## RS Aggarwal Solutions Class 10 Chapter 4 Triangles

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles

**Exercise 4A**

**Question 1:**

(i) In ∆ABC, DE || BC, AD = 3.6 cm, AB = 10 cm, AE = 4.5 cm

Hence, AC = 12.5 cm and EC = 8cm

(ii) In ∆ABC, DE || BC, AB = 13.3 cm, AC = 11.9 cm and EC = 5.1 cm

Hence, AD = 7.7 cm

(iii) In ∆ABC, DE || BC, AC = 6.6 cm,

Hence, AE = 2.4 cm

(iv) In ∆ABC, DE || BC, Given , EC=3.5cm

Hence AE = 4 cm

**Read More:**

- Angle Sum Property of a Triangle
- Median and Altitude of a Triangle
- The Angle of An Isosceles Triangle
- Areas of Two Similar Triangles
- Area of A Triangle
- To Prove Triangles Are Congruent
- Criteria For Similarity of Triangles
- Construction of an Equilateral Triangle

**Question 2:**

(i) D and E are points on the sides AB and AC respectively of a ∆ABC such that DE || BC, AD = x cm, DB = (x – 2) cmm, AE = (x + 2) cm, EC = (x – 1) cm

(ii) In ∆ABC, DE || BC, AD = 4 cm, DB = (x – 4) cm, AE = 8 cm, EC = (3x – 19) cm

Hence, x = 11

(iii) In ∆ABC, DE || BC, AD = (7x – 4) cm, AE = (5x – 2) cm, DB = (3x + 4)cm, EC = 3x cm

**More Resources**

**Question 3:**

Given: A ∆ABC, in which D and E are points on the sides AB and AC respectively.

To prove: DE ||BC

Proof:

(i) AD = 5.7 cm, DB = 9.5 cm, AE = 4.8 cm and EC = 8 cm

Since D and E are the points on AB and AC respectively.

Hence, by the converse of Thales theorem DE || BC

(ii) AB = 11.7 cm, AC = 11.2 cm, BD = 6.5 cm, AE = 4.2 cm

Since D and E are points on AB and AC respectively.

Hence, by the converse of Thales theorem DE is not parallel to BC.

(iii) AB = 10.8 cm, AD = 6.3 cm, AC = 9.6 cm, EC = 4 cm

Since D and E are the points on AB and AC respectively.

Therefore, (each is equal to 1.4)

Hence by the converse of Thales theorem DE || BC

(iv) AD = 7.2 cm, AE = 6.4 cm, AB = 12 cm, AC = 10 cm

Since D and E are points on the side AB and AC respectively.

Hence, by the converse of Thales theorem DB is not parallel to BC

**Question 4:**

(i) AB = 6.4 cm, AC = 8 cm, BD = 5.6 cm

Let BC = x

Now, DC = (BC – BD)

= (x – 5.6) cm

In ∆ABC, AD is the base for of A

So, by the angle bisector theorem, We have

Hence, BC = 12.6 cm and DC = (12.6 – 5.6) cm = 7 cm

(ii) AB = 10 cm, AC = 14 cm, BC = 6cm

Let BD = x,

DC = (BC – BD) = (6 – x) cm

In ∆ABC, AD is the bisector of A

So, By angle bisector theorem,

Hence, BD = 2.5 cm and DC = (6 – 2.5) cm = 3.5 cm

(iii) AB = 5.6 cm, BD = 3.2 cm and BC = 6 cm

DC = BC – BD = (6 – 3.2) cm = 2.8 cm

Let AC = x,

In ∆ABC, AD is the base for of A

So, by the angle bisector theorem we have

Hence, AC = 4.9 cm

(iv) AB = 5.6 cm, AC = 4 cm, DC = 3 cm

Let BD = x,

In ∆ABC, AD is the base for of A

So, by the angle bisector theorem we have

Hence, BD = 4.2 cm

So BC = BD + AC = (4.2 + 3) cm

BC = 7.2 cm

**Question 5:**

∆ABC and ∆PQR are similar triangles, therefore corresponding sides of both the triangles are proportional.

Hence, AB = 16 cm

**Question 6:**

∆ABC and ∆DEF are two similar triangles, therefore corresponding sides of both the triangles are proportional.

Hence,

Let perimeter of ∆ABC = x cm

Hence, perimeter of ∆ABC = 35 cm

**Question 7:**

We know that CD || AB in trap ABCD and its diagonals intersect at O.

Since the diagonals of a trapezium divides each other proportionally therefore, we have

**Question 8:**

Given ABC, the bisector of B meets AC at D, line PQ || AC meets AB, BC and BD at P, Q, R respectively.

To Prove : PR x BQ = QR x BP

Proof: In ∆BQP,

BR is the bisector of B

Therefore, by Basic proportionality theorem

**Question 9:**

Let ABCD be the trapezium and let E and F be the midpoints of AD and BC respectively.

Const: Produce AD and BC to meet at P

In ∆PAB, DC || AB

**Question 10:**

Given: ∆ABC and ∆DBC lie on the same side of BC. P is a point on BC, PQ || AB and PR || BD are drawn meeting AC at Q and CD at R respectively.

To Prove: QR || AD

Proof: In ∆ABC

Hence, in ∆ACD, Q and R the points in AC and CD such that

QR || AD (by the converse of Thales theorem)

Hence proved.

**Question 11:**

Given BD = CD and OD = DX

Join BX and CX

Thus, the diagonals of quad OBXC bisect each other

OBXC is a parallelogram

BX || CF and so, OF || BX

Similarly, CX || OE

In ∆ABX, OF || BX

**Question 12:**

Given: ABCD is a parallelogram in which P is the midpoint of DC and Q is a point on AC such that . PQ produced meets BC at R.

To prove: R is the midpoint of BC

Construction: Join BD

Proof: Since the diagonals of a || gm bisect each other at S such that

Q is the midpoint of CS

So, PQ || DS.

Therefore, QR || SB.

In ∆CSB, Q is the midpoint of CS and QR || SB.

So R is the midpoint of BC.

**Question 13:**

Given: ABCD is a quadrilateral in which AD = BC. P, Q , R, S are the midpoints of AB, AC, CD and BD.

To prove: PQRS is a rhombus

Proof: In ∆ABC,

Since P and Q are mid points of AB and AC

Therefore, PQ || BC and (Mid-point theorem)

Similarly,

SP || RQ and PQ || SR and PQ = RQ = SP = SR

Hence, PQRS is a rhombus.

**Question 14:**

Given: ABC is a triangle in which AB = AC. D and E are points on AB and AC respectively such that AD = AE

To prove: The points B, C, E and D are concyclic.

Proof: AB = AC (given)

Quad BCEA is cyclic

Hence, the point B, C, E, D are concyclic

### Exercise 4B

**Question 1:**

**Question 2:**

∆ODC ~ ∆OBC

∠BOC = 115

∠CDO = 70

_{(i) }DOC = (180 – ∠BOC)

= (180 – 115)

= 65

(ii) ∠OCD = 180 – ∠CDO – ∠DOC

∠OCD = 180 – (70 + 65)

= 45

(iii) Now, ∆ABO ~ ∆ODC

∠AOB = COD (vert. Opp s) = 65

∠OAB = ∠OCD = 45

∠OBA = ∠ODC(alternate angles) = 70

So, ∠OAB = 45 and ∠OBA = 70

**Question 3:**

Given: ∆OAB ∆OCD

AB = 8 cm, BO = 6.4 cm, CD = 5 cm, OC = 3.5 cm

**Question 4:**

Given: ADE = B,

AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm, BC = 4.2 cm

Proof:

In ∆ADE and ∆ABC,

A = A (common)

ADE = B (given)

Therefore, ∆ADE ∆ABC (AA Criterion)

Hence, DE = 2.8 cm

**Question 5:**

Given: ∆ABC ∆PQR in such a way that perimeter of respective ∆ABC = 36 cm and ∆PQR = 24 cm and PQ= 10 cm. Then, we have to find AB, Let AB = x

We know that the ratio of perimeters of two similar triangles is the same as the ratio of their corresponding sides.

Hence the corresponding side of the second triangle is 15 cm.

**Question 6:**

Given: AB = 100 cm, BC = 125 cm, AC = 75 cm

Proof:

In ∆BAC and ∆BDA

BAC = BDA = 90^{0}

B = B (common)

BAC ∆BDA (by AA similarities)

Therefore, AD = 60 cm

**Question 7:**

Given that AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm

In ∆CBA and ∆CDB

CBA = CDB = 90^{0}

C = C (Common)

Therefore, ∆CBA ∆CDB (by AA similarities)

Hence, BC = 8.1 cm

**Question 8:**

Given that BD = 8 cm, AD = 4 cm

In ∆DBA and ∆DCB, we have

BDA = CDB = 90^{0}

DBA= DCB [each = 90^{0} – A]

∆DBA ∆DCB (by AAA similarity)

Hence, CD = 16 cm

**Question 9:**

Given: P is a point on AB.

Then, AB = AP + PB = (2 + 4) cm = 6 cm

Also Q is a point on AC.

Then, AC = AQ + QC = (3 + 6) cm = 9 cm

Thus, in ∆APQ and ∆ABC

A = A (common)

∆APQ ∆ABC (by SAS similarity)

Hence proved.

**Question 10:**

Given: ABCD is a parallelogram and E is point on BC.

Diagonal DB intersects AE at F.

To Prove: AF x FB = EF x FD

Proof: In ∆AFD and ∆EFB

AFD = EFB (vertically opposite s)

DAF = BEF (Alternate s)

Hence proved.

**Question 11:**

In the given figure: DB ⊥ AB, AC ⊥ BC and DB || AC

DBC = ACB

AB is the transversal

DBE = BAC [Alternate s]

In ∆BDE and ∆ABC

∠DEB = ∠ACB = 90^{0}

∠DBE = ∠BAC

∆DBE ~ ∆ABC [By AA similarity]

⇒

Hence proved.

**Question 12:**

Let AB be the vertical stick and let AC be its shadow.

Then, AB = 7.5 m and AC = 5 m

Let DE be the vertical tower and let DF be its shadow

Then, DF = 24 m, Let DE = x meters

Now, in ∆BAC and ∆EDF,

∆BAC ∆EDF by SAS criterion

Therefore, height of the vertical tower is 36 m.

**Question 13:**

In ∆ACP and ∆BCQ

CA = CB

∠CAB = ∠CBA

∴ ∆ACP ~ ∆BCQ

**Question 14:**

∠1 = ∠2 (given)

Also, ∠2 = ∠1

Therefore, by SAS similarity criterion ∆ACB ~ ∆DCE

### Exercise 4C

**Question 1:**

Given: ∆ACB ~ ∆DEF,

area of ∆ABC = 64cm^{2} and area of ∆DEF = 121 cm^{2}

We know that the ratio of the area of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Hence, BC = 11.2 cm

**Question 2:**

Given: ∆ABC ~ ∆PQR,

area of ∆ABC = 9cm^{2} and area of ∆PQR = 16cm^{2}.

We know that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

Hence, QR = 6 cm

**Question 3:**

Given: ∆ABC ~ ∆PQR,

area of ∆ABC = 4 area of ∆PQR.

Let area of ∆PQR = x. Then area of ∆ABC = 4x.

We know that the ratio of the areas of two similar triangle is equal to the ratio of the square of their corresponding sides.

Hence. QR = 6 cm

**Question 4:**

Given: ∆ACB ~ ∆DEF such that ar(∆ABC) = 169cm^{2 }and ar(∆DEF) = 121cm^{2}

We know that the ratio of the area of similar triangles is equal to the ratio of the square of their corresponding sides.

Hence, the longest side of smallest triangle side is 22 cm.

**Question 5:**

Given: ∆ACB ~ ∆DEF

ar(∆ABC) = 100cm^{2 } and ar(∆DEF) = 49cm^{2 }

Let AL and DM be the corresponding altitude of ABC and DEF respectively such that AL = 5 cm and let DM = x cm

We know that the ratio of the area of two similar triangles is equal to the ratio of the square of corresponding altitudes.

Therefore, the required altitude is 3.5 cm

**Question 6:**

Given: ∆ACB ~ ∆DEF

Let AL and DM be the corresponding altitudes of ∆ABC and ∆DEF respectively such that AL = 6 cm and DM = 9 cm.

We know that the ratio of squares of altitudes of two similar triangles is equal to the ratio of the corresponding areas.

Hence, ratio of their areas = 4 : 9

**Question 7:**

Given: ∆ACB ~ ∆DEF such that

ar(∆ABC) = 81cm^{2} and ar(∆DEF) = 49cm^{2}

Let AL and DM be the corresponding altitudes of ∆ABC and ∆DEF respectively, such that AL = 6.3 cm and Let DM = x cm

We know that the ratio of the area of two similar triangles is equal to the ratio of the square of corresponding altitudes:

Hence, the required altitude 4.9 cm

**Question 8:**

Given: ∆ACB ~ ∆DEF such that ar(∆ABC) = 100 cm^{2} and ar(∆DEF) = 64cm^{2}

Let AP and DQ be the corresponding medians of ∆ABC and ∆DEF respectively such that DQ = 5.6cm.

Let AP = x cm.

We know that the ratio of the areas of two similar triangle is equal be the ratio of the squares of their corresponding medians.

Hence, AP = 7 cm

**Question 9:**

Given: AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm

AB = AP + PB = (1 + 3) cm = 4 cm

AC = AQ + QC = (1.5 + 4.5) cm = 6 cm

In ∆APQ and ∆ABC, we have

∠APQ = ∠ABC (corresponding ∠s)

∠AQP = ∠ACB (corresponding ∠s)

∆APQ ~ ∆ABC [by AA similarity]

Hence proved.

**Question 10:**

Given DE || BC

DE = 3 cm and BC = 6 cm

ar(∆ADE) = 15cm^{2}

In ∆ADE and ∆ABC, we have

**Question 11:**

In ∆BAC and ∆ADC, we have

∠BAC = ∠ADC = 90^{0 }(AD ⊥ BC)

∠ACB = ∠DCA (common)

∆BAC ~ ∆ADC

Therefore, the ratio of the areas of ∆ABC and ∆ADC = 169:25

**Question 12:**

Let DE = 3x and BC = 5x

In ∆ADE and ∆ABC, we have

∠ADE = ∠ABC (corres. ∠s)

∠AED = ∠ACB (corres. ∠s)

∆ADE ~ ∆ABC (by AA similarity)

Let, ar(∆ADE) = 9x^{2} units

Then, ar(∆ABC) = 25x^{2} units

Therefore, ratio of ar(∆ADE) to the ar(trap BCED) = 9:16

**Question 13:**

In ∆ABC, D and E are midpoint of AB and AC respectively.

So, DE|| BC and

Now, in ∆ADE and ∆ABC, we have

∠ADE = ∠ABC (corres. ∠s)

∠AED = ∠ACB (corres. ∠s)

∆ADE ~ ∆ABC (by AA similarity)

Let AD = x and AB = 2x

Therefore, the ratio of the areas of ∆ADE and ∆ABC = 1:4

### Exercise 4D

**Question 1:**

For a given triangle to be a right angled, the sum of the squares of the two sides must be equal to the square of the largest side.

(i) Let a = 9cm, b = 16 cm and c = 18 cm. Then

Hence the given triangle is not right angled.

(ii) Let a = 7cm, b = 24 cm and c = 25 cm, Then

Hence, the given triangle is a right triangle.

(iii) Let a = 1.4 cm, b = 4.8 cm, and c = 5 cm

Hence, the given triangle is a right triangle

(iv) Let a = 1.6 cm, b = 3.8 cm and c = 4 cm

Hence, the given triangle is not a right triangle

(v) Let p = (a – 1) cm, q = cm and r = (a + 1)cm^{2}

Hence, the given triangle is a right triangle

**Question 2:**

Starting from A, let the man goes from A to B and from B to C, as shown in the figure.

Then,

AB = 80 m, BC = 150 m and ∠ABC = 90^{0}

From right ∆ABC, we have

By Pythagoras theorem, we have

Hence, the man is 170m north-east from the starting point.

**Question 3:**

Starting from O, let the man goes from O to A and then A to B as shown in the figure.

Then,

OA = 10 m, AB = 24 m and ∠OAB = 90^{0}

Using Pythagoras theorem:

Hence, the man is 26 m south-west from the starting position.

**Question 4:**

Let AB be the building and CB be the ladder.

Then,

AB = 12 m, CB = 13 m and ∠CAB = 90^{0}

By Pythagoras theorem, we have

Hence, the distance of the foot of the ladder from the building is 5 m.

**Question 5:**

Let AB be the wall where window is at B, CB be the ladder and AC be the distance between the foot of the ladder and wall.

Then,

AB = 20 m, AC = 15 m, and ∠CAB = 90^{0}

By Pythagoras theorem, we have

Hence, the length of ladder is 25 m.

**Question 6:**

Let AB and CD be the given vertical poles.

Then,

AB = 9 m, CD = 14 m and AC = 12 m

Const: Draw, BE || AC.

Then,

CE = AB = 9m and BE = AC = 12 m

DE = (CD – CE)

= (14 – 9)

= 5 m

In right ∆BED, we have

Hence, the distance between their tops is 13 m.

**Question 7:**

In ∆PQR, ∠QPR = 90^{0}, PQ = 24 cm, and QR = 26cm^{2 }

In ∆POR, PO = 6 cm, QR = 8cm and ∠POR = 90^{0}

In ∆POR,

In ∆PQR,

By Pythagoras theorem, we have

(sum of square of two sides equal to square of greatest side)

Hence, ∆PQR is a right triangle which is right angled at P.

**Question 8:**

Given: ∆ABC is a right angled isosceles triangle in which ∠ACB = 90^{0}

**Question 9:**

Given: ∆ABC is an isosceles triangle with AC = BC and AB^{2}=2AC^{2}

∆ABC is a right triangle right angled at C.

(by converse of Pythagoras theorem)

**Question 10:**

Given: ∆ABC is an isosceles triangle with AB = AC = 13cm^{2 }

Const: Draw altitude from A to BC (AL ⊥ BC).

Now, AL = 5 cm

In ∆ALB,

∠ALB = 90^{0}

In ∆ALC,

**Question 11:**

Given: ∆ABC in which AB = AC = 2a units and BC = a units

Const: Draw AD ⊥ BC then D is the midpoint of BC.

In ∆ABC

**Question 12:**

In an equilateral triangle all sides are equal.

Then, AB = BC = AC = 2a units

Const: Draw an altitude AD ⊥ BC

Given BC = 2a. Then, BD = a

In ∆ABD,

∠ADB = 90^{0}

**Question 13:**

∆ABC is an equilateral triangle in which all side are equal.

Therefore, AB = BC = AC = 12 cm

If BC = 12 cm

Then, BD = DC = 6 cm

In ∆ADB,

Hence the height of the triangle is cm.

**Question 14:**

Let ABCD is the given rectangle, let BD is a diagonal making a ∆ADB.

⇒ ∠BAD = 90^{0}

Using Pythagoras theorem:

Hence, length of diagonal DB is 34 cm.

**Question 15:**

Let ABCD be the given rhombus whose diagonals intersect at O.

Then AC = 24 cm and BD = 10 cm

We know that the diagonals of a rhombus bisect each other at right angles.

From right ∆AOB, we have

Hence, each side of a rhombus 13 cm

**Question 16:**

Given: ∆ABC in which D is the midpoint of BC. AE ⊥ BC and AC > AB.

Then, BD = CD and ∠AED = 90^{0}

Then, ∠ADE < 90^{0} and ∠ADC > 90^{0}

In ∆AED,

Putting value of AE^{2} from (1) in (2), we get

**Question 17:**

Given: D is the midpoint of side BC, AE ⊥ BC, BC = a, AC = b, AB = c, ED = x, AD = p and AE = h

In ∆AEC, ∠AEC = 90^{0}

AD^{2 }= 2AE^{2 }+ ED^{2} (by Pythagoras theorem)

⇒ p^{2 }= h^{2 }+ x^{2}

(i) In ∆AEC, ∠AEC = 90^{0}

(ii) In ∆ABE, ∠ABE = 90^{0}

(iii) Adding (1) and (2), we get

(iv) Subtracting (2) from (1), we get

**Question 18:**

Const: Draw a perpendicular AE from A

Thus, AE ⊥ BC

Proof:

In ∆ABC, AB = AC

And AE is a bisector of BC

Then, BE = EC

In right angle triangles ∆AED and ∆ACE

Hence proved.

**Question 19:**

ABC is an isosceles triangle right angled at B,

Let AB = BC = x cm

By Pythagoras theorem,

Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles are helpful to complete your math homework.

If you have any doubts, please comment below. A Plus Topper try to provide online math tutoring for you.

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