Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Students can download Maths Chapter 4 Geometry Ex 4.7 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 4 Geometry Ex 4.7

Multiple Choice Questions

Question 1.
The exterior angle of a triangle is equal to the sum of two ……….
(a) Exterior angles
(b) Interior opposite angles
(c) Alternate angles
(d) Interior angles
Solution:
(b) Interior opposite angles

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 2.
In the quadrilateral ABCD, AB = BC and AD = DC Measure of ∠BCD is …………..
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 1
(a) 150°
(b) 30°
(c) 105°
(d) 72°
Solution:
(c) 105°
Hint:
Join BD
∠DBC = 54°; ∠BDC = 21°
∴ ∠BCD = 180° – (54° + 21°)
= 180° – 75°
= 105°

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 3.
ABCD is a square, diagonals AC and BD meet at O. The number of pairs of congruent triangles with vertex O are ……….
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 2
(a) 6
(b) 8
(c) 4
(d) 12
Solution:
(a) 6

Question 4.
In the given figure CE || DB then the value of x° is ………
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 3
(a) 45°
(b) 30°
(c) 75°
(d) 85°
∠B = 360° – (∠A + ∠D + ∠BCD)
= 360° – (110° + 75° + 60°)
= 360° – 245°
= 115°
∠DBC = 115° – 30°
= 85°
∴ ∠x = 85° (BD || CE) Alternate angles are equal

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 5.
The correct statement out of the following is ……..
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 4
(a) ΔABC ≅ ΔDEF
(b) ΔABC ≅ ΔEDF
(c) ΔABC ≅ ΔFDE
(d) ΔABC ≅ ΔFED
Solution:
(d) ΔABC ≅ ΔFED

Question 6.
If the diagonal of a rhombus are equal, then the rhombus is a ……..
(a) Parallelogram but not a rectangle
(b) Rectangle but not a square
(c) Square
(d) Parallelogram but not a square
Solution:
(c) Square

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 7.
If bisectors of ∠A and ∠B of a quadrilateral ABCD meet at O, then ∠AOB is ……..
(a) ∠C + ∠D
(b) \(\frac{1}{2}\) (∠C + ∠D)
(c) \(\frac{1}{2}\) ∠C + \(\frac{1}{3}\) ∠D
(d) \(\frac{1}{3}\) ∠C + \(\frac{1}{2}\) ∠D
Solution:
(b) \(\frac{1}{2}\) (∠C + ∠D)

Question 8.
The interior angle made by the side in a parallelogram is 90° then the parallelogram is a …….
(a) rhombus
(b) rectangle
(c) trapezium
(d) kite
Solution:
(b) rectangle
Hint:
Opposite sides are equal and each angle is 90° then it is a rectangle.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 9.
Which of the following statement is correct?
(a) Opposite angles of a parallelogram are not equal
(b) Adjacent angles of a parallelogram are complementary.
(c) Diagonals of a parallelogram are always equal.
(d) Both pairs of opposite sides of a parallelogram are always equal.
Solution:
(d) Both pairs of opposite sides of a parallelogram are always equal.

Question 10.
The angles of the triangle are 3x – 40, x + 20 and 2x – 10 then the value of x is ………
(a) 40°
(b) 35°
(c) 50°
(d) 45°
Solution:
(b) 35°
Hint:
3x – 40 + x + 20 + 2x – 10 = 180° (Sum of the angles of a triangle is 180°)
6x – 30 = 180°
6x = 180°+ 30°
x = \(\frac{210°}{6°}\)
= 35°

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 11.
PQ and RS are two equal chords of a circle with centre O such that ∠POQ = 70°, then ∠ORS = ……….
(a) 60°
(b) 70°
(c) 55°
(d) 80°
Solution:
(c) 55°
Hint:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 5
∠POQ = 70° (Vertically opposite Angle)
∠ORS and ∠OSR are equal. (OR = OS radius of the circle)
∠ORS + ∠OSR + ∠ROS = 180°
x° + x° + 70° = 180°
2x + 70° = 180°
2x = 180° – 70° = 110°
x = \(\frac{110°}{2}\) = 55°
∴ ∠ORS = 55°

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 12.
A chord is at a distance of 15cm from the centre of the circle of radius 25cm. The length of the chord is ……….
(a) 25cm
(b) 20cm
(c) 40cm
(d) 18cm
Solution:
(c) 40cm
Hint:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 6
In the right triangle OAC,
AC² = OA² – OC²
= 25² – 15²
= (25 + 15) (25 – 15)
= 40 × 10
AC² = 400
AC = \(\sqrt{400}\)
= 20
Length of the chord AB = 20 + 20 = 40 cm.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 13.
In the figure, O is the centre of the circle and ∠ACB = 40° then ∠AOB = ………
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 7
(a) 80°
(b) 85°
(c) 70°
(d) 65°
Solution:
(a) 80°
Hint:
∠AOB = 2∠ACB (The angle subtended by an arc of the circle at the centre is double the angle subtended at the remaining part of the circle.)
= 2 × 40° = 80°

Question 14.
In a cyclic quadrilaterals ABCD, ∠A = 4x, ∠C = 2x the value of x is ……..
(a) 30°
(b) 20°
(c) 15°
(d) 25
Solution:
(a) 30°
Hint:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 8
∠A + ∠C = 180° (Sum of the opposite angle of cyclic quadrilateral is 180°)
4x + 2x = 180°
x = \(\frac{180°}{6}\)
= 30°

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 15.
In the figure, O is the centre of a circle and diameter AB bisects the chord CD at a point E such that CE = ED = 8 cm and EB = 4cm. The radius of the circle is ………
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 9
(a) 8cm
(b) 4cm
(c) 6cm
(d) 10cm
Solution:
(i) 10cm
Hint:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 10
Let the radius OD be x.
OE = OB – BE
= x – 4 (OB radius of the circle)
In the ΔOED,
OD² = OE² + ED²
x² = (x – 4)² + 8²
x² = x² + 16 – 8x + 64
8x = 80
x = \(\frac{80}{8}\)
= 10 cm
Radius of the circle = 10 cm.

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 16.
In the figure, PQRS and PTVS are two cyclic quadrilaterals, If ∠QRS = 80°, then ∠TVS = ………
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 11
(a) 80°
(b) 100°
(c) 70°
(d) 90°
Solution:
(a) 80°
Hint:
∠SPQ = 180° – 80° (Opposite angles of a cyclic quadrilateral PQRS)
= 100°
In the cyclic quadrilateral PVTS,
∠V = 180° – 100° = 80° (Opposite angles of a cyclic quadrilateral)

Question 17.
If one angle of a cyclic quadrilateral is 75°, then the opposite angle is ………
(a) 100°
(b) 105°
(c) 85°
(d) 90°
Solution:
(b) 105°
Hint:
Opposite angles = 180° – 75°
= 105° (Sum of the opposite angle is 180°)

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 18.
In the figure, ABCD is a cyclic quadrilateral in which DC produced to E and CF is drawn parallel to AB such that ∠ADC = 80° and ∠ECF = 20°, then BAD = ?
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 12
(a) 100°
(b) 20°
(c) 120°
(d) 110°
Solution:
(c) 120°
Hint:
∠BAD = ∠ABC + ∠ECF (By exterior angle property)
= 100° + 20°
= 120°

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 19.
AD is a diameter of a circle and AB is a chord If AD = 30cm and AB = 24cm then the distance of AB from the centre of the circle is ………
(a) 10cm
(b) 9cm
(c) 8cm
(d) 6cm
Solution:
(b) 9cm
Hint:
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 13
In ΔAOC,
AO = 15 cm
AC = \(\frac{1}{2}\) AB
= \(\frac{1}{2}\) × 24
= 12 cm
In ΔAOC,
OC² = AO² – AC²
= 15² – 12²
= 225 – 144
= 81
OC = \(\sqrt{81}\)
= 9 cm

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Question 20.
In the given figure, If OP = 17cm, PQ = 30cm and OS is perpendicular to PQ, then RS is ……….
Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7 14
(a) 10cm
(b) 6cm
(c) 7cm
(d) 9cm
Solution:
(d) 9cm
Hint:
In ΔOPR, OP = 17cm, PR = 30/2 = 15 cm.
OR² = OP² – PR²
= 17² – 15²
= (17 + 15) (17 – 15)
= 32 × 2
OR² = \(\sqrt{64}\)
= 8 cm
RS = OS – OR
= 17 – 8
= 9 cm

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.7

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top