Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  Short Answer Questions: Triangles

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.
Class 10 Maths Chapter 6 Question Answers - Triangles

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

Class 10 Maths Chapter 6 Question Answers - Triangles
Q2: In the figure, ABC is a D such that BD ⊥ AC and CE ⊥ AB. Prove that:
 PD × BP = PC × EP

Class 10 Maths Chapter 6 Question Answers - Triangles

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,
Class 10 Maths Chapter 6 Question Answers - Triangles
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.
Class 10 Maths Chapter 6 Question Answers - Triangles

Sol: In Δ QOA and Δ POB,
        <QOA = <BOP [Vertically opposite angles]
         <QAO = <PBO [Each = 90°]
          ∴Using AA similarity, we have:
Class 10 Maths Chapter 6 Question Answers - Triangles
Q4: In the figure, ABCD is a trapezium in which AB y CD. If D BOC ~ D AOD, then prove that AD = BC.
Class 10 Maths Chapter 6 Question Answers - Triangles

Sol: We have a trapezium ABCD in which AB y DC.
Since Δ BOC   ~ Δ AOD [Given]
Class 10 Maths Chapter 6 Question Answers - Triangles (1)
In Δ ODC and ΔOBA,
      <COD = <AOB [Vertically opp. angles]
      <ODC = <OBA [Alt. angles]
∴ Using AA similarity, we have:
       Δ ODC ~ Δ OBA
Class 10 Maths Chapter 6 Question Answers - Triangles 
From (1) and (2)
= Class 10 Maths Chapter 6 Question Answers - Triangles
⇒ OB × OB = OA × OA
⇒ (OB)2 = (OA)2 
⇒  OA = OB  ...(3)
From (1) to (3), we have
Class 10 Maths Chapter 6 Question Answers - Triangles
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
           Class 10 Maths Chapter 6 Question Answers - Triangles
        Class 10 Maths Chapter 6 Question Answers - Triangles
         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
           Class 10 Maths Chapter 6 Question Answers - Triangles

 [Œ AB = 3 + 9 = 12 cm and
AC = 5 + 15 = 20 cm]
           Class 10 Maths Chapter 6 Question Answers - Triangles

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:

Class 10 Maths Chapter 6 Question Answers - Triangles

We have a rectangle PQRS such that
SO2 = PO· PQ
i.e., SO × SO = PO × PQ
Class 10 Maths Chapter 6 Question Answers - Triangles

  ...(1)
[Œ PQ = SR, opp. sides of rectangle PQRS]
Now, in Δ POS and Δ OSR, we have:
Class 10 Maths Chapter 6 Question Answers - Triangles
<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.
Class 10 Maths Chapter 6 Question Answers - Triangles

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
=   Class 10 Maths Chapter 6 Question Answers - Triangles 
Now in right Δ ABD, we have
AB2 = AD2 + BD2
⇒ AD2 = AB2 - BD2
= (AB + BD) (AB - BD) Class 10 Maths Chapter 6 Question Answers - TrianglesClass 10 Maths Chapter 6 Question Answers - Triangles
Q8: In an equilateral triangle with side ‘a’, prove that its area  Class 10 Maths Chapter 6 Question Answers - Triangles.
Class 10 Maths Chapter 6 Question Answers - Triangles

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)

Class 10 Maths Chapter 6 Question Answers - TrianglesNow, area of Δ  ABC =  1/2× Base × altitude
Class 10 Maths Chapter 6 Question Answers - TrianglesThus, the area of an equilateral triangle Class 10 Maths Chapter 6 Question Answers - Triangles

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.
Class 10 Maths Chapter 6 Question Answers - Triangles
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 = Class 10 Maths Chapter 6 Question Answers - Triangles  ...(1)
Now, in right Δ ADB, we have:
    AB2 = AD2 + BD2 [Using Pythagoras Theorem]
Class 10 Maths Chapter 6 Question Answers - TrianglesClass 10 Maths Chapter 6 Question Answers - Triangles
⇒ 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:
 =
Class 10 Maths Chapter 6 Question Answers - Triangles
Class 10 Maths Chapter 6 Question Answers - Triangles 

Sol: We have a right Δ ABC such that <C = 90°.
      Also, CD ⊥ AB
     Now, ar (Δ ABC) =1/2 × Base × Height Class 10 Maths Chapter 6 Question Answers - TrianglesAlso, Class 10 Maths Chapter 6 Question Answers - Triangles   
Class 10 Maths Chapter 6 Question Answers - Triangles  ... (2)
From (1) and (2), we have
= Class 10 Maths Chapter 6 Question Answers - Triangles
Dividing throughout by abp, we have:
 Class 10 Maths Chapter 6 Question Answers - Triangles
Squaring both sides,
= Class 10 Maths Chapter 6 Question Answers - Triangles     ...(3)
Now, In right D ABC,
     AB2 = AC2 + BC2 
⇒ c2 = b2 + a2         ...(4)
∴ From (3) and (4), we get
= Class 10 Maths Chapter 6 Question Answers - TrianglesClass 10 Maths Chapter 6 Question Answers - Triangles


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

Class 10 Maths Chapter 6 Question Answers - Triangles

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. 
Class 10 Maths Chapter 6 Question Answers - Triangles

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 + CF[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
          ⇒ Class 10 Maths Chapter 6 Question Answers - Triangles
             Thus OC = 5√3cm.

Q13: In the figure, if AD ⊥ Bc,  then prove that:
 AB
2 + CD2 = AC2 + BD2  
Class 10 Maths Chapter 6 Question Answers - Triangles

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:
Class 10 Maths Chapter 6 Question Answers - Triangles

Class 10 Maths Chapter 6 Question Answers - Triangles

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
Class 10 Maths Chapter 6 Question Answers - Triangles
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.
Class 10 Maths Chapter 6 Question Answers - Triangles

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:
Class 10 Maths Chapter 6 Question Answers - Triangles

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.
Class 10 Maths Chapter 6 Question Answers - Triangles

Sol: In Δ ABC and Δ ADE, we have:
            <A = <A [Common]
            <C = <E [Each = 90°]
∴ Δ ABC ~ Δ ADE [AA Similarity]
Class 10 Maths Chapter 6 Question Answers - Triangles  ...(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
Class 10 Maths Chapter 6 Question Answers - Triangles   

⇒ DE = = 2.77 cm and   Class 10 Maths Chapter 6 Question Answers - Triangles 

Class 10 Maths Chapter 6 Question Answers - Triangles


Q17: In the given figure, DEFG is a square and <BAC = 90°. Show that DE2 = BD × EC.
Class 10 Maths Chapter 6 Question Answers - Triangles

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:
= Class 10 Maths Chapter 6 Question Answers - Triangles

⇒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
Class 10 Maths Chapter 6 Question Answers - Triangles

Sol: Œ BD = 1/3 CD
       ⇒ 3 BD = CD
             ∴BC = BD + DC
          ⇒ BC = BD + 3 BD
          ⇒ BC = 4 BD ...(1)
And Class 10 Maths Chapter 6 Question Answers - TrianglesFrom (2) ...(4)

In right Δ ADC, Using Pythagoras theorem,
     CA2 = AD2 + DC
= Class 10 Maths Chapter 6 Question Answers - TrianglesFrom (3) ...(4)

In right Δ ADB, Using Pythagoras theorem,
     AD2 = AB2 - BD2
= Class 10 Maths Chapter 6 Question Answers - Triangles

Class 10 Maths Chapter 6 Question Answers - Triangles

Class 10 Maths Chapter 6 Question Answers - Triangles

Q19: If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.
Class 10 Maths Chapter 6 Question Answers - Triangles

Sol: We have a quadrilateral ABCD such that its diagonals intersect at O and
Class 10 Maths Chapter 6 Question Answers - Triangles

    <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°
Class 10 Maths Chapter 6 Question Answers - Triangles

           <1 = <2 [Vertically opp. angles]
∴ Using AA similarity, we have:
           Δ ABE ~ Δ DCE
⇒ Their corresponding sides are proportional.
Class 10 Maths Chapter 6 Question Answers - Triangles

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
Class 10 Maths Chapter 6 Question Answers - Triangles


Q22: In Δ ABC, if AD is the median, then show that AB2 + AC2 = 2 [AD2 + BD2]. 
Class 10 Maths Chapter 6 Question Answers - Triangles

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 + BD[BD = CD]
    = 2 [AD]2 + 2BD[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 AD- 3 AB2
 Sol:
D is the mid-point of BC.
            ∴ BC = 2 BD
Class 10 Maths Chapter 6 Question Answers - Triangles 

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

The document Class 10 Maths Chapter 6 Question Answers - Triangles is a part of the Class 10 Course Mathematics (Maths) Class 10.
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FAQs on Class 10 Maths Chapter 6 Question Answers - Triangles

1. What are the different types of triangles?
Ans. There are three types of triangles based on their sides: equilateral triangles (all sides are equal), isosceles triangles (two sides are equal), and scalene triangles (all sides are different).
2. How can we determine the type of triangle if only the angles are given?
Ans. If the angles of a triangle are all less than 90 degrees, it is an acute triangle. If one angle is exactly 90 degrees, it is a right triangle. And if one angle is greater than 90 degrees, it is an obtuse triangle.
3. How can we find the area of a triangle?
Ans. The area of a triangle can be found using the formula: Area = (base * height) / 2. The base and height are perpendicular to each other and the base can be any side of the triangle.
4. Can we determine the missing angle in a triangle if the other two angles are known?
Ans. Yes, the sum of the three angles in a triangle is always 180 degrees. So, if two angles are known, the missing angle can be found by subtracting the sum of the known angles from 180 degrees.
5. Can we determine the length of a side in a triangle if the angles and one side length are known?
Ans. Yes, using trigonometric ratios such as sine, cosine, and tangent, we can find the length of a side in a triangle if the angles and one side length are known. These ratios relate the angles of a triangle to the lengths of its sides.
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