Category: Class 9

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.

Practice Set 4.1 Geometry 9th Std Maths Part 2 Answers Chapter 4 Constructions of Triangles

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.

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

Problem Set 3 Geometry 9th Std Maths Part 2 Answers Chapter 3 Triangles

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)

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.

Practice Set 3.5 Geometry 9th Std Maths Part 2 Answers Chapter 3 Triangles

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

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 1.3 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 1 Basic Concepts in Geometry.

Practice Set 1.3 Geometry 9th Std Maths Part 2 Answers Chapter 1 Basic Concepts in Geometry

Question 1.
Write the following statements in ‘if-then’ form.
i. The opposite angles of a parallelogram are congruent.
ii. The diagonals of a rectangle are congruent.
iii. In an isosceles triangle, the segment joining the vertex and the midpoint of the base is perpendicular to the base.
Answer:
i. If a quadrilateral is a parallelogram, then its opposite angles are congruent.
ii. If a quadrilateral is a rectangle, then its diagonals are congruent.
iii. If a triangle is isosceles triangle, then segment joining the vertex of a triangle and midpoint of the base is perpendicular to the base.

Question 2.
Write converses of the following statements.
i. The alternate angles formed by two parallel lines and their transversal are congruent.
ii. If a pair of the interior angles made by a transversal of two lines are supplementary, then the lines are parallel.
iii. The diagonals of a rectangle are congruent.
Answer:
i. If the alternate angles made by two lines and their transversal are congruent, then the two lines are parallel.
ii. If two parallel lines are intersected by a transversal, then the interior angles formed bv the transversal are supplementary.
iii. If the diagonals of a quadrilateral are congruent, then that quadrilateral is a rectangle.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 1.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 1 Basic Concepts in Geometry.

Practice Set 1.2 Geometry 9th Std Maths Part 2 Answers Chapter 1 Basic Concepts in Geometry

Question 1.
The following table shows points on a number line and their co-ordinates. Decide whether the pair of segments given below the table are congruent or not.

i. seg DE and seg AB
ii. seg BC and seg AD
iii. seg BE and seg AD
Solution:
i. Co-ordinate of the point E is 9.
Co-ordinate of the point D is -7.
Since, 9 > -7
∴ d(D, E) = 9 – (-7) = 9 + 7 = 16
∴ l(DE) = 16 …(i)
Co-ordinate of the point A is -3.
Co-ordinate of the point B is 5.
Since, 5 > -3
∴ d(A, B) = 5 – (-3) = 5 + 3 = 8
∴ l(AB) = 8 …(ii)
∴ l(DE) ≠ l(AB) …[From (i) and (ii)]
∴ seg DE and seg AB are not congruent.

ii. Co-ordinate of the point B is 5.
Co-ordinate of the point C is 2.
Since, 5 > 2
∴ d(B, C) = 5 – 2 = 3
∴ l(BC) = 3 …(i)
Co-ordinate of the point A is -3.
Co-ordinate of the point D is -7.
Since, -3 > -7
∴ d(A, D) = -3 – (-7) = -3 + 7 = 4
∴ l(AD) = 4 . ..(ii)
∴ l(BC) ≠ l(AD) … [From (i) and (ii)]
∴ seg BC and seg AD are not congruent.

iii. Co-ordinate of the point E is 9.
Co-ordinate of the point B is 5.
Since, 9 > 5
∴ d(B, E) = 9 – 5 = 4
∴ l(BE) = 4 …(i)
Co-ordinate of the point A is -3.
Co-ordinate of the point D is -7.
Since, -3 > -7
∴ d(A, D) = -3 – (-7) = 4
∴ l(AD) = 4 …(ii)
∴ l(BE) =l(AD) …[From (i) and (ii)]
∴ seg BE and seg AD are congruent.
i.e, seg BE ≅ seg AD

Question 2.
Point M is the midpoint of seg AB. If AB = 8, then find the length of AM.
Solution:
Point M is the midpoint of seg AB and l(AB) = 8. …[Given]

Question 3.
Point P is the midpoint of seg CD. If CP = 2.5, find l(CD).
Solution:
Point P is the midpoint of seg CD and l(CP) = 2.5 …[Given]

∴ l(CD) = 2.5 x 2
∴ l(CD) = 5

Question 4.
If AB = 5 cm, BP = 2 cm and AP = 3.4 cm, compare the segments.
Solution:
Given, l(AB) = 5 cm, l(BP) = 2 cm,
l(AP) = 3.4 cm … [Given]
r Since, 2 < 3.4 < 5
∴ l(BP) < l(AP) < l(AB)
i.e., seg BP < seg AP < seg AB

Question 5.
Write the answers to the following questions with reference to the figure given below:

i. Write the name of the opposite ray of ray RP
ii. Write the intersection set of ray PQ and ray RP.
iii. Write the union set of ray PQ and ray QR.
iv. State the rays of which seg QR is a subset.
v. Write the pair of opposite rays with common end point R.
vi. Write any two rays with common end point S.
vii. Write the intersection set of ray SP and ray ST.
Answer:
i. Ray RS or ray RT
ii. Ray PQ
iii. Line QR
iv. Ray QR, ray QS, ray QT, ray RQ, ray SQ, ray TQ
v. Ray RP and ray RS, ray RQ and ray RT vi. Ray ST, ray SR
vii. Point S

Question 6.
Answer the questions with the help of figure given below.

i. State the points which are equidistant from point B.
ii. Write a pair of points equidistant from point iii. Find d(U,V), d(P,C), d(V,B), d(U, L).
Answer:
i. Points equidistant from point B are a. A and C, because d(B, A) = d(B, C) = 2 b. D and P, because d(B, D) = d(B, P) = 4
ii. Points equidistant from point Q are a. L and U, because d(Q, L) = d(Q, U) = 1 b. P and R, because d(P, Q) = d(Q, R) = 2
iii. a. Co-ordinate of the point U is -5. Co-ordinate of the point V is 5. Since, 5 > -5
∴ d(U, V) = 5 – (-5)
= 5 + 5
∴ d(U, V) = 10

b. Co-ordinate of the point P is -2.
Co-ordinate of the point C is 4.
Since, 4 > -2
∴ d(P, C) = 4 – (-2)
= 4 + 2
∴ d(P, C) = 6

c. Co-ordinate of the point V is 5.
Co-ordinate of the point B is 2.
Since, 5 > 2
∴ d(V, B) = 5 – 2
∴ d(V, B) = 3

d. Co-ordinate of the point U is -5.
Co-ordinate of the point L is -3.
Since, -3 > -5
∴ d(U, L) = -3 – (-5)
= -3 + 5
∴ d(U, L) = 2

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 1.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 1 Basic Concepts in Geometry.

Practice Set 1.1 Geometry 9th Std Maths Part 2 Answers Chapter 1 Basic Concepts in Geometry

Question 1.
Find the distances with the help of the number line given below.

i. d(B, E)
ii. d (J, J)
iii. d(P, C)
iv. d(J, H)
v. d(K, O)
vi. d(O, E)
vii. d(P, J)
viii. d(Q, B)
Solution:
i. Co-ordinate of the point B is 2.
Co-ordinate of the point E is 5.
Since, 5 > 2
∴ d(B, E) = 5 – 2
∴ d(B, E) = 3

ii. Co-ordinate of the point J is -2.
Co-ordinate of the point A is 1.
Since, 1 > -2
∴ d(J, A) = 1 – (-2)
= 1 + 2
∴ d(J, A) = 3

iii. Co-ordinate of the point P is -4.
Co-ordinate of the point C is 3.
Since, 3 > -4
∴ d(P,C) = 3 – (-4)
= 3 + 4
∴ d(P,C) = 7

iv. Co-ordinate of the point J is -2.
Co-ordinate of the point H is -1.
Since, -1 > -2
∴ d(J,H) = – 1 – (-2)
= -1 + 2
∴ d(J,H) = 1

v. Co-ordinate of the point K is -3.
Co-ordinate of the point O is 0.
Since,0 > -3
∴ d(K, O) = 0 – (-3)
= 0 + 3
∴ d(K, O) = 3

vi. Co-ordinate of the point O is 0.
∴ Co-ordinate of the point E is 5.
Since, 5 > 0
∴ d(O, E) = 5 – 0
∴ d(O, E) = 5

vii. Co-ordinate of the point P is -4.
Co-ordinate of the point J is -2.
Since -2 > -4
∴ d(P, J) = -2 – (-4)
= – 2+ 4
∴ d(P, J) = 2

viii. Co-ordinate of the point Q is -5.
Co-ordinate of the point B is 2.
Since,2 > -5
∴ d(Q,B) = 2 – (-5)
= 2 + 5
∴ d(Q, B) = 7

Question 2.
If the co-ordinate of A is x and that of B is . y, find d(A, B).
i. x = 1, y = 7
ii. x = 6, y = -2
iii. x = -3, y = 7
iv. x = -4, y = -5
v. x = -3, y = -6
vi. x = 4, y = -8
Solution:
i. Co-ordinate of point A is x = 1.
Co-ordinate of point B is y = 7
Since, 7 > 1
∴ d(A, B) = 7 – 1
∴ d(A, B) = 6

ii. Co-ordinate of point A is x = 6.
Co-ordinate of point B is y = -2.
Since, 6 > -2
∴ d(A, B) = 6 – ( -2) = 6 + 2
∴ d(A, B) = 8

iii. Co-ordinate of point A is x = -3.
Co-ordinate of point B is y = 7.
Since, 7 > -3
∴ d(A, B) = 7 – (-3) = 7 + 3
∴ d(A, B) = 10

iv. Co-ordinate of point A is x = -4.
Co-ordinate of point B is y = -5.
Since, -4 > -5
∴ d(A, B) = -4 – (-5)
= -4 + 5
∴ d(A, B) = 1

v. Co-ordinate of point A is x =-3.
Co-ordinate of point B is y = -6.
Since, -3 > -6
∴ d(A, B) = -3 – (-6)
= -3 + 6
∴ d(A, B) = 3

vi. Co-ordinate of point A is x = 4.
Co-ordinate of point B is y = -8.
Since, 4 > -8
∴ d(A, B) = 4 – (-8)
= 4 + 8
∴d(A, B) = 12

Question 3.
From the information given below, find which of the point is between the other two. If the points are not collinear, state so.
i. d(P, R) = 7, d(P, Q) = 10, d(Q, R) = 3
ii. d(R, S) = 8, d(S, T) = 6, d(R, T) = 4
iii. d(A, B) = 16, d(C, A) = 9, d(B, C) = 7
iv. d(L, M) =11, d(M, N) = 12, d(N, L) = 8
v. d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8
vi. d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
Solution:
i. Given, d(P, R) = 7, d(P, Q) = 10, d(Q, R) = 3
d(P, Q) = 10 …(i)
d(P, R) + d(Q, R) = 7 + 3 = 10 .. .(ii)
∴ d(P, Q) = d(P, R) + d(Q, R) …[From (i) and (ii)]
∴ Point R is between the points P and Q
i. e., P – R – Q or Q – R – P.
∴ Points P, R, Q are collinear.

ii. Given, d(R, S) = 8, d(S, T) = 6, d(R, T) = 4
d(R, S) = 8 …(i)
d(S, T) + d(R, T) = 6 + 4 = 10 …(h)
∴ d(R, S) ≠ d(S, T) + d(R, T) … [From (i) and (ii)]
∴ The given points are not collinear.

iii. Given, d(A, B) = 16, d(C, A) = 9, d(B, C) = 7
d(A, B) = 16 …(i)
d(C, A) + d(B, C) = 9 + 7 = 16 …(ii)
∴ d(A, B) = d(C, A) + d(B, C) …[From(i) and (ii)]
∴ Point C is between the points A and B.
i. e., A – C – B or B – C – A.
∴ Points A, C, B are collinear

iv. Given, d(L, M) = 11, d(M, N) = 12, d(N, L) = 8
d(M, N) = 12 …(i)
d(L, M) + d(N, L) = 11 + 8 = 19 …(ii)
∴d(M, N) + d(L, M) + d(N, L) … [From (i) and (ii)]
∴ The given points are not collinear.

v. Given, d(X, Y) = 15, d(Y, Z) = 7, d(X, Z) = 8
d(X, Y) = 15 …(i)
d(X,Z) + d(Y, Z) = 8 + 7= 15 …(ii)
∴ d(X, Y) = d(X, Z) + d(Y, Z) …[From (i) and (ii)]
∴ Point Z is between the points X and Y
i. e.,X – Z – Y or Y – Z – X.
∴ Points X, Z, Y are collinear.

vi. Given, d(D, E) = 5, d(E, F) = 8, d(D, F) = 6
d(E, F) = 8 …(i)
d(D, E) + d(D, F) = 5 + 6 = 11 …(ii)
∴ d(E, F) ≠ d(D, E) + d(D, F) … [From (i) and (ii)]
∴ The given points are not collinear.

Question 4.
On a number line, points A, B and C are such that d(A, C) = 10, d(C, B) = 8. Find d(A, B) considering all possibilities.
Solution:
Given, d(A, C) = 10, d(C, B) = 8.

Case I: Points A, B, C are such that, A – B – C.

∴ d(A, C) = d(A, B) + d(B, C)
∴ 10 = d(A, B) + 8
∴ d(A, B) = 10 – 8
∴ d(A, B) = 2

Case II: Points A, B, C are such that, A – C – B.

∴ d(A, B) = d(A, C) + d(C, B)
= 10 + 8
∴ d(A, B) = 18

Case III: Points A, B, C are such that, B – A – C.

From the diagram,
d (A, C) > d(B, C)
Which is not possible
∴ Point A is not between B and C.
∴ d(A, B) = 2 or d(A, B) = 18.

Question 5.
Points X, Y, Z are collinear such that d(X, Y) = 17, d(Y, Z) = 8, find d(X, Z).
Solution:
Given,d(X, Y) = 17, d(Y, Z) = 8
Case I: Points X, Y, Z are such that, X – Y – Z.

∴ d(X, Z) = d(X, Y) + d(Y, Z)
= 17 + 8
∴ d(X, Z) = 25

Case II: Points X, Y, Z are such that, X – Z – Y.

∴ d(X,Y) = d(X,Z) + d(Z,Y)
∴ 17 = d(X, Z) + 8
∴ d(X, Z) = 17 – 8
∴ d(X, Z) = 9

Case III: Points X, Y, Z are such that, Z – X – Y.

From the diagram,
d(X, Y) > d (Y, Z)
Which is not possible
∴ Point X is not between Z and Y.
∴ d(X, Z) = 25 or d(X, Z) = 9.

Question 6.
Sketch proper figure and write the answers of the following questions. [2 Marks each]
i. If A – B – C and l(AC) = 11,
l(BC) = 6.5, then l(AB) = ?
ii. If R – S – T and l(ST) = 3.7,
l(RS) = 2.5, then l(RT) = ?
iii. If X – Y – Z and l(XZ) = 3√7,
l(XY) = √7, then l(YZ) = ?
Solution:
i. Given, l(AC) =11, l(BC) = 6.5

l(AC) = l(AB) + l(BC) … [A – B – C]
∴ 11= l(AB) + 6.5
∴ l(AB) = 11 – 6.5
∴ l(AB) = 4.5

ii. Given, l(ST) = 3.7, l(RS) = 2.5

l(RT) = l(RS) + l(ST) … [R – S – T]
= 2.5 + 3.7
∴ (RT) = 6.2

iii. l(XZ) = 3√7 , l(XY) = √7,

l(XZ) = l(X Y) + l(YZ) … [X – Y – Z]
∴ 3 √7 ⇒ √7 + l(YZ)
∴ l(YZ)= 3√7 – √7
∴ l(YZ) = 2 √7

Question 7.
Which figure is formed by three non-collinear points?
Solution:
Three non-collinear points form a triangle.

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.

Practice Set 3.3 Geometry 9th Std Maths Part 2 Answers Chapter 3 Triangles

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

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.

Practice Set 3.1 Geometry 9th Std Maths Part 2 Answers Chapter 3 Triangles

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.

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.

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.

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°]

∴ ∠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°.

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.

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,

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]

∴ ∠EDF = ∠GFD [Alternate angles]
∴ ∠EDF = y° ….(iii) [From (ii)]
line DE || line GF and EM is their transversal. [Given]

∴ ∠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]

∴ ∠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)

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)

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)

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

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

Problem Set 2 Geometry 9th Std Maths Part 2 Answers Chapter 2 Parallel Lines

Question 1.
Select the correct alternative and fill in the blanks in the following statements.

i. If a transversal intersects two parallel lines then the sum of interior angles on the same side of the transversal is ____.
(A) 0°
(B) 90°
(C) 180°
(D) 360°
Answer:
(C) 180°

ii. The number of angles formed by a transversal of two lines is _____.
(A) 2
(B) 4
(C) 8
(D) 16
Answer:
(C) 8

iii. A transversal intersects two parallel lines. If the measure of one of the angles is 40°, then the measure of its corresponding angle is ______.
(A) 40°
(B) 140°
(C) 50°
(D) 180°
Answer:
(A) 40°

iv. In ∆ABC, ∠A = 76°, ∠B = 48°, then ∠C = _____.
(A) 66°
(B) 56°
(C) 124°
(D) 28°
Answer:
In ∆ABC, ∠A + ∠B + ∠C = 180°
∴ ∠C = 180° – 76° – 48° = 56°
(B) 56°

v. Two parallel lines are intersected by a transversal. If measure of one of the alternate interior angles is 75° then the measure of the other angle is _____.
(A) 105°
(B) 15°
(C) 75°
(D) 45°
Answer:
(C) 75°

Question 2.
Ray PQ and ray PR are perpendicular to each other. Points B and A are in the interior and exterior of ∠QPR respectively. Ray PB and ray PA are perpendicular to each other.
Draw a figure showing all these rays and write –
i. A pair of complementary angles
ii. A pair of supplementary angles
iii. A pair of congruent angles.
Solution:

i. Complementary angles:
∠RPQ = 90° [Ray PQ ⊥ ray PR]
∴ ∠RPB + ∠BPQ = 90° [Angle addition property]
∠RPB and ∠BPQ are pair of complementary angles
∠APB = 90° [Ray PA ⊥ ray PB]
∴ ∠APR + ∠RPB = 90°
∠APR and ∠RPB are pair of complementary angles.

ii. Supplementary angles:
∠APB + ∠RPQ = 90° + 90° = 180°
∴ ∠APB and ∠RPQ are a pair of supplementary angles.

iii. Congruent angles:
a. ∠APB = ∠RPQ [Each is of 90°]
b. ∠APB = ∠RPQ
∴ ∠APR + ∠RPB = ∠RPB + ∠BPQ [Angle addition property]
∴ ∠APR = ∠BPQ
∴ ∠APR ≅ ∠BPQ

Question 3.
Prove that, if a line is perpendicular to one of the two parallel lines, then it is perpendicular to the other line also.

Given: line AB || line CD and line EF intersects them at P and Q respectively.
line EF ⊥ line AB
To prove: line EF ⊥ line CD
Solution:
Proof:
line EF ⊥ line AB [Given]
∴ ∠APR = 90° ….(i)
line AB || line CD and line EF is their transversal.
∴ ∠EPB ≅ ∠PQD …..(ii) [Corresponding angles]
∴ ∠PQD = 90° [From (i) and (ii)]
∴ line EF ⊥ line CD

Question 4.
In the given figure, measures of some angles are shown. Using the measures find the measures of ∠x and ∠y and hence show that line l || line m.

Solution:
Proof:
∠x = 130°
∠y = 50° [Vertically opposite angles]
Here, m∠PQT + m∠QTS = 130° + 50° = 180°
But, ∠ PQT and ∠ QTS are a pair of interior angles on lines l and m when line n is the transversal,

∴ line l || line m [Interior angles test]

Question 5.
In the given figure, Line AB || line CD || line EF and line QP is their transversal. If y : z = 3 : 7 then find the measure of ∠x.

Solution:
y : z = 3 : 7 [Given]
Let the common multiple be m
∴ ∠j = 3m and ∠z = 7m ….(i)
line AB || line EF and line PQ is their transversal [Given]
∠x = ∠z
∴ ∠x = 7m …..(ii) [From (i)]
line AB || line CD and line PQ is their transversal [Given]
∠x + ∠y = 180°
∴ 7m + 3m = 180°
∴ 10m = 180°
∴ m = 18°
∴ ∠x = 7m = 7(18°) [From (ii)]
∴ ∠x = 126°

Question 6.
In the given figure, if line q || line r, line p is their transversal and if a = 80°, find the values of f and g.

Solution:
i. ∠a = 80° [Given]
∠g = ∠a [Alternate exterior angles]
∴ ∠g = 80° …..(i)

ii. Now, line q || line r and line p is their transversal.
∴ ∠f + ∠g = 180°
∴ ∠f + 80° = 180° [Interior angles]
∴ ∠f = 180° – 80° [From (i)]
∴ ∠f=100°

Question 7.
In the given figure, if line AB || line CF and line BC || line ED then prove that ∠ABC = ∠FDE.

Given: line AB || line CF and line BC || line ED
To prove: ∠ABC = ∠FDE
Solution:
Proof:
line AB || line PF and line BC is their transversal.
∴ ∠ABC = ∠BCD ….(i) [Alternate angles]
line BC || line ED and line CD is their transversal.
∴ ∠BCD = ∠FDE ….(ii) [Corresponding angles]
∴ ∠ABC = ∠FDE [From (i) and (ii)]

Question 8.
In the given figure, line PS is a transversal of parallel line AB and line CD. If Ray QX, ray QY, ray RX, ray RY are angle bisectors, then prove that □ QXRY is a rectangle.

Given: line AB || line CD
Rays QX, RX, QY, RY are the bisectors of ∠AQR, ∠QRC, ∠BQR and ∠QRD respectively.
To prove: □QXRY is a rectangle.
Proof:
∠XQA = ∠XQR = x° ……(i) [Ray QX bisects ∠AQR]
∠YQR = ∠YQB =y° …….(ii) [Ray QY bisects ∠BQR]
∠XRQ = ∠XRC = u° …….. (iii) [Ray RX bisects ∠CRQ]
∠YRQ = ∠YRD = v° ……(iv) [Ray RY bisects ∠DRQ]
line AB || line CD and line PS is their transversal.

∠AQR+ ∠CRQ= 180° [Interior angles]
(∠XQA + ∠XQR) + (∠XRQ + ∠XRC) = 180° [Angle addition property]
∴ (x + x) + (u + u) = 180° [From (i) and (ii)]
∴ 2x + 2u = 180°
∴ 2(x + u) = 180°
∴ x + u = 90° ……..(v)
In ∆XQR
∠XQR + ∠XRQ + ∠QXR = 180° [Sum of the measures of the angles of triangle is 180°]
∴ x + u + ∠QXR = 180° [From (i) and (iii)]
∴ 90 + ∠QXR = 180° [From (v)]
∴ ∠QXR = 180°- 90°
∴ ∠QXR = 90° …..(vi)
Similarily we can prove that,

∴ y + v = 90°
Hence ∠QYR = 90° ……(vii)
Now, ∠AQR + ∠BQR = 180° [Angles is linear pair]
(∠XQA + ∠XQR) + (∠YQR + ∠YQB) = 180° [Angle addition property]
∴ (x + x) + (y + y) = 180° [From (i) and (ii)]
∴ 2x + 2y = 180°
∴ 2(x+y) = 180°
∴ x +y = 90°
i.e. ∠XQR + ∠YQR = 90° [From (i) and (ii)]
∴ ∠XQY = 90° ….(viii) [Angle addition property]
Similarly we can prove that,
∠XRY = 90° …(ix)
In □QXRY
∠QXR = ∠QYR = ∠XQY = ∠XRY = 90° [From (vi), (vii), (viii) and (ix)]
∴□ QXRY is a rectangle.

Maharashtra Board Class 9 Maths Chapter 2 Parallel Lines Problem Set 2 Intext Questions and Activities

Question 1.
To verify the properties of angles formed by a transversal of two parallel lines. (Textbook pg. no. 14)
Take a piece of thick coloured paper. Draw a pair of parallel lines and a transversal on it. Paste straight sticks on the lines. Eight angles will be formed. Cut pieces of coloured paper, as shown in the figure, which will just fit at the comers of ∠1 and ∠2. Place the pieces near different pairs of corresponding angles, alternate angles and interior angles and verify their properties.

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 3 Triangles.

Practice Set 3.2 Geometry 9th Std Maths Part 2 Answers Chapter 3 Triangles

Question 1.
In each of the examples given below, a pair of triangles is shown. Equal parts of triangles in each pair are marked with the same signs. Observe the figures and state the test by which the triangles in each pair are congruent.

By SSS test
∆ABC ≅ ∆PQR

By SAS test
∆ XYZ ≅ ∆LMN

By ASA test
∆PRQ ≅ ∆STU

By hypotenuse side test
∆LMN ≅ ∆PTR

Question 2.
Observe the information shown in pairs of triangles given below. State the test by which the two triangles are congruent. Write the remaining congruent parts of the triangles.
Solution:

From the information shown in the figure,
In ∆ABC and ∆PQR,
∠ABC ≅ ∠PQR
seg BC ≅ seg QR
∠ACB ≅ ∠PRQ
∴ ∆ABC ≅ ∆PQR [ASA test]
∴ ∠BAC ≅ ∠QPR [Corresponding angles of congruent triangles]
seg AB ≅ segPQ and segAC ≅ seg PR [Corresponding sides of congruent triangles]

From the information shown in the figure,
In ∆PTQ and ∆STR,
seg PT ≅ seg ST
∠PTQ ≅ ∠STR [Vertically opposite angles]
seg TQ ≅ seg TR
∴ ∆PTQ ≅ ∆STR [SAS test]
∴ ∠TPQ ≅ ∠TSR and ∠TQP ≅ ∠TRS [Corresponding angles of congruent triangles]
seg PQ ≅ seg SR [Corresponding sides of congruent triangles]

Question 3.
From the information shown in the figure, state the test assuring the congruence of ∆ABC and ∆PQR. Write the remaining congruent parts of the triangles.

Solution:
In ∆BAC and ∆PQR,
seg BA ≅ seg PQ
seg BC ≅ seg PR
∠BAC ≅ ∠PQR = 90° [Given]
∴ ∆BAC ≅ ∆PQR [Hypotenuse side test]
∴ seg AC ≅ seg QR [c.s.c.t.]
∠ABC ≅ ∠QPR and ∠ACB ≅ ∠QRP [c.a.c.t.]

Question 4.
As shown in the adjoining figure, in ∆LMN and ∆PNM, LM = PN, LN = PM. Write the test which assures the congruence of the two triangles. Write their remaining congruent parts.

Solution:
In ∆LMN and ∆PNM,
seg LM ≅ seg PN
seg LN ≅ seg PM [Given]
seg MN ≅ seg NM [Common side]
∴ ∆LMN ≅ ∆PNM [SSS test]
∴ ∠LMN ≅ ∠PNM,
∴ ∠MLN ≅ ∠NPM, and ∠LNM ≅ ∠PMN [c.a.c.t.]

Question 5.
In the adjoining figure, seg AB ≅ seg CB and seg AD ≅ seg CD. Prove that ∆ABD ≅ ∆CBD.

Solution:
proof:
In ∆ABD and ∆CBD,
seg AB ≅ seg CB
seg AD ≅ seg CD [Given]
seg BD ≅ seg BD [Common side]
∴ ∆ABD ≅ ∆CBD [SSS test]

Question 6.
In the adjoining figure, ZP ≅ ZR, seg PQ ≅ seg RQ. Prove that APQT ≅ ARQS.

Proof:
In ∆PQT and ∆RQS,
∠P ≅ ∠R
seg PQ ≅ seg RQ [Given]
∠Q ≅ ∠Q [Common angle]
∴ ∆PQT ≅ ∆RQS [ASA test]