Class 10: Similarity – Exercise 16(a-1) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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:

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: Given $\angle GHE = \angle DFE= 90^o$ , $DH= 8, DF=12,$ $DG=3x-1$ 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: sm10 In 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:

(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: Given $ABCD$ 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 $\frac{GB}{AD}=\frac{BC}{ED}$

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) $DC:AD = BC:AB$

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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Class 10: Similarity – Exercise 16(a-1)

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