CBSE Class 10 Mathematics chapter 6 - Triangle Exercise 6.3

 

Q1.	State which pairs of triangles in Fig. 6.34 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)
Ans.	Given, in ΔABC and ΔPQR, ∠A = ∠P = 60° ∠B = ∠Q = 80° ∠C = ∠R = 40° As per AAA similarity criterion "If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar." ∴ ΔABC ~ ΔPQR

(ii)

Ans.

Given, in  ΔABC and ΔPQR, $$\frac{AB}{QR}=\frac{2}{4}=\frac{1}{2}$$ $$\frac{BC}{RP}=\frac{2.5}{5}=\frac{1}{2}$$ $$\frac{CA}{PA}=\frac{3}{6}=\frac{1}{2}$$ As per SSS similarity criterion "If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar." Therefore ΔABC ~ ΔQRP

(iii)
Ans.	According to figure, in ΔLMP and ΔDEF, LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6 $$\frac{MP}{DE}=\frac{2}{4}=\frac{1}{2}$$ $$\frac{PL}{DF}=\frac{3}{6}=\frac{1}{2}$$ $$\frac{LM}{EF}=\frac{2.7}{5}=\frac{27}{50}$$ Therefore $$\frac{MP}{DE}\neq\frac{PL}{DF}\neq\frac{LM}{EF}$$ Therefore, ΔLMP and ΔDEF are not similar.

(iv)
Ans.	According to figure, In ΔMNL and ΔQPR $$\frac{MN}{PQ}=\frac{2.5}{5}=\frac{25}{50}=\frac{1}{2}$$ $$\frac{ML}{QR}=\frac{5}{10}=\frac{1}{2}$$ ∠M = ∠Q = 70° As per SAS similarity criterion "If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar." Therefore,  ∴ ΔMNL ~ ΔQPR

(v)
Ans.	According to figure, In ΔABC and ΔDEF AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80° $$\frac{AB}{DF}=\frac{2.5}{5}=\frac{25}{50}=\frac{1}{2}$$ $$\frac{BC}{EF}=\frac{3}{6}=\frac{1}{2}$$ Therefore, ∠B ≠ ∠F Hence, ΔABC and ΔDEF are not similar.

(vi)

Ans.

We know that sum of angle in a triangle equal to 180°. so, In ΔDEF, $$\angle{D}+\angle{E}+\angle{F}=180^{0}$$ Putting value $$\Rightarrow\angle{70^{0}}+\angle{80^{0}}+\angle{F}=180^{0}$$ $$\Rightarrow\angle{F}=180^{0}-\angle{70^{0}}+\angle{80^{0}}$$ $$\Rightarrow\angle{F}=\angle{30^{0}}$$ Similarly, In ΔPQR, $$\angle{P}+\angle{Q}+\angle{R}=180^{0}$$ Putting value $$\Rightarrow\angle{P}+\angle{80^{0}}+\angle{30^{0}}=180^{0}$$ $$\Rightarrow\angle{P}=180^{0}-\angle{80^{0}}+\angle{30^{0}}$$ $$\Rightarrow\angle{P}=\angle{70^{0}}$$ Now, comparing both the triangles, ΔDEF and ΔPQR, we have ∠D = ∠P = 70° ∠F = ∠Q = 80° ∠F = ∠R = 30° Therefore, by AAA similarity criterion - "If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar." Hence, ΔDEF ~ ΔPQR

Q2.

In Fig. 6.35, ΔODC ~ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Ans.

Given: ∠BOC = 125° ∠CDO = 70° and  ΔODC ~ ΔOBA As per given figure, DOB is a straight line. Therefore $$\angle{DOC}+\angle{COB}=180^{0}$$ Putting Value $$\Rightarrow\angle{DOC}+\angle{125^{0}}=180^{0}$$ $$\Rightarrow\angle{DOC}=180^{0}-125^{0}$$ $$\Rightarrow\angle{DOC}=55^{0}$$ We know that sum of the measures of the angles of a triangle is 180º, therefore In ΔDOC, $$\angle{DCO}+\angle{CDO}+\angle{DOC}=180^{0}$$ $$\Rightarrow\angle{DCO}+70^{0}+55^{0}=180^{0}$$ $$\Rightarrow\angle{DCO}=55^{0}$$ It is given that, ΔODC ~ ΔOBA, Therefore, we can say ΔODC similar to ΔOBA. Hence, Corresponding angles are equal in similar triangles $$\angle{OAB}=\angle{OCD}$$ $$\Rightarrow\angle{OAB}=55^{0}$$

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 $$\frac{OA}{OC}=\frac{OB}{OD}$$
Ans.
	Given: Diagonals AC and BD and AB || DC Construct a trapezium as per given As per construction, In ΔDOC and ΔBOA, alternate interior angles will be equal because AB || DC Therefore ∠CDO = ∠ABO Similarly, ∠DCO = ∠BAO Triangles ΔDOC and ΔBOA are vertically opposite angles will be equal; Therefore, ∠DOC = ∠BOA Hence, by AAA similarity criterion, ΔDOC ~ ΔBOA

Q4.

In Fig. 6.36, $$\frac{QR}{QS}=\frac{QT}{PR}$$ , ∠1= ∠2, show that ΔPQS ~ΔTQR

Ans.

Given: In ΔPQR,∠PQR = ∠PRQ Given, $$\frac{QR}{QS}=\frac{QT}{PR}$$ We have to prove that ΔPQS ~ ΔTQR Proof: We are given that In ΔPQR,∠PQR = ∠PRQ Therefore,PQ = PR .........(i) As per given $$\frac{QR}{QS}=\frac{QT}{PR}$$ We can write it as by using equation (i) $$\frac{QR}{QS}=\frac{QT}{QP}.................(ii)$$ This means that ΔPQS ~ ΔTQR are similar by the SAS similarity criterion because we have: Side PQ = PR (Common side) Angle Q = Angle Q (by given proportionality) ∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]

Q5.	S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.
Ans.	Given: S and T are points on sides PR and QR of ΔPQR. ∠P=∠RTS We have to prove that:ΔRPQ ~ ΔRTS. Proof: We are given that In ΔRPQ and ΔRTS ∠P = ∠RTS Angle R is Common angle, therefore ∠R = ∠R Hence, by AA Similarity criterion :ΔRPQ ~ ΔRTS. Hence proved.

Q6.

In Fig. 6.37, if ΔABE ≅ ΔACD Show that ΔADE ~ ΔABC.

Ans.

Given: ΔABE ≅ ΔACD Proof: As per given ΔABE ≅ ΔACD We know that if two triangles are congruent, then their corresponding parts (angles and sides) are also congruent. Therefore, AB = AC .................. (i) and AD = AE .............  (ii) In ΔADE and ΔABC, Divide equation(ii) by equation (i), we get $$\frac{AD}{AB}=\frac{AE}{AC}$$ and ∠A = ∠A  (Common angle) Therefore by using SAS similarity criterion, we can say that ΔADE ~ ΔABC Hence proved.

Q7.	In Fig. 6.38, altitude AD and CE of ΔABC intersect each other at the point P. Show that: (i) ΔAEP ~ ΔCDP. (ii) ΔABD ~ ΔCBE. (iii) ΔAEP ~ ΔADB. (iv) ΔPDC ~ ΔBEC.
Ans.	Given: altitude AD and CE of ΔABC intersect each other at the point P

(i)
	In ΔAEP and ΔCDP, Both triangle share a common angle which formed by altitude ∠AEP = ∠CDP (both are right angles) ∠APE = ∠CPD (Vertically opposite angles) Hence, by using AA similarity criterion we can say that ΔAEP ~ ΔCDP Hence proved.

(ii)
	In ΔABD and ΔCEB, Both triangle share a common angle ∠ADB = ∠CEB (both are right angles) ∠ABD = ∠CBE (Common angles) Hence, by using AA similarity criterion we can say that ΔABD ~ ΔCEB Hence proved.

(iii)
	In ΔAEP ~ ΔADB, Both triangle share a common angle which formed by altitude ∠AEP = ∠ADB (both are right angles.) ∠PAE = ∠DAB (Common angles) Hence, by using AA similarity criterion we can say that ΔAEP ~ ΔADB Hence proved.

(iv)
	In ΔPDC ~ ΔBEC, Both triangle share a common angle ∠PDC = ∠BEC (both are right angles) ∠PCD = ∠BCE (Common angles) Hence, by using AA similarity criterion we can say that ΔPDC ~ ΔBEC Hence proved.

Q8.	E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
Ans.	Given: E is a point on the side AD BE intersects CD at F Proof: In ΔABE and ΔCFB, ∠EAB = ∠FCB Both angles are vertically opposite angles and are therefore congruent. ∠AEB = ∠CBF Both angles are alternate interior angles when AE || BC Therefore, by AA similarity criterion, we can say ΔABE ~ ΔCFB Hence proved.

Q9.

In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) ΔABC ~ ΔAMP (ii) CA/PA = BC/MP

Ans.

(i) Given:∠B = ∠M are right angle. ΔABC and ΔAMP are right angle triangle Proof: In given ΔABC and ΔAMP, ∠CAB and ∠MAP are common angles in both triangle, so ∠CAB = ∠MAP ∠ABC and ∠AMP are right angles, ∠ABC = ∠AMP = 90° Therefore, by AA similarity criterion, we can say ΔABC ~ ΔAMP Hence proved. (ii) As per AA similarity criterion"If two triangles are similar then the corresponding sides are always equal" So, in ΔABC and ΔAMP  ∠ABC = ∠AMP = 90° Hence, $$\frac{CA}{PA}=\frac{BC}{MP}$$ Its proved.

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) CD/GH = AC/FG (ii) ΔDCB ~ ΔHGE (iii) ΔDCA ~ ΔHGF

Ans.

(i) As per the given condition, ΔABC ~ ΔFEG. ∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE Since, ∠ACB = ∠FGE Therefore, by Angle bisector theorem We can say ∠ACD = ∠FGH And, ∠DCB = ∠HGE In ΔACD and ΔFGH, ∠A = ∠F ∠ACD = ∠FGH Therefore, by AA similarity criterion, we can say ΔACD ~ ΔFGH Therefore, $$\Rightarrow\frac{CD}{GH}=\frac{AC}{FG}$$ Hence proved. (ii) As per AA similarity criterion"If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar" In ΔDCB and ΔHGE, ∠DCB = ∠HGE (As Already proved) ∠B = ∠E (As already proved) Therefore,  ΔDCB ~ ΔHGE Hence proved. (iii) As per AA similarity criterion"If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar" In ΔDCA and ΔHGF, ∠ACD = ∠FGH (As already proved) ∠A = ∠F (As already proved) Therefore,  ΔDCA ~ ΔHGF Hence proved.

Q11.	In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF
Ans.	As per AA similarity criterion"If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar" Given Triangle is an isosceles triangle. Therefore, AB = AC ⇒ ∠ABD = ∠ECF In ΔABD and ΔECF, ∠ADB = ∠EFC (Each right angle) ∠BAD = ∠CEF Therefore,  ΔABD ~ ΔECF

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 Fig. 6.41). Show that ΔABC ~ ΔPQR.

Ans.

Given, in ΔABC and ΔPQR AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR. Proof: Let suppose that in ΔABC, $$AB=k_{1}$$ $$BC=k_{2}$$ $$AC=k_{3}$$ $$AD=m_{1}$$   and  Let suppose that in ΔPQR, $$AB=l_{1}$$ $$BC=l_{2}$$ $$AC=l_{3}$$ $$AD=m_{2}$$   As per given, we can express this relationship as follow- $$\frac{k_{1}}{l_{1}}=\frac{k_{2}}{l_{2}}=\frac{k_{3}}{l_{3}}=\frac{m_{1}}{m_{2}}$$ Take Ratio  $$\frac{k_{1}}{l_{1}}=\frac{k_{2}}{l_{2}}$$ since, they are equal. This mean - $$\frac{k_{1}}{k_{2}}=\frac{l_{1}}{l_{2}}$$ $$\frac{k_{2}}{k_{3}}=\frac{l_{2}}{l_{3}}$$ $$\frac{k_{3}}{k_{1}}=\frac{l_{3}}{l_{1}}$$ These relationships mean that the ratios of corresponding sides of ΔABC and ΔPQR are equal. Therefore, ΔABC ~ ΔPQR (by SSS similarity criterion) Hence Proved.

Q13.	D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2= CB.CD.
Ans.	Given: ∠ADC = ∠BAC D is a point on the side BC of a triangle ABC Proof: In ΔADC and ΔBAC, As per given ∠ADC = ∠BAC ∠BCA = ∠ACD (by common angles) Therefore, by AA similarity criterion, we can say ΔADC ~ ΔBAC We know that corresponding sides of similar triangles are in proportion. Therefore, $$\frac{CA}{CB}=\frac{CD}{CA}$$ by cross multiply, we get $$\Rightarrow CA^{2}=CB.CD$$ Hence, proved.

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 ΔABC ~ ΔPQR.

Ans.

Given: Two triangle - ΔABC in which AB and AC are sides and AD is median and in ΔPQR,  PQ and PR are sides and PM is median In ΔABC and ΔPQR, $$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$$ To Prove: We have to prove that ΔABC ~ ΔPQR Construction: Produce  AD to E such that AD=DE. Join CE. Produce PM to N such that PM=MN. Join RN. Proof: Step 1: In ΔABD and ΔCDE: By Construction we can write as AD = DE Since AD is the median, so BD = DC and  ∠ADB =∠CDE (Vertically Opposite Angles) By SAS criterion of congruence ΔABD ≅ ΔCDE By Corresponding parts of congruent triangles ⇒AB = CE Step 2: In ΔPQM and ΔMNR: By Construction we can write as PM = MN Since PM is the median, so QM = MR and  ∠PMQ =∠NMR (Vertically Opposite Angles) By SAS criterion of congruence ΔPQM ≅ ΔMNR By Corresponding parts of congruent triangles ⇒PQ = RN Step 3: From given  $$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$$ By using the result from step 1 and 2 $$\Rightarrow\frac{CE}{RN}=\frac{AC}{PR}=\frac{AD}{PM}$$ Since AD=DE and PM=MN), so $$\Rightarrow\frac{CE}{RN}=\frac{AC}{PR}=\frac{2AD}{2PM}$$ Since 2AD=AE and 2PM=PN), so $$\Rightarrow\frac{CE}{RN}=\frac{AC}{PR}=\frac{AE}{PN}$$ By SSS similarity criterion we can say ΔACE ~ ΔPRN Step4: Because ΔACE ~ ΔPRN, Therefore, Corresponding angles are equal ∠2 =∠4 ∠1 =∠3 Therefore, ∠1+∠2 =∠3 +∠4 ⇒∠A =∠P (from the similarity criteria ΔACE ~ ΔPRN) Hence ΔABC ~ ΔPQR are similar by SAS similarity criteria. Its Proved.

Q15.

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

Ans.

Given: Length (Height) of pole : 6 m Shadow of pole : 4 m Shadow of tower : 28 m Let height of tower : h m In ΔPQR and ΔABC ∠Q =∠B (are right angle) ∠R =∠C (by angular elevation of sum properties of triangle) we know that as per AA Similarity criteria If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar. Therefore, ΔPQR ~ ΔABC We also know that if two triangle are similar then their corresponding sides are prepositional. Therefore $$\Rightarrow\frac{PQ}{AB}=\frac{QR}{BC}$$ Putting value $$\Rightarrow\frac{6}{h}=\frac{4}{28}$$ $$\Rightarrow h=\frac{6\times28}{4}$$ $$\Rightarrow h=6\times7$$ $$\Rightarrow h=42m$$ Hence, Height of Tower is 42 m.

Q16.

If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, prove that $$\frac{AB}{PQ}=\frac{AD}{PM}$$

Ans.

Given: AD and PM are medians of triangles ABC and PQR. ΔABC ~ ΔPQR To Prove: $$\frac{AB}{PQ}=\frac{AD}{PM}$$ Proof: We know that the corresponding sides of similar triangle are in same ratio and their corresponding angles are similar, therefore $$\Rightarrow\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}................(i)$$ $$\angle{A}=\angle{P},\angle{B}=\angle{Q},\angle{C}=\angle{R}...........(ii)$$ Medians AD and PM divide their opposite sides, therefore $$\Rightarrow BD=\frac{BC}{2} and QM = \frac{QR}{2}................(iii)$$ From (i) and (iii) we get $$\Rightarrow \frac{AB}{PQ} and  \frac{BD}{QM}.................(iv)$$ From (iii) and (iv) $$\Rightarrow \frac{AB}{PQ} and  \frac{BD}{QM}$$ By SAS similarity criteria we can say ΔABD ~ ΔPQM Therefore, $$\frac{AB}{PQ}=\frac{BD}{QM}=\frac{DA}{PM}$$ Hance proved.