Samacheer Kalvi Class 9 Maths Chapter 4 Geometry Solutions
Welcome to NCTB Solutions. Here with this post we are going to help 9th class students by providing Solutions for Samacheer Kalvi Class 9 Maths chapter 4 Geometry. Here students can easily find all the solutions for Geometry Exercise 4.1, 4.2, 4.3, 4.4, 4.5, 4.6 and 4.7. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 4 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus.
Geometry Exercise 4.1 Solutions :
(1) In the figure, AB is parallel to CD, find x
Solution :
Figure – (i)
∠A = 140°
∠b = 180-140 = 40
∠c = 150°
∠3 = 180-150 = 30 °
∠x = 40+30
= 70°
Figure – (ii)
∠B = 48°
48°+24°+∠3 = 180°
∠3 = 180°
180°+∠5 = 180°
∠5 = 72°
∠3+∠4+∠5 = 72°+108°+108°
= 288°
Figure – (iii)
∠D = 53
So, ∠x = 180°- (38+53)
= 180°-91
= 89°
(2) The angles of a triangle are in the ratio 1: 2 : 3, find the measure of each angle of the triangle.
Solution :
Let, C.F = x
We know that,
x + 2x + 3x = 180°
or, 6x = 180°
or, x = 180°/6
∴ x = 30°
∴ Angles are,
= 30°
= 2 × 30 = 60°
= 3 × 30 = 90°
(3) Consider the given pairs of triangle and say whether each pair is that of congruent triangle. if the triangles are congruent, say how; if they are not congruent say why and also say why and also say if small modification would make them congruent
Solution :
Figure – (i)
AB=PQ
RQ=BC
∆ABC is not congruent to ∆PQR
Figure – (ii)
AD=BC and AB=DC and BD is common
So, ∆ABD ≅ BDC
Figure – (iii)
PY=PZ and XY=XZ
And PY is common
So, ∆PXZ ≅ ∆PXY.
Figure – (iv)
OA=OC ∠AB = ∠COD and obviously ∠BAO = ∠DOC
So, ∆ABO ≅ ∆COD
Figure – (v)
AO = OC
BO = BD
Obviously ∠ABO =∠COD
So ∆ABO = ∆DOC
Figure – (vi)
∠M = 90°
AB = AC
∆BAM ≅ AMC.
(4) ΔABC and ΔDEF are two triangles in which AB = DF, ∠ACB = 70°, ∠ABC = 60°; ∠DEF = 70° and ∠EDF = 60°. prove that the triangles are congruent
Solution :
∠ACB=70°, ∠ABC = 60°
So, ∠BAC = 180°-130° = 50
Similarly,
∠DFE = 50
So, ∠A = ∠E
Hence ∆ABC ≅ ∆DEF
(5) Find all the three angles of the ΔABC
Solution :
We know,
x+35°+ 2x-5 = 4x-15°
= 4x-3x = 30+15°
= x = 45°
∠A = 45° + 35 = 80°
∠B = 2x – 5° = 85 °
∠C = 4x -15° = 165°
Geometry Exercise 4.2 Solutions :
(1) The angles of a quadrilateral are in the ratio 2 : 4 : 5 : 7. Find all the angles.
Solution :
Let, the angle be 2x, 4x, 5x and 7x
So, 2x + 4x + 5x + 7x = 360°
= x = 20°
So, angle are 40°, 80°, 100°, 140°
(3) ABCD is a rectangle whose diagonals AC and BD intersect at O. If ∠OAB = 46°, find ∠OBC
Solution :
= ∠OAB + ∠OBC = 90°
= ∠OBC = 90°- 46°
= ∠OBC = 44°
(4) The lengths of the diagonals of a Rhombus are 12 cm and 16 cm. Find the side of the rhombus.
Solution :
OA = 16/2 = 8 cm
OB = 12/2 = 6 cm
So, AB² = 6² +8²
= 36 + 64
= AB² = 100
∴ AB = 10
Therefore, the side of the rhombus is 10.
(7) Iron rods a, b, c ,d, e, and f, are making a design in a bridge as shown in the figure if a || b ,c || d, e || f, find the marked angles between
(i) b and c
(ii) d and e
(iii) d and f
(iv) c and f
Solution :
∠1 = 30°
∠2 + 75° = 180°
= ∠2 = 105°
So, ∠2 + ∠3 = 180°
∠3 = 75°
So, ∠3 + ∠4 = 180°
=∠4 = 105°
Hence, (i) Angles between b and c = 30
(ii) Angles between d and e = 105 °
(iii) Angles between d and f = 75 °
(iv) Angles between c and f = 105°
(8) In the given Fig. 4.39, ∠A = 64° , ∠ABC = 58°. If BO and CO are the bisectors of ∠ABC and ∠ACB respectively of ΔABC, find x° and y°
Solution :
In ∆ABC,
∠C = 180° – (64 + 58)
= 58°
∠OBC = 58/22 = 29
Similarly, ∠OBC = 29°
In ∆OBC,
= x° + y° + ∠OBC = 180°
= x = 122°
x° + y° + ∠OBC = 180°
= 122° + y + 29° = 180°
= x = 180° – 122 – 29
= y = 58 – 29
= y = 29°
(9) In the given Fig. 4.40, if AB = 2, BC = 6, AE = 6, BF = 8, CE = 7, and CF = 7, compute the ratio of the area of quadrilateral ABDE to the area of ΔCDF. (Use congruent property of triangles).
Solution :
AB = 2, BC = 6, AE = 6, BF = 8, CE l= 7, EC = 7
In ΔAEC and ΔBCF
AC = 8 and BF = 8
AE = 6 and BC = 6
CE = 7 and CF = 7
So, ΔAEC ≅ ΔBCF
So, area of ΔAEC = area of ΔBCF
Subtract ΔBDC from both triangles
Then area of ABDE = Area of ΔCDF
i.e 1 = 1
Geometry Exercise 4.3 Solutions :
(1) The diameter of the circle is 52 cm and the length of one of its chord is 20 cm. Find the distance of the chord from the centre.
Solution :
So, OD² = AO² – AD²
= √26² – 101²
= √576
= 24 cm
Therefore, the distance of the chord from the centre is 24 cm.
(2) The chord of length 30 cm is drawn at the distance of 8 cm from the centre of the circle. Find the radius of the circle.
Solution :
Radius of the circle,
= √8² + 15²
= √289
= 17
Hence, the radius of the circle will be 17
(3) Find the length of the chord AC where AB and CD are the two diameters perpendicular to each other of a circle with radius 4√2 cm and also find ∠OAC and ∠OCA.
Solution :
∠OAC = 90°
Since, ∠AOC = 90°
= ∠OAC = 45°
Length of the chord
= √(4√2)² + (4√2)²
= √32 + 32
= 8
(4) A chord is 12 cm away from the centre of the circle of radius 15 cm. Find the length of the chord.
Solution :
BC = √15² – 12²
= √225 – 144
= 9
Length of the chord is
= 2 × 9
= 18 cm
Therefore, the length of the chord is 18 cm.
(5) In a circle, AB and CD are two parallel chords with centre O and radius 10 cm such that AB = 16 cm and CD = 12 cm determine the distance between the two chords?
Solution :
Given, AB = 16, CD = 12
FE = OE + OF
So, OE = √10² – 6²
= √64
= 8
OF = √10² – 8²
= √36
= 6
So, EF = 6 + 8
= 14 cm
Therefore, the distance between the two chords will be 14 cm.
(6) Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Solution :
OD = 4/2
= 2 cm
AD = BD
= √5² -4²
= 3
So, length of common chord AB = 6 cm
Hence, the length of the common chord will be 6 cm.
(7) Find the value of x° in the following figures :
Solution :
Figure – (i)
We know, ∠BAC = 1/2
∠BOC = 1/2 × 90
= 45°
Figure – (ii)
∠QPR = 1/2 × 80 = 40°
∠P + ∠Q + ∠R = 180°
= 40 + ∠Q + 80 = 180
= ∠Q = 60°
Given, 50 + x = 60
= x = 10°
Figure – (iii)
∠MPN = 90°
∠M + ∠P = 180° – 110° = 70°
∠MPO = 70/2 = 35
But, ∠MPN = 90°
So, ∠O x° = 90 – 35
= 55°
Figure – (iv)
∠YOZ = 120° × 2 = 240°
x + ∠YOZ = 360°
= x = 360° – 240°
= x = 120°
Figure – (v)
∠BOC = 360° – 240° = 120°
∴ ∠BAC = 120°/2
= 60°
(8) In the given figure, ∠CAB = 25°, find ∆BDC, ∆DBA and ∆COB
Solution :
Given, ∠CAB = 25°
∴ ∠BDC = 25°
Now, ∠DBA = 180° – (90° + 25)
= 180° – 115°
= 65°
∴ ∠COB = 2 × 25°
= 50°
Geometry Exercise 4.4 Solutions :
(1) Find the value of x in the given figure.
Solution :
∠ABC = 180° – 120° = 60°
∠BCA = 90°
So, ∠CAB
= 180° – (90° + 60°)
= 30°
Therefore, the value of x is 30°
(2) In the given figure, AC is the diameter of the circle with centre O. if ∠ADE = 30°; ∠DAC = 35° and ∠CAB = 40°
Find (i) ∠ACD (ii) ∠ACB (iii) ∠DAE
Solution :
(i) ∠ACD = 180° – (90° + 35°)
= 180° – 125°
= 55°
(ii) ∠ACB = 180° – (90° + 40)
= 180° – 135°
= 50°
(iii) ∠ADC = 90°
∠CAE = 180° – 120° = 60°
∴ ∠DAE = 60 – 35 = 25
(3) Find all the angles of the given cyclic quadrilateral ABCD in the figure.
Solution :
∠A + ∠C = 180°
= 2y + 4 + 4y – 4 = 180°
= 6y = 180°
= y = 30°
∠B + ∠D = 180°
= 6x – 4 + 7x + 2 = 180°
= 13x = 182
= x = 14°
∠A = 2 × 30° + 4° = 64°
∠B = 6 × 14 – 4 = 80°
∠C = 4 × 30 – 4 = 116°
∠D = 7 × 14 + 2 = 100°
(5) In the given figure, AB and CD are the parallel chords of a circle with centre O. such that AB = 8 cm and CD = 6 cm. if OM ⊥ AB and OL⊥CD distance between LM is 7 cm. find the radius of the circle
Solution :
Given, LM = 7
OM = 7x
MB = 4 cm
OB = √4² + (7-x)²
OD = √3² + x²
∴ √4² + (7 – x)² = √3² + x²
= 16 + (7 – x)² = 9 + x²
= 149 – 14x + x² = 9 + x²
= 14x = 56
= x = 4
OD = √3² + 4²
= √9 + 16
= 5
Therefore, the radius of the circle will be 5
(6) The arch of a bridge has dimensions as shown, where the arch measure 2 m at its highest point and its width is 6 m. What is the radius of the circle that contains the arch?
Solution :
Let, r be radius
OD = r – 2
AD = 3 cm
OD² = OD² + BD²
r² = (r-2)² + 3²
= r² = r² – 4r + 4 + 9
= 4r = 13
= r = 3.25
Hence, Radius of the figure will be 3.25 cm.
(7) In figure, ∠ABC = 120°, where A, B and C are points on the circle with centre O. find ∠OAC
Solution :
(∠AOC) = 2 × 120° = 240°
∠AOC = 360° – 240° = 120°
∠OAC + ∠OCA = 180° – 120° = 60°
= 2∠AOC = 60°
= ∠OAC = 30°
(8) A school wants to conduct tree plantation programme. For this a teacher allotted a circle of radius 6m ground to nineth standard students for planting sapplings. Four students plant trees at the points A, B, C and D as shown in figure. Here AB = 8 m, CD = 10 m and AB⊥CD. If another student places a flower pot at the point P, the intersection of AB and CD, then find the distance from the centre to P.
Solution :
As per the question,
OA = 6
AB = 8
CD = 10
OM² = √6² – 4² = √20
ON = √6² – 5² = √11
∴ OP = √6N² + NP²
= √(√11)² + (√20)²
= √11 + 20
= √31 cm.
(9) In the given figure, ∠POQ = 100° and ∠PQR = 30°, then find ∠RPO
Solution :
As per the given question,
∠POQ = 100°
∠PQR = 30°
In △PQO
∠OPQ = ∠POQ = 180°
= ∠OPQ + 100° = 180°
= ∠OPQ = 40°
∴ In △PRQ, ∠R + ∠P + ∠Q = 180°
= 50 + (40 + x) + 30° = 180°
= x = 180° + 30 – 40 – 50°
= 60°
∴ ∠RPO = 60°
Geometry Exercise 4.7 Solutions :
Multiple Choice Questions :
(1) The exterior angle of a triangle is equal to the sum of two
(1) Exterior angles
(2) Interior opposite angles
(3) Alternate angles
(4) Interior angles
Solution :
Correct option – (2)
The exterior angle of a triangle is equal to the sum of two Interior opposite angles.
(2) In the quadrilateral ABCD, AB = BC and AD = DC Measure of ∠BCD is
(1) 150°
(2) 30°
(3) 105°
(4) 72°
Solution :
Correct option – (3)
∠C = 72/2 + 138/2
= 36 + 69
= 150°
Therefore, Measure of ∠BCD is 150°
(3) ABCD is a square, diagonals AC and BD meet at O. The number of pairs of congruent triangles with vertex O are
(1) 6
(2) 8
(3) 4
(4) 12
Solution :
Correct option – (1)
The number of pairs of congruent triangles with vertex O are 6.
(4) In the given figure CE || DB then the value of x° is
(1) 45°
(2) 30°
(3) 75°
(4) 85°
Solution :
Correct option – (4)
So, the value of x° is 85°
(5) The correct statement out of the following is
(1) ΔABC ≅ ΔDEF
(2) ΔABC ≅ ΔDEF
(3) ΔABC ≅ ΔFDE
(4) ΔABC ≅ ΔFED
Solution :
Correct option – (4)
The correct statement out of the following is ΔABC ≅ ΔFED.
(6) If the diagonal of a rhombus are equal, then the rhombus is a
(1) Parallelogram but not a rectangle
(2) Rectangle but not a square
(3) Square
(4) Parallelogram but not a square
Solution :
Correct option – (3)
If the diagonal of a rhombus are equal, then the rhombus is a Square.
(7) If bisectors of ∠A and ∠B of a quadrilateral ABCD meet at O, then ∠AOB is
(1) ∠C + ∠D
(2) 1/2 (∠C + ∠D)
(3) 1/2∠C + 1/3∠D
(4) 1/3∠C + 1/2∠D
Solution :
Correct option – (2)
Then ∠AOB is 1/2 (∠C + ∠D)
(8) The interior angle made by the side in a parallelogram is 90° then the parallelogram is a
(1) rhombus
(2) rectangle
(3) trapezium
(4) kite
Solution :
Correct option – (2)
The interior angle made by the side in a parallelogram is 90° then the parallelogram is a Rectangle.
(9) Which of the following statement is correct?
(1) Opposite angles of a parallelogram are not equal.
(2) Adjacent angles of a parallelogram are complementary.
(3) Diagonals of a parallelogram are always equal.
(4) Both pairs of opposite sides of a parallelogram are always equal.
Solution :
Correct option – (4)
The correct statement is – Both pairs of opposite sides of a parallelogram are always equal.
(10) The angles of the triangle are 3x–40, x+20 and 2x–10 then the value of x is
(1) 40°
(2) 35°
(3) 50°
(4) 45°
Solution :
Correct option – (2)
Therefore, the value of x is 35°
(11) PQ and RS are two equal chords of a circle with centre O such that ∠POQ = 70°, then ∠ORS =
(1) 60°
(2) 70°
(3) 55°
(4) 80°
Solution :
Correct option – (3)
Hence, ∠ORS = 55°
(12) A chord is at a distance of 15 cm from the centre of the circle of radius 25 cm. The length of the chord is
(1) 25 cm
(2) 20 cm
(3) 40 cm
(4) 18 cm
Solution :
Correct option – (3)
Therefore, the length of the chord is 40 cm.
(13) In the figure, O is the centre of the circle and ∠ACB = 40° then ∠AOB =
(1) 80°
(2) 85°
(3) 70°
(4) 65°
Solution :
Correct option – (1)
Therefore, ∠AOB = 80°
(14) In a cyclic quadrilaterals ABCD, ∠A = 4x, ∠C = 2x the value of x is
(1) 30°
(2) 20°
(3) 15°
(4) 25°
Solution :
Correct option – (1)
So, the value of x is 30°
(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 = 4 cm. the radius of the circle is
(1) 8 cm
(2) 4 cm
(3) 6 cm
(4) 10 cm
Solution :
Correct option – (4)
Thus, the radius of the circle will be 10 cm.
(16) In the figure, PQRS and PTVS are two cyclic quadrilaterals, if ∠QRS = 100°, then ∠TVS =
(1) 80°
(2) 100°
(3) 70°
(4) 90°
Solution :
Correct option – (1)
Then ∠TVS = 80°
(17) If one angle of a cyclic quadrilateral is 75°, then the opposite angle is
(1) 100°
(2) 105°
(3) 85°
(4) 90°
Solution :
Correct option – (2)
If one angle of a cyclic quadrilateral is 75°, then the opposite angle is 105°
(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 =
(1) 100°
(2) 20°
(3) 120°
(4) 110°
Solution :
Correct option – (3)
Then ∠BAD = 120°
(19) AD is a diameter of a circle and AB is a chord If AD = 30 cm and AB = 24 cm then the distance of AB from the centre of the circle is
(1) 10 cm
(2) 9 cm
(3) 8 cm
(4) 6 cm
Solution :
Correct option – (2)
So, the distance of AB from the centre of the circle is 9 cm.
(20) In the given figure, if OP = 17 cm, PQ = 30 cm and OS is perpendicular to PQ, then RS is
(1) 10 cm
(2) 6 cm
(3) 7 cm
(4) 9 cm
Solution :
Here is the solution to your question :
OR = √17² – 15²
= 8
RS = 17 – 8
= 9
Therefore, option (4) 9 cm is the correct answer of this question.
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