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NCERT Solutions for Class 10th Maths Chapter 6 Triangles

Answers of Math NCERT solutions for class 10 Chapter 6 Triangles

Unit 6

Triangles 

Exercise 6.1:

Q1:

Fill in the blanks using correct word given in the brackets:-

  1.     All circles are ……… (congruent, similar)
  1.     All squares are …….. (similar, congruent)
  1.     All ……… triangles are similar. (isosceles, equilateral)
  1.     Two polygons of the same number of sides are similar, if their corresponding             angles are (a)……………… and their corresponding sides are (b)…………….(equal,         proportional)

Answer:

(i)     Similar

(ii)    Similar

(iii)   Equilateral 

(iv)    (a) Equal

(b) Proportional

Q2:

Give two different examples of pair of

(i)     Similar figures                 (ii)    Non-similar figures

Answer:

  1. Two equilateral triangles with sides 1 cm and 2 cmsjqcvP9 C5cVS0yT8yZN0Z1y4JHEhWYM2DX1FrFhONfV4UKSkfrAkjlXQPuC042uxmp9XiLuSUFUp7NR F2PNrAtF9v7PBNAPzvsmn6jv

Two squares with sides 1 cm and 2 cm

0jyRLiEWcO9GxNmqLS wysIqLkKtCqKSBI XCGucXois0IzMH7qOW60RXa7CvwN06ptjIaMOpIhO UGm xx1juJ Z5El3a8E3ckI19TFiWcrueU1PkFpUuy JCCuZ1y z6yZGzg
  1. Trapezium and square

Triangle and parallelogram

Q3:

State whether the following quadrilaterals are similar or not:

Answer:

Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional i.e., 1:2, but their corresponding angles are not equal.

Exercise 6.2 

Q1:

In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). 

(i) 

gV3HwrIFJbv7zeE4i6rV4

(ii)fgOFTtGuepyQmHz UOZLBznzqUjTW LceyhuFyEhBZ1Zn2 HhSDeLgKN59fRLCPN9AUFirCpmPqLY2dgNvwbuNfFASg06cUwMYOxXz3WdCx8otTG651N7p165

Answer:

(i)U20rgV2JfudLIpjfCsGN9G7hUJmKw2G4sfE6I80zEZRk3CxZo oWOZeXly7jVjFTk0W1Ce52OYLWCOHLTJic8HfyXv0Z4MzUp6m EL0yKO1e8tNdyjsM 8ltRTDyD1YEO53Mb3U

Let EC = x cm

It is given that DE || BC.

By using basic proportionality theorem, we obtain
GONoFdrYmqvd2S fPp5qotkq7ADvCbxubYxTu815XK4WqbnaJw8lQueJvaKnbGG0g03B0eQ 5Cwijcpn09FexdNiSIOwAxOawC5hgfIuHKct NQNNj4c6 LeoscO5VYjXKv4EhM

(ii)

LUotffkO1FCU jFMGz7LRzTsQJ8T36GdZUU7wkj 4M zypRsB0ejODHFNnjJ49Q8LgOccVz82y6ow0nsaAHH3n00hSaPheOStd OtEOlcsHtsRSQXqbsVCDP2fFNTv3sbOU5jZY

Let AD = x cm

It is given that DE || BC.

By using basic proportionality theorem, we obtain
fcOjYIetsV8Fe1q yR0CHnMhFCYks7z0KanZg8PIhNBRsAiqbLTHbVwkKDFc6DtRxyj42 5Etoza bIhUTz3RbGnXnlVcVySt 5UY dp VsKGU7qWaPy8WQPhPpSfRGq wRCqF0

Q2:

E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state

whether EF || QR.

  1.         PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
  1.     PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
  1. PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm

Answer:

(i)
onXpz4WNKSepiz5 ChquHBvLP XcK1iZ4LXRZz2XaVLBnrw9VUUy5zux7JiV5kpUXygelLb73xjDuaaXODmNepvlarMGb0Ma9W5F Tf1eZzi y0ehekzrtVt 3 NHjW8cBLnH Q

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm

v7G Y1n6by2Wi jUcxFrAwzyDSZhGfab3QJAbIX9wIzBGj4uMc8sRflZKWnEclHN0E6NxTjU3wGZukq8WVv6K7fGSx1jHcVwXX6517FH IN8rftrGa LnP ccMy6EWmP7YdeRIA

(ii)
pWyhc hHGi0u6ABHrhpwFVqp9vePEHs EsRZhNGq6HJuMxzToHkyn P19Mr6RSWcInskg 2R3qmTvw51W6TnA97oboPG3Ca5cSt3QPW3aZEoKj5u2LQfBfbJutKKrZyKo FN9Wk

PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm

(iii)

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
f McIN Wbz8 J5zB1SclE5Hu7mlWd onRsUnPvOK 5 Rz19viTZdAMbo7eNlAMT6JZbOhBZtWgHLxOJzJWuaZlrgqR UI2qGbeILcEuIgyuSoNCWIgKM3mcK 0s Pi NXMlb1Tk

Q3:

In the following figure, if LM || CB and LN || CD, prove that
BZoyjvr9So47P1fbAAh7IpTUdjbUOW7AoY3FfhKWBS y hyOUXJRJe pTJFovzqVMCNyQy6IgbIDvQ7iBtFwtXWodrJWFb20Mju4ZI5oYOF2uvsik0WA u 9OIcWvm0dld2cNRk

GIqFAZCyUsxHcide0J6j54ej7Yj VWiTNd81sFX6kVyY7TjcHRGeLlXF8czD919Ta DJysYzv3eNOSBJxSc5 qTGkyzs6FQ1H7AKMOytqR3XHNYEnoAStNTaNLxcy QpC AiuII

Answer:GIqFAZCyUsxHcide0J6j54ej7Yj VWiTNd81sFX6kVyY7TjcHRGeLlXF8czD919Ta DJysYzv3eNOSBJxSc5 qTGkyzs6FQ1H7AKMOytqR3XHNYEnoAStNTaNLxcy QpC AiuII

In the given figure, LM || CB

By using basic proportionality theorem, we obtain

4SOZ7KL9GVEjAiLDAbk5Fesml3rRxowrSkiCF54A dEk6j62AKM2fuqNy6rXAiCapWitbXpo8B2Rjd4utomfdIl8xgDv9JnHhUcbvTbzP2p81wG0Qgic23rvj3uFMO P0zocd18

Q4:

In the following figure, DE || AC and DF || AE. Prove that
TUCESwYjPJh2H2lYwIBqj8Z04h37ImM1pGurryY7YeZjP1QzyhMfnbVUdv li9bN9L5d5rJ6zle7FK g5FqQgNlQQu1U37aw7Fum2CtzE mDmCDRPOA 144z20KKFKzf61h3VYcPLFMA9L2v9pIB9lBtgYuwsJmK5zdYwHYXdXJF9Dvlqxio1HeI5HPnWB0Fk2xUcU8KnBcLPFC2ee M9ervnZdJZI9l56sWYZLuMTJqBJBGbJfVYuLd8XGpL4mZv1cw6W1K5T8hJs

Answer:

In ΔABC, DE || AC
wHuHl vPacYn9tmIm3vngmKCp7ciemqDKRkIIlfYuwL8uDyoQUPHPzX8tRamBzWy f2alYhKyMqYRiQuEdANGYQqutcom4pitwm99Q8sbVJ9i1NfJKmCZmVOZOGPghd OynvEJQ

UoYFmadf5ZPnZ1NgE1nx7r69Zu4lmlIbttA5ik i9UkddVK4m6TSyGaqatAV3cSD3VGlu006vh0PNFnNW35volwAXD1 f6od5Unm2w2tWaJibRKWacpeTu17NPF ApYTyo0OJzY

lY08jhwfo bANxJiA ShZHB2ShapprrCL6BoEZKoLUvVui

Q5:

In the following figure, DE || OQ and DF || OR, show that EF || QR.

Answer:nrf L4NJk0a6RI9U524rkfJQ1ByXaJE0GlJCvqL8CW7wXFxsWsjSkO6FFcFkZUuIqtyAMNGvIvtGUODjnD E 1rJE2Dx67f6 pBbeHb nvw6wXI

In Δ POQ, DE || OQ
IYR8W

FSbYpsCUuUHUSDSN6mDwNEzNAuoPfnnm20AFlMBY2
rsP0pRERoMRbS16PDEmiGNHiC6glA1hL
CgeUybF1lh6CuUvd3JyLAWEBq3rZed4CkAlVzqTjse 7OZUJtD2kmJZCL3RAQEFwyy NB43gGyY4ReXaJwTQssTZos1u0RNfSJL7Ydl56Z ea7nNtUC93OlmoTctgncFnU4lvc

Q6:

In the following figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.rPv1JOPlcArq fi18lLImpAdd1TNfT9XuFX9zAT2bF3fDcgVQZ3l3BvyBdJAiGnJVvgVlDK9YTaemon21xuAy5 IngqW OrXi21fQJz1OVPgbUhAenf0v2rqioY q0pzti7TA9g

Answer:

vVSzt1PW f5NMKUXUADOrWzmHUCZhWzCG6iP1MKKx gZh iph kOXnrRwy02 6Qo1CTCJIk2KgueVOra1sujeD2umaDHmKRivmuLUU 8 BDzaueZcLTLi8jmFOtK7fey4oAE2qA

In Δ POQ, AB || PQ
34d3zPSvX2oRncSgsNRqe3vH5pgaTJaZz6Wi2a1201ruXqFT6ZZc4iMCUN35ABKDGJGA1PNi640brAC9N V9 p2S6dfbxTqfvwM6cKv3s746z2uU BG0D23ihYA8yIQKNOxRRgCDxmKKrseA JuSrFIC7JKOa6BllB4cPWG3t0lwKzslS61M3ipRHj r1dELzJmfKPtYEDoOKXjPf6wVPg7zi21nkBcIns92LgJqw3d60KPfUQ7O4PQUnMGTpB9sJN6eM 4LfmtYgamP LyOOv8BxJAMKniHCMSO9yzuU5vz8UJLYU5HUnUaYw9yUsUcBGd AYQ ibU0H 7ki1t BTPu2OdaQ wwYYU1gh hIgTcuvY5oEe7MS66 vAgRhQ1Y

Ol6IOozf9ubtd2QLlvl1DhgDlKWOjfI dbOWsKdT2dRJ50Z7EV f4ix2Jzy AeGXruoZVi1TDXPwswANuH8C78RIUz4YpNow WDuHXgkgi5WjXwpyep KrkUpbg8cNGQiy 4 0

*Q7:

Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Answer:

Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such that4dJv2aYnWKX bfakFRXAUQf3A7x1aGQoEb fNtMrwa4s56Rp0U4meTzA77sAjot37 TyIGOItyCabPKf8XAHypX S5ulWfHHYYnMW7AZOsSz kYiRfGH btnz2puplf8070uDo85IzV mROkloawRGc0YW0cKKeJIWne9Pi4pEh7nuqCA1lXUXTdJ5URBeZblGZ yBfTMVnuqURUahlZDaTzm18 OKYjSrt2tkNdQmiYYr6WkX3mbQpKe89RPr24UnwzWNJP3nUciM

Or, Q is the mid-point of AC.

*Q8:

Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Answer:7 YE7YEhV3YC6GTlcn4fvFi0r69MAfilzUtSdx3Vp 57itISfxY3l1u9pau9gqMsSsbch08ME0ZoHWuxBuBrqJ4K7pBOpNGHqFT7mXVF5iAiJMFCn90k Fagwhfsm3Ots6atYw8

Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively.

i.e., AP = PB and AQ = QC

It can be observed that

liyZkbq CiKa7P 3YAEQXZGl1T97VrnxNtYOgADSLQj1Y2xpOmfc52J2Shxy0RoS44vmvayz8QTdT6L9 sgpChVHogqrssRhpQqHWsa YUfr5CGdOzqjdY8Nu82DoZmXeroLC48

Hence, by using basic proportionality theorem, we obtain
ql9y4kYMH1CWmZgZuAegdEGa EnSHd4PtOuiDLI2WJSwW8y LKXGwQp 71F7aHGAi QVMh 7fJqCVp 5YGjJJtR AfgfyMSkUvJcaxnv8YZ29 5gVSzFcHK2vndpv0zru8CgYk8

*Q9:

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show
KQsdq3GEACH9l dETkQ ThUzv deLnNSPcoyf8NuHEBzkITPvE akpB7dKB5icE 7lMnbudmgdy88AhOhmdAo22D Y17cC Mtn oSn Hlv

that

Answer:

Draw a line EF through point O, such thatTqC7m

In ΔADC,

By using basic proportionality theorem, we obtain
NKy6t4Y hZVqYkC 9Om1zU gW7enNs1iWw8ad7vPSIjY69U6i6GKJUZL 9VOKo8Ra3kqhWDlyY2AXnPO1JF2SS6OdUzpSdMOfBs6eniqYptV3Wzh4jw2T2v8CeISkUZ8ijZM yA
iOMbmfsjTJCes7MmnHL6fdnEBWCHzGgHmBfwQojY1lA CZF 1QihHxRJwSK04j6EHKT

In ΔABD,

So, by using basic proportionality theorem, we obtain

From equations (1) and (2), we obtain

qzdYg8heg8N3bv2Y7uEWTwb6kG 7z7pAvE2Ei87 cP3uO9HPJtVx4TjPbBfyRknt AtrARwJBYAIvtLrimCv7KL6zWwKM EjgQGt0hT5CiKhrdNo1w1sWaYScKTOAnSG0LQuEo

*Q10:

The diagonals of a quadrilateral ABCD intersect each other at the point O such that Show that ABCD is a trapezium.

Answer:

Let us consider the following figure for the given question.
lPgURENNkzVY2m5xYJwnqvfuef9r 0oOdm3eUp1EdQ0jlYTkyO9zX tZ5RCjf1SPQ1qZ3SC2 hLyx8XquZ3KrS c689oTRUFjfnFaSCm

Draw a line OE || AB
riGXKaZQcu BUuRVJx0p0NNKAYJLfIYFaclFc5Aox0iHgc47JodQKH21L4AxXnWzYZescQxbFa3dGN

In ΔABD, OE || AB

By using basic proportionality theorem, we obtainypnp0pR ltLaHuhIxfVkNCu8KCq4AiB91rMJS3Rmc5hTse1O0UW4RLxhMNfJXMWMbSomdJ3 nSW nN1miEOZlLAZt7PiNEMZlLGFB8

However, it is given that

1Rph3nqqXg6TTZJncBkTskU7IBNoqATQMGUTgN3MF0paJp1ZwM p530QQtNBEQXKfRf1HD4NF0tCj3nAx1 q YM79WEMrsI6FxT6DPhJfsu VUaJXDG

[By the converse of basic proportionality theorem]

iDJuW8sIEZ 7 BM142YW6gC8qwjpmWACZAUrSNDQ5Wkh Qi7go 0ZdaF8H9bvbovRCOdzMv1dcFQU4RYAHywAazWcx5vc8IgydX9GTS5qQJbuDe7P7 Pub0T0if0Q7kSmk JJ8

∴ ABCD is a trapezium.

Exercise 6.3

Q1:

State which pairs of triangles in the following figure are similar? Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

(i)

(ii)

B k65LeHo3xfokxJMfXH8lzyjD0LU6SJvigbJf fJkt2vQWgZLazLdXO7Tfdo03foTCzaIqb3sH3VkrkKVUx7Fvcs97 nNXct2dXsLspgculfVMOgYve9OFF5NB5hC5gWUB1Jb8

(iii)

JUyBnpsgp1vJLDVZ2b0mBIeeEkN0J4Lc CwL3tsK3G9WsErV9sSmkQwmhGD BuLONfK08EpSloAmTINnYyPfpW7FbLCrCb V XUXSqH sB2O3aY6FZwYTTMjsgDAiOSd3S39qA

(iv)    hRihPimj33CKuDHENC68QQHpCQvo1Vr9VccDO7nPVu4A XjuKlYwEdYm7SHZrqaakOGIZNz0ePWfFLIqbzfwzPUv0bkEkXHwcysh6F8oJ0bYBG5Y1lf5V

(v)   

YKjiDT2l6XzB3zy 8r43lBYR4QUFNIu1kbeqjCPd8B9pjIb4BYl7H7Gu2vXvdX9SaPOyYpwb jfTcXXkrvA1bcqv5rCFgFa4WjbAJX0M wM0tnSbmFKlILkSmq9ieyr72wkGIjc

(vi)   

Answer:

(i)     ∠A = ∠P = 60°

    ∠B = ∠Q = 80°

    ∠C = ∠R = 40°

    Therefore, [By AAA similarity criterion]

(ii) 

tBHMfaEWBAsoYd7lR 7qH0rhFtd5WMTjoW8BWDDpS47PCbLBdoams PQdOgDGaB1OmSR EhVz2eq0RZUyBDHeAO A0X 5rCyCgvmIX8CT UEH2JVFBt6PICNRCVS32hdDfiZqPs

(iii)        The given triangles are not similar as the corresponding sides are not proportional.

iw75aa3 Y2IkN4G tpH089oiEJdGWBh9 airbfc5h1kBVUKJ6 95pHuTa8fi4fK2 RAlTBQA49hn1MfFkQZ Zq28zcTO5WB Ejmkhe2TNdZK79wtcsv7Esx EKLGtzhOFgh9ydo

(iv)        In ≅ ΔMNL ≅  ΔQPR we observe that,

    MNQP = MLQR = 12

Q2:

In the following figure, ΔODC ~ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, 

∠ DCO and ∠ OAB 

sPgI6E736X1L26E5uvBd65kXXnsNiZQ

Answer:

DOB is a straight line.

∴ ∠ DOC + ∠ COB = 180°

⇒ ∠ DOC = 180° – 125°

= 55°

In ΔDOC,

∠ DCO + ∠ CDO + ∠ DOC = 180°

(Sum of the measures of the angles of a triangle is 180º.)

⇒ ∠ DCO + 70º + 55º = 180°

⇒ ∠ DCO = 55°

It is given that ΔODC ≅ΔOBA.

∴ ∠ OAB = ∠ OCD [Corresponding angles are equal in similar triangles.]

⇒ ∠ OAB = 55°

Q3:

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that
QUqUjtcC5TBTYPv5FbU8o7MHsVC82lAeodX 7gwVfRby2xURtIWX6eA8fP

Answer:

In ΔDOC and ΔBOA,

∠CDO = ∠ABO [Alternate interior angles as AB || CD]

∠DCO = ∠BAO [Alternate interior angles as AB || CD]

∠DOC = ∠BOA [Vertically opposite angles]

∴ ΔDOC ≅¼ ΔBOA [AAA similarity criterion]bBSP6trNvTusRtKU8pglWv3E2is3Mu9 DMosrh55s835V5BT7vnIURCpxQUwKvg oVUabzcHDwIhSG6iDsJzrQgt8tjemAZWEZDVrbVIrBquCUN2fTo1RyUv FBiP2UoJMAOrw

Q4:

In the following figure, and Show that

dSoyiXcJHWiwmvyKUjE kqAjAfWiea

Answer:

In ΔPQR, ∠PQR = ∠PRQ

∴ PQ = PR (i)

Given,

yGfVj

Q5:

S and T are point on sides PR and QR of ΔPQR such that ∠ P = ∠ RTS. Show that ΔRPQ ~ ΔRTS.zA8q lnjOkHbhdeYadi 1ZgSfi5VVO qcN

Answer:

In ΔRPQ and ΔRST,

∠ RTS = ∠ QPS (Given)

∠ R = ∠ R (Common angle)

(By AA similarity criterion)

Q6:

In the following figure, if ΔABE≅ΔACD, show that ΔADE ≅ ΔABC.

iOU bNviExE ThDXawweEBnEztQU6turUmLcRsepxMVNbFEyi5QI3 GF51H9OIWA9 K4razRk0Jz EkCHwM0Tp7agXv810h0Iivsl1Buqm3mz2sod6eRsKtwHrAzSByNLy59 tY

Answer:

It is given that ΔABE ≅ ΔACD.

∴ AB = AC [By CPCT] (1) And, AD = AE [By CPCT] (2) In zCELDEbOBL0 rzeWB3MTwaWOIpSh4PyX9Epe 4rzbmClHYhVsv2dWQYrolVNYpmO and

J8mutvItmckAb9wbkAvhPJgausS6bGDPK1J2EqIWRkQUKTYn0V2H9N0hOaoGK [Dividing equation (2) by (1)]

∠A = ∠A [Common angle]

∴ ΔADE ~ ΔABC [By SAS similarity criterion]

Q7:

In the following figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:JNCJBA azRrEBkz4FTr7umwIuHs3cj0Hj0A4Rc8Nm0 8RlprhcW1f9kAkDPov0R3FDKRWK6i1MTgxW VsCya0Jkcpid4RcnFUax FJI5cRvJsNYscr nkzKUSKVkXnbcFLHukZc

(i)   

(ii)   

(iii)   

(iv)       

Answer:

(i)   

DbhqcE 5RXkqLP YuyyvNFy5eOVgI6aeTSAj TbiI4AJVTrzt47wDqB0iGuO tbkASEkriSlpX eq 4k2hX84kROgl6SCO nbUU3gkM1j Re7IsUeIzyrowkmc Nc OUcMkBLls

    In ΔAEP and ΔCDP,

    ∠ AEP = ∠ CDP (Each 90°)

    ∠ APE = ∠ CPD (Vertically opposite angles) Hence, by using AA similarity criterion,

eVG8SiMzOObi1feuOwp0SY3dhFIN59fVZEc0hmTNADMk2iw7mtz nvdBF2B49wT2pzFf09vGYBuuuPhjcNTMD5C6ailXVjHPaHZTE62SqLlHR iMEjBFq4 BNvdq8q3kuWA

(ii)

j1 gU7Tfh6HDaVpNX72zp1kNcUjxonaNndU64ZJKxQ0dBRSi5K 53G G2vWjwfYoU1z5w8Pog6xCpcDiqJiTUV b2 XO 4xh3xFy zIdcdOb Yuy05

    In ΔABD and ΔCBE,

    ∠ ADB = ∠ CEB (Each 90°)

    ∠ ABD = ∠ CBE (Common)

    Hence, by using AA similarity criterion,

Kv8c8yfHvxkArwoxYfaV2N7ZiSZ lCkfg1p4DLxhMebhFI6OF PHAyzNxmXEmOIEtKBqXz4vq8hTU vPyJo3h96PT2w4Hm8Wpukvfs awWQmGGcVKx3yhPrDPNqUKBsA XmS0Dw

(iii)   

tIdQCUJH0DOdTWTfluMXKpaqttKNjBmt fOqa6YIPU32g8R3IlAOqqT3cZYyPyVhn8GGtCsoL8ACIolX1TuzMa4QFkVHt9vEQ4K ghjzM8UdPAIKXgObqPK3cEjqdzSj4ho4cJM

In ΔAEP and ΔADB,

∠ AEP = ∠ ADB (Each 90°)

∠ PAE = ∠ DAB (Common)

Hence, by using AA similarity criterion,

aaAK79w2Y8DhzSWxGNOo2eTmTTRBmDe1T4XaRIRMOWjdtduLS8O8 KDoDEzKKYUBf0JZ HFaUj5wDJWLSHCFPdkr n1BvqC9oIvqVspUXLOc2eqf377RYf0ICpgTQBwnQjZ4jjs

(iv)

kBF9rCgcZRaiXg6cBu5 mr3xiczqE1 K7X1r6iDosSvt0LLyWkAB9eCjmdiNSbQaw0vC zwmnnTWkvtO56bPBBQGKbaFHR1dyOKdP3Di cpS5yCn paulmOJOVRgL6h1 9r4iwU

In ΔPDC and ΔBEC,

∠ PDC = ∠ BEC (Each 90°)

∠ PCD = ∠ BCE (Common angle) 

Hence, by using AA similarity criterion, 

FjsLIv5q0C3DbkZ7OWHR5Ofg NaplfAuXL4TeGnI7P91yRlmjWfZA8se 9RejEWpbIol1

Q8:

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE

Answer:
Diba3os w1UOB7Y20plNJsO5

In ΔABE and ΔCFB,

∠ A = ∠ C (Opposite angles of a parallelogram)

∠ AEB = ∠ CBF (Alternate interior angles as AE || BC)

(By AA similarity criterion)

Q9:

In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:

tIXszGFpVdcSCmcJwymmdFWKxywVa5k368QA00Sdee1kWVCT U9icb3t90bUN0zyVXT3fKicIG6B7zSTTCmFDxX5TwtJkQ36BTtFOPCZUIF8 MMwTLP28ZQGbQe2KpqR rkdwA0
  1.         ΔABC ~ ΔAMP
  2. tQ6DYyIxEg2Kyc9q MgH2ElDlglgto3ELAbg IhG PnQD5ogzf4GiqDh7gTbYfMIu6ds7FHNTQEjXS6mkT0pkavKXWyW ERdaKj

Answer:

In ΔABC and ΔAMP,

∠ABC = ∠AMP (Each 90°)

∠A = ∠A (Common)

∴ ΔABC TIbJxLfxKmMKP8SpRL2NIU3nY 4skw6g0BWfjmld7YqGx0089NFNCMc6bw8lWgp6ocpouΔAMP (By AA similarity criterion)s rgT3v1ZLDvIY346GgPUnXeneQ5QhMgR5Iry3 9bTkrbheGXqskbQkWwvOORtj9M43igXj Ac8AswIDxgg G9LL y RWbfz8VhhK748KaT6vOQJcQhWgbzT6st9LDIVdNBjXIU

Q10:

CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of

ΔABC and ΔEFG respectively. If ΔABC ΔFEG show that:

(i)    

(ii)        ΔDCB ~ ΔHGE

(iii)        ΔDCA ~ ΔHGF

Answer:

It is given that ΔABC xAkQuX PVWS78gv u0Z9t0db3o83N NtF9DXqhbaknvg3WppDJMAWCPV7zxEU3gNvvEp3Qq CTGPEfUgAc2i WD5oaRwwso3dl4XaWaArfn9LI9OnQSbJbwod wkTSToWTFdd0ΔFEG.

∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE

∠ACB = ∠FGE

∴ ∠ACD = ∠FGH (Angle bisector) And, ∠DCB = ∠HGE (Angle bisector) In ΔACD and ΔFGH,

∠A = ∠F (Proved above)

∠ACD = ∠FGH (Proved above)

∴ ΔACDeaR0O5y9U2pFLc0VNjKBFeyRboq vkgLuKADCPu4Uvo9EFaQqNOy4y xWieHerjzJOl C1TMVgJfΔFGH (By AA similarity criterion)p7wRaxVgsDAEsOs1IPe8rKUuTViO76Gch3CcETOlG117 OVZdbTiwJPQ5FVFewsripOvdvQFU2XsfwKxl2nghxD4uGueEu4BFEbJSAq1P1RYudb aPw 3543pdErkklmUAfUcjk

In ΔDCB and ΔHGE,

∠DCB = ∠HGE (Proved above)

∠B = ∠E (Proved above)

∴ ΔDCB Tpa JZsX23ydKNo1tX0CaclTux1DSisz AzcUhdjqmQSD 4xaE85yzYv5E2N25qodsJUqTI4y1FJjNAc8xj8 3nm YKd4nsqjmuhylcOTnigCo97cggCVK7lM4F0lFrzkbf5TOMΔHGE (By AA similarity criterion) In ΔDCA and ΔHGF,

∠ACD = ∠FGH (Proved above)

∠A = ∠F (Proved above)

∴ ΔDCA 0EOjGRAIWLTgewf ftcΔHGF (By AA similarity criterion)

*Q11:

In the following figure, E is a point on side CB produced of an isosceles triangle ABC with 

oHg zBIfKbJDOf9QgjMItcrLH8OAFFnfzuIl6pz3Ep5i9ZucjktWhphJH8lP UCTXKsYM1zHF7ym13ifB3w1kqOmQ7mEXUTLObUu9z3Fuf51J8BuXkKVMT8Lb9LPvQWwWoVd9 s

AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that

Answer:

It is given that ABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ ABD = ∠ ECF

In ΔABD and ΔECF,

∠ADB = ∠ EFC (Each 90°)

∠ BAD = ∠ CEF (Proved above)

∴ ΔABD yUSp3uVXEzmTZ8IZnEDF5UbnOBWNQomJVZBpgxY KETBEr7tnNPH55Qw3ZTZgSrIgpwMRdq5Xz7plcOVayuo3tYnIwFXa4H9flq aΔECF  (By using AA similarity criterion)

*Q12:

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and

median PM of ΔPQR (see the given figure). Show that ΔABC ~ ΔPQR.

Answer:MStFJFN47FkwIPpHOy0TYhL7pQWfD46dUp3VrtjBzX8Bf8pNpvNEvRjahZJeTPmn1XkzENSZhkVL5f6cJCBTsqmKJgL1Nyegaf UjWBvPLVXVepbZq0jBqGeRLQt1BZAPUThjLA

Median divides the opposite side.

fGBGuCPAxItY tgzuRVHfUB1ovf4Hu 6tb4NywroUseqiYWA5tGcn6z5Go1F1IJqRheb82sbQ8cpIspIEsSNEivq9hhu4at88oXGlKBypX4kgqWiIPg1K2suifcw5 GT42zndEs

Given that,iazidK9DNG BIjcFhO4q

In ΔABD and ΔPQM,

uiW2O86hba5tvEbyz57p04CW dyOdYFT1momElrh4u xwgSu Utbr45SLwroMwOIcgUDtUPoz2lOlceToVL0KYes 1CJQc6FkT I9rky0qiG df2kw3PrBdI2oRAKfqPyDlJJ4 (Proved above)

∴ ΔABD yqgN34z21yR1YzJESyWqzLFLQ3yRdkjqnrMoWJKKN5i15fgIsswJH5LNVJbEszDhG2KOFxpLFXnFGAWibp4r bkXsJl 6Z4hmH HKoUEGUqE0Yqmhw Biir7cKNfhyyF3belYvYΔPQM (By SSS similarity criterion)

⇒ ∠ABD = ∠PQM (Corresponding angles of similar triangles)

In ΔABC and ΔPQR,

∠ABD = ∠PQM (Proved above)

4zsp6NT5vrjpnZuERpNsF6VRTRcti kTxxibmiceWlOH85j8sz97ed9NL7qTbU6HfnXZF5o7EwKSypd4cRZ8aBYX0WFpsxRkS Gc6YnkE6oJzzatYFxzxr55y1dy8 aCiuUnpgo

∴ ΔABC ΔPQR  (By SAS similarity criterion)

*Q13:

D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that  dZ1MZTgebm1ShlqeiA01d9u9us7fW24c6rRJaJw2QKx1cbAKnn0SbLV2aujCwgUuypb5VcFG3kkKE

Answer:

3miE14NiOpYrcCCUUTYbVxQiRGFQOf5pTiQyrJOdj42FK9 cI2upVimusdlL5nT7VgpsYcHe2CAg4A jxcdbEHOgVFlfulyF8TGM3I1jvxOnG imTgLOWnct5Ln 8C39RC6lcRs

In ΔADC and ΔBAC,

∠ADC = ∠BAC (Given)

∠ACD = ∠BCA (Common angle)

∴ ΔADC wjq2Aujer05hnp8O1fwqOZHZsR9KF4o1r8AtscfzAk18wKdOs2TCd DWzvfGmRylcp8PTrCxFHLcbKAgAdPYFzGQC0IS6RhBugKLUdfFWffZI8S1ΔBAC (By AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion.

CuACqjynuIACns8Xfi WrQhMjAUgOKaCX wu9M0nUQwpjG6ZM521ONAPV

*Q14:

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that

Answer:

fc hTJaBH51trsAjyQrWMdWQeGbXNheDxn1aBH24J0m6SCB2QpKm3he56L9Z IvzMGBQ4Ch KXVvaxVXUfAgOtBEtz3c 3 T1Hm3ts5SA6003siNhsCgwNidgzepjn7zHLSuIzg

Given that,MR t4 eW7LaFix aq7a 0dfifeWtqBzkvp WX75csped4gfumHNuy7mQr djTlQlcbRAoLDItoDceDlACKpC6EuhSwMbv9Mw2W7GOcu0GHrFZ7nuhg0u3hdYogrhpRogafNXAuQ

Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. Then, join B to E, C to E, Q to L, and R to L.
5GMNveN7LBTo5q8Yg rWKcjvB1LxG0mGk1fofbHo 4hdTKseG8bCnt5x3GUwePgm 7q9IzSFbKFpMmuAcBPOQM1VgFMveRqEuRA3BcDl3mfQU 1IS0lHC6hWogOzEh3CSGaThsQ

We know that medians divide opposite sides. Therefore, BD = DC and QM = MR

Also, AD = DE (By construction) And, PM = ML (By construction)

In quadrilateral ABEC, diagonals AE and BC bisect each other at point D. 

Therefore, quadrilateral ABEC is a parallelogram.

∴ AC = BE and AB = EC (Opposite sides of a parallelogram are equal)

Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR It was given that

4gVIrQ5nqe9KlsQfgn5vEI35lF 3meTy64kwwEui G PeAe7OyHsIL RpUPFYEzR7BxoQP75BVvGIBGm5bwdUnvlf qq5M4A0i7s Zp9KY HYzTU71oNv JEwdwymPR7DIUV1Kc

∴ ΔABE kVxq0M LIsL9HldFtNVMx 9ZZwΔPQL (By SSS similarity criterion)

We know that corresponding angles of similar triangles are equal.

∴ ∠BAE = ∠QPL … (1)

Similarly, it can be proved that ΔAEC y1i527eaR HrtwzFIhd2WFWR7bvBByGINFRKOsPaVXhb5fUfvFiekHErdgs8CSpmPFmj9AHo7ZTAIuVV1z5oeqGzoΔPLR and

∠CAE = ∠RPL … (2)

Adding equation (1) and (2), we obtain

∠BAE + ∠CAE = ∠QPL + ∠RPL

⇒ ∠CAB = ∠RPQ … (3) In ΔABC and ΔPQR,

PrAPXH3Amq17EdsJoBdWa6p5c23PgO4nmU1Hy4Eh9MNCRK1 sIgCSWe7oMcozNuZDZl4UzpDypo 9Yu7omnM4yNEN 1gIvALEl M8TmQHAh64y63a QVmHz DOZwE0KzCKau0VA

 (Given)

∠CAB = ∠RPQ [Using equation (3)]

∴ ΔABC ΔPQR (By SAS similarity criterion)

*Q15:

A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Answer:OX5juuaVJclie L WBvf UwqoBFnT7iGisHrLAP I9d2k8HSMGfWlfD4V8Lv8kwqDmtYOxi2r 9nz3UQkIDvi70wBthjB5FNXR8i7iagkLBTHSq9oPhg0xjBpK YzV24Q927jcQ

Let AB and CD be a tower and a pole respectively.

Let the shadow of BE and DF be the shadow of AB and CD respectively.

At the same time, the light rays from the sun will fall on the tower and the pole at the same angle. Therefore, ∠DCF = ∠BAE

And, ∠DFC = ∠BEA

∠CDF = ∠ABE (Tower and pole are vertical to the ground)

∴ ΔABE BVz6c2K Av gBDr9E3hTe3kRniPbBMmxgAOVcOkmB16jf02I9hBMTrqrSbAYslq4JpGvKp8iz5yPPXr5PR8dTq cW9by1iaD e3E739BG7bgodYIfXSCLyT2O0J WkGHkX4RoEU ΔCDF (AAA similarity criterion)

XnupsuCdY32AIrWLnGMjVDCuP1evWxJw

Therefore, the height of the tower will be 42 metres.

*Q16:

If AD and PM are medians of triangles ABC and PQR, respectively where SBpsHJRsFgYt6GT dlTG9WSLzbGzrRiFJPoBOw4ChRC1InccMfDn Y2rAZAUprhx2hzUZQ9JWCLzhopzKbMH76Yw ZgtYOGjWgyWubMkC4jpbzFhdAAH ApPiO0W7kbTTpQqdpM

5LDXFSg3i aNdEK6yBoiobM4VKnmi4JvTt8JwENWLqm01DHw1M31HgShNItjf5ZAwDJEa7HJ NmcIC5eKNcZp DazYZ23kAlUzUpabn1 jTg3YuQjUSGC5b80JilgT7o4BgqLkw

Answer:

It is given that ΔABC OqFDhK0V98GFcvG0PpAWK9oDjT5BcRw6MBtYkqpEeJSiwb GE 5n15Z71B0G9f7gBO7NjolO0t2krK qpRksDvgnBuZ5gKnvlRGgvKvgtΔPQR

We know that the corresponding sides of similar triangles are in proportion.

∴  qElyVlitzPllJ7T WugKlO4d9hfYbhO4eY7CLa RsFWDWvE o7Tk1eE … (1)

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)

Since AD and PM are medians, they will divide their opposite sides.

∴  fGBGuCPAxItY tgzuRVHfUB1ovf4Hu 6tb4NywroUseqiYWA5tGcn6z5Go1F1IJqRheb82sbQ8cpIspIEsSNEivq9hhu4at88oXGlKBypX4kgqWiIPg1K2suifcw5 GT42zndEs … (3)

From equations (1) and (3), we obtain

BJHaFrLp6CRMj1tFVJDUJospGiTDUtjiKocEJatf9chCXXY zYF ZKs86ty2du tSHp3hEBJIzr46u0 bAwKRzAEas EzndvG8 DoyFO898ZfIZJLXRdAXviOis5kjXke8s QeE … (4)

 In ΔABD and ΔPQM,

∠B = ∠Q [Using equation (2)]

[Using equation (4)]

∴ ΔABD Q1J2nz E1V4HTYLzrA3lAVTzNDG9SMrsqGq5QFyN28SOxoDXjelsH1 zeWTzXO94ndm8l0AdN3UtNXcky2ba6AzykoR 5WHs1M2XeuqH7ΔPQM (By SAS similarity criterion)

uiW2O86hba5tvEbyz57p04CW dyOdYFT1momElrh4u xwgSu Utbr45SLwroMwOIcgUDtUPoz2lOlceToVL0KYes 1CJQc6FkT I9rky0qiG df2kw3PrBdI2oRAKfqPyDlJJ4BJHaFrLp6CRMj1tFVJDUJospGiTDUtjiKocEJatf9chCXXY zYF ZKs86ty2du tSHp3hEBJIzr46u0 bAwKRzAEas EzndvG8 DoyFO898ZfIZJLXRdAXviOis5kjXke8s QeE

Exercise 6.4

Q1:

Let  and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC. 

Answer:
EJcFKTRiQGoM3dTISOXbZ6HdPwk 8cSE2CZ ZeUYezmg4K6GBAM CNX73A rjrk1lrdejzCDAldaUyneDNrW3Vakh1y6j Ft9SuDravkkRzkCzmUUnWRXKWwzM6P4i B6HJjptQ

luzzYfBSaIU3StapqKvLJss2apghk jp3JjFT04zoXaqqUdEarcur1cjfjHwcJp3Z0uI1e9VsvOZ nI37gNJwnyGP fQN6BS8ry7vS4t79MDpXyXkBmSfq1pW0MlLP2Tel I fA

8MxQJMZxAC fCZdw5LSl1JCsdu4mkum7kUpCbbC3N174rANDOtiCwT4dkMjicHtZ5Rt9vWfuZrQqZAiPVPZxFgMpZ QR3 uIceoDxExGdMtiBQ5MFaMIld7vTKDbZa CatyO1ww

Q2:

Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Answer:

xGrRlLJco6FCtqNunj7sA5JrQfA4zrIWiD klfh20wj17TGKscupwIdPvI9Vk1ybG WhwmQxp7AVUgZy0snCu3sqDYkWrK8zGvxeik 8JYnxwbu Fsvd0k43II s3qp0oZheP4Q

Since AB || CD,

∴ ∠OAB = ∠OCD and ∠OBA = ∠ODC (Alternate interior angles)

In ΔAOB and ΔCOD,

∠AOB = ∠COD (Vertically opposite angles)

∠OAB = ∠OCD (Alternate interior angles)

∠OBA = ∠ODC (Alternate interior angles)

∴ ΔAOB BiNaMRi82iHVdBwEtXSdQO7SZgMHZttT5D10g c3gyqioxl32riZ qqLoxYsrrJqRzCOBC Jq1bH3IlT 346fz6ObPI2WwR8wwgvCmD8mbn5tX9s8wVZQo2MLGsuLU9FhwC msMΔCOD (By AAA similarity criterion)SbQsdTwJhWMqbe04YDDlAkLc ldZwvHmZDnNV62VmByXKFKlJ1t7nPAFVHcdzzmw8rf96KmlOOJ1uZOSD46OXiTkPmIj9GZW4g68Vr0Hzv0camyW0o8IVazDA9o7KpqYhBIiTc

Q3:

In the following figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that

Answer:

Let us draw two perpendiculars AP and DM on line BC.

1 1hl25SdKGm8fa3ENg2hR8M940XC 823VqixUfiUZSbTOdHCEzu2qZW6m6AUGDvonKNkyQRk DBf8
WwFVonPJfQ9jdmwBld0740cKeGor I05Q TjlgP3b EOAbCAFT3Rp20O5A008OyC5T7SscKkEQJQ5RfnqJu2O

We know that area of a triangle =

.

In ΔAPO and ΔDMO,

∠APO = ∠DMO (Each = 90°)

∠AOP = ∠DOM (Vertically opposite angles)

∴ ΔAPO BAT3zFXF6MlSvABxtkr6xh8IO1RdF1HsL054Nd8ECOR1nZMyRjSmUxaKNoNhDw7N4uOXwqOqPXYHBh 2KA3t0jhc1ppUs3FIslFkIQKXof 7bPZO024IcFpL1UFbOZKce4iGwlUΔDMO (By AA similarity criterion)7Ex4up8zzU9xNC8kHRnfDQReQFN0lTP yd jh5 mDDy4G0nAueJXNiYzUQy fDSgOlyWIVNNSe3wD1XFUJ0AKI oWFJIo4C6uH3FMAv9VMS89nMTUeDOH2jdElNWH wlg 7g6Y

Q4:

If the areas of two similar triangles are equal, prove that they are congruent.

Answer:

Let us assume two similar triangles as ΔABC Ua3GAZaMKcl2kMAqIVxw3w 4nRckHu3uMGMeIxsJtinwePYwey2334ndzl27p6N7wIndKe xwgF jfAxeW7x6 Y7tLkL3BjbLE 8a ΔPQR.

jPgFct3EikRDryNpNEXyDSPcN wMns948VFd4aOhaRGAGwXO4oSg1ke9 ZC7UEzH GsCS3mAa01o l8RnVeOknR5gg6TowJaj9vnsGsHEe6

Q5:

D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.

Answer:ZC9bd0RTSf1QMvIyYTfwhG63EAbuRRTrZqYEwhTFTKMo e7GyoySFJ4gsOpm A2VZaGJkSTp fcMGzifKk yUasJ3ptrKZ0j 0j3KZCqvzLyt9G3wwAoXNnww8YsgB SLKNkrIA

D and E are the mid-points of ΔABC.

lhoqJ06 p7mJHbKg9y ngRBwwz1C66StY3o35CSJRoLx hLdQJuBI2Z2P3c VfYgp7bjdyjN5gYYVGZ4FNGYcCL30hkmRFHLkR61cIvEcMSMSVBki81mJ r

*Q6:

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Answer:v98himVfBNSQdFAbGZOiC JoPeFfrJmAVCKwEPEneMlnIn CUcm CIWI5uUzxPJhbAKl JXfF23 1THEII1hGgrlozi0hz8YyxWWuMvNBU2MGMms5

Let us assume two similar triangles as ΔABC H38hc2so AiGi1joPhvDp26OGs6 ΔPQR. Let AD and PS be the medians of these triangles. sniqNJjEx dg476INZzXNn TsfZu1TxaXJQqwxe1IMRnidnicVQsptbrj5Znh5TLjzFUAqNab FMcND1wVNJLn1XcX3s FuVn57USmmjKsMOWfPdgXocZ4rAWnKnRnGAUHWLMdo ΔABC ΔPQR

                    …(1)EQGvnopSnjQXtOesw7SNSJoffMgkDAAgpc7rRhLLdycwfXdf1XgyKN5QW57XeUmXpHp 7U4 badY4dkJXmrFNPX JuzbuRRqhipujc92zwCWdl6gAJ0rilliMOcqGnUZZlpmdXg

∠A = ∠P, ∠B = ∠Q, ∠C = ∠R     …(2)

Since AD and PS are medians,

and,

Equation (1) becomes

… (3)

In ΔABD and ΔPQS,

∠B = ∠Q [Using equation (2)]

And,  [Using equation (3)]

∴ ΔABD ΔPQS (SAS similarity criterion)

Therefore, it can be said that

zAhKFxDf9vD2xU1ME80xomxfCLpbGKqpSWVFEGaL6xIxCYaTvRH3MuQgfk 9btUrt51e R2TUbLvZHZF3cU5WVx5kXfyz    … (4)VgDhnuMU8 SMD1pn8UebnOuNG0oTMQebFhIGCWv12g3uoS5xYbcVvK6IlYN294 sPF09qaFVWoaTNwJcX8TFBm6klaTtyZz55FrfPDf2RvLb bC a8 VN7NCKyHD7bMubM1Z1dw

From equations (1) and (4), we may find that

IhYvHh0lSf6NfgrZ73l3kErOAV ZIqNh7T8zBAvmbBpvatD8p7bp745vuGr 1HTVsmtPofJW

And hence,kp85WLxU5ayPqT8PozjuDC MsLB5U bmuvnMmvBJeQmBKWQkAN9LLnA92FkaJyELnTfvNrn

*Q7:

Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Answer:
OzPD7N1uWly5XcUH709xKvx0UBtu E6jUjP79682 QfL5D6Hl94g

Let ABCD be a square of side a. Therefore, its diagonal OC74MtqQuPCdY GSnCSogQTesZTS4DeDlmA6 QmlFlGTZ12spiROi6oCtxeCrT2QwJnunRdvn4w3 h FKdbMTugrA

Two desired equilateral triangles are formed as ΔABE and ΔDBF.

Side of an equilateral triangle, ΔABE, described on one of its sides = a

Side of an equilateral triangle, ΔDBF, described on one of its diagonals OC74MtqQuPCdY GSnCSogQTesZTS4DeDlmA6 QmlFlGTZ12spiROi6oCtxeCrT2QwJnunRdvn4w3 h FKdbMTugrA

We know that equilateral triangles have all its angles as 60 º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.

gUtcQjFZacl8627i6Yao4T28 g12bna M8UG5WE4zCcqBiVB5pOR Kf9q PU0IiuS4gCNSzHnXr4IcXp8VlnjadFKhv QOCEbQlTDLA2vV jIP8FsublkgaZrSjp9pUt 5d5HU

*Q8:

ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is

(A) 2:1

(B) 1:2

(C) 4:1

(D) 1:4

Answer:

W Ap ROuU4YHJxftGFHtPUK6eI9sdB7Jxi8qgxYPg5rXcf sFo3axFt0sWRXyyWqHw f 1tgEe0obVkYT1rchNIUpwjFo1A3gMy52ZGg4BLXjvHjvPmho1t2NrniGqrq4UB ysY

We know that equilateral triangles have all its angles as 60 º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.

Let side of ΔABC = x

Therefore, side of mQspin8hEIwb4IO8aj4cAAIlO0LtJY9iIkUDd9fPoEIsiKr0quBDlhSOMwUPVo4ZEm727dcKvZkYZkPFNyCkrnZbD2fH 8ocwaOqzT37gS6oSQRuucZQXK8BV2b2EeoayVFGHdc

Hence, the correct answer is (C).

*Q9:

Sides of two similar triangles are in the ratio 4:9. Areas of these triangles are in the ratio

(A) 2:3            (B) 4:9         (C) 81:16         (D) 16:81

Answer:

If two triangles are similar to each other, then the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides of these triangles.

It is given that the sides are in the ratio 4:9.

Therefore, ratio between areas of these triangles

Hence, the correct answer is (D).

Exercise 6.5

Q1:

Sides of triangles are given below. Determine which of them are right triangles? In case of a right triangle, write the length of its hypotenuse.

  1.         7 cm, 24 cm, 25 cm
  2.         3 cm, 8 cm, 6 cm
  3.     50 cm, 80 cm, 100 cm
  4.     13 cm, 12 cm, 5 cm

Answer:

  1.     It is given that the sides of the triangle are 7 cm, 24 cm, and 25 cm. Squaring the     lengths of these sides, we will obtain 49, 576, and

    Or,

    The sides of the given triangle are satisfying Pythagoras theorem. Therefore, it is a     right triangle.

    We know that the longest side of a right triangle is the hypotenuse. Therefore, the     length of the hypotenuse of this triangle is 25 cm.

  1.     It is given that the sides of the triangle are 3 cm, 8 cm, and 6 cm. Squaring the             lengths of these sides, we will obtain 9, 64, and 36. However,

        Or,

    Clearly, the sum of the squares of the lengths of two sides is not equal to the     square of the length of the third side. Therefore, the given triangle is not satisfying     Pythagoras theorem.

    Hence, it is not a right triangle.

  1.     Given that sides are 50 cm, 80 cm, and 100 cm.

    Squaring the lengths of these sides, we will obtain 2500, 6400, and 10000.

    However,

    Or,

    Clearly, the sum of the squares of the lengths of two sides is not equal to the     square of the length of the third side. Therefore, the given triangle is not satisfying     Pythagoras theorem.

    Hence, it is not a right triangle.

  1.     Given that sides are 13 cm, 12 cm, and 5 cm.

    Squaring the lengths of these sides, we will obtain 169, 144, and 25. Clearly,    

    Or,

    The sides of the given triangle are satisfying Pythagoras theorem. Therefore, it is a     right triangle.

    We know that the longest side of a right triangle is the hypotenuse. Therefore, the     length of the hypotenuse of this triangle is 13 cm.

Q2:

PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM x MR.
Jeb8pwUvmBxAotcMv7YK6FOuE aZgYB1x1QihI0CiiumKgBM VaXEpT8B688E7AmM40sTeU5abs44GsaopVJUCricm0uvZvN40gupHGMeDCccumqSGKNR6dvYA3l5tHXnUk Cow

Answer:

Q-3: 

In the given figure, ABD is a triangle right angled at A and AC i. BD. Show that 

(i)   

(ii)   

(iii)   

Answer:

C:\Users\varun\AppData\Local\Microsoft\Windows\Temporary Internet Files\Content.Word\WhatsApp Image 2021-06-08 at 4.57.07 PM.JPEG

Q4:

ABC is an isosceles triangle right angled at C. prove that AB2 = 2 AC2.

Answer:QOs8jgP7LbQYklJ9yE7dhk8cRrbchPWRcp5aOJke67FKG67UUnBtZS4UAtrH

Given that ΔABC is an isosceles triangle.

∴ AC = CB

Applying Pythagoras theorem in ΔABC (i.e., right-angled at point C), we obtain

Hence proved.

Q5:

ABC is an isosceles triangle with AC = BC. If AB2 = 2 AC2, prove that ABC is a right triangle.

Answer:
jE zswDJWXgnadmjR8JQYUFDw6VKGpakn6aDoh

Given that,U2Hh Z 4VboO58LjeOGQF4czUGF ASGbSKT25wWlpAlikMrDNwA6Ggel98QC7NxkSXYfv8yGDN SFj tdEY5XHA1i8sLuRYyXT xcw WHaDvi8pEkvhE6Ihai1b8EeyHQT92X08

Q6:

ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Answer:mHH9ZL i0rHja9GXwAWM0 c7uxucsedz1DB2gLDjNfHRnsAgPD7hlW5SHue Jg76UjGfE2YQBHeRd

Let AD be the altitude in the given equilateral triangle, ΔABC.

We know that altitude bisects the opposite side.

∴ BD = DC = a

ts2OlqdaJ9ca1XzZi9PyMncHIR LvqAOrPfXH7tui7Xwhj9lJy bMWotYb5qsrWk7LIR1TcQsUKYXwg

In an equilateral triangle, all the altitudes are equal in length. Therefore, the length of each altitude will be

Q7:

Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.

Answer:Ib3CGgQDcnR8kwTw8qWb6WjGvI0UbYcdXbAggpAKjDyCEc1sbD4ODRgKl3iBNKYSBNj j2NMH5ClrBgC HXdJAfmkOpAVe1jo

In ΔAOB, ΔBOC, ΔCOD, ΔAOD,

Applying Pythagoras theorem, we obtain

tbjgmiFcTCQeN2rz4fZH3jDvls fOMfiGv9hrcAkDrSQVm20EHSZOLT3nUhmeDUXjUeOZh4XudaI6DdUOLfI1f90ly5sqeIube8JgR 6LvBJpgayNrWPHyChFcHHCGsbJfJ0Hqk

Hence proved.

Q8:

In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show thatPR8TpbTNMAX5dh0uCQnW8failss6QwTSTdvFd2AL0hYOy zGQX7EZUjPqcUjzu7meCIDkjt5YW07P S64DDSoqauPLlvtUtckkMwnUowCBu5u5mfHrfPH9SjiTd cZw3JLASH A

  1.         OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
  1.     AF2 + BD2 + CE2 = AE2 + CD2 + BF2

Answer:

Join OA, OB, and OC.

IjkOgcPvrYOyZt
  1.     Applying Pythagoras theorem in ΔAOF, we obtain
QnZbcUPJSzDxqNsKss0vNzJrms7ZTz07Wpu7U5t0BNAusFZQQ2XAzhuxQiebNeSuwopYxpJTpFL9bKz jfzk0EWmn2UJ2 Q XcA VqIKHAcQrdYckgCXVGfU Z5E MTY3DeVV c

Similarly, in ΔBOD,

FTZIJRe8GKbf7MDG3fiBFLPgsTICZsL fzcaOLQZ7XopkDWOWyYBAPMcO4Fm8kRqdgLxL4yKRSxsANX5wz7 lZHaEacPwSgvo3f3GYRBMSrSM p5EM88HUYyH49q FF9fTB1czA

Similarly, in ΔCOE,

NtXnO22xCXEs4L3fs7DqkfTgcMv3LhORvCvDF2QX6gH6ZKgVT1zDW3jOdB3XXRR

(ii)    From the above result,

hwBZnUsv HupnEs5DYtXd6uQZ3SD3W E 5jyLymTHJbxTRIm9Ieb

Q9:

A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Answer:a6hf1UwJEICkeXXUVVrmGsWzfaDUg MOeH 4sxg75EGWTw45r A8aOux A bUIq82wEsG wZbTPRgfFVRO9yI VCUfNXW5CwMrmAb0M6MGCjGgtn5RYbTWtz NP7w1VgjA7qCNM

Let OA be the wall and AB be the ladder.

Therefore, by Pythagoras theorem,cn45co4tJlmuzq2PO AiGCL10lR51wmH6s1ubnQTcHNnAE52Gga4Vm1aagIBg2tP3 xrYm89PGrQTvKNv25V5 M syfPm zk iVSHj6GDepocyrOwwuTwMMc9MaLr oA1ZWplcA

Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.

Q10:

A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Answer:VtRSafuCJ3ICvaeDeILBTbBrUyxsjbTF97fDEiNfTJLMBazyAA1LKoUhOeJX 9IjuF6dXFqiP66iKCZwH3uS90bnyIpIqI1Hn1wYGBX86LB9Lk8 bTGPE6KDD s2HRNuv5XuIeM

Let OB be the pole and AB be the wire. By Pythagoras theorem,

OA2+OB2=AB2

OA2+182=242

OA2=252

Therefore, the distance from the base is

Q11:

An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be

the two planes after hours?

Answer:LClBya40iWv0BEXPsiALtrnDNj1nfM6xHjN

Distance travelled by the plane flying towards north inSimilarly, distance travelled by the plane flying towards west in Let these distances be represented by OA and OB respectively.

Applying Pythagoras theorem,

Distance between these planes afterck 9zYr2WKlq9rAwcZapyv ypzNrLRscEu83WPALhmwIVzAPAVuN9pjeI0bsb0 MlDz6XK8 7G0Fp1d 9p9s632gw8CSBzMKl7aZu6NdTYZ OX18q6n51JT1AlTWbAz ayZP58gHdZRZrHaojBP0FBw yicjaGMEAu4ttlqbDZZAX5 qcWC6rmuJqXoC32pHEDIH0MRa eSohqCelJj3nS3lYGX RBxqZlNpwD7pQH s JazvOuxP CiLmOROm4397 LPZSKvDfJIs

Therefore, the distance between these planes will be    km after

Q12:

Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Answer:

Let CD and AB be the poles of height 11 m and 6 m. Therefore, CP = 11 – 6 = 5 m

From the figure, it can be observed that AP = 12m

Applying Pythagoras theorem for ΔAPC, we obtainm49FcVtGCH GA zlsgh0mEB079bQyN2bkGY7 5F6ZlyP99jI4N0kupMNLaht

Therefore, the distance between their tops is 13 m.

*Q13:

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2+ BD2 = AB2 + DE2

Answer:8ThVDCn6wwr mN0arbJAC5NK FUCUmwemxBzafkdFn B CkpbtGXCd5rtp0GG915KB0vyKfyCz5ZMrP5Shsi8Pw6NYdL2v

Applying Pythagoras theorem in ΔACE, we obtain

C:\Users\varun\Desktop\JPG.JPG
L9aAIvuwGxyXUXO8C4neN3Vnxe 6VWxRqPYvxR bQ0fTW4hKAVUlnCS53pPI0YwY0ecLVd6fS3ki0KAB s LOpfuRZVPOtVcI19kQeKeRjvOloSHlf3TXT7dlbl hTzBlcOOTs4

*Q14:

The perpendicular from A on side BC of a ΔABC intersect BC at D such that DB = 3 CD. Prove that 2 AB2 = 2 AC2 + BC2Ac7f7nBW6 wC0bdxkeyZzQVNTKhz1F2 nhOhgVjpn85TCdUopj8DzrWg9xGU05jfbmZun4tibJELH1udiG95r0XWLwof It7FSSwL5IiOPPmE8WJz17 zLR8utnEn2oTlKdhCck

Answer:

Applying Pythagoras theorem for ΔACD, we obtainC5Gx70rzv5bw67821ElJPrteyEnfsnPVdmd5Ht fXzjAMAMX9N l7kSVcTYecvojux4cloOEPjL73TJDMnP71MkVu5M4MOkwt4BGlf8A6VhpouzxwpC j0VWS7sptYGhEhhBBIY

Applying Pythagoras theorem in ΔABD, we obtain

Kd p UTMXx6DM058aMz mGe05hprl0cuEjdZunaAZa z1EEXKLxSabJ1KC3m7gM7VXJbT18QwgveKnXbhQcZLo7XSvM4WMP5JV10PEZsXJJZl76K24tC1CP vOo59PqydnXZcJI

Hence  proved.

*Q15:

In an equilateral triangle ABC, D is a point on side BC such thatBC. Prove that 9

Answer:

9GRJ2OjBc tSozrvIXfSozCdW6FbP45oxDl8eBA2l2GmglwIHcT0wpZhNcpfxQ7sWxi38tOJINWt 6qtP yaDQmDR6A Sr8vN8xlrE

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

and

Given that,
FEPs9A1POoftThefog8IJeNI6rGRY fNK5DSIWjZdrKxlmx L ED9mBZ22WiGvCe047VJymtzouJpj 1oIbPJH9DXYJi7bT0jscRycJ5mBZpkxlhS6RMRLl oP r1YXzN16BJPM

∴ BD =

IGb0mAHWB23uzJsVihm2FvTvmBdOd7ii7DIaBQTiOT vGSRxlu Td587AEfbjGzbl0AU2GEKhrl6kosC0rxVA 91ZSSUTDmDazk9

Applying Pythagoras theorem in ΔADE, we obtain

AD= AE+ DE2pLIGCM8H75ZPo4SL wyJSgmGX5DyCyRdP8okPEWlEJRf1LugQl3vRj9L2illVSqfWQHTHl3a0o7RBz mn4GEKrlaTgUSaB9Z5iHu oPO4K3kGtutXbFQ s776bkQXyoH30R3wdc

⇒ 9 AD= 7 AB2

Hence proved.

*Q16:

In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Answer:
PzXQ4UlqIwnSCU5tZMaNQuakrYzBGbFpBKEFrJTFwfKyRGT9kB zS 4PjCwtCWcu6KX3uXtmif7XdoHSfzbM0usXlLmhZTAnKoqBXCAMYkpRB9eGkIHPyhr1V7boKgSrn8RwnE8

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

ZihHVDL1SGAyfHJSUsK7vmuKcHk1GPUCJkIGodcWNGFRmL1OaJHlt7tA6ypA7HhTxItvrTZpmkXIOUgLxRWlPNkHGMFMAXkyZJuVmwL8Clfh4xvcl q0KZPML0wyAB9IPN V05Y

Applying Pythagoras theorem in ΔABE, we obtain

AB2 = AE2 + BE2

5kA3fIL6bA22t5ZdgZCjjDzyYiEsrcTIC2ybum8RkjEJ92L6v3M WUhYOhSMDO kueXEnb8OmsNFoqOPt83 w6rCO2mYjFzA1VwVkdCkU10H 1auB0kSKA1acQ0Y6ksOpI N 04

4AE2 = 3a2

⇒ 4 × (Square of altitude) = 3 × (Square of one side)

Hence proved.

*Q17:

Tick the correct answer and justify: In The angle B is:

(A) 120°         (B) 60°        (C) 90°         (D) 45°

Answer:

Given thatcm, AC = 12 cm, and BC = 6 cm It can be observed that

AB2 = 108

AC2 = 144 And, BC2 = 36

AB2 +BC2 = AC2

The given triangle, ΔABC, is satisfying Pythagoras theorem.

Therefore, the triangle is a right triangle, right-angled at B.

∴ ∠B = 90°

Hence, the correct answer is (C).

Exercise 6.6 

*Q1:

In the given figure, PS is the bisector of ∠QPR of ΔPQR. Prove that

Answer:shnCiagLj9VRu8SCMhbP3lgDdcMYleVzYSocDWG40K5wqtvKdJgEF0jKFMiKQhQNPG2zb1cQDdFTIge3wwS DTpUem0sySRXliNTXPqsfBHEsHBoLUamTAkSv8QkqbeNnJLAZfQ

L6VG4ONdp0J3Td2sViuRnIZL bkWGICCQOxSIatjOvGSgTkZiZ2v69RR11M3Lzsn3PNCSqT8kHnK

Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T. Given that, PS is the angle bisector of ∠QPR.

∠QPS = ∠SPR … (1)

By construction,

∠SPR = ∠PRT (As PS || TR) … (2)

∠QPS = ∠QTR (As PS || TR) … (3)

Using these equations, we obtain

∠PRT = ∠QTR

∴ PT = PR

By construction, PS || TR

By using basic proportionality theorem for ΔQTR,

QSSR=QPPT

⇒ QSSR

*Q2:

In the given figure, D is a point on hypotenuse AC of ΔABC, DM ⊥ BC and DN ⊥ AB, Prove that:

  1.         DM2 = DN.MC
  1.     DN2 = DM.AN156zHZRtXWKhCtT7JRzRILk0rgbyYvitqgv5toz6iNqiaKC 2JaWXP1Xe9 hHnOKHG7bHF6kaxKtW2UnFJokBeVkpZm88Yh4nB4J6HNtXNWBUOEnQaw60gKUq5 i314yuh2mSYY

Answer:

  1. Let us join DB.8SQ2YiGS azRSMgUu0AJdRfUQCJrUclIMRk YgcacsDdGqF AZo8JIiu6NG P21Xxo1BSSqXS2YIszQUBrUByFRzODTqQjcC

We have, DN || CB, DM || AB, and ∠B = 90°

∴ DMBN is a rectangle.

∴ DN = MB and DM = NB

The condition to be proved is the case when D is the foot of the perpendicular drawn from B to AC.

∴ ∠CDB = 90°

⇒ ∠2 + ∠3 = 90° … (1) In ΔCDM,

∠1 + ∠2 + ∠DMC = 180°

⇒ ∠1 + ∠2 = 90° … (2)

In ΔDMB,

∠3 + ∠DMB + ∠4 = 180°

⇒ ∠3 + ∠4 = 90° … (3)

From equation (1) and (2), we obtain

∠1 = ∠3

From equation (1) and (3), we obtain

∠2 = ∠4

In ΔDCM and ΔBDM,

∠1 = ∠3 (Proved above)

∠2 = ∠4 (Proved above)

∴ ΔDCM KKZtX28n BYtS3XgjMJkMcNiA715Hz7zA8fAIlmwQz7iczWBLXEAEOTLOZPLVhuZWvCbeYrLgNtACx9qECefkSKCev4o6x0ybCnNW3WagJFReC61OCxtj5 UsaUaQMrCc9CkM sΔBDM AA similarity criterion)

⇒ DM2 = DN × MC

Hence proved.

(ii) In right triangle DBN,

∠5 + ∠7 = 90° … (4)9 WP90DdaBIZtnglxEw 8z0o5YWtB3j2gKOSEI4opHgIwQ4YhZ7DjdyiHxwGFm7l4OC5yFn

In right triangle DAN,

∠6 + ∠8 = 90° … (5)

D is the foot of the perpendicular drawn from B to AC.

∴ ∠ADB = 90°

⇒ ∠5 + ∠6 = 90° … (6)

From equation (4) and (6), we obtain

∠6 = ∠7

From equation (5) and (6), we obtain

∠8 = ∠5

In ΔDNA and ΔBND,

∠6 = ∠7 (Proved above)

∠8 = ∠5 (Proved above)

∴ ΔDNA js4V4JJZWgjhLzoBRmobzjNuhm7uSXbiwCYtxuS6UHsnV92rSyEEEuzU2fLcnlLEpQzcXFN3N8brlISRYYFwt q3czxKx49nkyUgdwNqnZcp2AUxrfNcnWKrJu 1kpthU3w SXAΔBND (AA similarity criterion)Wfqt lBO JkULMgYOGZIV21v83dhxKt8WX 7kqFFmoChbaFdF1aYVZklSvp5u0RfebIecCK if81nEGWF Wkd XtI8aV0KldXEknyc7v72uu0OTfeUUwmyuesjIRqEL 5lF1vs

⇒ DN2 = AN × NB

⇒ DN2 = AN × DM (As NB = DM)

Hence proved.

*Q3:

In the given figure, ABC is a triangle in which ∠ ABC> 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2BC.BD.
MHJwXrw0PABNFoMYCRg

Answer:

Applying Pythagoras theorem in ΔADB, we obtain

AB2 = AD2 + DB2 … (1)

Applying Pythagoras theorem in ΔACD, we obtain

AC2 = AD2 + DC2

AC2 = AD2 + (DB + BC)2

AC2 = AD2 + DB2 + BC2 + 2DB x BC

AC2 = AB2 + BC2 + 2DB x BC [Using equation (1)]

*Q4:

In the given figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that Qp p3m5MurHmW0ZN3Sg0soaLSEpupFr6NpDmhVWOmyKDjdZaEj 0dJl W0jBpOLq5Rk5SQTvImucMTzxtcd6Vp29H9LGWX4TDfX5avxbzJFy2kAhjS9Iz6q6JrzlQj8MM s30sI

AC2 = AB2 + BC2 – 2BC.BD.

Answer:

Applying Pythagoras theorem in ΔADB, we obtain

AD2 + DB2 = AB2

⇒ AD2 = AB2 – DB2 … (1)

Applying Pythagoras theorem in ΔADC, we obtain

AD2 + DC2 = AC2

AB2 – BD2 + DC2 = AC2 [Using equation (1)] AB2 – BD2 + (BC – BD)2 = AC2

AC2 = AB2 – BD2 + BC2 + BD2 -2BC x BD

= AB2 + BC2 – 2BC x BD

Hence proved.

*Q5:

In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that: (i)   

(ii)   

(iii)   

Answer:
A26KkAHGZYYkN Du9RCJRbaMa4TpVsSHVAYnWCHQe vm8LR2R6ERlUWYIKwcB5GuQqf hqIiTCiZ TR SnktHhvf XEMoE38UbmMOn94ardzmZKDiGgx5hJCr5yinOXasI jY1Y

  1.         Applying Pythagoras theorem in ΔAMD, we obtain

    AM2 + MD2 = AD2 … (1)

    Applying Pythagoras theorem in ΔAMC, we obtain

    AM2 + MC2 = AC2

    AM2 + (MD + DC)2 = AC2

    (AM2 + MD2) + DC2 + 2MD.DC = AC2

    AD2 + DC2 + 2MD.DC = AC2 [Using equation (1)] Using the result, we     obtain

ixr0dRB4fgD T8hTiPp8E5R jV6Gl13egzsA94ODtu3J1TAOz BMZxVOPIl3IVF5pryiaoIbzPL0lH7nDf9gdA7McY2GuDlN2WQ6mR G z1cWIdz5kpsxb9G7eixvJiF4LHmS s
  1.     Applying Pythagoras theorem in ΔABM, we obtain 
qOpO FYtOaWXCIQz4Qu 4sX3o4QCP2i xO137rHCMl1gDq4A2rGreepqpLR42ZKKLKlHY omus9Tz696xec3Za Vg6QTMJPr7hXypNCQRDmZNiUOvZQ WeAMeAGXrdTJS9ilY
WFVSe0vG9eCD9nRo2p9cdwEfj2hiYnbdQOQ2tf1e91SY8Gh2ugdcC2zGJoFxkGeobm39OyRl lo2jFenWiuF0Jv CERR9Tyv AzqNfCvpwR0B3tF65hK5jH8QKf 8XHY84krPpA

       AB2 = AM2 + MB2

    = (AD2 – DM2) + MB2

    = (AD2 – DM2) + (BD – MD)2

    = AD2 – DM2 + BD2 + MD2 – 2BD × MD

    = AD2 + BD2 – 2BD × MDBtqK7r6LUWqAijUFHf2wUmH GiiSynM3kXPhTZrl7wS5q5zrbpgVWrVkeHw0zyF HzEZB8erjRvIwWF6cbfOAr yfA2 pgHQcKpO0vXTUxiNZgqRcyMfb9bUiOkjhG9Mi9 XLg

  1. Applying Pythagoras theorem in ΔABM, we obtain

    AM2 + MB2 = AB2 … (1)

    Applying Pythagoras theorem in ΔAMC, we obtain

    AM2 + MC2 = AC2 … (2)

    Adding equations (1) and (2), we obtain

    2AM2 + (BD – DM)2 + (MD + DC)2 = AB2 + AC2

    2AM2+BD2 + DM2 – 2BD.DM + MD2 + DC2 + 2MD.DC = AB2 + AC2

    2AM2 + 2MD2 + BD2 + DC2 + 2MD ( – BD + DC) = AB2 + AC2

bbw4yEH0c2c4CvKDNlsDI3VSzJxRRwGHjqDE0HK6FxbbWl8GBNOUAJDdlJpYUKYItBdjISxGAhZIQLDEsBqINqWVJktrDhLhn81CVfim27Tah1V pXmS6PQvpAJUWyoSqnaqiw

*Question 6:

Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Answer:

Let ABCD be a parallelogram.

Let us draw perpendicular DE on extended side AB, and AF on side DC. Applying Pythagoras theorem in ΔDEA, we obtain

DE2 + EA2 = DA2                                 … (i)

Applying Pythagoras theorem in ΔDEB, we obtain DE2 + EB2 = DB2

DE2 + (EA + AB)2 = DB2

            … (ii)

Applying Pythagoras theorem in ΔADF, we obtain AD2 = AF2 + FD2

Applying Pythagoras theorem in ΔAFC, we obtain

AC2 = AF2 + FC2

= AF2 + (DC − FD)2

= AF2 + DC2 + FD2 − 2DC × FD

        ….(iii)

Since ABCD is a parallelogram, AB = CD                 … (iv)

And, BC = AD                                 … (v) 

In ΔDEA and ΔADF,

∠DEA = ∠AFD (Both 90°)

∠EAD = ∠ADF (EA || DF)

AD = AD (Common)

∴ ΔEAD V4ExDJocbIozAMFC1uhY7Seo0yE67q20hZVbRw 71442k1jLIqeUWP1bJsWlno2JvFT54IyGEkcpD6b6 DaWpw8l6s0Y5eSNk7NGPKlas9VEV55zTG4835wHCaCRu9i1tRtCLf0ΔFDA (AAS congruence criterion)

EA = DF                                     … (vi)

Adding equations (i) and (iii), we obtain

563YcZyyZwAFafnMR2 Sga m56MZSN9HeuZqyLPCg3mSKzUh0rYAS8IDVO295QHORw89dXKzOONvmUVGRoQwh0w3di47CoUtrFkDEgs97c9Kj0buMwOe0rEHdeiMUls6DwVk4Kw

[Using equations (iv) and (vi)]

AB2 + BC2 + CD2 + DA2 = AC2 + BD2

Hence proved.

*Question 7:

In the given figure, two chords AB and CD intersect each other at the point P. prove that:

(i)   

(ii)       

ekyxXG4wKGkzJwJvJptYuGiHmwUwEZkq3sLZ sDmls5kXxkMy6uwqOJwODG5aqVXnwQGkxO18RLHTPBT7fhx22VfT8GGjZz0YpAvzkAQwO1rJs7 2v9pt94JZ K90DGlYzvhycM

Answer:

Let us join CB.

ytsLTGPn6Ri0iEUD6vAUvKpWhRKabIbVOGUo QR2w3Npr9C34cDdG5BtOaKA00JhzUlciYttSZOE5C A7p8nBggrZsB h0HQhBUJCrer5h5pDE6iUUHiA3mcPwVaBJa Sqz4sfY

  1.         In ΔAPC and ΔDPB,

    ∠APC = ∠DPB (Vertically opposite angles)

    ∠CAP = ∠BDP (Angles in the same segment for chord CB) ΔAPC ~ ΔDPB (By AA     similarity criterion)

    Hence proved

  1.     We have already proved that

    ΔAPC ~ ΔDPB

    We know that the corresponding sides of similar triangles are proportional.

    ∴ AP. PB = PC. DP

YEbI9mgmmVNTXczvDzyh2DAX5PohnPzbtfxtgiUCbJjlM

    Hence proved

*Question 8:

In the given figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that

(i)   

(ii)   
oiU67jWSft7YlGiUrLnXIm3tCLmZwoFEw d SM6 jcgHQTQrGyPeYMTMfiDLl0PvEA51o1E7f FlyUSelWsRzhaQkHMWLi8p dD4SL16GX7fgBey 6gChDRWwT3nZSruPJOR9Tc

Answer:

  1.         In ΔPAC and ΔPDB,

    ∠P = ∠P (Common)

    ∠PAC = ∠PDB (Exterior angle of a cyclic quadrilateral is ∠PCA = ∠PBD equal to     the opposite interior angle)

    ∴ ΔPAC ~ ΔPDB

    Hence proved.

  1. We know that the corresponding sides of similar triangles are proportional.            1xs598MtpSYgDA0PnsDmr03UL2zSW4soF5gH6DPm 8itc6kEvIWxrzG4xNvMVIiFH0qD0LsvPEaS74mshrMuuqr2MqVelxkgG KL2YHfKOSPN5Y6KgFJE vdWfEfvSMYsNVzcE

∴ PA.PB = PC.PD

Hence proved

*Question 9:

In the given figure, D is a point on side BC of ΔABC such that (i)   

(ii)    . Prove that AD is the bisector of ∠BAC.
LgbR19541yb0m3lQvlWYxoydCEQEDf2rKZcXCWSVz xmtnGj208IZ gDeMnD30yjsdSy t8h IgF8KAuDadtbk7pZ2ENCaSrWhPISMx0GJThas69JBh 5I

Answer:

Let us extend BA to P such that AP = AC. Join PC.

lXe9l82sQJfJLsqpHLMijYnQUdnd52ifxUYDHzWEllWFZf991uYKcipvePJbkZB0ogyztcXLhnwQYzdDJ76cVY0ymD8PhuIZxfjNXkRdFwdjr Zi3pmqKHIPKr9wKabtwQRN Wo

It is given that,

By using the converse of basic proportionality theorem, we obtain AD || PC

∠BAD = ∠APC (Corresponding angles) … (1)RAkttU1hA6lUY

And, ∠DAC = ∠ACP (Alternate interior angles) … (2)

bLfFHwFkw9bvO38pfpSCLUA Ufwv3mWRX XvWOwQ2XCPsDXqh3NtLgUVmgbODkFdWkbcFsJV54t2itIUu4fLDS3TbUF5ggUHRP2qeEOVjmkcKHJjCfZobCg

By construction, we have AP = AC

∠APC = ∠ACP … (3)hvDRqa WMScrbLSvb3T43Pio1ILrkzr5 VW5oBr49x8v3ErUK1OWWixjyJu28Qp8uO9Dt6oSN1mTT0qhvfmqT51X075yBPqoAthLECcql6oDWOJPcfU8c7NMJ z1TwIqqZhsYuQ

On comparing equations (1), (2), and (3), we obtain

∠BAD = ∠APC

AD is the bisector of the angle BAC.

*Question 10:

Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?
o 1ZMd zREUCOkxWwnoMsHuCmoEJiYJn1h5f7Tflbt rKUc47WaIkgJyybtSlOLd eKL9lT1HUnALrC1h8izB5cOnzj7lNivnHM67Wt2oaLYTwI 1woH8kYDnQKsyrfWmhJkG6I

Answer:

YETfXx1xZhHJhQdyBLkiikZJenNNlZ5AyRTr5mE08VNPves2 nsBULkt7bFDNXPZM2FAPZLkSyCLQd6aVp1qWdtYeIYLicVuQDtJC34MoGpxILj sKZDFhK7nGsYPJT wk7b0hM

Let AB be the height of the tip of the fishing rod from the water surface. 

Let BC be the horizontal distance of the fly from the tip of the fishing rod.

Then, AC is the length of the string.

AC can be found by applying Pythagoras theorem in ΔABC.

dmPP5kmCgk7ntSy0cfIjIn2rvDdHKD1PYHDEtm9hEn5xVoFKE6YVB4kCnSqxF7fVIhKQf6VRn dZWA
hg1hxh6trr C4XtXzEA rKtcIb fJLGxNELP88sSh2

Thus, the length of the string out is 3 m.

She pulls the string at the rate of 5 cm per second.

Therefore, string pulled in 12 seconds
Y9PDu6cCL1bJbIo4S0qPz9Zc6mVJDzNP kRZS1Hb3X5h a S0Uz6nJirFMUgaJtWDwB wb3RByWkARIn8RG B 8rrbG2TqOOxEDP9bREqYoI jc8qqfFv1I9mMhKFpp86OGzfs8

Let the fly be at point D after 12 seconds. 

Length of string out after 12 seconds is AD.

AD = AC − String pulled by Nazima in 12 seconds

= (3.00 − 0.6) m

= 2.4 m in ΔADB,

AB2 + BD2 = AD2

(1.8 m)2 + BD2 = (2.4 m)2

Horizontal distance of fly = BD + 1.2 m

= (1.587 + 1.2) m

= 2.787 m

= 2.79 m

NCERT Solutions for Class 10 Maths Chapter 6- Triangles

NCERT Solutions Class 10 Maths Chapter 6, Triangle, is part of the Unit Geometry which constitutes 15 marks of the total marks of 80. On the basis of the CBSE Class 10 syllabus, this chapter belongs to the unit that has the second-highest weightage. Hence, having a clear understanding of the concepts, theorems and the problem-solving methods in this chapter is mandatory to score well in the Board examination of class 10 maths.

Main topics covered in this chapter include:

6.1 Introduction

From your earlier classes, you are familiar with triangles and many of their properties. In Class 9, you have studied congruence of triangles in detail. In this chapter, we shall study about those figures which have the same shape but not necessarily the same size. Two figures having the same shape (and not necessarily the same size) are called similar figures. In particular, we shall discuss the similarity of triangles and apply this knowledge in giving a simple proof of Pythagoras Theorem learnt earlier.

6.2 Similar Figures

In Class 9, you have seen that all circles with the same radii are congruent, all squares with the same side lengths are congruent and all equilateral triangles with the same side lengths are congruent. The topic explains similarity of figures by performing relevant activity. Similar figures are two figures having the same shape but not necessarily the same size.

6.3 Similarity of Triangles

The topic recalls triangles and its similarities. Two triangles are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). It explains Basic Proportionality Theorem and different theorems are discussed performing various activities.

6.4 Criteria for Similarity of Triangles

In the previous section, we stated that two triangles are similar (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). The topic discusses the criteria for similarity of triangles referring to the topics we have studied in earlier classes. It also contains different theorems explained with proper examples.

6.5 Areas of Similar Triangles

You have learnt that in two similar triangles, the ratio of their corresponding sides is the same. The topic Areas of Similar Triangles consists of theorem and relatable examples to prove the theorem.

6.6 Pythagoras Theorem

You are already familiar with the Pythagoras Theorem from your earlier classes. You have also seen proof of this theorem in Class 9. Now, we shall prove this theorem using the concept of similarity of triangles. Hence, the theorem is verified through some activities and make use of it while solving certain problems.

6.7 Summary

The summary contains the points you have studied in the chapter. Going through the points mentioned in the summary will help you to recollect all the important concepts and theorems of the chapter.

Triangle is one of the most interesting and exciting chapters of the unit Geometry as it takes us through the different aspects and concepts related to the geometrical figure triangle. A triangle is a plane figure that has three sides and three angles. This chapter covers various topics and sub-topics related to triangles including the detailed explanation of similar figure, different theorems related to the similarities of triangles with proof, and the areas of similar triangles. The chapter concludes by explaining the Pythagoras theorem and the ways to use it in solving problems. Read and learn the Chapter 6 of NCERT textbook to learn more about Triangles and the concepts covered in it. Ensure to learn the NCERT Solutions for Class 10 effectively to score high in the board examination.

Key Features of NCERT Solutions for Class 10 Maths Chapter 6 – Triangles

  • Helps to ensure that the students use the concepts in solving the problems.
  • Encourages the children to come up with diverse solutions to problems.
  • Hints are given for those questions which are difficult to solve.
  • Helps the students in checking if the solutions they gave for the questions are correct or not.

Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 6

Why should we learn all the concepts present in NCERT Solutions for Class 10 Maths Chapter 6?

The concepts covered in Chapter 6 of NCERT Solution Maths provide questions based on the exam pattern and model question paper. So it is necessary to learn all the concepts present in NCERT Solutions for Class 10 Maths Chapter 6.

List out the important topics present in NCERT Solutions for Class 10 Maths Chapter 6?

The topics covered in the chapters are Introduction to the triangles, similar figures, similarity of triangles, criteria for similarity of triangles, areas of similar triangles and Pythagoras Theorem. These concepts are important from an exam perspective. It is strictly based on the latest syllabus of CBSE board, and also depends on CBSE question paper design and marking scheme.

How many exercises are there in NCERT Solutions for Class 10 Maths Chapter 6?

There are 6 exercises are there in NCERT Solutions for Class 10 Maths Chapter 6. First exercise contains 3 questions, second exercise contains 10 questions, third exercise has 16 questions, fourth exercise has 9 questions, fifth exercise has 17 questions and last or sixth exercise has ten questions based on the concepts of triangles.


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NCERT Solutions Class 10 Maths Chapters

  • Chapter 1 Real Numbers
  • Chapter 2 Polynomials
  • Chapter 3 Pair Of Linear Equations In Two Variables
  • Chapter 4 Quadratic Equations
  • Chapter 5 Arithmetic Progressions
  • Chapter 6 Triangles
  • Chapter 7 Coordinate Geometry
  • Chapter 8 Introduction To Trigonometry
  • Chapter 9 Some Applications Of Trigonometry
  • Chapter 10 Circles
  • Chapter 11 Constructions
  • Chapter 12 Areas Related To Circles
  • Chapter 13 Surface Areas And Volumes
  • Chapter 14 Statistics
  • Chapter 15 Probability

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