Question 1: State True or False:

(i) Two similar polygons are necessarily congruent – False

(ii) Two congruent polygons are necessarily similar – True

(iii) All equiangular triangles are similar – True

(iv) All isosceles triangles are similar – False

(v) Two isosceles right angles triangles are similar – True

(vi) Two isosceles triangles are similar, if an angle of one is congruent to the corresponding angle of the other – True

(vii) The diagonals of the trapezium, divide each into proportional segments – True

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Question 2: In \triangle ABC, DE \parallel BC , where D and E  are points on AB and AC  respectively. Prove that \triangle ADE \sim \triangle ABC . Also find the length of DE, \ if \ AD = 12 \ cm,  BD = 24 \ cm \ and \ BC = 8 \ cm .

Answer:s11.jpg

In \triangle ABC, DE \parallel BC

Consider \triangle ABC \ and\  \triangle ADE

\angle ABC = \angle ADE (alternate angles)

\angle ACB =\angle AED  (alternate angles)

Therefore \triangle ADE \sim \triangle ABC    (AAA postulate)

Hence \frac{AD}{AB} = \frac{DE}{BC} = \frac{AE}{AC} 

\frac{12}{12+24} = \frac{DE}{8}

\Rightarrow DE = \frac{12}{36} \times 8 = \frac{8}{3}

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Question 3: s1Given \angle GHE = \angle DFE= 90^o , DH= 8, DF=12, DG=3x-1  and DE=4x+2 . Find the length of the segments DG \ and \ DE

Answer:

Consider \triangle EFD and \triangle GHD

\angle GHE = \angle DFE= 90^o (Given)

\angle EDG = \angle HDG \ (\angle D \ is \ common) 

Therefore \triangle EFD \sim \triangle GHD    (AAA postulate)

\frac{DH}{DF}=\frac{DG}{DE} 

\frac{8}{12}=\frac{3x-1}{4x+2} 

32x+16 = 36x-12 

4x=28 \Rightarrow x = 7 

Substituting DG = 3 \times 7 -1 = 20 

DE = 4 \times 7 +2 = 30 

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Question 4: D is a point of side BC of  \triangle ABC such that \angle ADC = \angle BAC . Prove that CA^2=CB \times CD .

Answer:s12

Consider \triangle ABC \ and \  \triangle ADC

\angle BAC = \angle ADC (Given)

\angle ACB = \angle ACD \ (\angle C \ is \ common) 

Therefore \triangle ABC \ and \  \triangle ADC   (AAA postulate)

Therefore \frac{CA}{CB}=\frac{CD}{CA} 

\Rightarrow CA^2=CB \times CD

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Question 5:  sm10In the given figure, \triangle ABC and \triangle AMP are right angled at B and M  respectively.  Given AC=10 \ cm, AP = 15 \ cm  and PM= 12 \ cm . (i) Prove \triangle ABC \sim \triangle AMP (ii) Find AB and AC . [2012]

Answer:

In  \triangle ABC \ and \  \triangle AMP

\angle BAC = \angle PAM (common angle)

\angle ABC =\angle PMA

Therefore \triangle ABC \sim \triangle AMP (AAA postulate)

Since \triangle ABC \sim \triangle AMP

\frac{AP}{AM}= \frac{BC}{PM}= \frac{AC}{AP}

Given AC=10 cm, AP = 15 cm and PM= 12 cm

\Rightarrow AB = \frac{10}{15} \times 11 = 7.33 \ cm

\Rightarrow BC = \frac{10}{15} \times 12 = 8 \ cm

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Question 6: E \ and \  F are points in sides DC \ and \ AB respectively of parallelogram ABCD . If diagonal AC and segment EF intersect at G ; prove that: AG \times EG = FG \times CG .

Answer:s10

In  \triangle AGF \ and \  \triangle EGC

\angle AGF = \angle EGC (opposite angles)

\angle ECG =\angle FAG   (alternate angles)

Therefore \triangle AGF \sim \triangle ECG (AAA postulate)

Therefore   \frac{EG}{FG}=\frac{CG}{AG} 

 \Rightarrow AG \times EG = FG \times CG

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Question 7: Given RS \ and \  PT  are altitudes of  \triangle PQR  . Prove that (i) \triangle PQT \sim \triangle QRS  (ii) PQ \times QS = RQ \times QT 

Answer:s9

(i)   Consider \triangle PQT \ and \  \triangle QRS

\angle QTP = \angle QSR= 90^o (Given, altitudes)

\angle PQT = \angle SQR \ (\angle Q \ is \ common) 

Therefore \triangle PQT \sim \triangle QRS    (AAA postulate)

(ii)  \frac{PQ}{RQ}=\frac{QT}{QS} 

PQ \times QS = RQ \times QT 

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Question 8: s3Given ABCD  is a rhombus, DPR \ and \  CBR  are straight lines. Prove that DP \times CR = DC \times PR  .

Answer:

In  \triangle DPA \ and \  \triangle RPC

\angle DPA = \angle RPC (vertically opposite angles)

\angle PAD =\angle PCR   (alternate angles)

Therefore \triangle DPA \sim \triangle RPC (AAA postulate)

Therefore   \frac{DB}{PR}=\frac{AD}{CR} 

\frac{DP}{PR}=\frac{DC}{CR}  ( AD = DC \ as \ ABCD is a Rhombus)

 \Rightarrow DP \times CR = DC \times PR

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Question 9: Given FB=FD, AE \perp FD \ and \  FC \perp AD   . Prove \frac{GB}{AD}=\frac{BC}{ED}   s4

Answer:

FB=FD, AE \perp FD   and FC \perp AD   .

Therefore \angle FDB=\angle FBD 

Consider \triangle AED \ and \  \triangle FCB

\angle AED = \angle FCB= 90^o (Given)

\angle ADE = \angle FBC 

Therefore \triangle AED \sim \triangle FBC    (AAA postulate)

\frac{AD}{FB}=\frac{ED}{BC} 

\frac{FB}{AD}=\frac{BC}{ED} 

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Question 10: In \triangle ABC, \angle B = 2 \angle C   and the bisector of \angle B  meets CA  at point D  . Prove that  (i) \triangle ABC \sim \triangle ABD  (ii) DC:AD = BC:AB  s13.jpg

Answer:

Given \triangle ABC, \angle B = 2 \angle C  

\angle ABD = \angle DBC 

Therefore \angle ABD = \angle DBC = \angle ACB 

Consider \triangle ABC  \ and \  \triangle ABD

\angle BAC = \angle DAB  (Given)

\angle ACB = \angle ABD 

Therefore \triangle ABC \sim \triangle ABD    (AAA postulate)

\frac{BC}{BD}=\frac{AB}{AD} 

\frac{BC}{AB}=\frac{BD}{AD} 

\frac{BC}{AB}=\frac{DC}{AD}  ( \angle DBC = \angle DCB \Rightarrow DC = BD )

DC:AD = BC:AB 

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