Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

Class 10 Mathematics by VP Classes

Class 10 : Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

The document Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
All you need of Class 10 at this link: Class 10

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.
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

Sol. Since ABCD is a trapezium,
∴ AB y cΔ ⇒ <1 = <2 and <3 = <4 (Alternate angle)
∴  Δ APB ~ Δ CPD
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

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

Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

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,
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
⇒ BP × PD = EP × PC
[By cross multiplication]

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.
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

Sol. In Δ QOA and Δ POB,
<QOA = <BOP
[Vertically opposite angles]
<QAO = <PBO [Each = 90°]
∴Using AA similarity, we have:
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

Q4. In the figure, ABCD is a trapezium in which AB y CD. If D BOC ~ D AOD, then prove that AD = BC.
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

Sol. We have a trapezium ABCD in which AB y DC.
Since Δ BOC   ~ Δ AOD [Given]
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev .(1)
In Δ ODC and ΔOBA,
<COD = <AOB
[Vertically opp. angles]
  <ODC = <OBA [Alt. angles]
∴ Using AA similarity, we have:
Δ ODC ~ Δ OBA
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev  ...(2)
From (1) and (2)
= Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
⇒ OB × OB = OA × OA
⇒ OB2 = OA2  ⇒  OA = OB  ...(3)
From (1) to (3) we have
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

Q5. P and Q are points on 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
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
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
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
⇒ =
[ΠAB = 3 + 9 = 12 cm and
AC = 5 + 15 = 20 cm]
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

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.

Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

We have a rectangle PQRS such that
SO2 = PO· PQ
i.e., SO × SO = PO × PQ
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
⇒ =  ...(1)
[ΠPQ = SR, opp. sides of rectangle PQRS]
Now, in Δ POS and Δ OSR, we have:
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
<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.
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

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
=   Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
⇒   Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev 
Now in right Δ ABD, we have
AB2 = AD2 + BD2
⇒ AD2 = AB2 - BD2
= (AB + BD) (AB - BD)
= Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

Q8. In an equilateral triangle with side ‘a’, prove that its area  Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev.
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

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)
= Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
Now, area of Δ  ABC =  1/2× Base × altitude
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
Thus, the area of an equilateral triangle
=  Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

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.
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
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 = Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev  ...(1)
Now, in right Δ ADB, we have:
AB2 = AD2 + BD2 [Using Pythagoras Theorem]
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
⇒ 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:
 =
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev 

Sol. We have a right Δ ABC such that <C = 90°.
Also, CD ⊥ AB
Now, ar (Δ ABC) =1/2 × Base × Height
= Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev   ...(1)
Also, Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev   
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev . .. (2)
From (1) and (2), we have
= Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
Dividing throughout by abp, we have:
= Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
Squaring both sides,
= Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev  =       ...(3)
Now, In right D ABC,
AB2 = AC2 + BC2 
⇒ c2 = b2 + a2         ...(4)
∴ From (3) and (4), we get
= Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRevShort Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev


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

Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

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. 
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

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
[ΠBF = 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
⇒ Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRevThus OC = 5√3cm.
Q13. In the figure, if AD ⊥ Bc,  then prove that:
 AB
2 + CD2 = AC2 + BD2  
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

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:
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRevSol. 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 right Δ ADB
AD2 = AB2 - BD2  ...(2)
From (1) and (2),
CA2 = AB2 - BD2+ CD2
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRevQ15. In the given figure, M is the mid-point of side CD of parallelogram ABCD. The line BM is drawn intersecting AC at L, and AD produce D at E. Prove that EL = 2 BL.
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

Sol. We have parallelogram ABCD in which M is the mid point of CD.
In Δ EMD and Δ BMC
MD = MC [ΠM 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:

Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRevQ16. 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.
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRevSol.In Δ ABC and Δ ADE, we have:<A = <A [Common]
<C = <E [Each = 90°]
∴ Δ ABC ~ Δ ADE [AA Similarity]
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev  ...(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
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev   ⇒ DE = = 2.77 cmAnd = Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev ⇒ AE Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev


Q17. In the given figure, DEFG is a square and <BAC = 90°. Show that DE2 = BD × EC.
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

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:
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRevBD × EC = EF × DGBut 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
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

Sol. ΠBD = 1/3 CD
⇒ 3 BD = CD
∴BC = BD + DC
⇒ BC = BD + 3 BD
⇒ BC = 4 BD ...(1)
And Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRevFrom (2) ...(4)In right Δ ADC,
Using Pythagoras theorem,
CA2 = AD2 + DC2
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRevFrom (3) ...(4)In right Δ ADB
Using Pythagoras theorem,
AD2 = AB2 - BD2
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRevShort Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRevShort Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRevQ19.If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

Sol. We have a quadrilateral ABCD such that its diagonals intersect at O and
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev<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°
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

<1 = <2 [Vertically opp. angles]
∴ Using AA similarity, we have:
Δ ABE ~ Δ DCE
⇒ Their corresponding sides are proportional.
⇒ Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev⇒ AE· EC = BE· ED
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
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev


Q22. In Δ ABC, if AD is the median, then show that AB2 + AC2 = 2 [AD2 + BD2]. 
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRevSol. 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 mid point of BC. Prove that:
 AC2 = 4 AD- 3 AB2
 Sol.
D is the mid-point of BC.
∴ BC = 2 BD
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev Now, in Δ ABCAC2 = BC2 + AB2
= (2 BD)2 + AB2
= 4 BD2 + AB2 ...(1)
In 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



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