Class 10 Maths Chapter 6 Question Answers - Triangles

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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.

Sol: Since ABCD is a trapezium,
∴ AB y cΔ ⇒ <1 = <2 and <3 = <4 (Alternate angle)
∴  Δ APB ~ Δ CPD
⇒ ABCD = APCP
But CP : AP = 1 : 2 (Given)
∴ APCP = 21
⇒ ABCD = 21 ⇒ AB = 2 × CD

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,
BPPC = EPPD
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.

Sol: In Δ QOA and Δ POB,
<QOA = <BOP [Vertically opposite angles]
<QAO = <PBO [Each = 90°]
∴Using AA similarity, we have:
ar (Δ QOA)ar (Δ POB) = (OQ)(OP)²
ar (Δ QOA)250 = (8)(5)²
ar (Δ QOA)250 = 6425
ar (Δ QOA) = 250 × 6425 = 640 cm²

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]
OBOA = OCOD = BCAD (1)
In Δ ODC and Δ OBA,
∠COD = ∠AOB [Vertically opp. angles]
∠ODC = ∠OBA [Alt. angles]
∴ Using AA similarity, we have:
Δ ODC ~ Δ OBA
ODOB = OCOA ⇒ ODOC = OBOA ...(2)
From (1) and (2):
OBOA = OBOA
⇒ OB × OB = OA × OA
⇒ (OB)² = (OA)²
⇒ OA = OB   ...(3)
From (1) to (3), we have:
ADBC = 11
AD = BC

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
APPB = AQQC = 39 = 515 ⇒ 13
i.e., PQ divides AB and AC in the same ratio
∴ PQ || BC
Now, in Δ APQ and Δ ABC:
∠P = ∠B [Corresponding angles]
∠A = ∠A [Common]
∴ Using AA similarity,
Δ APQ ~ Δ ABC
APAB = AQAC = PQBC
312 = 520 = PQBC [∵ AB = 3 + 9 = 12 cm and AC = 5 + 15 = 20 cm]
14 = PQBC
BC = 4PQ

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
SOPO = PQSO
SOPO = SRSO ...(1) [∵ PQ = SR, opp. sides of rectangle PQRS]
Now, in Δ POS and Δ OSR, we have:
POSO = SOSR       [∵ SOPO = SRSO in (1)]
<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
ADAB = ACAB = DBDC
⇒ DBDC = 1 ⇒ DB = DC = 12a
Now in right Δ ABD, we have:
AB² = AD² + BD²
⇒ AD² = AB² − BD²
⇒ AD² = (AB + BD)(AB − BD)
= (2a + a2) × (2a − a2)
= 5a2 × 3a2 = 154a²
AD = √154a² = a2√15 cm

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)
AD = (a + a2) (a − a2)
= 32 a × 12 a = 34 a²
AD = √34 a² = √32 a
Now, area of Δ ABC = 1/2 × Base × Altitude
= 12 × BC × AD
= 12 × a × √32 a
= √34 a² square units
Thus, the area of an equilateral triangle = √34 a².

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 = 12 BC ...(1)
Now, in right Δ ADB, we have:
AB² = AD² + BD²   [Using Pythagoras Theorem]
AB² = AD² + [12 BC]²
[∵ BD = 12 BC, from (1)]
AB² = AD² + 14 BC²
⇒ AB² − 14 BC² = AD²
4 AB² − AB²4 = AD²
⇒ 34 AB² = AD²
⇒ 3 AB² = 4 AD²
⇒ 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:
 =1p² = 1a² + 1b².
 Sol: We have a right Δ ABC such that ∠C = 90°.
Also, CD ⊥ AB
Now, ar (Δ ABC) = 12 × Base × Height
= 12 × BC × AC
= 12 × a × b ...(1)
Also, ar (Δ ABC) = 12 × AB × CD ... (2)
From (1) and (2), we have:
12 ab = 12 cp
⇒ ab = cp
Dividing throughout by abp, we have:
ababp = cpabp
1p = cab
Squaring both sides:
1p² = c²a²b² ...(3)
Now, in right Δ ABC,
AB² = AC² + BC²
⇒ c² = b² + a²   ...(4)
∴ From (3) and (4), we get:
1p² = b² + a²a²b²
= b²a²b² + a²a²b²
= 1a² + 1b²
Thus, 1p² = 1a² + 1b².

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
⇒ OC = √75 − √(25 × 3) − √(3 × 5)
OC = 5√3 cm
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:
CA² = AB² + 12 BC²

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
= AB² − (14 BC)² + (34 BC)²
= AB² − BC²16 + 9 BC²16
= AB² + 816 BC²
= AB² + 12 BC²
Thus, CA² = AB² + 12 BC²

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 [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:
ELBL = AEBC = AD + DEBC
= BC + BCBC [From (1)]
= 2 BCBC − 2
⇒ ELBL = 2 ⇒ EL = 2 BL

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]

ABAD = BCDE = ACAE ...(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:
133 = 12DE = 5AE
⇒ DE12 = 12 × 313 = 2.77 cm
And
AE5 = 5 × 313 = 1.15 cm
⇒ DE = 2.77 cm and AE = 1.15 cm

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:
= BDDG = EFEC
⇒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 BD = 14 BC    and    DC = 3 BD = 34 BC ...(4)
In right Δ ADC, Using Pythagoras theorem,
CA² = AD² + DC²
AD² + (34 BC)² [From (4)]
In right Δ ADB, Using Pythagoras theorem,
AD² = AB² − BD²
= AB² − [14 BC]²
= AB² − BC²16
CA² = (AB² − BC²16) + (34 BC)²
= AB² − BC²16 + 916 BC²
= AB² + 12 BC²
CA² = 2 AB² + BC²

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
OAOC = OBOD
∠AOB = ∠COD
<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.
⇒ AEED = BEEC
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

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