically opposite angles
    Important Questions for Class 10 Maths Chapter 6 Triangles 27

    Triangles Class 10 Important Questions Short Answer-II (3 Marks)

    Question 18.
    In the given figure, CD || LA and DE || AC. Find the length of CL if BE = 4 cm and EC = 2 cm. (2012)
    Important Questions for Class 10 Maths Chapter 6 Triangles 28
    Solution:
    In ∆ABL, CD || LA
    Important Questions for Class 10 Maths Chapter 6 Triangles 29

    Question 19.
    If a line segment intersects sides AB and AC of a ∆ABC at D and E respectively and is parallel to BC, prove that \(\frac{A D}{A B}=\frac{A E}{A C}\). (2013)
    Solution:
    Given. In ∆ABC, DE || BC

    To prove. \(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}\)
    Proof.
    In ∆ADE and ∆ABC
    ∠1 = ∠1 … Common
    ∠2 = ∠3 … [Corresponding angles
    ∆ADE ~ ∆ABC …[AA similarity
    ∴ \(\frac{\mathbf{A D}}{\mathbf{A B}}=\frac{\mathbf{A} \mathbf{E}}{\mathbf{A C}}\)
    …[In ~∆s corresponding sides are proportional

    Question 20.
    In a ∆ABC, DE || BC with D on AB and E on AC. If \(\frac{A D}{D B}=\frac{3}{4}\) , find \(\frac{\mathbf{B} C}{\mathbf{D} \mathbf{E}}\). (2013)
    Solution:
    Given: In a ∆ABC, DE || BC with D on AB and E
    on AC and \(\frac{A D}{D B}=\frac{3}{4}\)
    To find: \(\frac{\mathrm{BC}}{\mathrm{DE}}\)
    Proof. Let AD = 3k,
    Important Questions for Class 10 Maths Chapter 6 Triangles 31
    DB = 4k
    ∴ AB = 3k + 4k = 7k
    In ∆ADE and ∆ABC,
    ∠1 = ∠1 …[Common
    ∠2 = ∠3 … [Corresponding angles
    ∴ ∆ADE ~ ∆ABC …[AA similarity
    Important Questions for Class 10 Maths Chapter 6 Triangles 32

    Question 21.
    In the figure, if DE || OB and EF || BC, then prove that DF || OC. (2014)
    Important Questions for Class 10 Maths Chapter 6 Triangles 32
    Solution:
    Given. In ∆ABC, DE || OB and EF || BC
    To prove. DF || OC
    Proof. In ∆AOB, DE || OB … [Given
    Important Questions for Class 10 Maths Chapter 6 Triangles 34

    Question 22.
    If the perimeters of two similar triangles ABC and DEF are 50 cm and 70 cm respectively and one side of ∆ABC = 20 cm, then find the corresponding side of ∆DEF. (2014)
    Solution:
    Important Questions for Class 10 Maths Chapter 6 Triangles 35
    Given. ∆ABC ~ ∆DEF,
    Perimeter(∆ABC) = 50 cm
    Perimeter(∆DEF) = 70 cm
    One side of ∆ABC = 20 cm
    To Find. Corresponding side of ∆DEF (i.e.,) DE. ∆ABC ~ ∆DEF …[Given
    Important Questions for Class 10 Maths Chapter 6 Triangles 36
    ∴ The corresponding side of ADEF = 28 cm

    Question 23.
    A vertical pole of length 8 m casts a shadow 6 cm long on the ground and at the same time a tower casts a shadow 30 m long. Find the height of tower. (2014)
    Solution:
    Important Questions for Class 10 Maths Chapter 6 Triangles 37
    Let BC be the pole and EF be the tower Shadow AB = 6 m and DE = 30 m.
    In ∆ABC and ∆DEF,
    ∠2 = ∠4 … [Each 90°
    ∠1 = ∠3 … [Sun’s angle of elevation at the same time
    ∆ABC ~ ∆DEF …[AA similarity
    \(\frac{A B}{D E}=\frac{B C}{E F}\) … [In -As corresponding sides are proportional
    ⇒ \(\frac{6}{30}=\frac{8}{\mathrm{EF}}\) ∴ EF = 40 m

    Question 24.
    In given figure, EB ⊥ AC, BG ⊥ AE and CF ⊥ AE (2015)
    Prove that:
    (a) ∆ABG ~ ∆DCB
    (b) \(\frac{\mathbf{B C}}{\mathbf{B D}}=\frac{\mathbf{B E}}{\mathbf{B A}}\)
    Important Questions for Class 10 Maths Chapter 6 Triangles 38
    Solution:
    Important Questions for Class 10 Maths Chapter 6 Triangles 39
    Given: EB ⊥ AC, BG ⊥ AE and CF ⊥ AE.
    To prove: (a) ∆ABG – ∆DCB,
    (b) \(\frac{B C}{B D}=\frac{B E}{B A}\)
    Proof: (a) In ∆ABG and ∆DCB,
    ∠2 = ∠5 … [each 90°
    ∠6 = ∠4 … [corresponding angles
    ∴ ∆ABG ~ ∆DCB … [By AA similarity
    (Hence Proved)
    ∴ ∠1 = ∠3 …(CPCT … [In ~∆s, corresponding angles are equal

    (b) In ∆ABE and ∆DBC,
    ∠1 = ∠3 …(proved above
    ∠ABE = ∠5 … [each is 90°, EB ⊥ AC (Given)
    ∆ABE ~ ∆DBC … [By AA similarity
    \(\frac{B C}{B D}=\frac{B E}{B A}\)
    … [In ~∆s, corresponding sides are proportional
    ∴ \(\frac{B C}{B D}=\frac{B E}{B A}\) (Hence Proved)

    Question 25.
    ∆ABC ~ ∆PQR. AD is the median to BC and PM is the median to QR. Prove that \(\frac{\mathbf{A B}}{\mathbf{P Q}}=\frac{\mathbf{A D}}{\mathbf{P M}}\). (2017D)
    Solution:
    Important Questions for Class 10 Maths Chapter 6 Triangles 40
    ∆ABC ~ ∆PQR … [Given
    ∠1 = ∠2 … [In ~∆s corresponding angles are equal
    Important Questions for Class 10 Maths Chapter 6 Triangles 41

    Question 26.
    State whether the given pairs of triangles are similar or not. In case of similarity mention the criterion. (2015)
    Important Questions for Class 10 Maths Chapter 6 Triangles 42
    Solution:
    Important Questions for Class 10 Maths Chapter 6 Triangles 43
    (b) In ∆PQR, ∠P + ∠Q + ∠ZR = 180° …[Angle-Sum Property of a ∆
    45° + 78° + ∠R = 180°
    ∠R = 180° – 45° – 78° = 57°
    In ∆LMN, ∠L + ∠M + ∠N = 180° …[Angle-Sum Property of a ∆
    57° + 45° + ∠N = 180°
    ∠N = 180° – 57 – 45° = 78°
    ∠P = ∠M … (each = 45°
    ∠Q = ∠N … (each = 78°
    ∠R = ∠L …(each = 57°
    ∴ ∆PQR – ∆MNL …[By AAA similarity theorem

    Question 27.
    In the figure of ∆ABC, D divides CA in the ratio 4 : 3. If DE || BC, then find ar (BCDE) : ar (∆ABC). (2015)
    Important Questions for Class 10 Maths Chapter 6 Triangles 44
    Solution:
    Given:
    D divides CA in 4 : 3
    CD = 4K
    DA = 3K
    DE || BC …[Given
    Important Questions for Class 10 Maths Chapter 6 Triangles 45
    In ∆AED and ∆ABC,
    ∠1 = ∠1 …[common
    ∠2 = ∠3 … corresponding angles
    ∴ ∆AED – ∆ABC …(AA similarity
    ⇒ \(\frac { ar\left( \triangle AED \right) }{ ar\left( \triangle ABC \right) } =\left( \frac { AD }{ AC } \right) ^{ 2 }\)
    … [The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides
    ⇒ \(\\frac { \left( 3K \right) ^{ 2 } }{ \left( 7K \right) ^{ 2 } } =\frac { { 9K }^{ 2 } }{ { 49K }^{ 2 } } =\frac { ar\left( \triangle AED \right) }{ ar\left( \triangle ABC \right) } =\frac { 9 }{ 49 } \)
    Let ar(∆AED) = 9p
    and ar(∆ABC) = 49p
    ar(BCDE) = ar (∆ABC) – ar (∆ADE)
    = 49p – 9p = 40p
    ∴ \(\frac { ar\left( BCDE \right) }{ ar\left( \triangle ABC \right) } =\frac { 40p }{ 49p } \)
    ∴ ar (BCDE) : ar(AABC) = 40 : 49

    Question 28.
    In the given figure, DE || BC and AD : DB = 7 : 5, find \frac { ar\left( \triangle DEF \right) }{ ar\left( \triangle CFB \right) } [/latex] (2017OD)
    Important Questions for Class 10 Maths Chapter 6 Triangles 46
    Solution:
    Given: In ∆ABC, DE || BC and AD : DB = 7 : 5.
    To find: \(\frac { ar\left( \triangle DEF \right) }{ ar\left( \triangle CFB \right) } \) = ?
    Important Questions for Class 10 Maths Chapter 6 Triangles 47
    Proof: Let AD = 7k
    and BD = 5k then
    AB = 7k + 5k = 12k
    In ∆ADE and ∆ABC,
    ∠1 = ∠1 …(Common
    ∠2 = ∠ABC … [Corresponding angles
    Important Questions for Class 10 Maths Chapter 6 Triangles 48

    Question 29.
    In the given figure, the line segment XY is parallel to the side AC of ∆ABC and it divides the triangle into two parts of equal areas. Find the ratio \(\frac{\mathbf{A} \mathbf{X}}{\mathbf{A B}}\). (2017OD)
    Important Questions for Class 10 Maths Chapter 6 Triangles 49
    Solution:
    We have XY || AC … [Given
    So, ∠BXY = ∠A and ∠BYX = ∠C …[Corresponding angles
    ∴ ∆ABC ~ ∆XBY …[AA similarity criterion
    Important Questions for Class 10 Maths Chapter 6 Triangles 50

    Question 30.
    In the given figure, AD ⊥ BC and BD = \(\frac{1}{3}\)CD. Prove that 2AC2 = 2AB2 + BC2. (2012)
    Important Questions for Class 10 Maths Chapter 6 Triangles 51
    Solution:
    BC = BD + DC = BD + 3BD = 4BD
    ∴ \(\frac{\mathrm{BC}}{4}\) = BD
    In rt. ∆ADB, AD2 = AB2 – BD2 ….(ii)
    In rt. ∆ADC, AD2 = AC2 – CD2 …(iii)
    From (ii) and (iii), we get
    AC2 – CD2 = AB2 – BD2
    AC2 = AB2 – BD2 + CD2
    Important Questions for Class 10 Maths Chapter 6 Triangles 52
    ∴ 2AC2 = 2AB2 + BC2 (Hence proved)

    Question 31.
    In the given figure, ∆ABC is right-angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE. (2012, 2017D)
    Important Questions for Class 10 Maths Chapter 6 Triangles 53
    Solution:
    Given: ∆ABC is rt. ∠ed at C and DE ⊥ AB.
    AD = 3 cm, DC = 2 cm, BC = 12 cm
    To prove:
    (i) ∆ABC ~ ∆ADE; (ii) AE = ? and DE = ?
    Proof. (i) In ∆ABC and ∆ADE,
    ∠ACB = ∠AED … [Each 90°
    ∠BAC = ∠DAE …(Common .
    ∴ ∆ABC ~ ∆ADE …[AA Similarity Criterion

    (ii) ∴ \(\frac{A B}{A D}=\frac{B C}{D E}=\frac{A C}{A E}\) … [side are proportional
    \(\frac{A B}{3}=\frac{12}{D E}=\frac{3+2}{A E}\)
    …..[In rt. ∆ACB, … AB2 = AC2 + BC2 (By Pythagoras’ theorem)
    = (5)2 + (12)2 = 169
    ∴ AB = 13 cm
    Important Questions for Class 10 Maths Chapter 6 Triangles 54

    Question 32.
    In ∆ABC, if AP ⊥ BC and AC2 = BC2 – AB2, then prove that PA2 = PB × CP. (2015)
    Solution:
    Important Questions for Class 10 Maths Chapter 6 Triangles 55
    AC2 = BC2 – AB2 …Given
    AC2 + AB2 = BC2
    ∴ ∠BAC = 90° … [By converse of Pythagoras’ theorem
    ∆APB ~ ∆CPA
    [If a perpendicular is drawn from the vertex of the right angle of a triangle to the hypotenuse then As on both sides of the perpendicular are similar to the whole triangle and to each other.
    ∴ \(\frac{\mathrm{AP}}{\mathrm{CP}}=\frac{\mathrm{PB}}{\mathrm{PA}}\) … [In ~∆s, corresponding sides are proportional
    ∴ PA2 = PB. CP (Hence Proved)

    Question 33.
    ABCD is a rhombus. Prove that AB2 + BC2 + CD2 + DA2 = AC2 + BD2. (2013)
    Solution:
    Given. In rhombus ABCD, diagonals AC and BD intersect at O.
    Important Questions for Class 10 Maths Chapter 6 Triangles 56
    To prove: AB2 + BC2 + CD2 + DA2 = AC2 + BD2
    Proof: AC ⊥ BD [∵ Diagonals of a rhombus bisect each other at right angles
    ∴ OA = OC and
    OB = OD
    In rt. ∆AOB,
    AB2 = OA2 + OB2 … [Pythagoras’ theorem
    AB2 = \(\left(\frac{A C}{2}\right)^{2}+\left(\frac{B D}{2}\right)^{2}\)
    AB2 = \(\left(\frac{A C}{2}\right)^{2}+\left(\frac{B D}{2}\right)^{2}\)
    4AB2 = AC2 + BD2
    AB2 + AB2 + AB2 + AB2 = AC2 + BD2
    ∴ AB2 + BC2 + CD2 + DA2 = AC2 + BD2
    …[∵ In a rhombus, all sides are equal

    Question 34.
    The diagonals of trapezium ABCD intersect each other at point o. If AB = 2CD, find the ratio of area of the ∆AOB to area of ∆COD. (2013)
    Solution:
    In ∆AOB and ∆COD, … [Alternate int. ∠s
    ∠1 = ∆3
    ∠2 = ∠4
    Important Questions for Class 10 Maths Chapter 6 Triangles 57

    Question 35.
    The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac{A O}{B O}=\frac{C O}{D O}\). Show that ABCD is a trapezium. (2014)
    Solution:
    1st method.
    Given: Quadrilateral ABCD in which
    AC and BD intersect each other at 0.
    Such that \(\frac{A O}{B O}=\frac{C O}{D O}\)
    To prove: ABCD is a trapezium
    Const.: From O, draw OE || CD.
    Important Questions for Class 10 Maths Chapter 6 Triangles 58
    Important Questions for Class 10 Maths Chapter 6 Triangles 59
    But these are alternate interior angles
    ∴ AB || DC Quad. ABCD is a trapezium.

    Triangles Class 10 Important Questions Long Answer (4 Marks).

    Question 36.
    In a rectangle ABCD, E is middle point of AD. If AD = 40 m and AB = 48 m, then find EB. (2014D)
    Solution:
    Important Questions for Class 10 Maths Chapter 6 Triangles 60
    E is the mid-point of AD …[Given
    AE = \(\frac{40}{2}\) = 20 m
    ∠A = 90° …[Angle of a rectangle
    In rt. ∆BAE,
    EB2 = AB2 + AE2 …[Pythagoras’ theorem
    = (48)2 + (20)2
    = 2304 + 400 = 2704
    ∴ EB = \(\sqrt{2704}\) = 52 m

    Question 37.
    Let ABC be a triangle and D and E be two points on side AB such that AD = BE. If DP || BC and EQ || AC, then prove that PQ || AB. (2013)
    Solution:
    Important Questions for Class 10 Maths Chapter 6 Triangles 60
    In ∆ABC,
    DP || BC
    and EQ || AC … [Given
    Important Questions for Class 10 Maths Chapter 6 Triangles 62
    Now, in ∆ABC, P and Q divide sides CA and CB respectively in the same ratio.
    ∴ PQ || AB

    Question 38.
    In the figure, ∠BED = ∠BDE & E divides BC in the ratio 2 : 1.
    Prove that AF × BE = 2 AD × CF. (2015)
    Important Questions for Class 10 Maths Chapter 6 Triangles 63
    Solution:
    Construction:
    Draw CG || DF
    Proof: E divides
    BC in 2 : 1.
    \(\frac{B E}{E C}=\frac{2}{1}\) …(i)
    Important Questions for Class 10 Maths Chapter 6 Triangles 64
    Important Questions for Class 10 Maths Chapter 6 Triangles 65

    Question 39.
    In the given figure, AD = 3 cm, AE = 5 cm, BD = 4 cm, CE = 4 cm, CF = 2 cm, BF = 2.5 cm, then find the pair of parallel lines and hence their lengths. (2015)
    Important Questions for Class 10 Maths Chapter 6 Triangles 66
    Solution:
    Important Questions for Class 10 Maths Chapter 6 Triangles 67
    Important Questions for Class 10 Maths Chapter 6 Triangles 68

    Question 40.
    If sides AB, BC and median AD of AABC are proportional to the corresponding sides PQ, QR and median PM of PQR, show that ∆ABC ~ ∆PQR. (2017OD)
    Solution:
    Important Questions for Class 10 Maths Chapter 6 Triangles 69

    Question 41.
    Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. (2012)
    Solution:
    Given: ∆ABC ~ ∆DEF
    Important Questions for Class 10 Maths Chapter 6 Triangles 70
    Important Questions for Class 10 Maths Chapter 6 Triangles 71
    Important Questions for Class 10 Maths Chapter 6 Triangles 72

    Question 42.
    State and prove converse of Pythagoras theorem. Using the above theorem, solve the following: In ∆ABC, AB = 6\(\sqrt{3}\) cm, BC = 6 cm and AC = 12 cm, find ∠B. (2015)
    Solution:
    Part I:
    Statement: Prove that, in a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
    Important Questions for Class 10 Maths Chapter 6 Triangles 73
    To prove: ∠ABC = 90°
    Const.: Draw a right angle ∆DEF in which DE = BC and EF = AB.
    Proof: In rt. ∆ABC,
    AB2 + BC2 = AC2 …(i) Given
    In rt. ∆DEF
    DE2 + EF2 = DF2 … [By Pythagoras’ theorem
    BC2 + AB2 = DF2…(ii)…[∵ DE = BC; EF = AB
    From (i) and (ii), we get
    AC2 = DF2 = AC = DF
    Now, DE = BC …[By construction
    EF = AB …[By construction
    DF = AC … [Proved above :
    ∴ ∆DEF = ∆ABC … (SSS congruence :
    ∴ ∠DEF = ∠ABC …[c.p.c.t.
    ∵ ∠DEF = 90° ∴ ∠ABC = 90°
    Given: In rt. ∆ABC,
    AB2 + BC2 = AC2
    AB2 + BC2 = (6\(\sqrt{3}\))2 + (6)2
    = 108 + 36 = 144 = (12)2
    AB2 + BC2 = AC2 ∴ ∠B = 90° … [Above theorem

    Question 43.
    In the given figure, BL and CM are medians of a triangle ABC, right angled at A. Prove that: 4(BL2 + CM2) = 5BC2 (2012)
    Important Questions for Class 10 Maths Chapter 6 Triangles 74
    Solution:
    Given: BL and CM are medians of ∆ABC, right angled at A.
    To prove: 4(BL2 + CM2) = 5 BC2
    Proof: In ∆ABC, BC2 = BA2 + CA2 …(i)
    In ∆BAL,
    BL2 = BA2 + AL2 …[Pythagoras’ theorem
    BL2 = BA2 + \(\left(\frac{\mathrm{CA}}{2}\right)^{2}\)
    BL2 = BA2+ \(\frac{\mathrm{CA}^{2}}{4}\)
    ⇒ 4BL2 = 4BA2 + CA2 …(ii)
    Now, In ∆MCA,
    MC2 = CA2 + MA2 …[Pythagoras’ theorem
    MC2 = CA22 + \(\left(\frac{\mathrm{BA}}{2}\right)^{2}\)
    MC2 = CA2 + \(\frac{\mathrm{BA}^{2}}{4}\)
    4MC2 = 4CA2 + BA2
    Adding (ii) and (iii), we get
    4BL2 + 4MC2 = 4BA2 + CA2 + 4CA2+ BA2 …[From (ii) & (iii)
    4(BL2 + MC2) = 5BA2 + 5CA2
    4(BL2 + MC2) = 5(BA2 + CA2)
    ∴ 4(BL2 + MC2) = 5BC2 … [Using (1)
    Hence proved.

    Question 44.
    In the given figure, AD is median of ∆ABC and AE ⊥ BC. (2013)
    Prove that b2 + c2 = 2p2 + \(\frac{1}{2}\) a2.
    Important Questions for Class 10 Maths Chapter 6 Triangles 75
    Solution:
    Important Questions for Class 10 Maths Chapter 6 Triangles 76
    Proof. Let ED = x
    BD = DC = \(\frac{B C}{2}=\frac{a}{2}\) = …[∵ AD is the median
    In rt. ∆AEC, AC2 = AE2 + EC2 …..[By Pythagoras’ theorem
    b2 = h2 + (ED + DC)2
    b2 = (p2 – x2) + (x = \(\frac{a}{2}\))2
    …[∵ In rt. ∆AED, x2 + h2 = p2 ⇒ h2 = p2 – x2 …(i)
    b2 = p2 – x2 + x2 + \(\left(\frac{a}{2}\right)^{2}\)2+ 2(x)\(\left(\frac{a}{2}\right)\)
    b2 = p2 + ax + \(\frac{a^{2}}{4}\) …(ii)
    In rt. ∆AEB, AB2 = AE2 + BE2 … [By Pythagoras’ theorem
    Important Questions for Class 10 Maths Chapter 6 Triangles 77

    Question 45.
    In a ∆ABC, the perpendicular from A on the side BC of a AABC intersects BC at D such that DB = 3 CD. Prove that 2 AB2 = 2 AC2 + BC2. (2013; 2017OD)
    Solution:
    In rt. ∆ADB,
    AD2 = AB2 – BD2 …(i) [Pythagoras’ theorem
    In rt. ∆ADC,
    AD2 = AC2 – DC2 …(ii) [Pythagoras’ theorem
    Important Questions for Class 10 Maths Chapter 6 Triangles 78
    From (i) and (ii), we get
    AB2 – BD2 = AC2 – DC2
    AB2 = AC2 + BD2 – DC2
    Now, BC = BD + DC
    = 3CD + CD = 4 CD …[∵ BD = 3CD (Given)
    ⇒ BC2 = 16 CD2 …(iv) [Squaring
    Now, AB2 = AC2 + BD2 – DC2 …[From (iii)
    = AC2 + 9 DC2 – DC2 ….[∵ BD = 3 CD ⇒ BD2 = 9 CD2
    = AC2 + 8 DC2
    = AC2 + \(\frac{16 \mathrm{DC}^{2}}{2}\)
    = AC2 + \(\frac{B C^{2}}{2}\) … [From (iv)
    ∴ 2AB2 = 2AC2 + BC2 … [Proved

    Question 46.
    In ∆ABC, altitudes AD and CE intersect each other at the point P. Prove that: (2014)
    (i) ∆APE ~ ∆CPD
    (ii) AP × PD = CP × PE
    (iii) ∆ADB ~ ∆CEB
    (iv) AB × CE = BC × AD
    Solution:
    Important Questions for Class 10 Maths Chapter 6 Triangles 79
    Given. In ∆ABC, AD ⊥ BC & CE ⊥ AB.
    To prove. (i) ∆APE ~ ∆CPD
    (ii) AP × PD = CP × PE
    (iii) ∆ADB ~ ∆CEB
    (iv) AB × CE = BC × AD
    Proof: (i) In ∆APE and ∆CPD,
    ∠1 = ∠4 …[Each 90°
    ∠2 = ∠3 …[Vertically opposite angles
    ∴ ∆APE ~ ∆CPD …[AA similarity
    (ii) \(\frac{\mathrm{AP}}{\mathrm{CP}}=\frac{\mathrm{PE}}{\mathrm{PD}}\) … [In ~ ∆s corresponding sides are proportional
    ∴ AP × PD = CP × PE
    (iii) In ∆ADB and ∆CEB,
    ∠5 = ∠7 …[Each 90°
    ∠6 = ∠6 …(Common
    ∴ ∆ADB ~ ∆CEB …[AA similarity
    (iv) ∴ \(\frac{A B}{C B}=\frac{A D}{C E}\) … [In ~ ∆s corresponding sides are proportional
    ∴ AB × CE = BC × AD

    Question 47.
    In the figure, PQR and QST are two right triangles, right angled at R and T resepctively. Prove that QR × QS = QP × QT. (2014)
    Important Questions for Class 10 Maths Chapter 6 Triangles 80
    Solution:
    Given: Two rt. ∆’s PQR and QST.
    Important Questions for Class 10 Maths Chapter 6 Triangles 81
    To prove: QR × QS = QP × QT
    Proof: In ∆PRQ and ∆STQ,
    ∠1 = ∠1 … [Common
    ∠2 = ∠3 … [Each 90°
    ∆PRQ ~ ∆STO …(AA similarity
    ∴ \(\frac{Q R}{Q T}=\frac{Q P}{Q S}\) ..[In -∆s corresponding sides are proportional
    ∴ QR × QS = QP × QT (Hence proved)

    Question 48.
    In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \(\frac { ar\left( ABC \right) }{ ar\left( DBC \right) } =\frac { AO }{ DO } \). (2012)
    Important Questions for Class 10 Maths Chapter 6 Triangles 82
    Solution:
    Given: ABC and DBC are two As on the same base BC. AD intersects BC at O.
    To prove:
    Important Questions for Class 10 Maths Chapter 6 Triangles 83
    Important Questions for Class 10 Maths Chapter 6 Triangles 84
    Important Questions for Class 10 Maths Chapter 6 Triangles 85

    Question 49.
    Hypotenuse of a right triangle is 25 cm and out of the remaining two sides, one is longer than the other by 5 cm. Find the lengths of the other
    two sides. (2013)
    Solution:
    Let Base, AB = x cm
    Then altitude, BC = (x + 5) cm
    In rt. ∆,
    By Pythagoras’ theorem
    Important Questions for Class 10 Maths Chapter 6 Triangles 86
    AB2 + BC2 = AC2
    ⇒ (x)2 + (x + 5)2 = 252
    ⇒ x22 + x2 + 10x + 25 – 625 = 0
    ⇒ 2x2 + 10x – 600 = 0
    ⇒ x2 + 5x – 300 = 0 … [Dividing both sides by 2
    ⇒ x2 + 20x – 15x – 300 = 0
    ⇒ x(x + 20) – 15(x + 20) = 0
    (x – 15)(x + 20) = 0
    x – 15 = 0 or x + 20 = 0
    x = 15 or x = -20
    Base cannot be -ve
    ∴ x = 15 cm
    ∴ Length of the other side = 15 + 5 = 20 cm
    Two sides are = 15 cm and 20 cm

    Question 50.
    In Figure, AB ⊥ BC, FG ⊥ BC and DE ⊥ AC. Prove that ∆ADE ~ ∆GCF. (2016 OD)
    Important Questions for Class 10 Maths Chapter 6 Triangles 87
    Solution:
    In rt. ∆ABC,
    ∠A + ∠C = 90° …(i)
    In rt. ∆AED,
    ∠A + ∠2 = 90°
    From (i) and (ii), ∠C = ∠2
    Similarly, ∠A = ∠1
    Now in ∆ADE & ∆GCF
    ∠A = 1 … [Proved
    ∠C = 2 … [Proved
    ∠AED = ∠GFC … [rt. ∠s
    ∴ ∆ADE – ∆GCF …(Hence Proved)

    Important Questions for Class 10 Maths

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