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Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

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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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

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

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

Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10
⇒ 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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

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

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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

Sol. We have a trapezium ABCD in which AB y DC.
Since Δ BOC   ~ Δ AOD [Given]
Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10 .(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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10  ...(2)
From (1) and (2)
= Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10
⇒ OB × OB = OA × OA
⇒ OB2 = OA2  ⇒  OA = OB  ...(3)
From (1) to (3) we have
Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10
Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10
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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10
⇒ =
[ΠAB = 3 + 9 = 12 cm and
AC = 5 + 15 = 20 cm]
Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

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

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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10
⇒   Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10 
Now in right Δ ABD, we have
AB2 = AD2 + BD2
⇒ AD2 = AB2 - BD2
= (AB + BD) (AB - BD)
= Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10
Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

Q8. In an equilateral triangle with side ‘a’, prove that its area  Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10.
Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

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

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

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


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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10Thus OC = 5√3cm.
Q13. In the figure, if AD ⊥ Bc,  then prove that:
 AB
2 + CD2 = AC2 + BD2  
Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10Sol. 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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10Q15. 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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10Q16. 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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10Sol.In Δ ABC and Δ ADE, we have:<A = <A [Common]
<C = <E [Each = 90°]
∴ Δ ABC ~ Δ ADE [AA Similarity]
Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10  ...(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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10   ⇒ DE = = 2.77 cmAnd = Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10 ⇒ AE Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10


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

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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10BD × 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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

Sol. ΠBD = 1/3 CD
⇒ 3 BD = CD
∴BC = BD + DC
⇒ BC = BD + 3 BD
⇒ BC = 4 BD ...(1)
And Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10From (2) ...(4)In right Δ ADC,
Using Pythagoras theorem,
CA2 = AD2 + DC2
Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10From (3) ...(4)In right Δ ADB
Using Pythagoras theorem,
AD2 = AB2 - BD2
Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10Q19.If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.
Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

Sol. We have a quadrilateral ABCD such that its diagonals intersect at O and
Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10<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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10

<1 = <2 [Vertically opp. angles]
∴ Using AA similarity, we have:
Δ ABE ~ Δ DCE
⇒ Their corresponding sides are proportional.
⇒ Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10⇒ 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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10


Q22. In Δ ABC, if AD is the median, then show that AB2 + AC2 = 2 [AD2 + BD2]. 
Short Answer Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10Sol. 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 Questions: Triangles Notes | Study Mathematics (Maths) Class 10 - Class 10 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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