# 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)
[Œ PQ = 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.
<B = <C [Opp. angles to equal sides of a D]
=
Now in right Δ ABD, we have
⇒ 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.
AC = AB [Each = a]
∴DC = DB
⇒ 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.
AB = AB [Given]
Using RHS congruency, we have
⇒ 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 + 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
⇒
Thus OC = 5√3cm.

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

∴ 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: Œ BD = 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,
= From (3) ...(4)

In right Δ ADB, Using Pythagoras theorem,
=

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].

∴  BD = DC
Let us draw AE ≌ BC
Now, in rt. Δs AEB and AEC, we have
AB2 = BE2 + AE2 ...(1)
AC2 = CE2 + AE2 ...(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]
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

Now, in Δ ABC, AC2 = BC2 + AB2
= (2 BD)2 + AB2
= 4 BD2 + AB2 ...(1)
In the right Δ ABD,
Using Pythagoras theorem,
⇒ 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.
All you need of Class 10 at this link: Class 10

## 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.

## Mathematics (Maths) Class 10

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