Category: Class 9

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 6 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 6 Circle.

Problem Set 6 Geometry 9th Std Maths Part 2 Answers Chapter 6 Circle

Question 1.
Choose correct alternative answer and fill in the blanks.

i. Radius of a circle is 10 cm and distance of a chord from the centre is 6 cm. Hence, the length of the chord is ____.
(A) 16 cm
(B) 8 cm
(C) 12 cm
(D) 32 cm
Answer:

∴ OA² = AC² + OC²
∴ 10² = AC² + 6²
∴ AC² = 64
∴ AC = 8 cm
∴ AB = 2(AC)= 16 cm
(A) 16 cm

ii. The point of concurrence of all angle bisectors of a triangle is called the ____.
(A) centroid
(B) circumcentre
(C) incentre
(D) orthocentre
Answer:
(C) incentre

iii. The circle which passes through all the vertices of a triangle is called ____.
(A) circumcircle
(B) incircle
(C) congruent circle
(D) concentric circle
Answer:
(A) circumcircle

iv. Length of a chord of a circle is 24 cm. If distance of the chord from the centre is 5 cm, then the radius of that circle is ____.
(A) 12 cm
(B) 13 cm
(C) 14 cm
(D) 15 cm
Answer:

OA² = AC² + OC²
∴ OA² = 12² + 5²
∴ OA² = 169
∴ OA = 13 cm
(B) 13 cm

v. The length of the longest chord of the circle with radius 2.9 cm is ____.
(A) 3.5 cm
(B) 7 cm
(C) 10 cm
(D) 5.8 cm
Answer:
Longest chord of the circle = diameter = 2 x radius = 2 x 2.9 = 5.8 cm
(D) 5.8 cm

vi. Radius of a circle with centre O is 4 cm. If l(OP) = 4.2 cm, say where point P will lie ____.
(A) on the centre
(B) inside the circle
(C) outside the circle
(D) on the circle
Answer:
l(OP) > radius
∴Point P lies in the exterior of the circle.
(C) outside the circle

vii. The lengths of parallel chords which are on opposite sides of the centre of a circle are 6 cm and 8 cm. If radius of the circle is 5 cm, then the distance between these chords is _____.
(A) 2 cm
(B) 1 cm
(C) 8 cm
(D) 7 cm
Answer:

PQ = 8 cm, MN = 6 cm
∴ AQ = 4 cm, BN = 3 cm
∴ OQ² = OA² + AQ²
∴ 5² = OA² + 4²
∴ OA² = 25 – 16 = 9
∴ OA = 3 cm
Also, ON² = OB² + BN²
∴ 5² = OB² + 3²
∴ OB = 4 cm
Now, AB = OA + OB = 3 + 4 = 7 cm

Question 2.
Construct incircle and circumcircle of an equilateral ADSP with side 7.5 cm. Measure the radii of both the circles and find the ratio of radius of circumcircle to the radius of incircle.
Solution:


Steps of construction:
i. Construct ∆DPS of the given measurement.
ii. Draw the perpendicular bisectors of side DP and side PS of the triangle.
iii. Name the point of intersection of the perpendicular bisectors as point C.
iv. With C as centre and CM as radius, draw a circle which touches all the three sides of the triangle.
v. With C as centre and CP as radius, draw a circle which passes through the three vertices of the triangle.

Radius of incircle = 2.2 cm and Radius of circumcircle = 4.4 cm

Question 3.
Construct ∆NTS where NT = 5.7 cm. TS = 7.5 cm and ∠NTS = 110° and draw incircle and circumcircle of it.
Solution:


Steps of construction:
For incircle:
i. Construct ∆NTS of the given measurement.
ii. Draw the bisectors of ∠T and ∠S. Let these bisectors intersect at point I.
iii. Draw a perpendicular IM on side TS. Point M is the foot of the perpendicular.
iv. With I as centre and IM as radius, draw a circle which touches all the three sides of the triangle.
For circumcircle:
i. Draw the perpendicular bisectors of side NT and side TS of the triangle.
ii. Name the point of intersection of the perpendicular bisectors as point C.
iii. Join seg CN
iv. With C as centre and CN as radius, draw a circle which passes through the three vertices of the triangle.

Question 4.
In the adjoining figure, C is the centre of the circle, seg QT is a diameter, CT = 13, CP = 5. Find the length of chord RS.

Given: In a circle with centre C, QT is a diameter, CT = 13 units, CP = 5 units
To find: Length of chord RS
Construction: Join points R and C.
Solution:

i. CR = CT= 13 units …..(i) [Radii of the same circle]
In ∆CPR, ∠CPR = 90°
∴ CR² = CP² + RP² [Pythagoras theorem]
∴ 13² = 5² + RP² [From (i)]
∴ 169 = 25 + RP² [From (i)]
∴ RP² = 169 – 25 = 144
∴ RP = \(\sqrt { 144 }\) [Taking square root on both sides]
∴ RP = 12 cm ….(ii)

ii. Now, seg CP _L chord RS [Given]
∴ RP = \(\frac { 1 }{ 2 }\) RS [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ 12 = \(\frac { 1 }{ 2 }\) RS [From (ii)]
∴ RS = 2 x 12 = 24
∴ The length of chord RS is 24 units.

Question 5.
In the adjoining figure, P is the centre of the circle. Chord AB and chord CD intersect on the diameter at the point E. If ∠AEP ≅ ∠DEP, then prove that AB = CD.

Given: P is the centre of the circle.
Chord AB and chord CD intersect on the diameter at the point E. ∠AEP ≅ ∠DEP
To prove: AB = CD
Construction: Draw seg PM ⊥ chord AB, A-M-B
seg PN ⊥ chord CD, C-N-D
Proof:

∠AEP ≅ ∠DEP [Given]
∴ Seg ES is the bisector of ∠AED.
PoInt P is on the bisector of ∠AED.
∴ PM = PN [Every point on the bisector of an angle is equidistant from the sides of the angle.]
∴ chord AB ≅ chord CD [Chords which are equidistant from the centre are congruent.]
∴ AB = CD [Length of congruent segments]

Question 6.
In the adjoining figure, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that ∆ABC is an isosceles triangle.

Given: O is the centre of the circle.
diameter CD ⊥ chord AB, A-E-B
To prove: ∆ABC is an isosceles triangle.
Proof:
diameter CD ⊥ chord AB [Given]
∴ seg OE ⊥ chord AB [C-O-E, O-E-D]
∴ seg AE ≅ seg BE ……(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord]
In ∆CEA and ∆CEB,
∠CEA ≅ ∠CEB [Each is of 90°]
seg AE ≅ seg BE [From (i)]
seg CE ≅ seg CE [Common side]
∴ ∆CEA ≅ ∆CEB [SAS test]
∴ seg AC ≅ seg BC [c. s. c. t.]
∴ ∆ABC is an isosceles triangle.

Maharashtra Board Class 9 Maths Chapter 6 Circle Problem Set 6 Intext Questions and Activities

Question 1.
Every student in the group should do this activity. Draw a circle in your notebook. Draw any chord of that circle. Draw perpendicular to the chord through the centre of the circle. Measure the lengths of the two parts of the chord. Group leader should prepare a table as shown below and ask other students to write their observations in it. Write the property which you have observed. (Textbook pg. no. 77)


Answer:
On completing the above table, you will observe that the perpendicular drawn from the centre of a circle on its chord bisects the chord.

Question 2.
Every student from the group should do this activity. Draw a circle in your notebook. Draw a chord of the circle. Join the midpoint of the chord and centre of the circle. Measure the angles made by the segment with the chord.
Discuss about the measures of the angles with your friends. Which property do the observations suggest ? (Textbook pg. no. 77)

Answer:
The meausure of the angles made by the drawn segment with the chord is 90°. Thus, we can conclude that, the segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord.

Question 3.
Draw circles of convenient radii. Draw two equal chords in each circle. Draw perpendicular to each chord from the centre. Measure the distance of each chord from the centre. What do you observe? (Textbook pg. no. 79)
Answer:
Congruent chords of a circle are equidistant from the centre.

Question 4.
Measure the lengths of the perpendiculars on chords in the following figures.

Did you find OL = OM in fig (i), PN = PT in fig (ii) and MA = MB in fig (iii)?
Write the property which you have noticed from this activity. (Textbook pg. no. 80)
Answer:
In each figure, the chords are equidistant from the centre. Also, we can see that the measures of the chords in each circle are equal.
Thus, we can conclude that chords of a circle equidistant from the centre of a circle are congruent.

Question 5.
Draw different triangles of different measures and draw in circles and circumcircles of them. Complete the table of observations and discuss. (Textbook pg. no. 85)
Answer:

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 6.3 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 6 Circle.

Practice Set 6.3 Geometry 9th Std Maths Part 2 Answers Chapter 6 Circle

Question 1.
Construct ∆ABC such that ∠B =100°, BC = 6.4 cm, ∠C = 50° and construct its incircle.
Solution:


Steps of construction:
i. Construct ∆ABC of the given measurement.
ii. Draw the bisectors of ∠B and ∠C. Let these bisectors intersect at point I.
iii. Draw a perpendicular IM on side BC. Point M is the foot of the perpendicular.
iv. With I as centre and IM as radius, draw a circle which touches all the three sides of the triangle.

Question 2.
Construct ∆PQR such that ∠P = 70°, ∠R = 50°, QR = 7.3 cm and construct its circumcircle.
Solution:
In ∆PQR,
m∠P + m∠Q + m∠R = 180° … [Sum of the measures of the angles of a triangle is 180°]
∴ 70° + m∠Q + 50° = 180°
∴ m∠Q = 180° – 70° + m∠Q + 50° = 180°
∴ m∠Q = 180° – 70° – 50°
∴ m∠Q = 60°


Steps of construction:
i. Construct A PQR of the given measurement.
ii. Draw the perpendicular bisectors of side PQ and side QR of the triangle.
iii. Name the point of intersection of the perpendicular bisectors as point C.
iv. Join seg CP
v. With C as centre and CP as radius, draw a circle which passes through the three vertices of the triangle.

Question 3.
Construct ∆XYZ such that XY = 6.7 cm, YZ = 5.8 cm, XZ = 6.9 cm. Construct its incircle.
Solution:


Steps of construction:
i. Construct ∆XYZ of the given measurement
ii. Draw the bisectors of ∠X and ∠Z. Let these bisectors intersect at point I.
iii. Draw a perpendicular IM on side XZ. Point M is the foot of the perpendicular.
iv. With I as centre and IM as radius, draw a circle which touches all the three sides of the triangle.

Question 4.
In ∆LMN, LM = 7.2 cm, ∠M = 105°, MN = 6.4 cm, then draw ∆LMN and construct its circumcircle.
Solution:


Steps of construction:
i. Construct ∆LMN of the given measurement.
ii. Draw the perpendicular bisectors of side MN and side ML of the triangle.
iii. Name the point of intersection of the perpendicular bisectors as point C.
iv. Join seg CM
v. With C as centre and CM as radius, draw a circle which passes through the three vertices of the triangle.

Question 5.
Construct ∆DEF such that DE = EF = 6 cm. ∠F = 45° and construct its circumcircle.
Solution:


Steps of construction:
i. Construct ∆DEF of the given measurement.
ii. Draw the perpendicular bisectors of side DE and side EF of the triangle.
iii. Name the point of intersection of perpendicular bisectors as point C.
iv. Join seg CE
v. With C as centre and CE as radius, draw a circle which passes through the three vertices of the triangle.

Maharashtra Board Class 9 Maths Chapter 6 Circle Practice Set 6.3 Intext Questions and Activities

Question 1.
Draw any equilateral triangle. Draw incircle and circumcircle of it. What did you observe while doing this activity? (Textbook pg. no. 85)
i. While drawing incircle and circumcircle, do the angle bisectors and perpendicular bisectors coincide with each other?
ii. Do the incentre and circumcenter coincide with each other? If so, what can be the reason of it?
iii. Measure the radii of incircle and circumcircle and write their ratio.
Solution:


Steps of construction:
i. Construct equilateral ∆XYZ of any measurement.
ii. Draw the perpendicular bisectors of side XY and side YZ of the triangle.
iii. Draw the bisectors of ∠X and ∠Z.
iv. Name the point of intersection of the perpendicular bisectors and angle bisectors as point I.
v. With I as centre and IM as radïus, draw a circle which touches all the three sides of the triangle.
vi. With I as centre and IZ as radius, draw a circle which passes through the three vertices of the triangle.
[Note: Here, point of intersection of perpendicular bisector and angle bisector is same.]

i. Yes.
ii. Yes.
The angle bisectors of the angles and the perpendicular bisectors of the sides of an equilateral triangle are coincedent. Hence, its incentre and circumcentre coincide.
iii. Radius of circumcircle = 3.6 cm,
Radius of incircle = 1.8 cm
\(\text { Ratio }=\frac{\text { Radius of circumcircle }}{\text { Radius of incircle }}=\frac{3.6}{1.8}=\frac{2}{1}=2 : 1\)

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 5 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 5 Quadrilaterals.

Problem Set 5 Geometry 9th Std Maths Part 2 Answers Chapter 5 Quadrilaterals

Question 1.
Choose the correct alternative answer and fill in the blanks.

i. If all pairs of adjacent sides of a quadrilateral are congruent, then it is called ____.
(A) rectangle
(B) parallelogram
(C) trapezium
(D) rhombus
Answer:
(D) rhombus

ii. If the diagonal of a square is 22√2 cm, then the perimeter of square is ____.
(A) 24 cm
(B) 24√2 cm
(C) 48 cm
(D) 48√2 cm
Answer:
In ∆ABC,
AC² = AB² + BC²

∴ (12² √2 )² = AB² + AB²
∴ \( A B^{2}=\frac{12^{2} \times 2}{2}=12^{2}\)
∴ AB = 12 cm
∴ Perimeter of □ABCD = 4 x 12 = 48 cm
(C) 48 cm

iii. If opposite angles of a rhombus are (2x)° and (3x – 40)°, then the value of x is ____.
(A) 100°
(B) 80°
(C) 160°
(D) 40°
Answer:
2x = 3x – 40 … [Pythagoras theorem]
∴ x = 40°
(D) 40°

Question 2.
Adjacent sides of a rectangle are 7 cm and 24 cm. Find the length of its diagonal.
Solution:

Let □ABCD be the rectangle.
AB = 7 cm, BC = 24 cm
In ∆ABC, ∠B = 90° [Angle of a rectangle]
AC² = AB² + BC² [Pythagoras theorem]
= 7² + 24²
= 49 + 576
= 625
AC = √625 [Taking square root of both sides]
= 25 cm
∴ The length of the diagonal of the rectangle is 25 cm.

Question 3.
If diagonal of a square is 13 cm, then find its side.
Solution:

Let □PQRS be the square of side x cm.
∴ PQ = QR = x cm …..(i) [Sides of a square]
∴ In ∆PQR, ∠Q = 90° [Angle of a square]
∴ PR² = PQ² + QR² [Pythagoras theorem]
∴ 13 = x + x [From (i)]
∴ 169 = 2x²

The length of the side of the square is 6.5√2 cm.

Question 4.
Ratio of two adjacent sides of a parallelogram is 3 : 4, and its perimeter is 112 cm. Find the length of its each side.
Solution:

Let □STUV be the parallelogram.
Ratio of two adjacent sides of a parallelogram is 3 : 4.
Let the common multiple be x.
ST = 3x cm and TU = 4x cm
∴ ST = UV = 3x cm
TU = SV = 4x cm …..(i) [Opposite sides of a parallelogram]
Perimeter of □STUV = 112 [Given]
∴ ST + TU + UV + SV = 112
∴ 3x + 4x + 3x + 4x = 112 [From (i)]
∴ 14x = 112
∴ x = \(\frac { 112 }{ 14 }\)
∴ x = 8
∴ ST = UV = 3x = 3 x 8 = 24 cm
∴ TU = SV = 4x = 4 x 8 = 32 cm [From (i)]
∴ The lengths of the sides of the parallelogram are 24 cm, 32 cm, 24 cm and 32 cm.

Question 5.
Diagonals PR and QS of a rhombus PQRS are 20 cm and 48 cm respectively. Find the length of side PQ.
Solution:

□PQRS is a rhombus. [Given]
PR = 20 cm and QS = 48 cm [Given]
∴ PT = \(\frac { 1 }{ 2 }\) PR [Diagonals of a rhombus bisect each other]
= \(\frac { 1 }{ 2 }\) x 20 = 10 cm
Also, QT = \(\frac { 1 }{ 2 }\) QS [Diagonals of a rhombus bisect each other]
= \(\frac { 1 }{ 2 }\) x 48 = 24 cm

ii. In ∆PQT, ∠PTQ = 90° [Diagonals of a rhombus are perpendicular to each other]
∴ PQ² = PT² + QT² [Pythagoras- theorem]
= 10² + 24²
= 100 + 576
∴ PQ² = 676
∴ PQ = \(\sqrt {676 }\) [Taking square root of both sides]
= 26 cm
∴ The length of side PQ is 26 cm.

Question 6.
Diagonals of a rectangle PQRS are intersecting in point M. If ∠QMR = 50°, then find the measure of ∠MPS.
Solution:

□PQRS is a rectangle.
∴ PM = \(\frac { 1 }{ 2 }\) PR …(i)
MS = \(\frac { 1 }{ 2 }\) QS …(ii) [Diagonals of a rectangle bisect each other]
Also, PR = QS …..(iii) [Diagonals of a rectangle are congruent]
∴ PM = MS ….(iv) [From (i), (ii) and (iii)]
In ∆PMS,
PM = MS [From (iv)]
∴ ∠MSP = ∠MPS = x° …..(v) [Isosceles triangle theorem]
∠PMS = ∠QMR = 50° ……(vi) [Vertically opposite angles]
In ∆MPS,
∠PMS + ∠MPS + ∠MSP = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ 50° +x + x = 180° [From (v) and (vi)]
∴ 50° + 2x= 180
∴ 2x= 180-50
∴ 2x= 130
∴ x = \(\frac { 130 }{ 2 }\) = 65°
∴ ∠MPS = 65° [From (v)]

Question 7.
In the adjoining figure, if seg AB || seg PQ , seg AB ≅ seg PQ, seg AC || seg PR, seg AC ≅ seg PR, then prove that seg BC || seg QR and seg BC ≅ seg QR.

Solution:
Given: seg AB || seg PQ , seg AB ≅ seg PQ,
seg AC || seg PR, seg AC ≅ seg PR
To prove: seg BC || seg QR, seg BC ≅ seg QR
Proof:
Consider □ABQP,
seg AB || seg PQ [Given]
seg AB ≅ seg PQ [Given]
∴ □ABQP is a parallelogram. [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]
∴ segAP || segBQ …..(i)
∴ seg AP ≅ seg BQ …..(ii) [Opposite sides of a parallelogram]
Consider □ACRP,
seg AC || seg PR [Given]
seg AC ≅ seg PR [Given]
∴ □ACRP is a parallelogram. [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]
∴ seg AP || seg CR …(iii)
∴ seg AP ≅ seg CR …….(iv) [Opposite sides of a parallelogram]
Consider □BCRQ,
seg BQ || seg CR
seg BQ ≅ seg CR
∴ □BCRQ is a parallelogram. [A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and congruent]
∴ seg BC || seg QR
∴ seg BC ≅ seg QR [Opposite sides of a parallelogram]

Question 8.
In the adjoining figure, □ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively. Then prove that PQ || AB and PQ = \(\frac { 1 }{ 2 }\) ( AB + DC).

Given : □ ABCD is a trapezium.
To prove:
Construction: Join points A and Q. Extend seg AQ and let it meet produced DC at R.
Proof:

seg AB || seg DC [Given]
and seg BC is their transversal.
∴ ∠ABC ≅ ∠RCB [Alternate angles]
∴ ∠ABQ ≅ ∠RCQ ….(i) [B-Q-C]
In ∆ABQ and ∆RCQ,
∠ABQ ≅∠RCQ [From (i)]

seg BQ ≅ seg CQ [Q is the midpoint of seg BC]
∠BQA ≅ ∠CQR [Vertically opposite angles]
∴ ∆ABQ ≅ ∆RCQ [ASA test]
seg AB ≅ seg CR …(ii) [c. s. c. t.]
seg AQ ≅ seg RQ [c. s. c. t.]
∴ Q is the midpoint of seg AR. ….(iii)

In ∆ADR,
Points P and Q are the midpoints of seg AD and seg AR respectively. [Given and from (iii)]

∴ seg PQ || seg DR [Midpoint theorem]
i.e. seg PQ || seg DC ……..(iv) [D-C-R]
But, seg AB || seg DC …….(v) [Given]
∴ seg PQ || seg AB [From (iv) and (v)]
In ∆ADR,

Question 9.
In the adjoining figure, □ABCD is a trapezium. AB || DC. Points M and N are midpoints of diagonals AC and DB respectively, then prove that MN || AB.

Solution:
Given: □ABCD is a trapezium. AB || DC.
Points M and N are midpoints of diagonals AC and DB respectively.
To prove: MN || AB
Construction: Join D and M. Extend seg DM to meet seg AB at point E such that A-E-B.
Proof:
seg AB || seg DC and seg AC is their transversal. [Given]
∴ ∠CAB ≅ ∠ACD [Alternate angles]
∴ ∠MAE ≅ ∠MCD ….(i) [C-M-A, A-E-B]
In ∆AME and ∆CMD,

∠AME ≅ ∠CMD [Vertically opposite angles]
seg AM ≅ seg CM [M is the midpoint of seg AC]
∠MAE ≅∠MCD [From (i)]
∴ ∆AME ≅ ∆CMD [ASA test]
∴ seg ME ≅ seg MD [c.s.c.t]
∴ Point M is the midpoint of seg DE. …(ii)
In ∆DEB,

Points M and N are the midpoints of seg DE and seg DB respectively. [Given and from (ii)]
∴ seg MN || seg EB [Midpoint theorem]
∴ seg MN || seg AB [A-E-B]

Maharashtra Board Class 9 Maths Chapter 5 Quadrilaterals Problem Set 5 Intext Questions and Activities

Question 1.
Draw five parallelograms by taking various measures of lengths and angles. (Textbook page no. 59)

Question 2.
Draw a parallelogram PQRS. Draw diagonals PR and QS. Denote the intersection of diagonals by letter O. Compare the two parts of each diagonal with a divider. What do you find? (Textbook page no. 60)

Answer:
seg OP = seg OR, and seg OQ = seg OS
Thus we can conclude that, point O divides the diagonals PR and QS in two equal parts.

Question 3.
To verify the different properties of quadrilaterals.

Material: A piece of plywood measuring about 15 cm x 10 cm, 15 thin screws, twine, scissor.
Note: On the plywood sheet, fix five screws in a horizontal row keeping a distance of 2 cm between any two adjacent screws. Similarly make two more rows of screws exactly below the first one. Take care that the vertical distance between any two adjacent screws is also 2 cm.
With the help of the screws, make different types of quadrilaterals of twine. Verify the properties of sides and angles of the quadrilaterals. (Textbook page no. 75)

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 6.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 6 Circle.

Practice Set 6.2 Geometry 9th Std Maths Part 2 Answers Chapter 6 Circle

Question 1.
Radius of circle is 10 cm. There are two chords of length 16 cm each. What will be the distance of these chords from the centre of the circle ?
Given: In a circle with centre O,
OR and OP are radii and RS and PQ are its congruent chords.
PQ = RS= 16 cm,
OR = OP = 10 cm
seg OU ⊥ chord PQ, P-U-Q
seg OT ⊥ chord RS, R-T-S
To find: Distance of chords from centre of the circle.
Solution:

i. PU = \(\frac { 1 }{ 2 }\)(PQ) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ PU= \(\frac { 1 }{ 2 }\) x 16 = 8 cm …(i)

ii. In ∆OUP, ∠OUP = 90°
∴ OP² = OU² + PU² [Pythagoras theorem]
∴ 10² = OU² + 8² [From (i)]
∴ 100 = OU² + 64
∴ OU² = 100 – 64 = 36
∴ OU = √36 [Taking square root on both sides]
∴ OU = 6 cm

iii. Now, OT = OU [Congruent chords of a circle are equidistant from the centre.]
∴ OT = OU = 6cm
∴ The distance of the chords from the centre of the circle is 6 cm.

Question 2.
In a circle with radius 13 cm, two equal chords are at a distance of 5 cm from the centre. Find the lengths of chords.
Given: In a circle with cente O,
OA and OC are the radii and AB and CD are its congruent chords,
OA = OC = 13cm
0E = OF = 5 cm
seg 0E ⊥ chord CD, C-E-D
seg OF ⊥ chord AB. A-F-B
To find: length of the chords
Solution:

i. In ∆AFO, ∠AFO = 90°
∴ AO² = AF² + FO² [Pythagoras theorem]
∴ 13² = AF² + 5²
∴ 169 = AF² + 25
∴ AF² = 169-25
∴ AF² = 144
∴ AF = \(\sqrt { 144 }\) [Taking square root on both sides]
∴ AF = 12 cm …..(i)

ii. Now AF = \(\frac { 1 }{ 2 }\)AB [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ 12 = \(\frac { 1 }{ 2 }\) (AB) [From (i)]
∴ AB = 12 x 2 = 24 cm
∴ CD = AB = 24 cm [chord AB ≅ chord CD]
∴ The lengths of the two chords are 24 cm each.

Question 3.
Seg PM and seg PN are congruent chords of a circle with centre C. Show that the ray PC is the bisector of ∠NPM.
Given: Point C is the centre of the circle.
chord PM ≅ chord PN
To prove: Ray PC is the bisector of ∠NPM.
Construction: Draw seg CR ⊥ chord PN, P-R-N
seg CQ ⊥ chord PM, P-Q-M
Proof:

chord PM chord PN [Given]
seg CR ⊥ chord PN
seg CQ ⊥ chord PM [Construction]
∴ segCR ≅ segCQ ….(i) [Congruent chords are equidistant from the centre]
In ∆PRC and ∆PQC,
∠PRC ≅ ∠PQC [Each is of 90°]
segCR ≅ segCQ [From (i)]
seg PC ≅ seg PC [Common side]
∴ ∆PRC ≅ ∆PQC [Hypotenuse side test]
∴ ∠RPC ≅ ∠QPC [c. a. c. t.]
∴ ∠NPC ≅ ∠MPC [N- R-P, M-Q-P]
∴ Ray PC is the bisector of ∠NPM.

Maharashtra Board Class 9 Maths Chapter 6 Circle Practice Set 6.2 Intext Questions and Activities

Question 1.
Prove the following two theorems for two congruent circles. (Textbook pg. no. 81)
i. Congruent chords in congruent circles are equidistant from their respective centres.
ii. Chords of congruent circles which are equidistant from their respective centres are congruent.
Write ‘Given’. ‘To prove’ and the proofs of these theorems.
Solution:
(i) Congruent chords in congruent circles are equidistant from their respective centres.
Given: Point P and point Q are the centres of congruent circles.
chord AB ≅ chord CD
seg PM ⊥ chord AB, A-M-B
seg QN ⊥ chord CD, C-N-D
To prove: PM = QN

Construction: Draw seg PA and seg QC.
Proof:
seg PM ⊥ chord AB, seg QN ⊥ chord CD [Given]
∴ AM = \(\frac { 1 }{ 2 }\)(AB) ………(i) [Perpendicular drawn from the centre of the circle to the
∴ CN = \(\frac { 1 }{ 2 }\)(CD) ……..(ii) chord bisects the chord.]
But, AB = CD ………(iii) [Given]
∴ AM = CN [From (i), (ii) and (iii)]
i.e., segAM ≅ segCN ….(iv) [Segments of equal lengths]
In ∆PMA and ∆QNC,
∠PMA ≅ ∠QNC [Each is of 90°]
hypotenuse PA ≅ hypotenuse QC [Radii of congruent circles]
seg AM ≅ seg CN [From (iv)]
∴ ∆PMA ≅ ∆QNC [Hypotenuse side test]
∴ segPM ≅ segQN [c. s. c. t.]
∴ PM ≅ QN [Length of congruent segments]

(ii) Chords of congruent circles which are equidistant from their respective centres are congruent.

Given: Point P and point Q are the centres of congruent circles.
seg PM ⊥ chord AB, A-M-B
seg QN ⊥ chord CD, C-N-D
PM = QN
To prove: chord AB ≅ chord CD
Construction: Draw seg PA and seg QC.
Proof:
In ∆PMA and ∆QNC,
∴ ∠PMA ≅ ∠QNC [Each is of 90°]
seg PM ≅ seg QN [Given]
hypotenuse PA ≅ hypotenuse QC [Radii of the congruent circles]
∴ ∆PMA ≅ ∆QNC [Hypotenuse side test]
∴ seg AM ≅ seg CN [c. s. c. t.]
∴ AM = CN ….(i) [Length of congruent segments]
Now, seg PM ⊥ chord AB, and seg QN ⊥ chord CD
∴ AM = \(\frac { 1 }{ 2 }\)(AB) …(ii)
∴ CN = \(\frac { 1 }{ 2 }\) (CD) ..(iii) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
∴ AB = CD [From (i), (ii) and (ii)]
∴ chord AB ≅ chord CD [Segments of equal lengths]

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 5.5 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 5 Quadrilaterals.

Practice Set 5.5 Geometry 9th Std Maths Part 2 Answers Chapter 5 Quadrilaterals

Question 1.
In the adjoining figure, points X, Y, Z are the midpoints of of ∆ABC respectively, cm. Find the lengths of side AB, side BC and side AC AB = 5 cm, AC = 9 cm and BC = 11c.m. Find the lengths of XY, YZ, XZ.

Solution:
i. AC = 9 cm [Given]
Points X and Y are the midpoints of sides AB and BC respectively. [Given]
∴ XY = \(\frac { 1 }{ 2 }\) AC [Midpoint tfyeprem]
= \(\frac { 1 }{ 2 }\) x 9 = 4.5 cm

ii. AB = 5 cm [Given]
Points Y and Z are the midpoints of sides BC and AC respectively. [Given]
∴ YZ = \(\frac { 1 }{ 2 }\) AB [Midpoint theorem]
= \(\frac { 1 }{ 2 }\) x 5 = 2.5 cm

iii. BC = 11 cm [Given]
Points X and Z are the midpoints of sides AB and AC respectively. [Given]
∴ XZ = \(\frac { 1 }{ 2 }\) BC [Midpoint theorem]
= \(\frac { 1 }{ 2 }\) x 11 = 5.5 cm
l(XY) = 4.5 cm, l(YZ) = 2.5 cm, l(XZ) = 5.5 cm

Question 2.
In the adjoining figure, □PQRS and □MNRL are rectangles. If point M is the midpoint of side PR, then prove that,
i. SL = LR
ii. LN = \(\frac { 1 }{ 2 }\) SQ.

Given: □PQRS and □MNRL are rectangles. M is the midpoint of side PR.
Solution:
Toprove:
i. SL = LR
ii. LN = \(\frac { 1 }{ 2 }\) (SQ)
Proof:
i. □PQRS and □MNRL are rectangles. [Given]
∴ ∠S = ∠L = 90° [Angles of rectangles]
∠S and ∠L form a pair of corresponding angles on sides SP and LM when SR is their transversal.

∴eg ML || seg PS …(i) [Corresponding angles test]
In ∆PRS,
Point M is the midpoint of PR and seg ML || seg PS. [Given] [From (i)]
∴ Point L is the midpoint of seg SR. ……(ii) [Converse of midpoint theorem]
∴ SL = LR

ii. Similarly for ∆PRQ, we can prove that,

Point N is the midpoint of seg QR. ….(iii)
In ∆RSQ,
Points L and N are the midpoints of seg SR and seg QR respectively. [From (ii) and (iii)]
∴ LN = \(\frac { 1 }{ 2 }\)SQ [Midpoint theorem]

Question 3.
In the adjoining figure, ∆ABC is an equilateral triangle. Points F, D and E are midpoints of side AB, side BC, side AC respectively. Show that ∆FED is an equilateral triangle.

Given: ∆ABC is an equilateral triangle.
Points F, D and E are midpoints of side AB, side BC, side AC respectively.
To prove: ∆FED is an equilateral triangle.
Solution:
Proof:
∆ABC is an equilateral triangle. [Given]
∴ AB = BC = AC ….(i) [Sides of an equilateral triangle]
Points F, D and E are midpoints of side AB and BC respectively.

∴ FD = \(\frac { 1 }{ 2 }\)AC …..(ii) [Midpoint theorem]
Points D and E are the midpoints of sides BC and AC respectively.

∴ DE = \(\frac { 1 }{ 2 }\)AB …..(iii) [Midpoint theorem]
Points F and E are the midpoints of sides AB and AC respectively.
∴ FE = \(\frac { 1 }{ 2 }\)BC
∴ FD = DE = FE [From (i), (ii), (iii) and (iv) ]
∴ ∆FED is an equilateral triangle.

Question 4.
In the adjoining figure, seg PD is a median of ∆PQR. Point T is the midpoint of seg PD. Produced QT intersects PR at M. Show that \(\frac { PM }{ PR }\) = \(\frac { 1 }{ 3 }\). [Hint: Draw DN || QM]

Solution:
Given: seg PD is a median of ∆PQR. Point T is the midpoint of seg PD.
To Prove: \(\frac { PM }{ PR }\) = \(\frac { 1 }{ 3 }\)
Construction: Draw seg DN ||seg QM such that P-M-N and M-N-R.
Proof:
In ∆PDN,
Point T is the midpoint of seg PD and seg TM || seg DN [Given]
∴ Point M is the midpoint of seg PN. [Construction and Q-T-M]
∴ PM = MN [Converse of midpoint theorem]
In ∆QMR,
Point D is the midpoint of seg QR and seg DN || seg QM [Construction]

∴ Point N is the midpoint of seg MR. [Converse of midpoint theorem]
∴ RN = MN …..(ii)
∴ PM = MN = RN …..(iii) [From (i) and (ii)]
Now, PR = PM + MN + RN [ P-M-R-Q-T-M]
∴ PR = PM + PM + PM [From (iii) ]
∴ PR = 3PM
\(\frac { PM }{ PR }\) = \(\frac { 1 }{ 3 }\)

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 5.1 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 5 Quadrilaterals.

Practice Set 5.1 Geometry 9th Std Maths Part 2 Answers Chapter 5 Quadrilaterals

Question 1.
Diagonals of a parallelogram WXYZ intersect each other at point O. If ∠XYZ∠ = 135°, then measure of ∠XWZ and ∠YZW? If l(OY) = 5 cm, then l(WY) = ?
Solution:

i. ∠XYZ = 135°
□WXYZ is a parallelogram.
∠XWZ = ∠XYZ
∴ ∠XWZ = 135° …..(i)

ii. ∠YZW + ∠XYZ = 180° [Adjacent angles of a parallelogram are supplementary]
∴ ∠YZW + 135°= 180° [From (i)]
∴ ∠YZW = 180°- 135°
∴ ∠YZW = 45°

iii. l(OY) = 5 cm [Given]
l(OY) = \(\frac { 1 }{ 2 }\) l(WY) [Diagonals of a parallelogram bisect each other]
∴ l(WY) = 2 x l(OY)
= 2 x 5
∴ l(WY) = 10 cm
∴∠XWZ = 135°, ∠YZW = 45°, l(WY) = 10 cm

Question 2.
In a parallelogram ABCD, if ∠A = (3x + 12)°, ∠B = (2x – 32)°, then liptl the value of x and the measures of ∠C and ∠D.
Solution:

□ABCD is a parallelogram. [Given]
∴ ∠A + ∠B = 180° [Adjacent angles of a parallelogram are supplementary],
∴ (3x + 12)° + (2x-32)° = 180°
∴ 3x + 12 + 2x – 32 = 180
∴ 5x – 20 = 180
∴ 5x= 180 + 20
∴ 5x = 200
∴ x = \(\frac { 200 }{ 5 }\)
∴ x = 40

ii. ∠A = (3x + 12)°
= [3(40) + 12]°
=(120 +12)°= 132°
∠B = (2x – 32)°
= [2(40) – 32]°
= (80 – 32)° = 48°
∴ ∠C = ∠A = 132°
∠D = ∠B = 48° [Opposite angles of a parallelogram]
∴ The value of x is 40, and the measures of ∠C and ∠D are 132° and 48° respectively.

Question 3.
Perimeter of a parallelogram is 150 cm. One of its sides is greater than the other side by 25 cm. Find the lengths of all sides.
Solution:

i. Let □ABCD be the parallelogram and the length of AD be x cm.
One side is greater than the other by 25 cm.
∴ AB = x + 25 cm
AD = BC = x cm
AB = DC = (x + 25) cm [Opposite angles of a parallelogram]

ii. Perimeter of □ABCD = 150 cm [Given]
∴ AB + BC + DC + AD = 150
∴ (x + 25) +x + (x + 25) + x – 150
∴ 4x + 50 = 150
∴ 4x = 150 – 50
∴ 4x = 100
∴ x = \(\frac { 100 }{ 4 }\)
∴ x = 25

iii. AD = BC = x = 25 cm
AB = DC = x + 25 = 25 + 25 = 50 cm
∴ The lengths of the sides of the parallelogram are 25 cm, 50 cm, 25 cm and 50 cm.

Question 4.
If the ratio of measures of two adjacent angles of a parallelogram is 1 : 2, find the measures of all angles of the parallelogram.
Solution:

i. Let □ABCD be the parallelogram.
The ratio of measures of two adjacent angles of a parallelogram is 1 : 2.
Let the common multiple be x.
∴ ∠A = x° and ∠B = 2x°
∠A + ∠B = 180° [Adjacent angles of a parallelogram are supplementary]
∴ x + 2x = 180
∴ 3x = 180
∴ x = \(\frac { 180 }{ 3 }\)
∴ x = 60

ii. ∠A = x° = 60°
∠B = 2x° = 2 x 60° = 120°
∠A = ∠C = 60°
∠B = ∠D= 120° [Opposite angles of a parallelogram]
∴ The measures of the angles of the parallelogram are 60°, 120°, 60° and 120°.

Question 5.
Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO show that □ABCD is a rhombus.

Given: AO = 5, BO = 12 and AB = 13.
To prove: □ABCD is a rhombus.
Solition:
Proof:
AO = 5, BO = 12, AB = 13 [Given]
AO² + BO² = 5² + 12²
= 25 + 144
∴ AO² + BO² = 169 …..(i)
AB² = 13² = 169 ….(ii)
∴ AB² = AO² + BO² [From (i) and (ii)]
∴ ∆AOB is a right-angled triangle. [Converse of Pythagoras theorem]
∴ ∠AOB = 90°
∴ seg AC ⊥ seg BD …..(iii) [A-O-C]
∴ In parallelogram ABCD,
∴ seg AC ⊥ seg BD [From (iii)]
∴ □ABCD is a rhombus. [A parallelogram is a rhombus perpendicular to each other]

Question 6.
In the adjoining figure, □PQRS and □ABCR are two parallelograms. If ∠P = 110°, then find the measures of all the angles of □ABCR.
Solution:

□PQRS is a parallelogram. [Given]
∴ ∠R = ∠P [Opposite angles of a parallelogram]
∴ ∠R = 110° …..(iii)
□ABCR is a parallelogram. [Given]
∴ ∠A + ∠R= 180° [Adjacent angles of a parallelogram are supplementary]
∴ ∠A+ 110°= 180° [From (i)]
∴ ∠A= 180°- 110°
∴ ∠A = 70°
∴ ∠C = ∠A = 70°
∴ ∠B = ∠R= 110° [Opposite angles of a parallelogram]
∴ ∠A = 70°, ∠B = 110°,
∴ ∠C = 70°, ∠R = 110°

Question 7.
In the adjoining figure, □ABCD is a parallelogram. Point E is on the ray AB such that BE = AB, then prove that line ED bisects seg BC at point F.

Given: □ABCD is a parallelogram.
BE = AB
To prove: Line ED bisects seg BC at point F i.e. FC = FB
Solution:
Proof:
□ABCD is a parallelogram. [Given]
∴ seg AB ≅ seg DC …….(i) [Opposite angles of a parallelogram]
seg AB ≅ seg BE ……..(ii) [Given]
seg DC ≅ seg BE ……..(iii) [From (i) and (ii)]
side DC || side AB [Opposite sides of a parallelogram]
i.e. side DC || seg AE and seg DE is their transversal. [A-B-E]
∴ ∠CDE ≅ ∠AED
∴ ∠CDF ≅ ∠BEF …..(iv) [D-F-E, A-B-E]
In ∆DFC and ∆EFB,
seg DC = seg EB [From (iii)]
∠CDF ≅ ∠BEF [From (iv)]
∠DFC ≅ ∠EFB [Vertically opposite angles]
∴ ∆DFC ≅ ∆EFB [SAA test]
∴ FC ≅ FB [c.s.c.t]
∴ Line ED bisects seg BC at point F.

Maharashtra Board Class 9 Maths Chapter 5 Quadrilaterals Practice Set 5.1 Intext Questions and Activities

Question 1.
Write the following pairs considering □ABCD. (Textbook pg. no 57)

Pairs of adjacent sides:
i. AB, AD
ii. AD, DC
iii. DC, BC
iv. BC, AB

Pairs of adjacent angles:
i. ∠A, ∠B
ii. ∠C, ∠D
iii. ∠B, ∠C
iv. ∠D, ∠A

Pairs of opposite sides:
i. AB, DC
ii. AD, BC

Pairs of opposite angles:
i. ∠A, ∠C
ii. ∠B, ∠D

Question 2.
Complete the following tree diagram. (Textbook pg. no 57)

Question 3.
In the above theorem, to prove ∠DAB ≅ ∠BCD, is any change in the construction needed? If so, how will you write the proof making the change? (Textbook pg. no. 60)

Solution:
Yes
Construction: Draw diagonal BD.
Proof:
side AB || side CD and diagonal BD is their transversal. [Given]
∴ ∠ABD ≅ ∠CDB ……..(i) [Alternate angles]
side BC || side AD and diagonal BD is their transversal. [Given]
∴ ∠ADB ≅ ∠CBD ……..(ii) [Alternate angles]
In ∆DAB and ∆BCD,
∠ABD ≅ ∠CDB [From (i)]
seg BD ≅ seg DB [Common side]
∴ ∠ADB ≅ ∠CBD [From (ii)]
∴ ∆DAB ≅ ∆BCD [ASA test]
∴ ∠DAB ≅ ∠BCD [c.a.c.t.]
Note: ∠DAB s ∠BCD can be proved using the same construction as in the above theorem.
∠BAC ≅ ∠DCA …..(i)
∠DAC ≅ ∠BCA ……(ii)
∴ ∠BAC + ∠DAC ≅ ∠DCA + ∠BCA [Adding (i) and (ii)]
∴ ∠DAB ≅ ∠BCD [Angle addition property]

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 5.2 Geometry 9th Class Maths Part 2 Answers Solutions Chapter 5 Quadrilaterals.

Practice Set 5.2 Geometry 9th Std Maths Part 2 Answers Chapter 5 Quadrilaterals

Question 1.
In the adjoining figure, □ABCD is a parallelogram, P and Q are midpoints of sides AB and DC respectively, then prove □APCQ is a parallelogram.

Given: □ABCD is a parallelogram. P and Q are the midpoints of sides AB and DC respectively.
To prove: □APCQ is a parallelogram.
Solution:
Proof:
AP = \(\frac { 1 }{ 2 }\) AB …..(i) [P is the midpoint of side AB]
QC = \(\frac { 1 }{ 2 }\) DC ….(ii) [Q is the midpoint of side CD]
□ABCD is a parallelogram. [Given]
∴ AB = DC [Opposite sides of a parallelogram]
∴ \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) DC [Multiplying both sides by \(\frac { 1 }{ 2 }\)]
∴ AP = QC ….(iii) [From (i) and (ii)]
Also, AB || DC [Opposite angles of a parallelogram]
i.e. AP || QC ….(iv) [A – P – B, D – Q – C]
From (iii) and (iv),
□APCQ is a parallelogram. [A quadrilateral is a parallelogram if its opposite sides is parallel and congruent]

Question 2.
Using opposite angles test for parallelogram, prove that every rectangle is a parallelogram.

Given:
□ABCD is a rectangle.
To prove: Rectangle ABCD is a parallelogram.
Solution:
Proof:
□ABCD is a rectangle.
∴ ∠A ≅ ∠C = 90° [Given]
∠B ≅ ∠D = 90° [Angles of a rectangle]
∴ Rectangle ABCD is a parallelogram. [A quadrilateral is a parallelogram, if pairs of its opposite angles are congruent]

Question 3.
In the adjoining figure, G is the point of concurrence of medians of ADEF. Take point H on ray DG such that D-G-H and DG = GH, then prove that □GEHF is a parallelogram.

Given: Point G (centroid) is the point of concurrence of the medians of ADEF.
DG = GH
To prove: □GEHF is a parallelogram.
Solution:
Proof:

Let ray DH intersect seg EF at point I such that E-I-F.
∴ seg DI is the median of ∆DEF.
∴ El = FI ……(i)
Point G is the centroid of ∆DEF.
∴ \(\frac { DG }{ GI }\) = \(\frac { 2 }{ 1 }\) [Centroid divides each median in the ratio 2:1]
∴ DG = 2(GI)
∴ GH = 2(GI) [DG = GH]
∴ GI + HI = 2(GI) [G-I-H]
∴ HI = 2(GI) – GI
∴ HI = GI ….(ii)
From (i) and (ii),
□GEHF is a parallelogram [A quadrilateral is a parallelogram, if its diagonals bisect each other]

Question 4.
Prove that quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle.

Given: □ABCD is a parallelogram.
Rays AS, BQ, CQ and DS bisect ∠A, ∠B, ∠C and ∠D respectively.
To prove: □PQRS is a rectangle.
Solution:
Proof:
∠BAS = ∠DAS = x° …(i) [ray AS bisects ∠A]
∠ABQ = ∠CBQ =y° ….(ii) [ray BQ bisects ∠B]
∠BCQ = ∠DCQ = u° …..(iii) [ray CQ bisects ∠C]
∠ADS = ∠CDS = v° ….(iv) [ray DS bisects ∠D]
□ABCD is a parallelogram. [Given]
∴ ∠A + ∠B = 180° [Adjacent angles of a parallelogram are supplementary]
∴ ∠BAS + ∠DAS + ∠ABQ + ∠CBQ = 180° [Angle addition property]
∴ x°+x°+ v° + v° = 180 [From (i) and (ii)]
∴ 2x° + 2v° =180
∴ x + y = 90° ……(v) [Dividing both sides by 2]
Also, ∠A + ∠D= 180° [Adjacent angles of a parallelogram are supplementary]
∴ ∠BAS + ∠DAS + ADS + ∠CDS = 180° [Angle addition property]
∴ x° + x° + v° + v° = 180°
∴ 2x° + 2v° = 180°
∴ x° + v° = 90° …..(vi) [Dividing both sides by 2]
In ∆ARB,
∠RAB + ∠RBA + ∠ARB = 180° [Sum of the measures of the angles of a triangle is 180°]
∴ x° + y° + ∠SRQ = 180° [A – S – R, B – Q – R]
∴ 90° + ∠SRQ = 180° [From (v)]
∴ ∠SRQ = 180°- 90° = 90° …..(vi)
Similarly, we can prove
∠SPQ = 90° …(viii)
In ∆ASD,
∠ASD + ∠SAD + ∠SDA = 180° [Sum of the measures of angles a triangle is 180°]
∴ ∠ASD + x° + v° = 180° [From (vi)]
∴ ∠ASD + 90° = 180°
∴∠ASD = 180°- 90° = 90°
∴ ∠PSR = ∠ASD [Vertically opposite angles]
∴ ∠PSR = 90° …..(ix)
Similarly we can prove
∠PQR = 90° ..(x)
∴ In □PQRS,
∠SRQ = ∠SPQ = ∠PSR = ∠PQR = 90° [From (vii), (viii), (ix), (x)]
∴ □PQRS is a rectangle. [Each angle is of measure 90°]

Question 5.
In the adjoining figure, if points P, Q, R, S are on the sides of parallelogram such that AP = BQ = CR = DS, then prove that □PQRS is a parallelogram.

Given: □ABCD is a parallelogram.
AP = BQ = CR = DS
To prove: □PQRS is a parallelogram.
Solution:
Proof:
□ABCD is a parallelogram. [Given]
∴ ∠B = ∠D ….(i) [Opposite angles of a parallelogram]
Also, AB = CD [Opposite sides of a parallelogram]
∴ AP + BP = DR + CR [A-P-B, D-R-C]
∴ AP + BP = DR + AP [AP = CR]
∴ BP = DR ….(ii)
In APBQ and ARDS,
seg BP ≅ seg DR [From (ii)]
∠PBQ ≅ ∠RDS [From (i)]
seg BQ ≅ seg DS [Given]
∴ ∆PBQ ≅ ∆RDS [SAS test]
∴ seg PQ ≅ seg RS …..(iii) [c.s.c.t]
Similarly, we can prove that
∆PAS ≅ ∆RCQ
∴ seg PS ≅ seg RQ ….(iv) [c.s.c.t]
From (iii) and (iv),
□PQRS is a parallelogram. [A quadrilateral is a parallelogram, if pairs of its opposite angles are congruent]

Maharashtra Board Class 9 Maths Chapter 5 Quadrilaterals Practice Set 5.2 Intext Questions and Activities

Question 1.
Points D and E are the midpoints of side AB and side AC of ∆ABC respectively. Point F is on ray ED such that ED = DF. Prove that □AFBE is a parallelogram. For this example write ‘given’ and ‘to prove’ and complete the proof. (Text book pg. no. 66)

Given: D and E are the midpoints of side AB and side AC respectively.
ED = DF
To prove: □AFBE is a parallelogram.
Solution:
Proof:
seg AB and seg EF are the diagonals of □AFBE.
seg AD ≅ seg DB [Given]
seg DE ≅ seg DF [Given]
∴ Diagonals of □AFBE bisect each other.
∴ □AFBE is a parallelogram. [ By test of parallelogram]

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

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

Question 1.
Construct ∆XYZ, such that XY + XZ = 10.3 cm, YZ = 4.9 cm, ∠XYZ = 45°.
Solution:

As shown in the rough figure draw segYZ = 4.9cm
Draw a ray YT making an angle of 45° with YZ
Take a point W on ray YT, such that YW= 10.3 cm
Now,YX + XW = YW [Y-X-W]
∴ YX + XW=10.3cm …..(i)
Also, XY + X∠10.3cm ……(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 point X.

Steps of construction:
i. Draw seg YZ of length 4.9 cm.
ii. Draw ray YT, such that ∠ZYT = 75°.
iii. Mark point W on ray YT such that l(YW) = 10.3 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 2.
Construct ∆ABC, in which ∠B = 70°, ∠C = 60°, AB + BC + AC = 11.2 cm.
Solution:

i. As shown in the figure, take point D and E on line BC, such that
BD = AB and CE = AC ……(i)
BD + BC + CE = DE [D-B-C, B-C-E]
∴ AB + BC + AC = DE …..(ii)
Also,
AB + BC + AC= 11.2 cm ….(iii) [Given]
∴ DE = 11.2 cm [From (ii) and (iii)]

ii. In ∆ADB
AB = BD [From (i)]
∴ ∠BAD = ∠BDA = x° ….(iv) [Isosceles triangle theorem]
In ∆ABD, ∠ABC is the exterior angle.
∴ ∠BAD + ∠BDA = ∠ABC [Remote interior angles theorem]
x + x = 70° [From (iv)]
∴ 2x = 70° x = 35°
∴ ∠ADB = 35°
∴ ∠D = 35°
Similarly, ∠E = 30°

iii. Now, in ∆ADE
∠D = 35°, ∠E = 30° and DE = 11.2 cm
Elence, ∆ADE can be drawn.

iv. Since, AB = BD
∴ Point B lies on perpendicular bisector of seg AD.
Also AC = CE
∴ Point C lies on perpendicular bisector of seg AE.
∴ Points B and C can be located by drawing the perpendicular bisector of AD and AE respectively.
∴ ∆ABC can be drawn.

Steps of construction:
i. Draw seg DE of length 11.2 cm.
ii. From point D draw ray making angle of 35°.
iii. From point E draw ray making angle of 30°.
iv. Name the point of intersection of two rays as A.
v. Draw the perpendicular bisector of seg DA and seg EA intersecting seg DE in B and C respectively.
vi. Join AB and AC.
Hence, ∆ABC is the required triangle.

Question 3.
The perimeter of a triangle is 14.4 cm and the ratio of lengths of its side is 2 : 3 : 4. Construct the triangle.
Solution:

Let the common multiple be x
∴ In ∆ABC,
AB = 2x cm, AC = 3x cm, BC = 4x cm
Perimeter of triangle = 14.4 cm
∴ AB + BC + AC= 14.4
∴ 9x = 14.4
∴ x = \(\frac { 14.4 }{ 9 }\)
∴ x = 1.6
∴ AB = 2x = 2x 1.6 = 3.2 cm
∴ AC = 3x = 3 x 1.6 = 4.8 cm
∴ BC = 4x = 4 x 1.6 = 6.4 cm

Question 4.
Construct ∆PQR, in which PQ – PR = 2.4 cm, QR = 6.4 cm and ∠PQR = 55°.
Solution:

Here, PQ – PR = 2.4 cm
∴ PQ > PR
As shown in the rough figure draw seg QR = 6.4 cm
Draw a ray QT making on angle of 55° with QR
Take a point S on ray QT, such that QS = 2.4 cm.
Now, PQ – PS = QS [Q-S-P]
∴ PQ – PS = 2.4 cm …(i)
Also, PQ – PR = 2.4 cm ….(ii) [Given]
∴ PQ – PS = PQ – PR [From (i) and (ii)]
∴ 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 6.4 cm.
ii. Draw ray QT, such that ∠RQT = 55°.
iii. Take point S on ray QT such that l(QS) = 2.4 cm.
iv. Join the points S and R.
v. Draw perpendicular bisector of seg SR intersecting ray QT.
Name that point as P.
vi. Join the points P and R.
Hence, ∆PQR is the required triangle.

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

Practice Set 4.3 Geometry 9th Std Maths Part 2 Answers Chapter 4 Constructions of Triangles

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

Practice Set 4.2 Geometry 9th Std Maths Part 2 Answers Chapter 4 Constructions of Triangles

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

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

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

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

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

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

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

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

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