ML Aggarwal Triangles Chapter Test for ICSE Class 9 Maths Solutions Ch-10. Step by Step Answer of Ch-Test Questions of Triangles for ML Aggarwal ICSE Class 9th Mathematics. Visit official website CISCE for detail information about ICSE Board Class-9.

## ML Aggarwal Triangles Chapter Test for ICSE Class 9 Maths Solutions Ch-10

Board | ICSE |

Subject | Maths |

Class | 9th |

Chapter-10 | Triangles |

Topics | Solution of Ch-Test Questions |

Academic Session | 2024-2025 |

### Solution of Ch-Test Questions

ML Aggarwal Triangles for ICSE Class 9 Maths Solutions Ch-10

**Question 1. ****In triangles ABC and DEF, ∠A = ∠D, ∠B = ∠E and AB = EF. Will the two triangles be congruent? Give reasons for your answer.**

**Answer : **In ∆ABC and ∆DEF

∠A = ∠D

∠B = ∠E

AB = EF

In ∆ABC, two angles and included side is

Given but in ∆DEF, corresponding angles are

Equal but side is not included of there angle.

Triangles cannot be congruent.

**Question 2. ****In the adjoining figure, ABCD is a square. P, Q and R are points on the sides AB, BC and CD respectively such that AP= BQ = CR and ∠PQR = 90°. Prove that**

(a) ∆PBQ ≅ ∆QCR

(b) PQ = QR

(c) ∠PRQ = 45°

**Answer : **From the figure, ABCD is a square

P,Q and R are the Points on the sides AB,

BC and CD respectively such that

AP = BQ = CR, ∠PQR = 90^{o}

To prove: (a) ∆PBQ = ∆QCR

(b) PQ = QR

(c) ∠PQR = 45^{o}

Proof:

AB = BC = CD **(Sides of Square)**

And AP = BQ = CR **(Given)**

Subtracting, we get

AB – AP = BC – BQ = CD – CR **(PB = QC = RD)**

Now in ∆PBQ and ∠QCR

PB = QC **(Proved)**

BQ = CR **(Given)**

∠B = ∠C **(Each 90 ^{o})**

∆PBQ ≅ ∆QCR

PQ = QR

But ∠PQR = 90^{o}

∠RPQ = ∠PQR **(Angles opposite to equal angles)**

But,

∠RPQ + ∠PRQ = 90^{o}

⇒ ∠RPQ = ∠PQR = 90^{o}

⇒ ∠RPQ = ∠PRQ = 90^{o}/2 = 45^{o}

**Question 3. ****In the given figure, OA ⊥ OD, OC X OB, OD = OA and OB = OC. Prove that AB = CD.**

**Answer : **From the figure OA ⊥ OD, OC ⊥ OB.

∠AOD = ∠COB **(each 90 ^{o})**

Adding ∠AOC

∠AOD + ∠AOC = ∠AOC + ∠COB

⇒ ∠COD = ∠AOB

Now, in ∆AOB and ∆DOC

OA = OD **(Given)**

OB = OC **(Given)**

∠AOB = ∠COD **(Proved)**

∆AOB ≅ ∆DOC

AB = CD

**Question 4. ****In the given figure, PQ || BA and RS CA. If BP = RC, prove that:**

(i) ∆BSR ≅ ∆PQC

(ii) BS = PQ

(iii) RS = CQ.

**Answer :**

PQ ∥ BA, RS ∥ CA

BP = RC

To prove :

(i) ∆BSR ≅ ∆PQC

(ii) RS = CQ

Proof:

BP = RC

⇒ BC – RC = BC – BP

⇒ BR = PC

Now, in ∆BSR and ∆PQC

∠B = ∠P **(Corresponding angles)**

∠R = ∠C **(Corresponding angles)**

BR = PC **(Proved)**

∆BSR ≅ ∆PQC

BS = PQ

RS = CQ

**Question 5. ****In the given figure, AB = AC, D is a point in the interior of ∆ABC such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC of ∆ABC**.

**Answer : **Since AB = AC

∠ABC = ∠ACB

But ∠DBC = ∠DBC

⇒ ∠ABD = ∠ACD

Now in ΔABD and ΔADC

AB = AC

AD = AD

∠ABD = ∠ACD

Therefore, ΔABD ≅ ΔADC …(SSA criteria)

Hence, ∠BAD = ∠CAD

Thus, AD bisects ∠BAC.

**Question 6. ****In the adjoining figure, AB || DC. CE and DE bisects ∠BCD and ∠ADC respectively. Prove that AB = AD + BC.**

**Answer : **DE Bisect ∠D

∠EDA = ∠EDC = ∠D/2

AB ║ CD & DE is line cutting these parallel lines

∠DEA = ∠EDC = ∠D/2

in Δ ADE

∠EDA & ∠DEA = ∠D/2

Hence AD = AE

Similarly

in Δ BCE

∠ECB = ∠CEB = ∠C/2

BE = BC

**AB = AE + BE = AD + BC**

**AB = AD + BC**

**Question 7. ****In ∆ABC, D is a point on BC such that AD is the bisector of ∠BAC. CE is drawn parallel to DA to meet BD produced at E. Prove that ∆CAE is isosceles.**

**Answer : **

∠BAC + ∠EAC = 180° ( Straight line)

in ΔACE

∠ACE + ∠AEC + ∠EAC = 180° ( sum of angles of Triangle)

Equating both

∠BAC + ∠EAC = ∠ACE + ∠AEC + ∠EAC

**∠BAC = ∠ACE + ∠AEC Eq 1**

∠BAD = (1/2) ∠BAC

∠BAD = ∠AEC ( AD ║ CE)

**∠AEC = (1/2) ∠BAC**

putting this in eq 1

∠BAC = ∠ACE + (1/2) ∠BAC

**∠ACE = (1/2) ∠BAC**

∠AEC = ∠ACE = (1/2) ∠BAC

AC = AE

Hence Δ CAE is an isosceles triangle

**Question 8. ****In the figure (ii) given below, ABC is a right angled triangle at B, ADEC and BCFG are squares. Prove that AF = BE.**

**Answer : **To prove : AE = BE

angle ACE = angle BCF

adding angle ACB both side

angle ACE +angle ACB = angle BCF + angle ACB

angle BCE = angle ACF

in ∆ BCE and ∆ACF

BC = CF ( square sides)

CE = AF ( square sides)

angle BCE = angle ACF

∆ BCD and ∆ ACF are congruent triangle

BE= AF

Hence Proved

**Question 9. ****In the given figure, BD = AD = AC. If ∠ABD = 36°, find the value of x .**

**Answer :**

Since x is an exterior angle

So we can say,

x = **∠**ABD + **∠** ACD

AC = AD = AC

AB = AC

**∠**ABD = **∠**ACD ( 36° = 36°)

x = 36° + 36°

x = 72°

Hence, the value of x is 72° .

**Question 10 ****In the adjoining figure, TR = TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that RB = SA.**

**Answer : ****Given:** In the figure , RST is a triangle

*RT *= *T**S ……..(1)*

∠1 = 2∠2 ……..(2)

And∠4 = √2 …………….(3)

And given to prove Δ*RBT≅ *Δ*SAT*

Let the point of intersection of RB and SA be denoted by O Since RB and SA intersect at O.

∴∠*AOR *= ∠*BOS [Vertically opposite angles]*

⇒∠1 = ∠4

⇒ 2∠2 = 2∠3 [From (2) and (3)

∠2 = ∠3 ……..(4)

Now we have *RT *= *T**S * ∠∠*T**R**S*

∴Δ*T**RS *is an isosceles triangle

∴∠*T**R**S *= ∠*T**SR ……..(5) [Angles opposite to equal sides are equal] *

But we have

* ∠T**RS *= ∠*T**RB *+ ∠2 ………(6)

And ∠*T**SR *= ∠*T**SA *+ ∠3 ……….(7)

Putting (6) and (7) in (5) we get

*∠T**RB *+∠2 = ∠*T**SA *+ ∠*B *

⇒∠ *TRB *=∠*TSA *[∵ *From *(4)]

Now considerΔ*RBT *and Δ*SA**T *

*RT *= *ST …. [From (1)] *

*∠T**RB *= ∠*T**SA …. [From (4)] *

*∠RTB *=*∠ **STA [Common angle] *

From ASA criterion of congruence, we have *ΔRB**T ≅**Δ**SAT *

**Question 11 .**

**(a) In the figure (1) given below, find the value of x.**

**(b) In the figure (2) given below, AB = AC and DE || BC. Calculate**

(i)x

(ii) y

(iii) ∠BAC

**(c) In the figure (1) given below, calculate the size of each lettered angle.**

**Answer: **Update soon

**Question 12.**

**(a) In the figure (1) given below, AD = BD = DC and ∠ACD = 35°. Show that**

(i) AC > DC (ii) AB > AD.

**(b) In the figure (2) given below, prove that**

(i) x + y = 90° (ii) z = 90° (iii) AB = BC

**Answer: **Update soon

**Question 13. ****In the adjoining figure, ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that**

(i) ∆ABD ≅ ∆ACD

(ii) ∆ABP ≅ ∆ACP

(iii) AP bisects ∠A as well as ∠D

(iv) AP is the perpendicular bisector of BC.

**Answer : ****(i) In △ABD and △ACD,**

AB=AC ….(since △ABC is isosceles)

AD=AD ….(common side)

BD=DC ….(since △BDC is isosceles)

ΔABD≅ΔACD …..SSS test of congruence,

∴∠BAD=∠CAD i.e. ∠BAP=∠PAC …..[c.a.c.t]……(i)

**(ii) In △ABP and △ACP,**

AB=AC …(since △ABC is isosceles)

AP=AP …(common side)

∠BAP=∠PAC ….from (i)

△ABP≅△ACP …. SAS test of congruence

∴BP=PC …[c.s.c.t]…..(ii)

∠APB=∠APC ….c.a.c.t.

**(iii) Since △ABD≅△ACD**

∠BAD=∠CAD ….from (i)

So, AD bisects ∠A

i.e. AP bisects ∠A…..(iii)

In △BDP and △CDP,

DP=DP …common side

BP=PC …from (ii)

BD=CD …(since △BDC is isosceles)

△BDP≅△CDP ….SSS test of congruence

∴∠BDP = ∠CDP ….c.a.c.t.

∴ DP bsects ∠D

So, AP bisects ∠D ….(iv)

From (iii) and (iv),

AP bisects ∠A as well as ∠D.

**(iv) We know that**

∠APB+∠APC=180 ….(angles in linear pair)

Also, ∠APB=∠APC …from (ii)

∴∠APB=∠APC = =90°

BP = PC and ∠APB = ∠APC = 90°

Hence, AP is perpendicular bisector of BC.

**Question 14. ****In the given figure, AP ⊥ l and PR > PQ. Show that AR > AQ.**

**Answer : **Given: PR > PQ

Apply Pythagoras theorem in triangle APQ,

AP² = AQ² – PQ² … (1)

Apply Pythagoras theorem in triangle APR,

AP² = AR² – PR² … (2)

From equation (1) and (2),

AR² – PR² = AQ² – PQ²

Given PR > PQ, So,

AR² – PQ² > AQ² – PQ²

AR² > AQ²

AR > AQ

Hence Proved.

**Question 15. ****If O is any point in the interior of a triangle ABC, show that**

OA + OB + OC > 1/2 (AB + BC + CA).

**Answer : **According to triangle inequality, sum of any two sides is greater than the third side. Therefore,

OA+OB > AB ……(1)

OB+OC > BC ……(2)

OC+OA > AC ……(3)

Adding (1), (2) and (3), we get

2(OA+OB+OC) > (AB+BC+AC)

Dividing both sides by 2, we get

(OA+OB+OC) > 1/2(AB+BC+AC).

— : End of ML Aggarwal Triangles Chapter Test Class 9 ICSE Maths Solutions :–

Return to :- ** ML Aggarawal Maths Solutions for ICSE Class-9**

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