Practice Set 4.2 Geometry 9th Standard Maths Part 2 Chapter 4 Constructions of Triangles Solutions Maharashtra Board

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:
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.2 1
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.
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.2 2

Question 2.
Construct ∆PQR, such that QR = 6.5 cm, ∠PQR = 60° and PQ – PR = 2.5 cm.
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.2 3
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.
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.2 4

Question 3.
Construct ∆ABC, such that BC = 6 cm, ∠ABC = 100° and AC – AB = 2.5 cm.
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.2 5
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.
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.2 6

Class 9 Maths Digest

Practice Set 14.1 Class 8 Answers Chapter 14 Compound Interest Maharashtra Board

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
Maharashtra Board Class 8 Maths Solutions Chapter 14 Compound Interest Practice Set 14.1 1
= 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
Maharashtra Board Class 8 Maths Solutions Chapter 14 Compound Interest Practice Set 14.1 2
∴ 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
Maharashtra Board Class 8 Maths Solutions Chapter 14 Compound Interest Practice Set 14.1 3
∴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
Maharashtra Board Class 8 Maths Solutions Chapter 14 Compound Interest Practice Set 14.1 4
= 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
Maharashtra Board Class 8 Maths Solutions Chapter 14 Compound Interest Practice Set 14.1 5
Maharashtra Board Class 8 Maths Solutions Chapter 14 Compound Interest Practice Set 14.1 6
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 4.1 Geometry 9th Standard Maths Part 2 Chapter 4 Constructions of Triangles Solutions Maharashtra Board

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:
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.1 1
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.
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.1 2

Question 2.
Construct ∆XYZ, in which YZ = 6 cm, XY + XZ = 9 cm, ∠XYZ = 50°.
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.1 3
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.
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.1 4

Question 3.
Construct ∆ABC, in which BC = 6.2 cm, ∠ACB = 50°, AB + AC = 9.8 cm.
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.1 5
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.
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.1 6

Question 4.
Construct ∆ABC, in which BC = 3.2 cm, ∠ACB = 45° Solution:and perimeter of AABC is 10 cm.
Solution:
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.1 7
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.
Maharashtra Board Class 9 Maths Solutions Chapter 4 Constructions of Triangles Practice Set 4.1 8

Class 9 Maths Digest

Problem Set 3 Geometry 9th Standard Maths Part 2 Chapter 3 Triangles Solutions Maharashtra Board

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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 1
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,
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 2Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 3
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.

Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 4
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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 5
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
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 6
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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 7
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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 8
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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 9
∴ ∠QPT = ∠PSR …(ii) [Corresponding angles]
seg PT || seg SR [Construction]
and seg PR is their transversal.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 10
∴ ∠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).
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 11
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 Solutions Chapter 3 Triangles Problem Set 3 12

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)
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 13

Question 2.
Check the congruence of triangles.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 14
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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 15
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 16
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)
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 17
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)
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 18
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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 19
Number of triangle: 1
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 20
Number of triangles: 4
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 21
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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Problem Set 3 22
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 13.1 Class 8 Answers Chapter 13 Congruence of Triangles Maharashtra Board

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.
Maharashtra Board Class 8 Maths Solutions Chapter 13 Congruence of Triangles Practice Set 13.1 1
ii.
Maharashtra Board Class 8 Maths Solutions Chapter 13 Congruence of Triangles Practice Set 13.1 2
iii.
Maharashtra Board Class 8 Maths Solutions Chapter 13 Congruence of Triangles Practice Set 13.1 3
iv.
Maharashtra Board Class 8 Maths Solutions Chapter 13 Congruence of Triangles Practice Set 13.1 4
v.
Maharashtra Board Class 8 Maths Solutions Chapter 13 Congruence of Triangles Practice Set 13.1 5
Solution:
i.
Maharashtra Board Class 8 Maths Solutions Chapter 13 Congruence of Triangles Practice Set 13.1 6
The two triangles are congruent by SAS test in the correspondence XWZ ↔ YWZ.

ii.
Maharashtra Board Class 8 Maths Solutions Chapter 13 Congruence of Triangles Practice Set 13.1 7
The two triangles are congruent by hypotenuse-side test in the correspondence KJI ↔ LJI.

iii.
Maharashtra Board Class 8 Maths Solutions Chapter 13 Congruence of Triangles Practice Set 13.1 8
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.
Maharashtra Board Class 8 Maths Solutions Chapter 13 Congruence of Triangles Practice Set 13.1 9
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.
Maharashtra Board Class 8 Maths Solutions Chapter 13 Congruence of Triangles Practice Set 13.1 10

  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)
Maharashtra Board Class 8 Maths Solutions Chapter 13 Congruence of Triangles Practice Set 13.1 11
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 3.5 Geometry 9th Standard Maths Part 2 Chapter 3 Triangles Solutions Maharashtra Board

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]
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.5 1
∴ 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:
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.5 2
∆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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.5 3
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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.5 4
Consider square ABCD and rectangle STUV.
∠A = ∠S, ∠B = ∠T, ∠C = ∠U, ∠D = ∠V
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.5 5

Class 9 Maths Digest

Practice Set 3.4 Geometry 9th Standard Maths Part 2 Chapter 3 Triangles Solutions Maharashtra Board

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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.4 1
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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.4 2
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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.4 3
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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.4 4
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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.4 5
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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.4 6
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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.4 7
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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.4 8
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)
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.4 9
Answer:
From the given figure, ∠Z = 25° and ∠Y = 51°
∴ ∠Y is greater.

Class 9 Maths Digest

Practice Set 3.3 Geometry 9th Standard Maths Part 2 Chapter 3 Triangles Solutions Maharashtra Board

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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.3 1
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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.3 2
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).
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.3 3
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.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.3 4
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
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.3 5
∴ 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)
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.3 6
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)
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.3 7
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 12.2 Class 8 Answers Chapter 12 Equations in One Variable Maharashtra Board

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 12.2 8th Std Maths Answers Solutions Chapter 12 Equations in One Variable.

Equations in One Variable Class 8 Maths Chapter 12 Practice Set 12.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 12.2 Chapter 12 Solutions Answers

Equation In One Variable Practice Set 12.2 Question 1.
Mother is 25 years older than her son. Find son’s age, if after 8 years ratio of son’s age to mother’s age will be \(\frac { 4 }{ 9 }\).
Solution:
Let the son’s present age be x years.
∴ Mother’s present age = (x + 25) years
After 8 years,
Son’s age = (x + 8) years
Mother’s age = (x + 25 + 8) = (x + 33) years
Since, the ratio of the son’s age to mother’s age after 8 years is \(\frac { 4 }{ 9 }\).
∴ \(\frac{x+8}{x+33}=\frac{4}{9}\)
∴ 9 (x + 8) = 4 (x + 33)
∴ 9x + 72 = 4x + 132
∴ 9x – 4x = 132 – 72
∴ 5x = 60
∴ x = \(\frac { 60 }{ 5 }\)
∴ x = 12
∴ Son’s present age is 12 years.

8th Std Maths Practice Set 12.2 Question 2.
The denominator of a fraction is greater than its numerator by 12. If the numerator is decreased by 2 and the denominator is increased by 7, the new fraction is equivalent to \(\frac { 1 }{ 2 }\) . Find the fraction.
Solution:
Let the numerator of the fraction be x.
The denominator of a fraction is greater than its numerator by 12.
∴ Denominator of the fraction = (x + 12)
∴ The required fraction = \(\frac { x }{ x+12 }\)
For the new fraction,
numerator is decreased by 2.
∴ The new numerator = (x – 2)
Also, denominator is increased by 7.
∴ The new denominator = (x + 12) + 7
= (x + 19)
Since, the new fraction is equivalent to \(\frac { 1 }{ 2 }\).
∴ \(\frac{x-2}{x+19}=\frac{1}{2}\)
∴ 2(x – 2) = 1(x + 19)
∴ 2x – 4 = x + 19
∴ 2x – x = 19 + 4
∴ x = 23
∴ The required fraction = \(\frac{x}{x+12}=\frac{23}{23+12}=\frac{23}{35}\)
∴ The required fraction is \(\frac { 23 }{ 35 }\)

Practice Set 12.2 Class 8 Question 3.
The ratio of the weights of copper and zinc in brass is 13:7. Find the weight of zinc in a brass utensil weighing 700 gm.
Solution:
Let the weight of zinc in the brass utensil be x gm.
Since, the ratio of the weights of copper to zinc in brass is 13:7.
Maharashtra Board Class 8 Maths Solutions Chapter 12 Equations in One Variable Practice Set 12.2 1
∴ Weight of copper in the brass utensil = \(\left(\frac{13}{7} x\right)\) gm
The weight of the brass utensil = 700 gm
∴ \(\frac { 13 }{ 7 }x+x=700\)
∴ \(\frac { 13 }{ 7 }x\) x × 7 + x × 7 = 700 × 7
∴ 13x + 7x = 4900
∴ 20x = 4900
∴ \(x=\frac { 4900 }{ 20 }\)
∴ x = 245
∴ The weight of zinc in the brass utensil is 245 gm.

Practice Set 12.2 8th Class Question 4.
Find three consecutive whole numbers whose sum is more than 45 but less than 54.
Solution:
Let the three consecutive whole numbers be (x – 1), x and (x + 1).
∴ Sum of the three numbers
= (x – 1) + x + (x + 1)
= 3x
Given that, the sum of the three numbers is greater than 45 and less than 54.
When the sum of the three numbers is 45,
3x = 45
∴ x = \(\frac { 45 }{ 3 }\)
∴ x = 15
When the sum of the three numbers is 54,
∴ 3x = 54
∴ x = \(\frac { 54 }{ 3 }\)
∴ x = 18
∴ the value of x is greater than 15 and less than 18.
∴ the value of x is either 16 or 17

Case I:
If the value of x is 16, then the three consecutive whole numbers are
(16 – 1), 16,(16 + 1)i.e., 15, 16, 17

Case II:
If the value of x is 17, then the three consecutive whole numbers are (17 – 1), 17, (17 + 1) i.e., 16, 17, 18.
∴ The three consecutive whole numbers are 15, 16, 17 or 16, 17, 18.

Practice Set 12.2 8th Standard Question 5.
In a two-digit number, digit at the ten’s place is twice the digit at unit’s place. If the number obtained by interchanging the digits is added to the original number, the sum is 66. Find the number.
Solution:
Let the digit at unit’s place be x.
The digit at the ten’s place is twice the digit at unit’s place.
∴ The digit at ten’s place = 2x

Digit in units place Digit in tens place Number
Original Number x 2x (2x × 10) + x = 20x + x = 21x
New Number 2x x (x × 10) + 2x = 10x + 2x = 12x

Since, the sum of the original number and the new number is 66.
∴ 21x + 12x = 66
∴ 33x = 66
∴ x = \(\frac { 66 }{ 33 }\)
∴ x = 2
∴ Original number = 21x = 21 × 2 = 42
∴ the original number is 42.

8th Standard Maths Practice Set 12.2 Question 6.
Some tickets of Rs 200 and some of Rs 100, of a drama in a theatre were sold. The number of tickets of Rs 200 sold was 20 more than the number of tickets of Rs 100. The total amount received by the theatre by sale of tickets was Rs 37000. Find the number of Rs 100 tickets sold.
Solution:
Let the number of tickets sold of Rs 100 be x.
The number of tickets of Rs 200 sold was 20 more than the number of tickets of Rs 100.
∴ Number of tickets sold of Rs 200 = (x + 20)
∴ Total amount received by the theatre through the sale of tickets = 100 × x + 200 × (x + 20)
= 100x + 200x + 4000
= 300x + 4000
Since, the total amount received by the theatre through the sale of tickets = Rs 37000
∴ 300x + 4000 = 37000
∴ 300x = 37000 – 4000
∴ 300x = 33000
∴ \(x=\frac { 33000 }{ 300 }\)
∴ x = 110
∴ 110 tickets of Rs 100 were sold.

8th Maths Practice Set 12.2 Question 7.
Of the three consecutive natural numbers, five times the smallest number is 9 more than four times the greatest number, find the numbers.
Solution:
Let the three consecutive natural numbers be (x – 1), x and (x + 1).
Here, the smallest number is (x – 1) and the greatest number is (x + 1).
Since, five times the smallest number is 9 more than four times the greatest number.
∴ 5 × (x – 1) = [4 × (x + 1)] + 9
∴ 5x – 5 = 4x + 4 + 9
∴ 5x – 5 = 4x + 13
∴ 5x – 4x = 13 + 5
∴ x = 18 .
∴ the three numbers are (18 – 1), 18, (18 + 1)
i. e., 17, 18, 19
∴ The three consecutive natural numbers are 17,18 and 19.

Raju Sold A Bicycle to Amit at 8 Question 8.
Raju sold a bicycle to Amit at 8% profit. Amit repaired it spending Rs 54. Then he sold the bicycle to Nikhil for Rs 1134 with no loss and no profit. Find the cost price of the bicycle for which Raju purchased it.
Solution:
Let the cost price at which Raju purchased the bicycle be Rs x.
Since, Raju sold the bicycle at 8% profit to Amit.
∴ Selling price of bicycle for Raju = x + 8% of x
Maharashtra Board Class 8 Maths Solutions Chapter 12 Equations in One Variable Practice Set 12.2 2
Since, Amit spent Rs 54 on repairing the bicycle and then sold it to Nikhil for Rs 1134, at no loss and no profit.
∴ Selling price of bicycle + repairing cost = Rs 1134
Maharashtra Board Class 8 Maths Solutions Chapter 12 Equations in One Variable Practice Set 12.2 3
∴ The cost price of the bicycle at which Raju purchased it is Rs 1000.

Class 8 Maths Practice Set 12.2 Question 9.
A cricket player scored 180 runs in the first match and 257 runs in the second match. Find the number of runs he should score in the third match so that the average of runs in the three matches be 230.
Solution:
Let the number of runs required by the cricket player to score in the third match be x.
Number of runs scored by the player in first match = 180
Number of runs scored in second match = 257
∴ Total runs scored by the player = 180 + 257 + x = 437 + x
Average of runs in the three matches = \(\frac { 437+x }{ 3 }\)
Since, the average of runs should be 230.
\(\frac { 437+x }{ 3 }=230\)
∴ 437 + x = 230 × 3
∴ 437 + x = 690
∴ x = 690 – 437
∴ x = 253
∴ The cricket player should score 253 runs in the third match.

8th Class Math Practice Set 12.2 Question 10.
Sudhir’s present age is 5 more than three times the age of Viru. Anil’s age is half the age of Sudhir. If the ratio of the sum of Sudhir’s and Viru’s age to three times Anil’s age is 5:6, then find Viru’s age.
Solution:
Let Viru’s present age be x years.
Sudhir’s present age is 5 more than three times the age of Viru.
∴ Sudhir’s present age = (3x + 5) years
Anil’s age is half the age of Sudhir.
∴ Anil’s present age = \(\left(\frac{3 x+5}{2}\right)\) years
Since, the ratio of the sum of Sudhir’s and Viru’s age to three times Anil’s age is 5:6.
Maharashtra Board Class 8 Maths Solutions Chapter 12 Equations in One Variable Practice Set 12.2 4
∴ 2 × (24x + 30) = 45x + 75
∴ 48x + 60 = 45x + 75
∴ 48x – 45x = 75 – 60
∴ 3x = 15
∴ x = \(\frac { 15 }{ 3 }\)
∴ x = 5
∴ Viru’s present age is 5 years.

Maharashtra Board Class 8 Maths Chapter 12 Equations in One Variable Practice Set 12.2 Intext Questions and Activities

8th Math Practice Set 12.2 Question 1.
Write correct numbers in the boxes given. (Textbook pg. no. 78)
length is 3 times the breadth
Maharashtra Board Class 8 Maths Solutions Chapter 12 Equations in One Variable Practice Set 12.2 5
Perimeter of the rectangle = 40
2(__x + __x) = 40
2 × __ x = 40
__ x = 40
x = __
∴ Breadth of rectangle = __ cm and Length of rectangle = __ cm
Solution:
length is 3 times the breadth
Maharashtra Board Class 8 Maths Solutions Chapter 12 Equations in One Variable Practice Set 12.2 6
Perimeter of the rectangle = 40
∴ 2(3x + 1x) = 40
∴ 2 × 4x = 40
∴ 8x = 40
∴ x = 5
∴ Breadth of rectangle = 5 cm and Length of rectangle = 15 cm

Std 8 Maths Digest

Practice Set 3.1 Geometry 9th Standard Maths Part 2 Chapter 3 Triangles Solutions Maharashtra Board

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 3 Triangles.

9th Standard Maths 2 Practice Set 3.1 Chapter 3 Triangles Textbook Answers Maharashtra Board

Class 9 Maths Part 2 Practice Set 3.1 Chapter 3 Triangles Questions With Answers Maharashtra Board

Practice Set 3.1 Geometry 9th Standard Question 1.
In the adjoining figure, ∠ACD is an exterior angle of ∆ABC. ∠B = 40°, ∠A = 70°. Find the measure of ∠ACD.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.1 1
Solution:
∠A = 70° , ∠B = 40° [Given]
∠ACD is an exterior angle of ∆ABC. [Given]
∴ ∠ACD = ∠A + ∠B
= 70° + 40°
∴ ∠ACD = 110°

Question 2.
In ∆PQR, ∠P = 70°, ∠Q = 65°, then find ∠R.
Solution:
∠P = 70°, ∠Q = 65° [Given]
In ∆PQR,
∠P + ∠Q + ∠R = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 70° + 65° + ∠R = 180°
∴ ∠R = 180° – 70° – 65°
∴ ∠R = 45°

Practice Set 3.1 Geometry 9th Question 3.
The measures of angles of a triangle are x°, (x – 20)°, (x – 40)°. Find the measure of each angle.
Solution:
The measures of the angles of a triangle are x°, (x – 20)°, (x – 40)°. [Given]
∴ x°+ (x – 20)° + (x – 40)° = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 3x – 60 = 180
∴ 3x = 180 + 60
∴ 3x = 240
∴ x = 240
∴ x = \(\frac { 240 }{ 3 }\)
∴ x = 80°
∴ The measures of the remaining angles are
x – 20° = 80° – 20° = 60°,
x – 40° = 80° – 40° = 40°
∴ The measures of the angles of the triangle are 80°, 60° and 40°.

9th Class Geometry Practice Set 3.1 Question 4.
The measure of one of the angles of a triangle is twice the measure of its smallest angle and the measure of the other is thrice the measure of the smallest angle. Find the measures of the three angles.
Solution:
Let the measure of the smallest angle be x°.
One of the angles is twice the measure of the smallest angle.
∴ Measure of that angle = 2x°
Another angle is thrice the measure of the smallest angle.
∴ Measure of that angle = 3x°
∴ The measures of the remaining two angles are 2x° and 3x°.
Now, x° + 2x° + 3x° = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 6x = 180
∴ x = 180
∴ x = \(\frac { 180 }{ 6 }\)
∴ x° = 30°
The measures of the remaining angles are 2x° = 2 x 30° = 60°
3x° = 3 x 30° = 90°
The measures of the three angles of the triangle are 30°, 60° and 90°.

Question 5.
In the adjoining figure, measures of some angles are given. Using the measures, find the values of x, y, z.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.1 2
Solution:
i. ∠NET = 100° and ∠EMR = 140°
∠EMN + ∠EMR = 180°
∴ z +140° =180°
∴ z = 180° -140°
∴ z = 40°

ii. Also, ∠NET + ∠NEM = 180° [Angles in a linear pair]
∴ 100° + y = 180°
∴ y = 180° – 100°
∴ y = 80°

iii. In ∆ENM,
∴ ∠ENM + ∠NEM + ∠EMN = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ x +80°+ 40°= 180°
∴ x = 180° – 80° – 40°
∴ x = 60°
∴ x = 60°, = 80°, z = 40°

Question 6.
In the adjoining figure, line AB || line DE. Find the measures of ∠DRE and ∠ARE using given measures of some angles.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.1 3
Solution:
i. ∠B AD = 70°, ∠DER = 40° [Given]
line AB || line DE and seg AD is their transversal.
∴ ∠EDA = ∠BAD [Alternate Angles]
∴ ∠EDA = 70° ….(i)
In ∆DRE,
∠EDR + ∠DER + ∠DRE = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 70°+ 40° +∠DRE = 180° [From (i) and D – R – A]
∴ ∠DRE = 180° -70° -40°
∴ ∠DRE = 70°

ii. ∠DRE + ∠ARE = 180° [Angles in a linear pair]
∴ 70° + ∠ARE = 180°
∴ ∠ARE = 180°-70°
∴ ∠ARE =110°
∴ ∠DRE = 70°, ∠ARE = 110°

Triangles Class 9 Practice Set 3.1 Question 7.
In ∆ABC, bisectors of ∠A and ∠B intersect at point O. If ∠C = 70°, find the measure of ∠AOB.
Solution:
∠OAB = ∠OAC = – ∠BAC ….(i) [Seg AO bisects ∠BAC]
∠OBA = ∠OBC = – ∠ABC …..(ii) [Seg RO bisects ∠ABC]
In AABC,
∠BAC + ∠ABC + ∠ACB = 180° [Sum of the measures of the angles of a triangle is 180°]
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.1 4
∴ ∠BAC + ∠ABC + 70° = 180°
∴ ∠BAC + ∠ABC = 180°- 70°
∴ ∠BAC + ∠ABC = 110°
∴ \(\frac { 1 }{ 2 }\)(∠BAC) + \(\frac { 1 }{ 2 }\) (∠ABC) = \(\frac { 1 }{ 2 }\) x 110° [MuItiplying both sides by \(\frac { 1 }{ 2 }\)]
∴ ∠OAB + ∠OBA = 55° ….(iii) [From (i) and (ii)]
In AOAB,
∠OAB + ∠OBA + ∠AOB = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 55° + ∠AOB = 180° [From (iii)]
∴ ∠AOB = 180°- 55°
∴ ∠AOB = 125°

Question 8.
In the adjoining figure, line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of ∠BPQ and ∠PQD respectively. Prove that m ∠PTQ = 90°.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.1 5
Given: line AB || line CD and line PQ is the transversal.
ray PT and ray QT are the bisectors of ∠BPQ and ∠PQD respectively.
To prove: m∠PTQ = 90°
Solution:
Proof:
∠TPB = ∠TPQ = \(\frac { 1 }{ 2 }\)∠BPQ …(i) [Ray PT bisects ∠BPQ]
∠TQD = ∠TQP = \(\frac { 1 }{ 2 }\)∠PQD ….(ii) [Ray QT bisects ∠PQD]
line AB || line CD and line PQ is their transversal. [Given]
∴∠BPQ + ∠PQD = 180° [Interior angles]
∴ \(\frac { 1 }{ 2 }\) (∠BPQ) + \(\frac { 1 }{ 2 }\) (∠PQD) = \(\frac { 1 }{ 2 }\) x 180° [Multiplying both sides by \(\frac { 1 }{ 2 }\)]
∠TPQ + ∠TQP = 90°
In ∆PTQ,
∠TPQ + ∠TQP + ∠PTQ = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 90° + ∠PTQ = 180° [From (iii)]
∴ ∠PTQ = 180° – 90°
= 90°
∴ m∠PTQ = 90°

Triangle Practice Set 3.1 Question 9.
Using the information in the adjoining figure, find the measures of ∠a, ∠b and ∠c.
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.1 6
Solution:
i. ∠c + 100° = 180° [Angles in a linear pair]
∴ ∠c = 180° – 100°
∴ ∠c = 80°

ii. ∠b = 70° [Vertically opposite angles]
iii. ∠a + ∠b +∠c = 180° [Sum of the measures of the angles of a triangle is 180°]
∠a + 70° + 80° = 1800
∴ ∠a = 180° – 70° – 80°
∴ ∠a = 30°
∴ ∠a = 30°, ∠b = 70°,∠ c = 80°

Practice Set 3.1 Geometry Question 10.
In the adjoining figure, line DE || line GF, ray EG and ray FG are bisectors of ∠DEF and ∠DFM respectively. Prove that,
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.1 7
i. ∠DEG = \(\frac { 1 }{ 2 }\) ∠EDF
ii. EF = FG
Solution:
i. ∠DEG = ∠FEG = x° ….(i) [Ray EG bisects ∠DEF]
∠GFD = ∠GFM = y° …..(ii) [Ray FG bisects ∠DFM]
line DE || line GF and DF is their transversal. [Given]
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.1 8
∴ ∠EDF = ∠GFD [Alternate angles]
∴ ∠EDF = y° ….(iii) [From (ii)]
line DE || line GF and EM is their transversal. [Given]
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.1 9
∴ ∠DEF = ∠GFM [Corresponding angles]
∴ ∠DEG + ∠FEG = ∠GFM [Angle addition property]
∴ x°+ x° = y° [From (i) and (ii)]
∴ 2x° = y°
∴ x° = \(\frac { 1 }{ 2 }\)y°
∴ ∠DEG = \(\frac { 1 }{ 2 }\)∠EDF [From (i) and (iii)]

ii. line DE || line GF and GE is their transversal. [Given]
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.1 11
∴ ∠DEG = ∠FGE …(iv) [Alternate angles]
∴ ∠FEG = ∠FGE ….(v) [From (i) and (iv)]
∴ In ∆FEG,
∠FEG = ∠FGE [From (v)]
∴ EF = FG [Converse of isosceles triangle theorem]

Maharashtra Board Class 9 Maths Chapter 3 Triangles Practice Set 3.1 Intext Questions and Activities

Class 9 Geometry Practice Set 3.1 Question 1. Can you give an alternative proof of the above theorem by drawing a line through point R and parallel to seg PQ in the above figure? (Textbook pg. no. 25)
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.1 12
Solution:
Yes.
Construction: Draw line RM parallel to seg PQ through a point R.
Proof:
seg PQ || line RM and seg PR is their transversal. [Construction]
∴ ∠PRM = ∠QPR ……..(i) [Alternate angles]
seg PQ || line RM and seg QR is their transversal. [Construction]
∴ ∠SRM = ∠PQR ……..(ii) [Corresponding angles]
∴ ∠PRM + ∠SRM = ∠QPR + ∠PQR [Adding (i) and (ii)]
∴ ∠PRS = ∠PQR + ∠QPR [Angle addition property]

3 Triangles Question 2. Observe the given figure and find the measures of ∠PRS and ∠RTS. (Textbook pg. no.25)
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.1 13
Solution:
∠PRS is an exterior angle of ∆PQR.
So from the theorem of remote interior angles,
∠PRS = ∠PQR + ∠QPR
= 40° + 30°
∴ ∠PRS = 70°
∴ ∠TRS=70° …[P – T – R]
In ∆RTS,
∠TRS + ∠RTS + ∠TSR = 180° …[Sum of the measures of the angles of a triangle is 180°]
∴ 70° + ∠RTS + 20° = 180°
∴ ∠RTS + 90° = 180°
∴ ∠RTS = 180°
∴ ∠RTS = 90°

9th Class Geometry Triangles Question 3. In the given figure, bisectors of ∠B and ∠C of ∆ABC intersect at point P. Prove that ∠BPC = 90° + \(\frac { 1 }{ 2 }\)∠BAC.
Complete the proof by filling in the blanks. (Textbook pg. no.27)
Maharashtra Board Class 9 Maths Solutions Chapter 3 Triangles Practice Set 3.1 14
Solution:
Proof:
In ∆ABC,
∠BAC + ∠ABC + ∠ACB = 180° …[Sum of the measures of the angles of a triangle is 180°]
∴ ∠BAC + – ∠ABC + ∠ACB = 180 … [Multiplying each term by \(\frac { 1 }{ 2 }\)]
∴ ∠BAC + ∠PBC + ∠PCB = 90°
∴ ∠PBC + ∠PCB = 90° – 1 ∠BAC ………(i)
In∆BPC,
∠BPC + ∠PBC + ∠PCB = 180° …….[Sum of measures of angles of a triangle]
∴ ∠BPC + 90° – \(\frac { 1 }{ 2 }\)∠BAC = 180° ……[From (i)]
∴ ∠BPC = 180° – 90°\(\frac { 1 }{ 2 }\)∠BAC
= 180°- 90°+ \(\frac { 1 }{ 2 }\)∠BAC
= 90°+ \(\frac { 1 }{ 2 }\)∠BAC

Class 9 Maths Digest