Prasanna

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.3 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 4 Constructions of Triangles.

9th Standard Maths 2 Practice Set 4.3 Chapter 4 Constructions of Triangles Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 4.3 Chapter 4 Constructions of Triangles Questions With Answers Maharashtra Board

Question 1.
Construct ∆PQR, in which ∠Q = 70°, ∠R = 80° and PQ + QR + PR = 9.5 cm.
Solution:

i. As shown in the figure, take point T and S on line QR, such that
QT = PQ and RS = PR ….(i)
QT + QR + RS = TS [T-Q-R, Q-R-S]
∴ PQ + QR + PR = TS …..(ii) [From (i)]
Also,
PQ + QR + PR = 9.5 cm ….(iii) [Given]
∴ TS = 9.5 cm

ii. In ∆PQT
PQ = QT [From (i)]
∴ ∠QPT = ∠QTP = x° ….(iv) [Isosceles triangle theorem]
In ∆PQT, ∠PQR is the exterior angle.
∴ ∠QPT + ∠QTP = ∠PQR [Remote interior angles theorem]
∴ x + x = 70° [From (iv)]
∴ 2x = 70° x = 35°
∴ ∠PTQ = 35°
∴ ∠T = 35°
Similarly, ∠S = 40°

iii. Now, in ∆PTS
∠T = 35°, ∠S = 40° and TS = 9.5 cm Hence, ∆PTS can be drawn.

iv. Since, PQ = TQ,
∴ Point Q lies on perpendicular bisector of seg PT.
Also, RP = RS
∴ Point R lies on perpendicular bisector of seg PS.
Points Q and R can be located by drawing the perpendicular bisector of PT and PS respectively.
∴ ∆PQR can be drawn.

Steps of construction:
i. Draw seg TS of length 9.5 cm.
ii. From point T draw ray making angle of 35°.
iii. From point S draw ray making angle of 40°.
iv. Name the point of intersection of two rays as P.
v. Draw the perpendicular bisector of seg PT and seg PS intersecting seg TS in Q and R respectively.
vi. Join PQ and PR.
Hence, ∆PQR is the required triangle.

Question 2.
Construct ∆XYZ, in which ∠Y = 58°, ∠X = 46° and perimeter of triangle is 10.5 cm.
Solution:

i. As shown in the figure, take point W and V on line YX, such that
YW = ZY and XV = ZX ……(i)
YW + YX + XV = WV [W-Y-X, Y-X-V]
∠Y + YX + ∠X = WV ……(ii) [From (i)]
Also,
∠Y + YX + ∠X = 10.5 cm …..(iii) [Given]
∴ WV = 10.5 cm [From (ii) and (iii)]

ii. In ∆ZWY
∠Y = YM [From (i)]
∴ ∠YZW = ∠YWZ = x° …..(iv) [Isosceles triangle theorem]
In ∆ZYW, ∠ZYX is the exterior angle.
∴ ∠YZW + ∠YWZ = ∠ZYX [Remote interior angles theorem]
∴ x + x = 58° [From (iv)]
∴ 2x = 58°
∴ x = 29°
∴ ∠ZWY = 29°
∴ ∠W = 29°
∴ Similarly, ∠V = 23°

iii. Now, in ∆ZWV
∠W = 29°, ∠V = 23° and
WV= 10.5 cm
Hence, ∆ZWV can be drawn.

iv. Since, ZY = YW
∴ Point Y lies on perpendicular bisector of seg ZW.
Also, ZX = XV
∴ Point X lies on perpendicular bisector of seg ZV.
∴ Points Y and X can be located by drawing the perpendicular bisector of ZW and ZV respectively.
∴ ∆XYZ can be drawn.

Steps of construction:
i. Draw seg WV of length 10.5 cm.
ii. From point W draw ray making angle of 29°.
iii. From point V draw ray making angle of 23°.
iv. Name the point of intersection of two rays as Z.
v. Draw the perpendicular bisector of seg WZ and seg VZ intersecting seg WV in Y and X respectively.
vi. Join XY and XX.
Hence, ∆XYX is the required triangle

Question 3.
Construct ∆LMN, in which ∠M = 60°, ∠N = 80° and LM + MN + NL = 11 cm.
Solution:

i. As shown in the figure, take point S and T on line MN, such that
MS = LM and NT = LN …..(i)
MS + MN + NT = ST [S-M-N, M-N-T]
∴ LM + MN + LN = ST …..(ii)
Also,
LM + MN + LN = 11 cm ….(iii)
∴ ST = 11 cm [From (ii) and (iii)]

ii. In ∆LSM
LM = MS
∴ ∠MLS = ∠MSL = x° …..(iv) [isosceles triangle theorem]
In ∆LMS, ∠LMN is the exterior angle.
∴ ∠MLS + ∠MSL = ∠LMN [Remote interior angles theorem]
∴ x + x = 60° [From (iv)]
∴ 2x = 60°
∴ x = 30°
∴ ∠LSM = 30°
∴ ∠S = 30°
Similarly, ∠T = 40°

iii. Now, in ∆LST
∠S = 30°, ∠T = 40° and ST = 11 cm
Hence, ALST can be drawn.

iv. Since, LM = MS
∴ Point M lies on perpendicular bisector of seg LS.
Also LN = NT
∴ Point N lies on perpendicular bisector of seg LT.
∴ Points M and N can be located by drawing the perpendicular bisector of LS and LT respectively.
∴ ∆LMN can be drawn.

Steps of construction:
i. Draw seg ST of length 11 cm.
ii. From point S draw ray making angle of 30°.
iii. From point T draw ray making angle of 40°.
iv. Name the point of intersection of two rays as L.
v. Draw the perpendicular bisector of seg LS and seg LT intersecting seg ST in M and N respectively.
vi. Join LM and LN.
Hence, ∆LMN is the required triangle.

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Problem Set 3 Class 9 Answers
• Practice Set 4.1 Class 9 Answers
• Practice Set 4.2 Class 9 Answers
• Practice Set 4.3 Class 9 Answers
• Problem Set 4 Class 9 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 13.2 8th Std Maths Answers Solutions Chapter 13 Congruence of Triangles.

Congruence of Triangles Class 8 Maths Chapter 13 Practice Set 13.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 13.2 Chapter 13 Solutions Answers

Congruence of Triangles Class 8th Practice Set 13.2 Question 1.
In each pair of triangles given below, parts shown by identical marks are congruent. State the test and the one-to-one correspondence of vertices by which triangles in each pair are congruent. Also state the remaining congruent parts.


Solution:
i. In ∆MST and ∆TBM,
∴ side MS ≅ side TB … [Given]
m∠MST = m∠TBM = 90° … [Given]
hypotenuse MT ≅ hypotenuse MT
…[Common side]
∴ ∆MST ≅ ∆TBM …[by hypotenuse-side test]
∴ side ST ≅ side BM …[Corresponding sides of congruent triangles]
∠SMT ≅ ∠BTM …[Corresponding sides of congruent triangles]
∠STM ≅ ∠BMT …[Corresponding sides of congruent triangles]

ii. In ∆PRQ and ∆TRS,
side PR ≅ side TR … [Given]
∠PRQ ≅ ∠TRS …[Vertically opposite angles]
side RQ ≅ side RS … [Given]
∴ ∆PRQ ≅ ∆TRS …[by SAS test]
∴ side PQ ≅ side TS …[Corresponding sides of congruent triangles]
∠RPQ ≅ ∠RTS …[Corresponding sides of congruent triangles]
∠PQR ≅ ∠TSR …[Corresponding sides of congruent triangles]

iii. In ∆DCH and ∆DCF,
∠DCH ≅ ∠DCF …[Given]
∠DHC ≅ ∠DFC …[Given]
side DC ≅ side DC …[Common side]
∴ ∆DCH ≅ ∆DCF …[by AAS test]
∴ side HC ≅ side FC …[Corresponding sides of congruent triangles]
side DH ≅ side DF…[Corresponding sides of congruent triangles]
∠HDC ≅ ∠FDC ….[Corresponding sides of congruent triangles]

Congruence of Triangles Practice Set 13.2 Question 2.
In the given figure, seg AD ≅ seg EC. Which additional information is needed to show that ∆ABD and ∆EBC will be congruent by AAS test?

Solution:
In ∆ABD and ∆CBE,
∴ seg AD ≅ seg CE …[Given]
∠ABD ≅ ∠CBE …[Vertically opposite angles]
∴ The necessary condition for the two triangles to be congruent by AAS test is
∠ADB ≅ ∠CEB, or
∠DAB ≅ ∠ECB

Maharashtra Board Class 8 Maths Chapter 13 Congruence of Triangles Practice Set 13.2 Intext Questions and Activities

Practice Set 13.2 Class 8 Question 1.
Draw ∆ABC and ∆LMN such that two pairs of their sides and the angles included by them are congruent.
Draw ∆ABC and ∆LMN, l(AB) = l(LM), l(BC) = l(MN), m∠ABC = m∠LMN.
Copy ∆ABC on a tracing paper. Place the paper on ∆LMN in such a way that point A coincides with point L, side AB overlaps side LM. What do you notice?(Textbook pg. no. 83)

Solution:
We notice that ∆ABC ≅ ∆LMN.

Congruence of Triangles Class 8 Solutions Question 2.
Draw ∆PQR and ∆XYZ such that l(PQ) = l(X Y), l(Q R) = l(YZ), l(RP) = l(ZX). Copy ∆PQR on a tracing paper. Place it on ∆XYZ observing the correspondence P ↔ X, Q ↔ Y, R ↔ Z. What do you notice? (Textbook pg. no. 84)

Solution:
We notice that ∆PQR ≅ ∆XYZ.

Congruence of Triangles Class 8 Question 3.
Draw ∆XYZ and ∆DEF such that, l(XZ) = l(DF), ∠X ≅ ∠D and ∠Z ≅ ∠F.
Copy ∆XYZ on a tracing paper and place it over ∆DEF. What do you notice?(Textbook pg. no. 84)

Solution:
We notice that ∆XYZ ≅ ∆DEF in the correspondence X ↔ D, Y ↔ E, Z ↔ F.

Question 4.
Draw two right angled triangles such that a side and the hypotenuse of one is congruent with the corresponding parts of the other. Copy one triangle on tracing paper and place it over the other. What do you notice? (Textbook pg. no. 84)

Solution:
We notice that the two triangles are congruent.
(Students should draw figures and verify the answers.)

Std 8 Maths Digest

• Practice Set 11.1 Class 8 Answers
• Practice Set 11.2 Class 8 Answers
• Practice Set 11.3 Class 8 Answers
• Practice Set 12.1 Class 8 Answers
• Practice Set 12.2 Class 8 Answers
• Practice Set 13.1 Class 8 Answers
• Practice Set 13.2 Class 8 Answers
• Practice Set 14.1 Class 8 Answers
• Practice Set 14.2 Class 8 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 4 Constructions of Triangles.

9th Standard Maths 2 Practice Set 4.2 Chapter 4 Constructions of Triangles Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 4.2 Chapter 4 Constructions of Triangles Questions With Answers Maharashtra Board

Question 1.
Construct ∆XYZ, such that YZ = 7.4 cm, ∠XYZ = 45° and XY – XZ = 2.7 cm.
Solution:

Here, XY – XZ = 2.7 cm
∴ XY > XZ
As shown in the rough figure draw seg YZ = 7.4 cm
Draw a ray YP making an angle of 45° with YZ
Take a point W on ray YP, such that
YW = 2.7 cm.
Now, XY – XW = YW [Y-W-X]
∴ XY – XW = 2.7 cm ….(i)
Also, XY – XZ = 2.7 cm ….(ii) [Given]
∴ XY – XW = XY – XZ [From (i) and (ii)]
∴ XW = XZ
∴ Point X is on the perpendicular bisector of seg ZW
∴ Point X is the intersection of ray YP and the perpendicular bisector seg ZW

Steps of construction:
i. Draw seg YZ of length 7.4 cm.
ii. Draw ray YP, such that ∠ZYP = 45°.
iii. Mark point W on ray YP such that l(YW) = 2.7 cm.
iv. Join points W and Z.
v. Join the points X and Z.
Hence, ∆XYZ is the required triangle.

Question 2.
Construct ∆PQR, such that QR = 6.5 cm, ∠PQR = 60° and PQ – PR = 2.5 cm.
Solution:

Here, PQ – PR = 2.5 cm
∴ PQ > PR
As shown in the rough figure draw seg QR = 6.5 cm
Draw a ray QT making on angle of 60° with QR
Take a point S on ray QT, such that QS = 2.5 cm.
Now, PQ – PS = QS [Q-S-T]
∴ PQ – PS = 2.5 cm ……(i) [Given]
Also, PQ – PR = 2.5 cm …..(ii) [From (i) and (ii)]
∴ PQ – PS = PQ – PR
∴ PS = PR
∴ Point P is on the perpendicular bisector of seg RS
∴ Point P is the intersection of ray QT and the perpendicular bisector of seg RS

Steps of construction:
i. Draw seg QR of length 6.5 cm.
ii. Draw ray QT, such that ∠RQT = 600.
iii. Mark point S on ray QT such that l(QS) = 2.5 cm.
iv. Join points S and R.
v. Draw perpendicular bisector of seg SR intersecting ray QT. Name the point as P.
vi. Join the points P and R.
Hence, ∆PQR is the required triangle.

Question 3.
Construct ∆ABC, such that BC = 6 cm, ∠ABC = 100° and AC – AB = 2.5 cm.

Solution:
Here, AC – AB = 2.5 cm
∴ AC > AB
As shown in the rough figure draw seg BC = 6 cm
Draw a ray BT making an angle of 100° with BC.
Take a point D on opposite ray of BT, :
such that BD 2.5 cm.
Now, AD – AB = BD [A-B-D]
∴ AD – AB = 2.5cm …..(i)
Also, AC – AB = 2.5 cm …..(ii) [Given]
∴ AD – AB = AC – AB [From (i) and (ii)]
∴ AD = AC
∴ Point A is on the perpendicular bisector of seg DC
∴ Point A is the intersection of ray BT and the perpendicular bisector of seg DC

Steps of construction:
i. Draw seg BC of length 6 cm.
ii. Draw ray BT, such that ∠CBT = 100°.
iii. Take point D on opposite ray of BT such that l(BD) = 2.5 cm.
iv. Join the points D and C.
v. Draw the perpendicular bisector of seg DC intersecting ray BT. Name the point as A.
vi. Join the points A and C.
Hence, ∆ABC is the required triangle.

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Problem Set 3 Class 9 Answers
• Practice Set 4.1 Class 9 Answers
• Practice Set 4.2 Class 9 Answers
• Practice Set 4.3 Class 9 Answers
• Problem Set 4 Class 9 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 14.1 8th Std Maths Answers Solutions Chapter 14 Compound Interest.

Compound Interest Class 8 Maths Chapter 14 Practice Set 14.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 14.1 Chapter 14 Solutions Answers

Practice Set 14.1 Class 8 Question 1.
Find the amount and the compound interest.

No	Principal (Rs)	Rate (p.c.p.a.)	Duration (years)
i.	2000	5	2
ii.	5000	8	3
iii.	4000	7.5	2

Solution:
i. Here P = Rs 2000, R = 5 p.c.p.a. and N = 2 years

= 5 × 441
∴ A = Rs 2205
I = Amount (A) – Principal (P)
= 2205 – 2000
= Rs 205
∴ The amount is Rs 2205 and the compound interest is Rs 205.

ii. Here, P = Rs 5000, R = 8 p.c.p.a. and N = 3 years

∴ A = Rs 6298.56
I = Amount (A) – Principal (P)
= 6298.56 – 5000
= Rs 1298.56
∴ The amount is Rs 6298.56 and the compound interest is Rs 1298.56.

iii. Here, P = Rs 4000, R = 7.5 p.c.p.a. and N = 2 years

∴A = Rs 4622.50
I = Amount (A) – Principal (P)
= 4622.50 – 4000
= Rs 622.50
∴The amount is Rs 4622.50 and the compound interest is Rs 622.50.

Compound Interest Practice Set 14.1 Question 2.
Sameerrao has taken a loan of Rs 12500 at the rate of 12 p.c.p.a. for 3 years. If the interest is compounded annually then how many rupees should he pay to clear his loan?
Solution:
Here, P = Rs 12,500, R = 12 p.c.p.a. and
N = 3 years

= 0.8 × 28 × 28 × 28
= Rs 17,561.60
Sameerrao should pay Rs 17,561.60 to clear his loan.

8th Standard Maths Practice Set 14.1 Question 3.
To start a business Shalaka has taken a loan of Rs 8000 at a rate of \(10\frac { 1 }{ 2 }\) p.c.p.a. After two years how much compound interest will she have to pay?
Solution:
Here, P = Rs 8000, N = 2 years and


I = Amount (A) – Principal (P)
= 9768.20 – 8000
= Rs 1768.20
∴ After two years Shalaka will have to pay Rs 1768.20 as compound interest.

Std 8 Maths Digest

• Practice Set 11.1 Class 8 Answers
• Practice Set 11.2 Class 8 Answers
• Practice Set 11.3 Class 8 Answers
• Practice Set 12.1 Class 8 Answers
• Practice Set 12.2 Class 8 Answers
• Practice Set 13.1 Class 8 Answers
• Practice Set 13.2 Class 8 Answers
• Practice Set 14.1 Class 8 Answers
• Practice Set 14.2 Class 8 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 4 Constructions of Triangles.

9th Standard Maths 2 Practice Set 4.1 Chapter 4 Constructions of Triangles Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 4.1 Chapter 4 Constructions of Triangles Questions With Answers Maharashtra Board

Question 1.
Construct APQR, in which QR = 4.2 cm, m∠Q = 40° and PQ + PR = 8.5 cm.
Solution:

As shown in the rough figure draw seg QR = 4.2 cm
Draw a ray QT making an angle of 40° with QR
Take a point S on ray QT, such that QS = 8.5 cm
Now, QP + PS = QS [Q-P-S]
∴ QP + PS = 8.5 cm …….(i)
Also, PQ + PR = 8.5 cm ……(ii) [Given]
∴ QP + PS = PQ + PR [From (i) and (ii)]
∴ PS = PR
∴ Point P is on the perpendicular bisector of seg SR
∴ The point of intersection of ray QT and perpendicular bisector of seg SR is point P.

Steps of construction:
i. Draw seg QR of length 4.2 cm.
ii. Djraw ray QT, such that ∠RQT = 40°.
iii. Mark point S on ray QT such that l(QS) = 8.5 cm.
iv. Join points R and S.
v. Draw perpendicular bisector of seg RS intersecting ray QT.
Name the point as P.
vi. Join the points P and R.
Hence, ∆PQR is the required triangle.

Question 2.
Construct ∆XYZ, in which YZ = 6 cm, XY + XZ = 9 cm, ∠XYZ = 50°.
Solution:

As shown in the rough figure draw seg YZ = 6 cm
Draw a ray YT making an angle of 50° with YZ
Take a point W on ray YT, such that YW = 9 cm
Now, YX + XW = YW [Y-X-W]
∴ YX + XW = 9 cm ….(i)
Also, XY + XZ = 9 cm ….(ii) [Given]
∴ YX + XW = XY + XZ [From (i) and (ii) ]
∴ XW = XZ
∴ Point X is on the perpendicular bisector of seg WZ
∴ The point of intersection of ray YT and perpendicular bisector of seg WZ is j point X.

Steps of construction:
i. Draw seg YZ of length 6 cm.
ii. Draw ray YT, such that ∠ZYT = 50°.
iii. Mark point W on ray YT such that l(YW) = 9 cm.
iv. Join points W and Z.
v. Draw perpendicular bisector of seg WZ intersecting ray YT. Name the point as X.
vi. Join the points X and Z.
Hence, ∆XYZ is the required triangle.

Question 3.
Construct ∆ABC, in which BC = 6.2 cm, ∠ACB = 50°, AB + AC = 9.8 cm.
Solution:

As shown in the rough figure draw seg CB = 6.2 cm
Draw a ray CT making an angle of 50° with CB
Take a point D on ray CT, such that
CD = 9.8 cm
Now, CA + AD = CD [C-A-D]
∴ CA + AD = 9.8 cm …….(i)
Also, AB + AC = 9.8 cm ……(ii) [Given]
∴ CA + AD = AB + AC [From (i) and (ii)]
∴ AD = AB
∴ Point A is on the perpendicular bisector of seg DB
∴ The point of intersection of ray CT and perpendicular bisector of seg DB is point A.

Steps of construction:
i. Draw seg BC of length 6.2 cm.
ii. Draw ray CT, such that ∠BCT = 50°.
iii. Mark point D on ray CT such that l(CD) = 9.8 cm.
iv. Join points D and B.
v. Draw perpendicular bisector of seg DB intersecting ray CT. Name the point as A.
vi. Join the points A and B.
Hence, ∆ABC is the required triangle.

Question 4.
Construct ∆ABC, in which BC = 3.2 cm, ∠ACB = 45° Solution:and perimeter of AABC is 10 cm.
Solution:

Perimeter of ∆ABC = AB + BC + AC
∴ 10 = AB + 3.2 + AC
∴ AB + AC = 10 – 3.2
∴ AB + AC = 6.8 cm
Now, In ∆ABC
BC = 3.2 cm, ∠ACB = 45° and AB + AC = 6.8 cm ….(i)
As shown in the rough figure draw j seg BC = 3.2 cm
Draw a ray CT making an angle of 45° with CB
Take a point D on ray CT, such that
CD = 6.8 cm
Now, CA + AD = CD [C-A-D]
∴ CA + AD = 6.8 cm …(ii)
Also, AB + AC = 6.8 cm ….(iii) [From (i)]
∴ CA + AD = AB + AC [From (ii) and (iii)]
∴ AD = AB
∴ Point A is on the perpendicular bisector of seg DB
∴ The point of intersection of ray CT and perpendicular bisector of seg DB is point A.

Steps of construction:
i. Draw seg BC of length 3.2 cm.
ii. Draw ray CT, such that ∠BCT = 45°.
iii. Mark point D on ray CT such l(CD) = 6.8 cm. that
iv. Join points D and B.
V. Draw perpendicular bisector of seg DB intersecting ray CT. Name the point as A.
vi. Join the points A and B.
Hence, ∆ABC is the required triangle.

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Problem Set 3 Class 9 Answers
• Practice Set 4.1 Class 9 Answers
• Practice Set 4.2 Class 9 Answers
• Practice Set 4.3 Class 9 Answers
• Problem Set 4 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 3 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 3 Triangles.

9th Standard Maths 2 Problem Set 3 Chapter 3 Triangles Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Problem Set 3 Chapter 3 Triangles Questions With Answers Maharashtra Board

Question 1.
Choose the correct alternative answer for the following questions.

i. If two sides of a triangle are 5 cm and 1.5 cm, the length of its third side cannot be ____.
(A) 3.7 cm
(B) 4.1 cm
(C) 3.8 cm
(D) 3.4 cm
Answer:
Sum of the lengths of two sides of a triangle > length of the third side
Here, 1.5 cm + 3.4 cm = 4.9 cm < 5 cm
∴ Third side ≠ 3.4 cm
(D) 3.4 cm

ii. In ∆PQR, if ∠R > ∠Q, then _____ .
(A) QR > PR
(B) PQ > PR
(C) 3.8 cm
(D) 3.4 cm
Answer:
(B) PQ > PR

iii. In ∆TPQ, if ∠T = 65°, ∠P = 95° , Which of the following is a true statement?
(A) PQ < TP
(B) PQ < TQ
(C) TQ < TP < PQ
(D) PQ < TP < TQ
Answer:
∠Q = 180° – (95° + 65°) = 20°
∴ ∠Q < ∠T < ∠P
∴ PT < PQ < TQ
(B) PQ < TQ

Question 2.
∆ABC is isosceles in which AB = AC. Seg BD and seg CE are medians. Show that BD = CE.

Given: In isosceles ∆ABC, AB = AC. seg BD and seg CE are the medians of ∆ABC.
To prove: BD = CE
Proof: AE = \(\frac { 1 }{ 2 }\) AB …..(i) [E is the midpoint of side AB]
AD = \(\frac { 1 }{ 2 }\) AC ….(ii) [D is the midpoint of side AC]
Also, AB = AC ……(iii) [Given]
∴ AE = AD ….(iv) [From (i), (ii) and (iii)]
In ∆ADB and ∆AEC,

seg AB ≅ seg AC ∠BAD ≅ ∠CAE
seg AD ≅ seg AE
∴ ∆ADB ≅ ∆AEC
∴ seg BD ≅ seg CE
∴ BD = CE

Question 3.
In ∆PQR, if PQ > PR and bisectors of ∠Q and ∠R intersect at S. Show that SQ > SR.

Given: In APQR, PQ > PR and bisectors of ∠Q and ∠R intersect at S.
To prove: SQ > SR
Solution:
Proof:
∠SQR = \(\frac { 1 }{ 2 }\) ∠PQR ….(i) [Ray QS bisects ∠PQR]
∠SRQ = \(\frac { 1 }{ 2 }\) ∠PRQ ….(ii) [Ray RS bisects ∠PRQ]
In ∆PQR,
PQ > PR [Given]
∴ ∠R > ∠Q [Angle opposite to greater side is greater.]
∴ \(\frac { 1 }{ 2 }\)(∠R) > \(\frac { 1 }{ 2 }\)(∠Q) [Multiplying both sides by \(\frac { 1 }{ 2 }\) ]
∴ ∠SRQ > ∠SQR ….(iii) [From (i) and (ii)]
In ∆SQR,
∠SRQ > ∠SQR [From (iii)]
∴ SQ > SR [Side opposite to greater angle is greater]

Question 4.
In the adjoining figure, points D and E are on side BC of ∆ABC, such that BD = CE and AD AE. Show that ∆ABD ≅ ∆ACE.

Given: Points D and E are on side BC of ∆ABC,
such that BD = CE and AD = AE.
To prove: ∆ABD ≅ ∆ACE
Proof:
In ∆ADE,
seg AD = seg AE [Given]
∴ ∠AED = ∠ADE …(i) [Isosceles triangle theorem]
Now, ∠ADE + ∠ADB = 180° …(ii) [Angles in a linear pair]
∴ ∠AED + ∠AEC = 180° ….(iii) [Angles in a linear pair]
∴ ∠ADE + ∠ADB = ∠AED + ∠AEC [From (ii) and (iii)]
∴ ∠ADE + ∠ADB = ∠ADE + ∠AEC [From (i)]
∴ ∠ADB = ∠AEC ….(iv) [Eliminating ∠ADE from both sides]
In ∆ABD and ∆ACE,
seg BD ≅ seg CE [Given]
∠ADB = ∠AEC [From (iv)]
seg AD ≅ seg AE [Given]
∴ ∆ABD ≅ ∆ACE [SAS test]

Question 5.
In the adjoining figure, point S is any point on side QR of ∆PQR. Prove that: PQ + QR + RP > 2PS

Proof:
In ∆PQS,
PQ + QS > PS …..(i) [Sum of any two sides of a triangle is greater than the third side]
Similarly, in ∆PSR,
PR + SR > PS …(ii) [Sum of any two sides of a triangle is greater than the third side]
∴ PQ + QS + PR + SR > PS + PS
∴ PQ + QS + SR + PR > 2PS
∴ PQ + QR + PR > 2PS [Q-S-R]

Question 6.
In the adjoining figure, bisector of ∠B AC intersects side BC at point D. Prove that AB > BD.

Given: Bisector of ∠BAC intersects side BC at point D.
To prove: AB > BD
Solution:
Proof:
∠BAD ≅ ∠DAC ….(i) [Seg AD bisects ∠BAC]
∠ADB is the exterior angle of ∆ADC.
∴ ∠ADB > ∠DAC ….(ii) [Property of exterior angle]
∴ ∠ADB > ∠BAD ….(iii) [From (i) and (ii)]
In AABD,
∠ADB > ∠BAD [From (iii)]
∴ AB > BD [Side opposite to greater angle is greater]

Question 7.
In the adjoining figure, seg PT is the bisector of ∠QPR. A line through R intersects ray QP at point S. Prove that PS = PR.

Given: Seg PT is the bisector of ∠QPR.
To prove: PS = PR
Construction: Draw seg SR || seg PT.
Solution:
Proof:
seg PT is the bisector of ∠QPR. [Given]
∴ ∠QPT = ∠RPT ….(i)
seg PT || seg SR [Construction]
and seg QS is their transversal.

∴ ∠QPT = ∠PSR …(ii) [Corresponding angles]
seg PT || seg SR [Construction]
and seg PR is their transversal.

∴ ∠RPT = ∠PRS …..(iii) [Alternate angles]
∴ ∠PRS = ∠PSR …(iv) [From (i), (ii) and (iii)]
In ∆PSR,
∠PRS = ∠PSR [From (iv)]
∴ PS = PR [Converse of isosceles triangle theorem]

Question 8.
In the adjoining figure, seg AD ⊥ seg BC. Seg AE is the bisector of ∠CAB and B – E – C. Prove that ∠DAE = \(\frac { 1 }{ 2 }\) (∠c – ∠B).

Given: seg AD ⊥ seg BC
seg AE is the bisector of ∠CAB.
To prove: ∠DAE = \(\frac { 1 }{ 2 }\) (∠C – ∠B) [∵ AD ⊥ BC]
∴ ∠DAE = 180° – 90° – ∠AED
∴ ∠DAE = 90° – ∠AED  ….(ii)
Proof:
∴ ∠CAE = \(\frac { 1 }{ 2 }\)∠A ….(i) [seg AE is the bisector of ∠CAB]
In ∆DAE,
∠DAE + ∠ADE + ∠AED = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ ∠DAE + 90° + ∠AED = 180° [∵ AD ⊥ BC]
∴ ∠DAE = 180° – 90° – ∠AED
∴ ∠DAE = 90° – ∠AED ….(ii)
In ∆ACE,
∴ ∠ACE + ∠CAE + ∠AEC = 180° [Sum of the measures of the angles of a triangle is 180°]
∠C + -∠A + ∠AED = 180° [From (i) and C-D-E]
∴ ∠AED = 180° – ∠C – \(\frac { 1 }{ 2 }\)∠A ……(iii)
∴ ∠DAE = 90° – 180°- ∠C+ \(\frac { 1 }{ 2 }\) ∠A [Substituting (iii) in (ii)]
∴ ∠DAE = ∠C + \(\frac { 1 }{ 2 }\)∠A – 90° …..(iv)
In ∆ABC,
∠A + ∠B + ∠C = 180°

Maharashtra Board Class 9 Maths Chapter 3 Triangles Problem Set 3 Intext Questions and Activities

Question 1.
Draw a triangle of any measure on a thick paper. Take a point T on ray QR as shown in the figure given below. Cut two pieces of thick paper which will exactly fit the comers of ∠P and ∠Q. See that the same two jpieces fit exactly at the comer of ∠PRT as shown in the figure. (Textbook pg. no. 24)

Question 2.
Check the congruence of triangles.

Draw ∆ABC of any measure on a card-sheet and cut it out. Place it on a card-sheet. Make a copy of it by drawing its border. Name it as ∆A₁B₁C₁. Now slide the ∆ABC which is the cut out of a triangle to some distance and make one more copy of it. Name it ∆A₂B₂C₂. Then rotate the cut out of triangle ABC a little, as shown in the figure, and make another copy of it. Name the copy as ∆A₃B₃C₃ . Then flip the triangle ABC, place it on another card-sheet and make a new copy of it. Name this copy as ∆A₄B₄C₄ . Have you noticed that each of ∆A₁B₁C₁, ∆A₂B₂C₂, ∆A₃B₃C₃ and ∆A₄B₄C₄ is congruent with ∆ABC ? Because each of them fits exactly with ∆ABC. Let us verify for ∆A₃B₃C₃. If we place ∠A upon ∠A₃, ∠B upon ∠B3 and ∠C upon ∠C3, then only they will fit each other and we can say that ∆ABC = ∆A₃B₃C₃. We also have AB = A₃B₃, BC = B₃C₃, CA = C₃A₃ . Note that, while examining the congruence of two triangles, we have to write their angles and sides in a specific order, that is with a specific one-to-one correspondence. If ∆ABC ≅ ∆PQR, then we get the following six equations:
∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ……..(i)
and AB = PQ, BC = QR, CA = RP …….(ii)
This means, with a one-to-one correspondence between the angles and the sides of two triangles, we get hree pairs of congruent angles and three pairs of congruent sides. (Textbook pg. no. 29)

Question 3.
Every student in the group should draw a right angled triangle, one of the angles measuring 30°. The choice of lengths of sides should be their own. Each one should measure the length of the hypotenuse and the length of the side opposite to 30° angle.
One of the students in the group should fill in the following table.


Did you notice any property of sides of right angled triangle with one of the angles measuring 30°? (Textbook pg. no. 34)
Answer:
We observe that the length of the side opposite to 30° is half the length of the hypotenuse.

Question 4.
The measures of angles of a set square in your compass box are 30°, 60° and 90°. Verify the property of the sides of the set square. (Textbook pg. no. 34)
[Students should attempt the above activity on their own.]

Question 5.
Draw a triangle ABC. Draw medians AD, BE and CF of the triangle. Let their point of concurrence be G, which is called the centroid of the triangle. Compare the lengths of AG and GD with a divider. Verify that the length of AG is twice the length of GD. Similarly, verify that the length of BG is twice the length of GE and the length of CG is twice the length of GF. Name the property of medians of a triangle observed here. (Textbook pg. no. 37)

Answer:
The point of concurrence of medians of the triangle divides each median in the ratio 2 : 1.

Question 6.
Draw a triangle ABC on a cardboard. Draw its medians and denote their point of concurrence as G. Cut out the triangle. Now take a pencil. Try to balance the triangle on the flat tip of the pencil. The triangle is balanced only when the point G is on the flat tip of the pencil. This activity shows an important property of the centroid (point of concurrence of the medians) of the triangle. Point it out. (Textbook pg. no. 37)

Answer:
The centroid of the triangle is the triangle’s centre of gravity. Hence, the triangle in the experiment remains balanced.

Question 7.
Take a photograph on a mobile or a computer. Recall what you do to reduce it or to enlarge it. Also recall what you do to see a part of the photograph in detail. (Textbook pg. no, 45 )

Question 8.
On a card-sheet, draw a triangle of sides 4 cm, 3 cm and 2 cm. Cut it out. Make 13 more copies of the triangle and cut them out from the card sheet. Note that all these triangular pieces are congruent. Arrange them as shown in the following figure and make three triangles out of them.

Number of triangle: 1

Number of triangles: 4

Number of triangles: 9
∆ABC and ∆DEF are similar in the correspondence ABC ↔ DEF.
∠A ≅ ∠D, ∠B ≅ ∠E, ∠C ≅ ∠F
and \(\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{4}{8}=\frac{1}{2} ; \frac{\mathrm{BC}}{\mathrm{EF}}=\frac{3}{6}=\frac{1}{2} ; \frac{\mathrm{AC}}{\mathrm{DF}}=\frac{2}{4}=\frac{1}{2}\) …the corresponding sides are in proportion.
Similarly, consider ∆DEF and ∆PQR. Are their angles congruent and sides proportional in the correspondence DEF ↔ PQR? (Textbook pg. no. 45)
Answer:
Yes.
∠D ≅ ∠P, ∠E ≅ ∠Q, ∠F ≅ ∠R
\(\frac{\mathrm{DE}}{\mathrm{PQ}}=\frac{8}{12}=\frac{2}{3} ; \frac{\mathrm{EF}}{\mathrm{QR}}=\frac{6}{9}=\frac{2}{3} ; \frac{\mathrm{DF}}{\mathrm{PR}}=\frac{4}{6}=\frac{2}{3}\)

Question 9.
Draw a triangle ∆A₁B₁C₁ on a card-sheet and cut it out. Measure ∠A₁, ∠B₁, ∠C₁ Draw two more triangles AA₂B₂C₂ and AA₃B₃C₃ such that
∠A₁ = ∠A₂ = ∠A₃, ∠B₁ = ∠B₂ = ∠B₃, ∠C₁ = ∠c₂ = ∠c₃
and B₁C₁ > B₂C₂ > B₃C₃. Now cut these two triangles also. Measure the lengths of the three triangles. Arrange the triangles in two ways as shown in the figure.

Check the ratios \(\frac{A_{1} B_{1}}{A_{2} B_{2}}, \frac{B_{1} C_{1}}{B_{2} C_{2}}, \frac{A_{1} C_{1}}{A_{2} C_{2}}\). You will notice that the ratios are equal.
Similarly, see whether the ratios \(\frac{A_{1} C_{1}}{A_{3} C_{3}}, \frac{B_{1} C_{1}}{B_{3} C_{3}}, \frac{A_{1} B_{1}}{A_{3} B_{3}}\) are equal. What do you observe? (Texthook pg. no. 46)
Answer:
From the activity we observe that, when corresponding angles of two triangles are equal, the ratios of their corresponding sides are also equal i.e., their corresponding sides are in the same proportion.

Question 10.
Prepare a map of road surrounding your school or home, upto a distance of about 500 metre. How will you measure the distance between two spots on a road? While walking, count how many steps cover a distance of about two metre. Suppose, your three steps cover a distance of 2 metre. Considering this proportion 90 steps means 60 metre. In this way you can judge the distances between different spots on roads and also the lengths of roads. You have to judge the measures of angles also where two roads meet each other. Choosing a proper scale for lengths of roads, prepare a map. Try to show shops, buildings, bus stops, rickshaw stand etc. in the map. (Textbook pg. no. 48)

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Problem Set 3 Class 9 Answers
• Practice Set 4.1 Class 9 Answers
• Practice Set 4.2 Class 9 Answers
• Practice Set 4.3 Class 9 Answers
• Problem Set 4 Class 9 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 13.1 8th Std Maths Answers Solutions Chapter 13 Congruence of Triangles.

Congruence of Triangles Class 8 Maths Chapter 13 Practice Set 13.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 13.1 Chapter 13 Solutions Answers

Congruence of Triangles Practice Set 13.1 Question 1.
In each pair of triangles in the following figures, parts bearing identical marks are congruent. State the test and correspondence of vertices by which triangles in each pair are congruent.
i.

ii.

iii.

iv.

v.

Solution:
i.

The two triangles are congruent by SAS test in the correspondence XWZ ↔ YWZ.

ii.

The two triangles are congruent by hypotenuse-side test in the correspondence KJI ↔ LJI.

iii.

The two triangles are congruent by SSS test in the correspondence HEG ↔ FGE.

iv.
The two triangles are congruent by ASA test is the correspondence SMA ↔ OPT.

v.

The two triangles are congruent by ASA test or SAS test or SAA test in the correspondence MTN ↔ STN.

Maharashtra Board Class 8 Maths Chapter 13 Congruence of Triangles Practice Set 13.1 Intext Questions and Activities

Practice Set 13.1 Question 1.
Write answers to the following questions referring to the given figure.

1. Which is the angle opposite to the side DE?
2. Which is the side opposite to ∠E?
3. Which angle is included by side DE and side DF?
4. Which side is included by ∠E and ∠F?
5. State the angles adjacent to side DE. (Textbook pg, no. 81)

Solution:

1. ∠DFE i.e. ∠F is the angle opposite to side DE.
2. Side DF is the side opposite to ∠E.
3. ∠EDF i.e. ∠D is included by side DE and side DF.
4. Side EF is included by ∠E and ∠F.
5. ∠DEF and ∠EDF i.e. ∠E and ∠D are adjacent to side DE.

Congruence of Triangles Class 8th Practice Set 13.1 Question 2.
In the given figure, parts of triangles indicated by identical marks are congruent.
a. Identify the one-to-one correspondence of vertices in which the two triangles are congruent and write the congruence.
b. State with reason, whether the statement, ∆XYZ ≅ ∆STU is right or wrong. (Textbook pg. no. 82)

Solution:
a. From the figure,
S ↔ X, T ↔ Z, U ↔ Y i.e.,
STU ↔ XZY, or SUT ↔ XYZ, or
TUS ↔ ZYX, or TSU ↔ ZXY, or
UTS ↔ YZX, or UST ↔ YXZ

∴ ∆STU ≅ ∆XZY, or ∆SUT ≅ ∆XYZ, or
∆TUS ≅ ∆ZYX, or ∆TSU ≅ ∆ZXY, or
∆UTS ≅ ∆YZX, or ∆UST ≅ ∆YXZ

b. If ∆XYZ ≅ ∆STU, then
∠Y ≅ ∠T, ∠Z ≅ ∠U,
seg XY ≅ seg ST, seg XZ ≅ seg SU
∴ But, all the above statements are wrong. The statement AXYZ ≅ ASTU is wrong.

Std 8 Maths Digest

• Practice Set 11.1 Class 8 Answers
• Practice Set 11.2 Class 8 Answers
• Practice Set 11.3 Class 8 Answers
• Practice Set 12.1 Class 8 Answers
• Practice Set 12.2 Class 8 Answers
• Practice Set 13.1 Class 8 Answers
• Practice Set 13.2 Class 8 Answers
• Practice Set 14.1 Class 8 Answers
• Practice Set 14.2 Class 8 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.5 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 3 Triangles.

9th Standard Maths 2 Practice Set 3.5 Chapter 3 Triangles Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 3.5 Chapter 3 Triangles Questions With Answers Maharashtra Board

Question 1.
If ∆XYZ ~ ∆LMN, write the corresponding angles of the two triangles and also write the ratios of corresponding sides.
Solution:
∆XYZ ~ ∆LMN [Given]
∴ ∠X ≅ ∠L
∠Y ≅ ∠M >
∠Z ≅ ∠N [Corresponding angles of similar triangles]
\( \frac{\mathrm{XY}}{\mathrm{LM}}=\frac{\mathrm{YZ}}{\mathrm{MN}}=\frac{\mathrm{XZ}}{\mathrm{LN}}\) [Corresponding sides of similar triangles]

Question 2.
In ∆XYZ, XY = 4 cm, YZ = 6 cm, XZ = 5 cm. If ∆XYZ ~ ∆PQR and PQ = 8 cm, then find the lengths of remaining sides of ∆PQR.
Solution:
∆XYZ ~ ∆PQR [Given]
∴ \( \frac{\mathrm{XY}}{\mathrm{PQ}}=\frac{\mathrm{YZ}}{\mathrm{QR}}=\frac{\mathrm{XZ}}{\mathrm{PR}}\) [Corresponding sides of similar triangles]

∴ PR = 10 cm
∴ QR = 12 cm, PR = 10cm

Question 3.
Draw a sketch of a pair of similar triangles. Label them. Show their corresponding angles by the same signs. Show the lengths of corresponding sides by numbers in proportion.
Solution:

∆GHI ~ ∆STU

Maharashtra Board Class 9 Maths Chapter 3 Triangles Practice Set 3.5 Intext Questions and Activities

Question 1.
We have learnt that if two triangles are equiangular then their sides are in proportion. What do you think if two quadrilaterals are equiangular? Are their sides in proportion? Draw different figures and verify. Verify the same for other polygons. (Textbook pg no 50)
Answer:
If two quadrilaterals are equiangular then their sides will not necessarily be in proportion.
Case 1: The two quadrilaterals are of the same type.

Consider squares ABCD and PQRS.
∠A = ∠P, ∠B = ∠Q, ∠C = ∠R, ∠D = ∠S
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{CD}}{\mathrm{RS}}=\frac{\mathrm{AD}}{\mathrm{PS}}\)

Case 2: The two quadrilaterals are of different types.

Consider square ABCD and rectangle STUV.
∠A = ∠S, ∠B = ∠T, ∠C = ∠U, ∠D = ∠V

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Problem Set 3 Class 9 Answers
• Practice Set 4.1 Class 9 Answers
• Practice Set 4.2 Class 9 Answers
• Practice Set 4.3 Class 9 Answers
• Problem Set 4 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.4 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 3 Triangles.

9th Standard Maths 2 Practice Set 3.4 Chapter 3 Triangles Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 3.4 Chapter 3 Triangles Questions With Answers Maharashtra Board

Question 1.
In the adjoining figure, point A is on the bisector of ∠XYZ. If AX = 2 cm, then find AZ.

Solution:
AX = 2 cm [Given]
Point A lies on the bisector of ∠XYZ. [Given]
Point A is equidistant from the sides of ∠XYZ. [Every point on the bisector of an angle is equidistant from the sides of the angle]
∴ A Z = AX
∴ AZ = 2 cm

Question 2.
In the adjoining figure, ∠RST = 56°, seg PT ⊥ ray ST, seg PR ⊥ ray SR and seg PR ≅ seg PT. Find the measure of ∠RSP.
State the reason for your answer.

Solution:
seg PT ⊥ ray ST, seg PR ⊥ ray SR [Given]
seg PR ≅ seg PT
∴ Point P lies on the bisector of ∠TSR [Any point equidistant from the sides of an angle is on the bisector of the angle]
∴ Ray SP is the bisector of ∠RST.
∠RSP = 56° [Given]
∴ ∠RSP = \(\frac { 1 }{ 2 }\)∠RST
= \(\frac { 1 }{ 2 }\) x 56°
∴ ∠RSP = 28°

Question 3.
In ∆PQR, PQ = 10 cm, QR = 12 cm, PR triangle. 8 cm. Find out the greatest and the smallest angle of the triangle.

Solution:
In ∆PQR,
PQ = 10 cm, QR = 12 cm, PR = 8 cm [Given]
Since, 12 > 10 > 8
∴ QR > PQ > PR
∴ ∠QPR > ∠PRQ > PQR [Angle opposite to greater side is greater]
∴ In ∆PQR, ∠QPR is the greatest angle and ∠PQR is the smallest angle.

Question 4.
In ∆FAN, ∠F = 80°, ∠A = 40°. Find out the greatest and the smallest side of the triangle. State the reason.

Solution:
In ∆FAN,
∠F + ∠A + ∠N = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 80° + 40° + ∠N = 180°
∴ ∠N = 180° – 80° – 40°
∴∠N = 60°
Since, 80° > 60° > 40°
∴ ∠F > ∠N > ∠A
∴  AN > FA > FN [Side opposite to greater angle is greater]
∴  In ∆FAN, AN is the greatest side and FN is the smallest side.

Question 5.
Prove that an equilateral triangle is equiangular.

Given: ∆ABC is an equilateral triangle.
To prove: ∆ABC is equiangular
i.e. ∠A ≅ ∠B ≅ ∠C …(i) [Sides of an equilateral triangle]
In ∆ABC,
seg AB ≅ seg BC [From (i)]
∴ ∠C = ∠A (ii) [Isosceles triangle theorem]
In ∆ABC,
seg BC ≅ seg AC [From (i)]
∴ ∠A ≅ ∠B (iii) [Isosceles triangle theorem]
∴ ∠A ≅ ∠B ≅ ∠C [From (ii) and (iii)]
∴ ∆ABC is equiangular.

Question 6.
Prove that, if the bisector of ∠BAC of ∆ABC is perpendicular to side BC, then AABC is an isosceles triangle.

Given: Seg AD is the bisector of ∠BAC.
seg AD ⊥ seg BC
To prove: AABC is an isosceles triangle.
Proof.
In ∆ABD and ∆ACD,
∠BAD ≅ ∠CAD [seg AD is the bisector of ∠BAC]
seg AD ≅ seg AD [Common side]
∠ADB ≅ ∠ADC [Each angle is of measure 90°]
∴ ∆ABD ≅ ∆ACD [ASA test]
∴ seg AB ≅ seg AC [c. s. c. t.]
∴ ∆ABC is an isosceles triangle.

Question 7.
In the adjoining figure, if seg PR ≅ seg PQ, show that seg PS > seg PQ.

Solution:
Proof.
In ∆PQR,
seg PR ≅ seg PQ [Given]
∴ ∠PQR ≅ ∠PRQ ….(i) [Isosceles triangle theorem]
∠PRQ is the exterior angle of ∆PRS.
∴ ∠PRQ > ∠PSR ….(ii) [Property of exterior angle]
∴ ∠PQR > ∠PSR [From (i) and (ii)]
i.e. ∠Q > ∠S ….(iii)
In APQS,
∠Q > ∠S [From (iii)]
∴ PS > PQ [Side opposite to greater angle is greater]
∴ seg PS > seg PQ

Question 8.
In the adjoining figure, in AABC, seg AD and seg BE are altitudes and AE = BD. Prove that seg AD = seg BE.

Solution:
Proof:
In ∆ADB and ∆BEA,
seg BD ≅ seg AE [Given]
∠ADB ≅ ∠BEA = 90° [Given]
seg AB ≅ seg BA [Common side]
∴ ∆ADB ≅ ∆BEA [Hypotenuse-side test]
∴ seg AD ≅ seg BE [c. s. c. t.]

Maharashtra Board Class 9 Maths Chapter 3 Triangles Practice Set 3.4 Intext Questions and Activities

Question 1.
As shown in the given figure, draw ∆XYZ such that side XZ > side XY. Find which of ∠Z and ∠Y is greater. (Textbook pg. no. 41)

Answer:
From the given figure, ∠Z = 25° and ∠Y = 51°
∴ ∠Y is greater.

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Problem Set 3 Class 9 Answers
• Practice Set 4.1 Class 9 Answers
• Practice Set 4.2 Class 9 Answers
• Practice Set 4.3 Class 9 Answers
• Problem Set 4 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.3 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 3 Triangles.

9th Standard Maths 2 Practice Set 3.3 Chapter 3 Triangles Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 3.3 Chapter 3 Triangles Questions With Answers Maharashtra Board

Question 1.
Find the values of x and y using the information shown in the given figure. Find the measures of ∠ABD and ∠ACD.

Solution:
i. ∠ACB = 50° [Given]
In ∆ABC, seg AC ≅ seg AB [Given]
∴ ∠ABC ≅ ∠ACB [Isosceles triangle theorem]
∴ x = 50°

ii. ∠DBC = 60° [Given]
In ABDC, seg BD ≅ seg DC [Given]
∴ ∠DCB ≅ ∠DBC [Isosceles triangle theorem]
∴ y = 60°

iii. ∠ABD = ∠ABC + ∠DBC [Angle addition property]
= 50° + 60°
∴ ∠ABD = 110°

iv. ∠ACD = ∠ACB + ∠DCB [Angle addition property]
= 50° + 60°
∴ ∠ACD = 110°
∴ x = 50°, y = 60°,
∠ABD = 110°, ∠ACD = 110°

Question 2.
The length of hypotenuse of a right angled triangle is 15. Find the length of median on its hypotenuse.

Solution:
Length of hypotenuse = 15 [Given]
Length of median on the hypotenuse = \(\frac { 1 }{ 2 }\) x length of hypotenuse [In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse]
= \(\frac { 1 }{ 2 }\) x 15 = 7.5
∴ The length of the median on the hypotenuse is 7.5 units.

Question 3.
In ∆PQR, ∠Q = 90°, PQ = 12, QR = 5 and QS is a median. Find l(QS).

Solution:
i. PQ = 12, QR = 5 [Given]
In APQR, ∠Q = 90° [Given]
∴ PR² = QR² + PQ² [Pythagoras theorem]
= 25 + 144
∴ PR² =169
∴ PR = 13 units [Taking square root of both sides]

ii. In right angled APQR, seg QS is the median on hypotenuse PR.
∴ QS = \(\frac { 1 }{ 2 }\)PR [In a right angled triangle, the length of the median on the hypotenuse is half the length of the hypotenuse]
= \(\frac { 1 }{ 2 }\) x 13
∴ l(QS) = 6.5 units

Question 4.
In the given figure, point G is the point of concurrence of the medians of ∆PQR. If GT = 2.5, find the lengths of PG and PT.

Solution:
i. In ∆PQR, G is the point of concurrence of the medians. [Given]
The centroid divides each median in the ratio 2 : 1.
PG : GT = 2 : 1

∴ PG = 2 x 2.5
∴ PG = 5 units

ii. Now, PT = PG + GT [P – G – T]
= 5 + 2.5
∴ l(PG) = 5 units, l(PT) = 7.5 units

Maharashtra Board Class 9 Maths Chapter 3 Triangles Practice Set 3.3 Intext Questions and Activities

Question 1.
Can the theorem of isosceles triangle be proved by doing a different construction? (Textbook pg. no.34)

Solution:
Yes
Construction: Draw seg AD ⊥ seg BC.
Proof:
In ∆ABD and ∆ACD,
seg AB≅ seg AC [Given]
∠ADB ≅ ∠ADC [Each angle is of measure 90°]
seg AD ≅ seg AD [Common side]
∴ ∆ABD ≅ ∆ACD [Hypotenuse side test]
∴ ∠ABD ≅ ∠ACD [c.a.c.t.]
∴ ∠ABC ≅ ∠ACB [B-D-C]

Question 2.
Can the theorem of isosceles triangle be proved without doing any construction? (Textbook pg, no.34)

Solution:
Yes
Proof:
In ∆ABC and ∆ACB,
seg AB ≅ seg AC [Given]
∠BAC ≅ ∠CAB [Common angle]
seg AC ≅ seg AB [Given]
∴ ∆ABC ≅ ∆ACB [SAS test]
∴ ∠ABC ≅ ∠ACB [c. a. c. t.]

Question 3.
In the given figure, ∆ABC is a right angled triangle, seg BD is the median on hypotenuse. Measure the lengths of the following segments.
i. AD
ii. DC
iii. BD
From the measurements verify that BD = \(\frac { 1 }{ 2 }\)AC. (Textbook pg. no. 37)
Solution:
AD = DC = BD= 1.9 cm
AC = AD + DC [A – D – C]
= 1.9 + 1.9
= 2 x 1.9 cm
∴ AC = 2 x BD
∴ BD = \(\frac { 1 }{ 2 }\) AC

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Problem Set 3 Class 9 Answers
• Practice Set 4.1 Class 9 Answers
• Practice Set 4.2 Class 9 Answers
• Practice Set 4.3 Class 9 Answers
• Problem Set 4 Class 9 Answers