You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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
Samacheer Kalvi 10th Maths Chapter 4 Geometry Unit Exercise 4 1
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 AA-criterion of similarity, we have ∆AEC ~ ∆ADB
(ii) We have
∆AEC ~ ∆ADB [As proved above]
⇒ \(\frac{C A}{B A}=\frac{E C}{D B} \Rightarrow \frac{C A}{A B}=\frac{C E}{D B}\)
Hence proved.

Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Unit Exercise 4

Question 2.
In the given figure AB||CD || EF . If AB = 6 cm, CD = x cm, EF = 4 cm, BD = 5 cm and DE = y cm. Find x and y.
Samacheer Kalvi 10th Maths Chapter 4 Geometry Unit Exercise 4 2
Solution:
In the given figure, ∆AEF, and ∆ACD are similar ∆s.
∠AEF = ∠ACD = 90°
∠A = ∠A (common)
∴ ∆AEF ~ ∆ACD (By AA criterion of similarity)
Samacheer Kalvi 10th Maths Chapter 4 Geometry Unit Exercise 4 3 Samacheer Kalvi 10th Maths Chapter 4 Geometry Unit Exercise 4 4
Subtstituting x = 2.4 cm in (3)
Samacheer Kalvi 10th Maths Chapter 4 Geometry Unit Exercise 4 5

Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Unit Exercise 4

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.
Samacheer Kalvi 10th Maths Chapter 4 Geometry Unit Exercise 4 6
In ∆BOC, OE is the bisector of ∠BOC
∴ \(\frac{\mathrm{OB}}{\mathrm{OC}}=\frac{\mathrm{BE}}{\mathrm{EC}}\) …………. (2)
In ∆COA, OF is the bisector of ∠COA.
∴ \(\frac{\mathrm{OC}}{\mathrm{OA}}=\frac{\mathrm{CF}}{\mathrm{FA}}\) …………… (3)
Multiplying the corresponding sides of (1), (2) and (3), we get
Samacheer Kalvi 10th Maths Chapter 4 Geometry Unit Exercise 4 7
⇒ 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.
Samacheer Kalvi 10th Maths Chapter 4 Geometry Unit Exercise 4 8
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°.
Samacheer Kalvi 10th Maths Chapter 4 Geometry Unit Exercise 4 9
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)
⇒ \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\) ⇒ DE||BC
∠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.

Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Unit Exercise 4

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.
Samacheer Kalvi 10th Maths Chapter 4 Geometry Unit Exercise 4 10

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) b2 = p2 + ax + \(\frac{a^{2}}{4}\)
(ii) c2 = p2 – ax + \(\frac{a^{2}}{4}\)
(iii) b2 + c2 = 2p2 + \(\frac{a^{2}}{2}\)
Solution:
From the figure, D is the mid point of BC.
Samacheer Kalvi 10th Maths Chapter 4 Geometry Unit Exercise 4 11
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.
AC2 = AD2 + DC2 + 2DC × DE
⇒ AC2 = AD2 + \(\frac{1}{2}\) BC2 + 2 . \(\frac{1}{2}\) BC . DE
⇒ AC2 = AD2 + \(\frac{1}{4}\) BC2 + BC . DE
⇒ AC2 = AD2 + BC . DE + \(\frac{1}{4}\) BC2
⇒ b2 = p2 + ax + \(\frac{1}{4}\) a2
Hence proved.

(ii) In ∆ABD, ∠ADE is an acute angle.
AB2 = AD2 + BD2 – 2BD . DE
⇒ AB2 = AD2 + \(\left(\frac{1}{2} \mathrm{BC}\right)^{2}\) – 2 × \(\frac{1}{2}\) BC . DE
⇒ AB2 = AD2 + \(\frac{1}{4}\) BC2 – BC . DE
⇒ AB2 = AD2 – BC . DE + \(\frac{1}{4}\) BC2
⇒ c2 = p2 – ax + \(\frac{1}{4}\) a2
Hence proved.

(iii) From (i) and (ii) we get .
AB2 + AC2 = 2AD2 + \(\frac{1}{2}\) BC2
i.e. c2 + b2 = 2p2 + \(\frac{a^{2}}{2}\)
Hence it is proved.

Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Unit Exercise 4

Question 7.
A man whose eye-level 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.
Samacheer Kalvi 10th Maths Chapter 4 Geometry Unit Exercise 4 12
∴ 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
Samacheer Kalvi 10th Maths Chapter 4 Geometry Unit Exercise 4 13

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)
Samacheer Kalvi 10th Maths Chapter 4 Geometry Unit Exercise 4 14
∴ ∠PAB = ∠PDC
∠BPY = ∠PDC
XY is parallel of CD.
Hence proved.

Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Unit Exercise 4

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:
Samacheer Kalvi 10th Maths Chapter 4 Geometry Unit Exercise 4 15
Samacheer Kalvi 10th Maths Chapter 4 Geometry Unit Exercise 4 16

Leave a Reply

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