Samacheer Kalvi 10th Maths Chapter 4 Geometry Notes PDF Download: Tamil Nadu STD 10th Maths Chapter 4 Geometry Notes 
Samacheer Kalvi 10th Maths Chapter 4 Geometry Notes PDF Download: Students of class can download the Samacheer Kalvi 10th Maths Chapter 4 Geometry Notes PDF Download from our website. We have uploaded the Samacheer Kalvi 10th Maths Chapter 4 Geometry notes according to the latest chapters present in the syllabus. Download Samacheer Kalvi 10th Maths Chapter 4 Geometry Chapter Wise Notes PDF from the links provided in this article.
Samacheer Kalvi 10th Maths Chapter 4 Geometry Notes PDF Download
We bring to you specially curated Samacheer Kalvi 10th Maths Chapter 4 Geometry Notes PDF which have been prepared by our subject experts after carefully following the trend of the exam in the last few years. The notes will not only serve for revision purposes, but also will have several cuts and easy methods to go about a difficult problem.
Board 
Tamilnadu Board 
Study Material 
Notes 
Class 
Samacheer Kalvi 10th Maths 
Subject 
10th Maths 
Chapter 
Chapter 4 Geometry 
Format 

Provider 
How to Download Samacheer Kalvi 10th Maths Chapter 4 Geometry Notes PDFs?
 Visit our website  https://www.samacheerkalvibook.com/
 Click on the Samacheer Kalvi 10th Maths Notes PDF.
 Look for your preferred subject.
 Now download the Samacheer Kalvi 10th Maths Chapter 4 Geometry notes PDF.
Download Samacheer Kalvi 10th Maths Chapter 4 Geometry Chapterwise Notes PDF
Students can download the Samacheer Kalvi 10th Maths Chapter 4 Geometry Notes PDF from the links provided in this article.
Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Unit Exercise 4
Question 1.
In the figure, if BD⊥AC and CE ∠ AB, prove that
Solution:
In the figure’s ∆AEC and ∆ADB.
We have ∠AEC =∠ADB = 90 (∵ CE ∠AB and BD ∠AC)
and ∠EAC =∠DAB
[Each equal to ∠A]
Therefore by AAcriterion of similarity, we have ∆AEC ~ ∆ADB
(ii) We have
∆AEC ~ ∆ADB [As proved above]
⇒ CABA=ECDB⇒CAAB=CEDB
Hence proved.
Question 2.
In the given figure ABCD  EF . If AB = 6 cm, CD = x cm, EF = 4 cm, BD = 5 cm and DE = y cm. Find x and y.
Solution:
In the given figure, ∆AEF, and ∆ACD are similar ∆^{s}.
∠AEF = ∠ACD = 90°
∠A = ∠A (common)
∴ ∆AEF ~ ∆ACD (By AA criterion of similarity)
Subtstituting x = 2.4 cm in (3)
Question 3.
O is any point inside a triangle ABC. The bisector of ∠AOB , ∠BOC and ∠COA meet the sides AB, BC and CA in point D, E and F respectively.
Show that AD × BE × CF = DB × EC × FA
Solution:
In ∆AOB, OD is the bisector of ∠AOB.
In ∆BOC, OE is the bisector of ∠BOC
∴ OBOC=BEEC …………. (2)
In ∆COA, OF is the bisector of ∠COA.
∴ OCOA=CFFA …………… (3)
Multiplying the corresponding sides of (1), (2) and (3), we get
⇒ DB × EC × FA = AD × BE × CF
Hence proved.
Question 4.
In the figure, ABC is a triangle in which AB = AC . Points D and E are points on the side AB and AC respectively such that AD = AE . Show that the points B, C, E and D lie on a same circle.
Solution:
In order to prove that the points B, C, E and D are concyclic, it is sufficient to show that ∠ABC + ∠CED = 180° and ∠ACB + ∠BDE = 180°.
In ∆ABC, we have AB = AC and AD = AE.
⇒ AB – AD = AC – AE
⇒ DB = EC
Thus we have AD = AE and DB = EC. (By the converse of Thale’s theorem)
⇒ ADDB=AEEC ⇒ DEBC
∠ABC = ∠ADE (corresponding angles)
⇒ ∠ABC + ∠BDE = ∠ADE + ∠BDE (Adding ∠BDE on both sides)
⇒ ∠ABC + ∠BDE = 180°
⇒ ∠ACB + ∠BDE = 180° (∵ AB = AC ∴ ∠ABC = ∠ACB)
Again DE  BC
⇒ ∠ACB = ∠AED
⇒ ∠ACB + ∠CED = ∠AED + ∠CED (Adding ∠CED on both sides).
⇒ ∠ACB + ∠CED = 180° and
⇒ ∠ABC + ∠CED = 180° (∵ ∠ABC = ∠ACB)
Thus BDEC is a quadrilateral such that
⇒ ∠ACB + ∠BDE = 180° and
⇒ ∠ABC + ∠CED = 180°
∴ BDEC is a cyclic quadrilateral. Hence B, C, E, and D are concyclic points.
Question 5.
Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels at a speed of 20 km/hr and the second train travels at 30 km/hr. After 2 hours, what is the distance between them?
Solution:
After 2 hours, let us assume that the first train is at A and the second is at B.
Question 6.
D is the mid point of side BC and AE⊥BC. If i BC a, AC = b, AB = c, ED = x, AD = p and AE = h , prove that
(i) b^{2} = p^{2} + ax + a24
(ii) c^{2} = p^{2} – ax + a24
(iii) b^{2} + c^{2} = 2p^{2} + a22
Solution:
From the figure, D is the mid point of BC.
We have ∠AED = 90°
∴ ∠ADE < 90° and ∠ADC > 90°
i.e. ∠ADE is acute and ∠ADC is obtuse,
(i) In ∆ADC, ∠ADC is obtuse angle.
AC^{2} = AD^{2} + DC^{2} + 2DC × DE
⇒ AC^{2} = AD^{2} + 12 BC^{2} + 2 . 12 BC . DE
⇒ AC^{2} = AD^{2} + 14 BC^{2} + BC . DE
⇒ AC^{2} = AD^{2} + BC . DE + 14 BC^{2}
⇒ b^{2} = p^{2} + ax + 14 a^{2}
Hence proved.
(ii) In ∆ABD, ∠ADE is an acute angle.
AB^{2} = AD^{2} + BD^{2} – 2BD . DE
⇒ AB^{2} = AD^{2} + (12BC)2 – 2 × 12 BC . DE
⇒ AB^{2} = AD^{2} + 14 BC^{2} – BC . DE
⇒ AB^{2} = AD^{2} – BC . DE + 14 BC^{2}
⇒ c^{2} = p^{2} – ax + 14 a^{2}
Hence proved.
(iii) From (i) and (ii) we get .
AB^{2} + AC^{2} = 2AD^{2} + 12 BC^{2}
i.e. c^{2} + b^{2} = 2p^{2} + a22
Hence it is proved.
Question 7.
A man whose eyelevel is 2 m above the ground wishes to find the height of a tree. He places a mirror horizontally on the ground 20 m from the tree and finds that if he stands at a point C which is 4 m from the mirror B, he can see the reflection of the top of the tree. How height is the tree?
Solution:
From the figure; ∆DAC, ∆FBC are similar triangles and ∆ACE & ∆ABF are similar triangles.
∴ height of the tree h = 6x = 10 m.
Question 8.
An emu which is 8 ft tall standing at the foot of a pillar which is 30 ft height. It walks away from the pillar. The shadow of the emu falls beyond emu. What is the relation between the length of the shadow and the distance from the emu to the pillar?
Solution:
Let OA (emu shadow) the x and AB = y.
⇒ pillar’s shadow = OB = OA + AB
⇒ OB = x + y
Question 9.
Two circles intersect at A and B. From a point P on one of the circles lines PAC and PBD are drawn intersecting the second circle at C and D. Prove that CD is parallel to the tangent at P.
Solution:
Let XY be the tangent at P.
TPT: CD is  to XY.
Construction: Join AB.
ABCD is a cyclic quadilateral.
∠BAC + ∠BDC= 180° ………… (1)
∠BDC = 180° – ∠BAC …………. (2)
Equating (1) and (2)
we get ∠BDC = ∠PAB
Similarly we get ∠PBA = ∠ACD
as XY is tangent to the circle at ‘P’
∠BPY = ∠PAB (by alternate segment there)
∴ ∠PAB = ∠PDC
∠BPY = ∠PDC
XY is parallel of CD.
Hence proved.
Question 10.
Let ABC be a triangle and D, E, F are points on the respective sides AB, BC, AC (or their extensions). Let AD: DB = 5 : 3, BE : EC = 3 : 2 and AC = 21. Find the length of the line segment CF.
Solution:
Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Additional Questions
Question 1.
In figure if PQ  RS, Prove that ∆POQ ~ ∆SOR
Solution:
PQ  RS
So, ∠P = ∠S (A Hernate angles)
and ∠Q = ∠R
Also, ∠POQ = ∠SOR (vertically opposite angle)
∴ ∆POQ ~ ∆SOR (AAA similarity criterion)
Question 2.
In figure OA . OB = OC . OD Show that ∠A = ∠C and ∠B = ∠D
Solution:
OA . OB = OC . OD (Given)
Also we have ∠AOD = ∠COB
(vertically opposite angles) …………. (2)
From (1) and (2)
∴ ∆AOD ~ ∆COB (SAS similarity criterion)
So, ∠A = ∠C and ∠B = ∠D
(corresponding angles of similar triangles)
Question 3.
In figure the line segment XY is parallel to side AC of ∆ABC and it divides the triangle into two parts of equal areas. Find the ratio AXAB
Solution:
Given XYAC
Question 4.
In AD⊥BC, prove that AB^{2} + CD^{2} = BD^{2} + AC^{2}.
Solution:
From ∆ADC, we have
AC^{2} = AD^{2} + CD^{2} …………… (1)
(Pythagoras theorem)
From ∆ADB, we have
AB^{2} = AD^{2} + BD^{2} …………. (2)
(Pythagoras theorem)
Subtracting (1) from (2) we have,
AB^{2} – AC^{2} = BD^{2} – CD^{2}
AB^{2} + CD^{2} = BD^{2} + AC^{2}
Question 5.
BL and CM are medians of a triangle ABC right angled at A.
Prove that 4(BL^{2} + CM^{2}) = 5BC^{2}.
Solution:
BL and CM are medians at the ∆ABC in which
A = ∠90°.
From ∆ABC
BC^{2} = AB^{2} + AC^{2} …………… (1)
(Pythagoras theorem)
4CM^{2} = 4AC^{2} + AB^{2} …………… (3)
Adding (2) and (3), we have
4(BL^{2} + CM^{2}) = 5(AC^{2} + AB^{2})
4(BL^{2} + CM^{2}) = 5BC^{2} [From (1)]
Question 6.
Prove that in a right triangle, the square of the hypotenure is equal to the sum of the squares of the others two sides.
Solution:
Proof:
We are given a right triangle ABC right angled at B.
We need to prove that AC^{2} = AB^{2} + BC^{2}
Let us draw BD ⊥ AC
Now, ∆ADB ~ ∆ABC
ADAB=ABAC (sides are proportional)
AD . AC = AB^{2} …………. (1)
Also, ∆BDC ~ ∆ABC
CDBC=BCAC
CD . AC = BC^{2} …………. (2)
Adding (1) and (2)
AD . AC + CD . AC = AB^{2} + BC^{2}
AC(AD + CD) = AB^{2} + BC^{2}
AC . AC = AB^{2} + BC^{2}
AC^{2} = AB^{2} + BC^{2}
Question 7.
A ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall and its top reaches a window 6 m above the ground. Find the length of the ladder.
Solution:
Let AB be the ladder and CA be the wall with the window at A.
Also, BC = 2.5 m and CA = 6 m
From Pythagoras theorem,
AB^{2} = BC^{2} + CA^{2}
= (2.5)^{2} + (6)^{2}
= 42.25
AB = 6.5
Thus, length at the ladder is 6.5 m.
Question 8.
In figure O is any point inside a rectangle ABCD. Prove that OB^{2} + OD^{2} = OA^{2} + OC^{2}.
Solution:
Through O, draw PQBC so that P lies on AB and Q lies on DC.
Now, PQBC
PQ⊥AB and PQ⊥DC (∵ ∠B = 90° and ∠C = 90°)
So, ∠BPQ = 90° and ∠CQP = 90°
Therefore BPQC and APQD are both rectangles.
Now from ∆OPB,
OB^{2} = BP^{2} + OP^{2} ……………. (1)
Similarly from ∆OQD,
OD^{2} = OQ^{2} + DQ^{2} …………. (2)
From ∆OQC, we have
OC^{2} = OQ^{2} + CQ^{2} ……….. (3)
∆OAP, we have
OA^{2} = AP^{2} + OP^{2} …………. (4)
Adding (1) and (2)
OB^{2} + OD^{2} = BP^{2} + OP^{2} + OQ^{2} + DQ^{2} (As BP = CQ and DQ = AP)
= CQ^{2} + OP^{2} + OQ^{2} + AP^{2}
= CQ^{2} + OQ^{2} + OP^{2} + AP^{2}
= OC^{2} + OA^{2} [From (3) and (4)]
Question 9.
In ∠ACD = 90° and CD⊥AB. Prove that BC2AC2=BDAD
Solution:
∆ACD ~ ∆ABC
So,
Question 10.
The perpendicular from A on side BC at a ∆ABC intersects BC at D such that DB = 3 CD. Prove that 2AB^{2} = 2AC^{2} + BC^{2}.
Solution:
We have DB = 3 CD
Since ∆ABD is a right triangle (i) right angled at D
AB^{2} – AD^{2} + BD^{2} …………. (ii)
By ∆ACD is a right triangle right angled at D
AC^{2} = AD^{2} + CD^{2} ………… (iii)
Subtracting equation (iii) from equation (ii),
we got
How to Prepare using Samacheer Kalvi 10th Maths Chapter 4 Geometry Notes PDF?
Students must prepare for the upcoming exams from Samacheer Kalvi 10th Maths Chapter 4 Geometry Notes PDF by following certain essential steps which are provided below.
 Use Samacheer Kalvi 10th Maths Chapter 4 Geometry notes by paying attention to facts and ideas.
 Pay attention to the important topics
 Refer TN Board books as well as the books recommended.
 Correctly follow the notes to reduce the number of questions being answered in the exam incorrectly
 Highlight and explain the concepts in details.
Samacheer Kalvi 10th Maths All Chapter Notes PDF Download
 Samacheer Kalvi 10th Maths Chapter 1 Relations and Functions Notes PDF Download: Tamil Nadu STD 10th Maths Chapter 1 Relations and Functions Notes
 Samacheer Kalvi 10th Maths Chapter 2 Numbers and Sequences Notes PDF Download: Tamil Nadu STD 10th Maths Chapter 2 Numbers and Sequences Notes
 Samacheer Kalvi 10th Maths Chapter 3 Algebra Notes PDF Download: Tamil Nadu STD 10th Maths Chapter 3 Algebra Notes
 Samacheer Kalvi 10th Maths Chapter 4 Geometry Notes PDF Download: Tamil Nadu STD 10th Maths Chapter 4 Geometry Notes
 Samacheer Kalvi 10th Maths Chapter 5 Coordinate Geometry Notes PDF Download: Tamil Nadu STD 10th Maths Chapter 5 Coordinate Geometry Notes
 Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Notes PDF Download: Tamil Nadu STD 10th Maths Chapter 6 Trigonometry Notes
 Samacheer Kalvi 10th Maths Chapter 7 Mensuration Notes PDF Download: Tamil Nadu STD 10th Maths Chapter 7 Mensuration Notes
 Samacheer Kalvi 10th Maths Chapter 8 Statistics and Probability Notes PDF Download: Tamil Nadu STD 10th Maths Chapter 8 Statistics and Probability Notes
0 comments:
Post a Comment