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Triangles Test - 6

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Triangles Test - 6
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  • Question 1
    1 / -0
    In the given triangle ABC, AB = AC and DE bisects ∠BDC. DE is parallel to AB. What is the measure of ∠BED?
    Solution
    In △ABC and △DEC
    ∠ABC = ∠DEC (AB ∥ DE)
    ∠ACB = ∠DCE (common)
    △ABC ～ △DEC
    Therefore, ∠BAC = ∠EDC = ∠BDE = 20° (DE bisects ∠BDC)

    ∠ABC = ∠ACB (AB = AC)

    ∠ABC + ∠ACB + ∠BAC = 180° ( Angle sum property)

    2∠ABC = 180° - ∠BAC = 180° - 20° = 160°
    ∠ABC = 80°

    ∠BED = 180° - 80° ( As ∠ABC = ∠DEC)
    = 100°
  • Question 2
    1 / -0
    PQR is an equilateral triangle. Another triangle QRT is attached to it. What is the measure of QTR, if TQR = 35°?

    Solution
    PQR is an equilateral triangle.
    ∴ PRQ = 60°
    In ΔQRT,
    QRP = RQT + RTQ (Exterior angle property of triangle)
    60° = 35° + RTQ
    ∴ RTQ = 25°
    Or, QTR = 25°
  • Question 3
    1 / -0
    In the given figure PU = UR and TU = ST. If the length of QR is 44 cm, then what is the length of SQ?

    Solution


    If V is the point on PQ such that VU II SR, then
    ∠PRS = ∠QVU and ∠USR = ∠VUS (Alternate angles on the parallel lines)
    Also, TU = ST (given)
    So, △UTV ～ △STQ
    Hence, SQ = UV

    SQ =QR = 44 ÷ 2 = 22 cm
  • Question 4
    1 / -0
    In the given triangle, if ∠PQU = ∠QRT = ∠QPS, then what is the ratio of the lengths UT, TS and US, respectively?

    Solution


    Since ∠UTS is an exterior angle of triangle QTR
    ∠PQU = ∠TRQ (Given)
    ∠UTS = ∠QRT + ∠TQR
    = ∠PQU + ∠TQR
    = ∠PQR

    Similarly, we also have ∠UST = ∠PRQ, which implies that △UTS ～△PQR (By AA similarity)

    So, UT : TS : US = 16 : 19 : 5
  • Question 5
    1 / -0
    In the isosceles △XYZ, XZ = ZY, what is the approximate length of MY, if MX, MN and MO are 10 cm, 16 cm and 6 cm, respectively?

    Solution
    In △XYZ,
    XZ = ZY (Given)
    ∠ZXY = ∠ZYX (Angles opposite to equal sides are equal)
    In △XOM and △MNY,
    ∠ZXY = ∠ZYX (Proved above)
    ∠MOX = ∠MNY (Each 90°)
    Thus, by AA similarity criterion,
    △XOM is similar to △MNY.

    ==

    =

    MY = = = 26.67 cm approximately.
  • Question 6
    1 / -0
    AB and CD are two parallel lines and AD and CB intersect each other at point O. If the lengths of the sides OB, OC and AB are 2 cm, 3 cm and 4 cm, respectively and perimeter of △OCD is 12 cm, then what is the length of OA?
    Solution
    The required figure is as follows:


    Since, ∠AOB = ∠COD (Vertically opposite angles)
    ∠ABC = ∠BAD (As AB∥CD, so ∠ABC and ∠BAD are alternate interior angles)
    Thus, △AOB and △COD are similar. (By AA similarity rule)
    Therefore, =
    =
    CD = (4 × 3) ÷ 2
    CD == 6 cm
    As, perimeter of △OCD = 12 cm
    So, OD = 12 – (6 + 3)
    OD = 12 – 9 = 3 cm
    In △OBA and △OCD,
    =
    =
    OA = 3 ×(2/3) = 2 cm
  • Question 7
    1 / -0
    In the given triangle, AM and CS are perpendicular to BC and AB, respectively. If the lengths of CS and BS are 4 cm and 3 cm, respectively and BM is 2 cm less than BC, then what is the value of AB × BS?
    Solution
    In △ABM and △CBS,
    ∠BMA = 90° = ∠CSB (Each 90°)
    ∠B is common in both the triangles.
    Therefore, by AA similarity criteria, △ABM is similar to △CBS.
    So, =
    AB × BS = BC × BM --- (i)
    In △CBS, by Pythagoras theorem,
    BC² = CS² + BS²
    BC² = 4² + 3²
    BC² = 16 + 9 = 25
    BC = 5 cm
    Thus, BC = 5 cm and BM = 5 – 2 = 3 cm (Because it is given that BM is 2 cm less than BC) --- (ii)
    Putting the values of BC and BM in equation (i), we get
    AB × BS = 5 × 3 = 15 cm²
  • Question 8
    1 / -0
    ABCD is a rhombus whose diagonals AC and BD intersect at point E. If AB = 10 cm and diagonal BD = 16 cm, then find the length of diagonal AC.
    Solution
    Diagonals of a rhombus bisect each other at right angles.

    BE = (BD) = ( × 16) cm = 8 cm, AB = 10 cm and AEB = 90°
    From right ΔEAB, we get
    AB2 = AE2 + BE2 (Pythagoras theorem)
    AE2 = (AB2 - BE2) = {(10)2 - (8)2} cm2
    = (100 - 64) cm2 = 36 cm2
    AE = cm = 6 cm
    AC = 2 × AE = 2 × 6 cm = 12 cm (Since AE is half of AC)
  • Question 9
    1 / -0
    In a triangle ABC, if angle B = 90° and D is a point on BC such that BD = 2DC then _____.
    Solution


    BD = 2DC
    AC2 = AB2 + BC2

    In ΔABD,
    AD2 = AB2 + BD2
    Or AB2 = AD2 - BD2
    AC2 = AD2 - BD2 + BC2
    = AD2 - (2DC)2 + (3DC)2
    [ BD = 2 DC and BC = 3 DC]
    = AD2 - 4DC2 + 9DC2 = AD2 + 5CD2
  • Question 10
    1 / -0
    An electrical pole is placed next to a tree. The height of the tree is 12 m and the shadow cast by it is 16 m. Ratio between PT and TA is 1 : 1. Find the height of the electrical pole and distance between the pole and the tree.

    Solution


    Given:

    Length of the tree (TB) = 12 m

    Shadow cast by the tree (AB) = 16 m

    PT : TA = 1 : 1

    Draw a line TD that touches PC.

    Let x be the difference between the height of the tree and the electric pole.

    Height of the electric pole = (12 + x) m.

    Distance between the tree and the pole = y m

    So, in △PCA and △TBA

    ∠B = ∠C = 90° (Each 90°)

    ∠PAC = ∠TAB (Common angle)

    △PCA ~ △TBA (By AA similarity)

    So,





    Firstly consider,




    12 + x = 24

    x = 24 - 12 = 12 cm

    So, height of the electric pole = 12 + 12 = 24 m

    Secondly consider,



    16 + y = 32

    y = 32 - 16 = 16 m

    So, the distance between the tree and the pole is 16 m.
  • Question 11
    1 / -0
    A floodlight is 24 m away from a building which is in triangular shape. The distance between their tops is 26 m. If the floodlight is 28 m above ground level. Find the height of the building.
    Solution


    Given:
    Length of the floodlight (AB) = 28 m
    Distance between the floodlight and the building (BE) = 24 m
    Draw a straight line from D that meets line AB at point F. So,
    BE = FD = 24 m
    ∠B = 90°
    So, FDEB is a rectangle.
    Opposite sides are equal in a rectangle.
    Let the height of the building be x metres.
    In, △AFD, ∠F = 90°
    Using Pythagoras theorem,
    AF2 + FD2 = AD2
    AF2 = AD2 - FD2
    AF2 = (26)2 - (24)2
    = 676 - 576 = 100
    AF2 = 100
    AF = ±10 m
    AF = 10 m (Because distance cannot be negative)
    DE = FB (FDEB is a rectangle)
    DE = AB - AF = 28 - 10 = 18 m
    Height of the building is 18 m
  • Question 12
    1 / -0
    Sides of which three lengths cannot form a triangle?
    Solution
    For triangle:
    Sum of smaller sides > Larger side
    Only in option 3:
    a = 3 cm, b = 3.5 cm and c = 7 cm
    Now,
    a + b = (3 + 3.5) cm = 6.5 cm
    a + b is not greater than c.
    So, a triangle with these three side lengths cannot be formed.
  • Question 13
    1 / -0
    Two individuals A and B walk towards the South and East directions, respectively with speeds of 13.2 kmph and 22.4 kmph. After travelling for 2.5 hours B stops and A starts travelling towards B. What is the minimum distance that A would have to travel to meet B?
    Solution

    From the given information we can get the following arrangements of A and B direction wise.
    The distance that A would have to travel = PR + RQ
    PQ = Distance travelled by B after 2.5 hours = 22.4 × 2.5 = 56 km
    PR = Distance travelled by A after 2.5 hours = 13.2 × 2.5 = 33 km
    Distance between A and B after 2.5 hours = QR
    Pythagoras theorem
    QR² = PQ² + PR²
    = (56)² + (33)²
    = 3136 + 1089
    = 4225
    QR = = 65 km
    The total distance that A would have to travel = 33 + 65 = 98 km
  • Question 14
    1 / -0
    During a military training, each cadet had to start from A and reach B covering 17 km and then reach E which is 8 km from B. From E, the cadets had to travel to C and then to point D and then finally back to A through point E. The distance from E to C was 4 km and that of E to D was 3 km.

    (1) Find the distance from A to E.
    (2) Find the total distance covered while moving from A to D via B and C.


    Solution


    (I) In triangle AEB
    ∠E = 90°
    AB² = BE² + AE²
    (17)² = (8)² + (AE)²
    289 = 64 + AE²
    225 = AE²
    AE = 15 km

    (II) Distance covered while moving from A to D via B and C.
    Using Pythagoras theorem in Triangle ECD, we get
    CD² = (3)² + (4)²
    CD² = 9 + 16
    CD = 5
    Total distance
    = AB + BE + EC + CD
    = 17 + 8 + 4 + 5
    = 34 km
  • Question 15
    1 / -0
    In triangle ABC, ∠C and in triangle PQR, ∠R are 90°. If A = P, DQ = 2.5 cm and AD is the perpendicular bisector of QP, then find the length of AB.

    Solution
    In triangle ABC and triangle PQR,
    QRP = ACB = 90° (Each 90°)
    CAB = RPQ (given)

    Triangle ABC and triangle PQR are similar to each other by AA similarity criteria,



    (PQ = QD + DP, i.e. PQ = 2.5 + 2.5 = 5 cm (Because AD is the perpendicular bisector of PQ)



    AB = 20 cm
  • Question 16
    1 / -0
    The area of triangle MNO is 196 cm2, which is similar to another triangle PQR having an area 225 cm2. In triangle MNO, DM, which is 14 cm in length, is perpendicular to NO and in triangle PQR, PE is perpendicular to QR. Also E divides QR, which is 30 cm in length, in the ratio 4 : 11. What is the length of PQ?

    Solution
    Ratio of areas of two similar triangles is square of the ratio of their corresponding sides.
    Ratio of the areas of triangles MNO and PQR is 196 : 225 or 142 : 152
    Hence, the ratio of corresponding sides of triangles MNO and PQR is 14 : 15
    As DM is 14 cm, then PE is 15 cm.
    QR = 30
    QE : ER = 4 : 11
    Hence, QE = 8 cm and ER = 22 cm
    In right-angled triangle PEQ
    PQ2 = PE2 + QE2 = 152 + 82 = 225 + 64
    PQ2 = 289
    PQ = 17 cm
  • Question 17
    1 / -0
    In the given figure, PQRS is a parallelogram. T is a point on side PS such that the ratio of lengths of PT and TS is 2 : 5. Find the length of RU and QU, respectively if the lines TQ and PR intersects at U.

    Solution
    In △QUR and △TPU
    ∠QRU = ∠TPU (Alternate angles)
    ∠TUP = ∠ QUR (Vertically opposite angles)
    Hence,
    △QRU ～ △TPU (By AA similarity)
    Therefore,

     .....eq (i)

    As

    2TS = 5PT
    TS = 2.5PT

    PS = PT + TS = PT + 2.5PT
    PS = 3.5PT

    As PS = QR
    QR = 3.5PT

    Hence, equation (i) becomes



    Therefore, RU = 3.5UP
    QU = 3.5UT
  • Question 18
    1 / -0
    Find RT in the given figure.

    Solution
    In triangle PTR and triangle QTS,
    ∠PRT = ∠TSQ = 90°
    And ∠PTR = ∠QTS (Vertically opposite angles)
    So, △PTR ～△QTS (By AA similarity)
    Hence,



    RT =
  • Question 19
    1 / -0
    In the given triangle DE || AB, find the value of DE in terms of m, n, p.

    Solution
    In △ABC and △DEC
    ∠CBA = ∠CED (As DE || AB)
    ∠ACB = ∠DCE (Common)
    △ABC ～ △DEC (By AA similarity)



    DE =
  • Question 20
    1 / -0
    In the given figure, PT × TS = QT × TR. PT × SR equal to
    Solution

    In triangles PTS and QTR
    ∠PTS = ∠QTR (Vertically opposite angles)
    Also PT × TS = QT × TR (Given)

    So,

    ∴

    So PTS ~ QTR (By AA similarity)

    ∴



    PT × SR = TR × PQ
  • Question 21
    1 / -0
    In the given figure, ABC is an isosceles triangle right angled at A where AB = AC = x cm and area of triangle ABC right angled at A is x cm2. Area of right triangle ADE is cm2. Find the length of DE.
    Solution
    In triangle ABC,
    BC2 = AB2 + AC2
    BC2 = x2 + x2
    BC2 = 2x2
    BC = cm
    Area of triangle ABC =× AD × BC
    x =× AD ×
    AD = cm

    In triangle ADE,
    Area of triangle ADE =× AD × AE
    =× × AE

    AE = cm
    DE2 = AD2 + AE2 = ()2 + ()2
    DE2 = 2 + 7
    DE2 = 9
    DE = 3 cm
  • Question 22
    1 / -0
    For each of the following statements, mention T for true and F for false.

    i. Sides of two similar triangles are in the ratio 2 : 1, so the areas of these triangles would be in ratio 6 : 1.
    ii. Three sides of a right angled triangle are 20 cm, 99 cm and 102 cm.
    iii. In triangle ABC, AB = 3 cm, BC = 4 cm, AC = 5 cm and ∠ACB = 90°.
    iv. If two sides of two triangles are proportional, then the third side is also proportional and the triangles are similar.
    Solution
    i. If the sides of two similar triangles are in ratio 2 : 1, then the ratio of their areas would be 22 : 12 i.e 4 : 1.
    Hence, i is false.

    ii. If the triangle is a right angled triangle, then,
    1022 must be equal to 992 + 202
    1022 = 10404
    992 + 202 = 9801 + 400 = 10201
    Hence, ii is false

    iii. Sides of the triangle are 3 cm, 4 cm and 5 cm which makes Pythagorean triplet and hence, the triangle is right angled and AC = 5 cm is the hypotenuse as it is the largest side. Hence, ∠ACB cannot be a right angle.
    Hence, iii is false.

    iv. For a fixed value of two sides of a triangle, third side of the triangle can vary depending upon the angle between the other sides. Hence, if two sides of the triangles are proportional, the third side need not to be proportional.
    Hence, iv is false.

    Hence, option 1 is correct.
  • Question 23
    1 / -0
    Match the columns:

    Column I Column II
    A
    Find DE (in cm)    		 i 7.5 cm
    B
    Area of triangle ABC = 168 cm2. Find the value of X.    		 ii 4 cm
    C
    △ABC ～ △DEF Find EF in (cm)    		 iii 7 cm
    D
    Find IJ (in cm).    		 iv 5 cm
    Solution
    A: △ABC ～△ADE







    DE = 80 ÷ 20 = 4 cm

    B:
    Area of triangle =× CD × AB
    168 =× 24 × AB
    AB = 168 ÷ 12 = 14 cm

    AC = BC
    So, D is the midpoint of AB.
    AD = 14 ÷ 2 = 7 cm
    ADC is a right angled triangle
    AC2= AD2+ DC2= 72+ 242= 49 + 576 = 625
    AC = 25 cm
    x2 = 25 cm
    x = 5 cm

    C:
    In triangle ABC, BC2= AC2– AB2
    BC2= 52– 42= 25 – 16 = 9
    BC = 3 cm
    ABC ～⊿DEF
    Hence, AC is proportional to DF.
    Proportionality factor = 12.5 ÷ 5 = 2.5
    Side EF is proportional to BC.
    EF = 3 × 2.5 = 7.5 cm

    D:
    Angles of both the triangles are 40°, 60°, 80°
    △IJK ～RPQ (∠I = ∠R = 80°, ∠J = ∠P = 40°, ∠K = ∠Q = 60°)
    Hence, IK is proportional to RQ.
    Proportionality factor = 7.2 ÷ 6 = 1.2
    IJ is proportional to RP.
    IJ = 8.4 ÷ 1.2 = 7 cm

    Hence, option 3 is correct.
  • Question 24
    1 / -0
    In triangle ABC, BD is 1 cm less than the length of AC. Find the lengths of BD and QR.

    Solution
    In triangle BAC,
    AB = CB
    Hence, D is the midpoint of AC.
    Let AD be x and BD be 2x - 1.
    AB2 = AD2 + BD2
    172 = x2 + (2x – 1)2
    289 = x2 + 4x2 + 1 – 4x
    5x2 – 4x – 288 = 0
    5x2 – 40x + 36x – 288 = 0
    5x(x – 8) + 36(x – 8) = 0
    (5x + 36) (x – 8) = 0
    x cannot be negative.
    Hence, x = AD = 8 cm
    BD = 2 × 8 – 1 = 15 cm
    AC = 16 cm
    In triangle PQR, ∠Q = 70°, ∠P = 40°, ∠R = 180° - 110° = 70°
    In triangle ABC, ∠A = 70°, ∠C = 70° (AB = BC), ∠ABC = 180° - 140° = 40°

    Hence, △PQR ～ △BAC

    PQ / AB = QR/ AC (Corresponding sides of similar triangles are proportional)

    4.25 /17 = QR/ 16

    QR = (16 × 4.25)/17 = 4 cm


  • Question 25
    1 / -0
    Which of the following statements is/are correct, according to the given triangle, where XY || PQ?



    A.
    B. △PQR ～△XYR
    C. If PQ = 8 cm, XY = 5cm and QR = 6 cm, then XR = 3.75 cm.
    D. 2(∠Q + ∠R) = 360 - (∠Y + ∠P)
    Solution
    A is not correct as
    B is not correct as △PQR ～ △YXR because ∠Y = ∠ P as PQ || XY

    C. as △PQR ～ △YXR

    1.6 = 1.6
    Hence, C is correct.

    D. ∠ P = ∠ Y ( corresponding angles)
    So, 2(∠Q + ∠R) = 360 - (∠Y + ∠P)
    2(∠Q + ∠R) = 360 - (2∠P)
    ∠Q + ∠R = 180° - ∠P
    ∠Q + ∠R + ∠P = 180°
    Hence, D is correct.
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