Class 10 Maths Chapter 6 Question Answers - Triangles

Q1: If one diagonal of a trapezium divides the other diagonal in the ratio of 1: 2. Prove that one of the parallel sides is double the other.

Sol: Since ABCD is a trapezium,
      ∴ AB y cΔ ⇒ <1 = <2 and <3 = <4 (Alternate angle)
      ∴  Δ APB ~ Δ CPD

Q2: In the figure, ABC is a D such that BD ⊥ AC and CE ⊥ AB. Prove that:
 PD × BP = PC × EP

Sol: In Δ BEP and Δ CPD,

 we have: <BPE = <CPD [Vertically opp. angles]
                     <BEP = <CDP [Each = 90°]
            ∴Using AA similarity, we have
                      Δ BEP ~ Δ CDP
          ∴Their corresponding sides are proportional,

Q3: AB is a line segment. PB ⊥ AB and QA ⊥ AB are such that PO = 5 cm and QO = 8 cm. If ar (D POB) = 250 cm2, then find the area of D QOA.

Sol: In Δ QOA and Δ POB,
        <QOA = <BOP [Vertically opposite angles]
         <QAO = <PBO [Each = 90°]
          ∴Using AA similarity, we have:

Q4: In the figure, ABCD is a trapezium in which AB y CD. If D BOC ~ D AOD, then prove that AD = BC.

Sol: We have a trapezium ABCD in which AB y DC.
Since Δ BOC   ~ Δ AOD [Given]
 (1)
In Δ ODC and ΔOBA,
      <COD = <AOB [Vertically opp. angles]
      <ODC = <OBA [Alt. angles]
∴ Using AA similarity, we have:
       Δ ODC ~ Δ OBA
 
From (1) and (2)
=
⇒ OB × OB = OA × OA
⇒ (OB)² = (OA)² 
⇒  OA = OB  ...(3)
From (1) to (3), we have

Q5: P and Q are points on the sides of AB and AC respectively of Δ ABC. If AP = 3 cm, PB= 9 cm, AQ = 5 cm, and QC = 15 cm, then show that BC = 4 PQ.

Sol: We have Δ ABC in which P and Q are such that
           AP = 3 cm,  PB = 9 cm
            AQ = 5 cm,  QC = 15 cm
           
       
         i.e., PQ divides AB and AC in the same ratio
         ∴ PQ y BC
         Now, in Δ APQ and Δ ABC
                <P = <B [Corresponding angles]
                <A = <A [Common]
  ⇒  Using AA similarity,
           Δ APQ ~ Δ ABC
           

 [Œ AB = 3 + 9 = 12 cm and
AC = 5 + 15 = 20 cm]
           

Q6: On one of the longer sides PQ of a rectangle PQRS, a point O is taken such that
 SO² = PO· PQ
 Prove that: Δ POS ~ Δ OSR.
 Sol:

We have a rectangle PQRS such that
SO² = PO· PQ
i.e., SO × SO = PO × PQ

  ...(1)
[ΠPQ = SR, opp. sides of rectangle PQRS]
Now, in Δ POS and Δ OSR, we have:

<1 = <2 [OE PQ y SR,
opp. sides of a rectangle]
⇒ Using SAS similarity, we have
Δ POS ~ Δ OSR

Q7: Determine the length of the altitude AD of an isosceles D ABC in which AB = AC = 2a cm and BC = a cm.

Sol: We have Δ ABC in AD ⊥ BC and AB = AC = 2a. Also BC = a.
          In Δ  ADB and Δ  ADC
               <ADB = <ADC   [Each = 90°]
              <B = <C [Opp. angles to equal sides of a D]
       ∴Δ  ADB ~ Δ  ADC
=    
Now in right Δ ABD, we have
AB² = AD² + BD²
⇒ AD² = AB² - BD²
= (AB + BD) (AB - BD) 
Q8: In an equilateral triangle with side ‘a’, prove that its area  .

Sol: We have Δ ABC such that
          AB = BC = AC = a
Let us draw altitude AD ⊥ BC.
In Δ ADC and Δ ADB,
         AD = AD [Common]
       AC = AB [Each = a]
     <ADC = <ADB [Each = 90°]
     ∴Δ ADC ≌ Δ ADB [RHS congruency]
     ∴DC = DB
Now, in right Δ  ADB,
      AB² = AD² + DB²
⇒ AD² = AB² - DB²
= (AB + DB) (AB - DB)

Now, area of Δ  ABC =  1/2× Base × altitude
Thus, the area of an equilateral triangle 

Q9: In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Sol: We have Δ ABC in which AB = AC = CA and an altitude AD ⊥ BC.
In Δ ADB and Δ ADC
        <ADB = <ADC [Each = 90°]
           AB = AB [Given]
           AD = AD [Common]
Using RHS congruency, we have
        < ADB ≌ < ADC
        ⇒ DB = DC =   ...(1)
Now, in right Δ ADB, we have:
    AB² = AD² + BD² [Using Pythagoras Theorem]

⇒ 3 [Side of the equilateral triangle]
= 4 [Altitude]2.

Q10: ABC is a right triangle in which <C = 90° and CD ⊥ AB. If BC = a, CA = b, AB = c and CD = p, then prove that:
 =
 

Sol: We have a right Δ ABC such that <C = 90°.
      Also, CD ⊥ AB
     Now, ar (Δ ABC) =1/2 × Base × Height Also,    
  ... (2)
From (1) and (2), we have
=
Dividing throughout by abp, we have:
 
Squaring both sides,
=      ...(3)
Now, In right D ABC,
     AB² = AC² + BC² 
⇒ c² = b² + a²         ...(4)
∴ From (3) and (4), we get
=

Q11: ABC is a right triangle, right-angled at A, and D is the mid-point of AB. Prove that
 BC² = CD² + 3 BD². 

Sol: We have a right Δ ABC in which <A = 90°
      ∴Using Pythagoras Theorem, we have:
            BC² = AB² + AC²   ...(1)
        Again, Δ ACD is right D, <A = 90°
        ∴CD² = AD² + AC²...(2) [Using Pythagoras Theorem]
       Subtracting (2) from (1), we get
          BC² - CD² = AB² - AD² ...(3)
      Since D is the mid-point of AB
      ∴2 BD = AB  and  AD = BD ...(4)
      From (3) and (4), we have:
         ∴BC² - CD² = (2 BD)² - (BD)² 
     = 4 BD² - BD²
             BC² = CD² + 3 BD²

Q12: In the figure, O is any point inside a rectangle ABCD such that OB = 6 cm, OD = 8 cm, and OA = 5 cm. Find the length of OC. 

Sol: Let us draw EOF y AB ⇒ OE ⊥ AD and OF ⊥ BC
        In Δ OFB, Œ <F = 90°
∴Using Pythagoras theorem, we have:
                 CB² = OF² + BF² ...(1)
        In Δ OED, <E = 90°
∴Using Pythagoras theorem, we have:
                OD² = OE² + DE² ...(2)
   Adding (1) and (2), we get
                OB² + OD² = OF² + BF² + OE² + DE²
             = OF² + AE² + OE² + CF² [Œ BF = AE and CF = DE]
            = (OF² + CF²) + (OE² + AE²)
             = OC² + OA²
              = OC² + 52
             ⇒ 62 + 82 = OC² + 52
             ⇒ 36 + 64 = OC² + 25
            ⇒ OC² = 36 + 64 - 25 = 75
          ⇒
             Thus OC = 5√3cm.

Q13: In the figure, if AD ⊥ Bc,  then prove that:
 AB² + CD² = AC² + BD²  

Sol: In D ADC, <ADC = 90°
         ∴ AD² = AC² - CD² .....(1) (Using Pythagoras Theorem)
Similarly, in D AbD,
     ⇒ AD² = AB² - DB².....(2)
From (1) and (2), we have
        AB² - DB² = AC² - CD²
  ⇒ AB² + CD² = AC² + BD²

Q14: In the given figure, AD ⊥ BC and BD =  CD. Prove that:

Sol: BD = 1/2 CD
  ∴ 3 BD = CD
Since BD + DC = BC
        ∴ BD + 3 BD = BC
      ⇒ 4 BD = BC
      ⇒ BD = 1/4 BC
     ⇒ CD = 3/4 BC
Now, in right Δ ADC, < D = 90°
By Pythagoras theorem, we get
        CA² = AD² + CD²  ...(1)
Also in the right Δ ADB
     AD² = AB² - BD²  ...(2)
From (1) and (2),
    CA² = AB² - BD²+ CD²

Q15: In the given figure, M is the mid-point of the side CD of parallelogram ABCD. The line BM is drawn intersecting AC at L, and AD produces D at E. Prove that EL = 2 BL.

So: We have parallelogram ABCD in which M is the midpoint of CD.
      In Δ EMD and Δ BMC
          MD = MC [Œ M is mid-point of CD]
       <EMD = <CMB [Vertically opposite angles]
       <MED = <MBC [Alternate interior angles]
 ∴  Δ BMC ≌   Δ EMD     [AAS congruency]
     ⇒ BC = ED  ⇒  AD = ED    ...(1)
[Œ BC = AD,  opposite sides of parallelogram]
Now, in   Δ AEL and   Δ CBL
      <AEL = <CBL [Alternate interior angles]
     <ALE = <CLB [Vertically opposite angles]
∴ By AA similarity, we have:

Q16. In the given figure, Δ ABC is right-angled at C and DE ^ AB. Prove that Δ ABC ~ Δ ADE and hence find the length of AE and DE.

Sol: In Δ ABC and Δ ADE, we have:
            <A = <A [Common]
            <C = <E [Each = 90°]
∴ Δ ABC ~ Δ ADE [AA Similarity]
  ...(1)

In right Δ ABC, <C = 90°
Using Pythagoras theorem, we have:
      AB² = BC² + AC²
  = 122 + 52
  = 144 + 25 = 169
  ⇒AB = √169 = 13 cm
Now, from (1), we get
   

⇒ DE = = 2.77 cm and    

Q17: In the given figure, DEFG is a square and <BAC = 90°. Show that DE² = BD × EC.

Sol: In Δ DBG and Δ ECF
          <3 + <1 = 90° = <3 + <4
        ∴<3 + <1 = <3 + <4
      ⇒ <1 = <4
           <D = E = 90°
∴Using AA similarity, we have:
=

⇒BD × EC = EF × DG  But DG = EF = DE
∴BD × EC = DE × DE
⇒ BD × EC = DE²
Thus, DE² = BD × EC

Q18: In the figure, AD ⊥ BC and BD =  CD. Prove that 2 CA² = 2 AB² + BC². 

Sol: ΠBD = 1/3 CD
       ⇒ 3 BD = CD
             ∴BC = BD + DC
          ⇒ BC = BD + 3 BD
          ⇒ BC = 4 BD ...(1)
And From (2) ...(4)

In right Δ ADC, Using Pythagoras theorem,
     CA² = AD² + DC² 
= From (3) ...(4)

In right Δ ADB, Using Pythagoras theorem,
     AD² = AB² - BD²
=

Q19: If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.

Sol: We have a quadrilateral ABCD such that its diagonals intersect at O and

    <AOB = <COD [Vertically opposite angles]
∴ Using SAS similarity, we have
      Δ AOB ~ Δ COD
⇒ Their corresponding angles are equal i.e., <1 = <2
     But they form a pair of int. alt. angles.
⇒ AB y DC
⇒ ABCD is a trapezium.

Q20: Two triangles ABC and DBC are on the same base BC and on the same side of BC in which
 <A = <D = 90°. If CA and BD meet each other at E, show that
 AE· EC = BE· ED
 Sol: We have right Δ ABC and right Δ DBC on the same base BC such that
            <A = <D = 90°
         In Δ ABE and Δ DCE
             <A = <D = 90°

           <1 = <2 [Vertically opp. angles]
∴ Using AA similarity, we have:
           Δ ABE ~ Δ DCE
⇒ Their corresponding sides are proportional.
⇒ 

Q21: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that
 Δ ABE ~ Δ CFB
 Sol: We have parallelogram ABCD
    In Δ ABE and Δ CFB, we have
                 <A = <C [Opposite angles of parallelogram]
                 <AEB = <EBC [Alternate angles, AD y BC]
∴ Using AA similarity, we get
                Δ ABE ~ Δ CFB

Q22: In Δ ABC, if AD is the median, then show that AB² + AC² = 2 [AD² + BD²]. 

Sol:  AD is a median,
         ∴  BD = DC
Let us draw AE ≌ BC
Now, in rt. Δs AEB and AEC, we have
          AB² = BE² + AE² ...(1)
         AC² = CE² + AE² ...(2)
Adding (1) and (2),
         AB² + AC² = BE² + AE² + CE² + AE²
    = (BD - ED)2 + AE2 + (CD + DE)2 + AE2
    = 2AE² + 2ED² + BD2 + CD²
    = 2 [AE² + ED²] + BD² + BD² [BD = CD]
    = 2 [AD]2 + 2BD² [AE² + ED² = AD²]
   = 2 [AD² + BD²]
Thus,  AB² + AC² = 2 [AD² + BD²]

Q23: Triangle ABC is right-angled at B and D is the midpoint of BC. Prove that:
 AC² = 4 AD² - 3 AB²
 Sol: D is the mid-point of BC.
            ∴ BC = 2 BD
 

Now, in Δ ABC, AC² = BC² + AB²
                                              = (2 BD)² + AB²
                                              = 4 BD² + AB² ...(1)
In the right Δ ABD,
              Using Pythagoras theorem,
               AD² = AB² + BD²
         ⇒ BD² = AD² - AB² ...(2)
From (1) and (2), we get
              AC² = 4 [AD² - AB²] + AB²
         ⇒ AC² = - 4 AB² + 4 AD² + AB²
          ⇒ AC² = - 3 AB² + 4 AD²
          or AC² = 4 AD² - 3 AB²