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Triangles EX 6.3 Solutions

Question - 1 : -
State which pairs of triangles in the given figures 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 :

Answer - 1 : -


Question - 2 : - In the given figure, ∆ODC ~ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Answer - 2 : -

Question - 3 : - 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 AO/OC = OB/OD

Answer - 3 : -

In ΔDOC and ΔBOA,

AB || CD, thus alternate interior angles willbe equal,

∴∠CDO = ∠ABO

Similarly,

∠DCO = ∠BAO

Also, for the two triangles ΔDOC and ΔBOA,vertically opposite angles will be equal;

∴∠DOC = ∠BOA

Hence, by AAA similarity criterion,

ΔDOC ~ ΔBOA

Thus, the corresponding sides areproportional.

DO/BO = OC/OA

⇒OA/OC = OB/OD

Hence, proved.

Question - 4 : -

In the fig.6.36,QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.

Answer - 4 : -

Solution:

In ΔPQR,

∠PQR = ∠PRQ

∴ PQ = PR ………………………(i)

Given,

QR/QS = QT/PRUsing equation (i), we get

QR/QS = QT/QP……………….(ii)

In ΔPQS and ΔTQR, by equation (ii),

QR/QS = QT/QP

∠Q = ∠Q

∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]

Question - 5 : - S and T are point on sides PR and QR of ΔPQRsuch that ∠P= ∠RTS. Show that ΔRPQ ~ ΔRTS.

Answer - 5 : -

Question - 6 : -

In the figure, ifΔABE ≅ ΔACD, show thatΔADE ~ ΔABC.

Answer - 6 : -

Question - 7 : -

In the figure,altitudes AD and CE of ΔABC intersect each other at the point P. Show that:

Answer - 7 : -

(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC


Question - 8 : - E is a point on the side AD produced of aparallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.

Answer - 8 : -

Given, E is a point on the side AD produced ofa parallelogram ABCD and BE intersects CD at F. Consider the figure below,.

In ΔABE and ΔCFB,

∠A = ∠C (Opposite angles ofa parallelogram)

∠AEB = ∠CBF (Alternateinterior angles as AE || BC)

∴ ΔABE ~ ΔCFB (AA similarity criterion)

Question - 9 : -

In the figure, ABCand AMP are two right triangles, right angled at B and M respectively, provethat:

Answer - 9 : -

(i) ΔABC ~ ΔAMP

(ii) CA/PA = BC/MP

Solution

Question - 10 : -

CD and GH arerespectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFGrespectively. If ΔABC ~ ΔFEG, Show that:

Answer - 10 : - (i) CD/GH = AC/FG
(ii) ΔDCB ~ ΔHGE
(iii) ΔDCA ~ ΔHGF


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