NCERT Solutions For Class 10 Math Chapter – 6 Exercise – 6.3

NCERT Solutions For Class 10 Math Chapter – 6 Exercise – 6.3

Q1. State which pairs of triangles in 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:

Solution:-

∆ABC ∆ [AAA similarity]
AB/QR = BC/PR = AC/PQ = 1/2

∆ABC ∆QRP. [ SSS Similarity]

Not similar as sides are not in proportional.
Not similar SAS not satisfied.
Not similar.
Similar as ∆DEF ∆PQR. [ AAA Similarity]

Q2. In figure 6.35 ∆ODC ∆OBA , ∟BOC = 125⁰ and ∟CDO = 70⁰ Find ∟DOC , ∟DCO , ∟OAB.

Solution:-

In the given figure

∟DOC = 180⁰ – ∟COB [ ∟DOC and ∟COB from a linear pair]

∟DOC = 180⁰ – 125⁰

∟DOC = 55⁰

In ∆ODC

∟DCO = 180⁰ – (∟DOC + ∟ODC )

∟DCO = 55⁰

In ∆ODC and ∆OBA

∆ODC – ∆OBA

= ∟DCO – ∟OAB

= ∟OAB = 55⁰

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 OA/OC = OB/OD.

Solution:-

In ∆AOB and ∆COD

∟AOB = ∟COD (Vertically opposite angles)

∟BAO = ∟DCO (alternate interior angle)

∆AOB ∆COD (AA criterion)

Hence , OA/OC = OB/OD

Q4. In figures QR/QS = QT/PR and ∟1 = ∟2 Show that ∆PQS ∆TQR.

Solution:-

In ∆PQR

∟1 = ∟2

= PR = PQ (In a triangle side opposite to equal angles are equal)

In ∆PQS and ∆TQR

∟PQS = ∟TQR = ∟1 (Same angle)

QR/QS = QT/PQ (PR = PQ)

= ∆PQS ∆TQR (SAS criterion)

Q5. S and T are points on sides PR and QR of ∆PQR such that ∟P = ∟RTS Show that ∆RPQ ∆RTS.

Solution:-

In ∆RPQ , ∆RTS

∟RPQ = ∟RTS (Given)

∟PRQ = ∟TRS (Common angle)

= ∆RPQ ∆RTS (AA criterion)

Q6. In Figure If ∆ABE ∆ACD show that ∆ABE ∆ ABC.

Solution:-

In ∆ABE and ∆ABC

AE = AD

AB = AC

Now Consider ∆ADE , ∆ABC

AD/AB = AE/AC

And ∟DAE = ∟BAC (Common angle)

= ∆ADE ∆ABC (SAS criterion)

Q7. In figure 6.38 altitudes AD and CE of ∆ABC intersect each other at the point P. Show that:

∆AEP ∆CDP
∆ABD ∆CBE
∆AEP ∆ADB
∆PDC ∆BEC

Solution:-

In ∆AEP and ∆CDP

∟AEP = ∟CDP = 90⁰

∟APE = ∟CPD (Vertically opposite angles)

= ∆AEP ∆CPD (AA criterion)

(ii) In ∆ABD and ∆CBE

∟ADB = ∟CEB = 90⁰

∟ABD = ∟CBE ( Common angles)

= ∆ABD ∆ CBE ( AA criterion)

(iii) In ∆AEP and ∆ADB

∟AEP = ∟ADB =90⁰

∟PAE = ∟BAD (Common angle)

= ∆AEP ∆ADB

In ∆PDC and ∆BEC

∟PDC = ∟BEC = 90⁰

∟PCD = ∟BCE ( Common angle)

= ∆PDC ∆BEC

Q8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ∆CFB.

Solution:-

In ∆ABE and ∆CFB

∟BAE = ∟FCB (Opposite angles of a parallelogram)

∟AEB = ∟FBC

= ∆ABE ∆CFE (AA Criterion)

Q9. In figure 6.39 ABC and AMP are two right triangles right angled at B and M respectively.

∆ABC ∆AMP
CA/PA = BC/MP

Solution:-

In ∆ABC and ∆AMP

∟ABC = ∟AMP = 90⁰

∟BAC = ∟MAP (Common angles)

= ∆ABC ∆AMP

(II) In ∆ABC and ∆AMP

CA/PA = BC/MP

Q10. CD and GH are respectively the bisectors of ∟ACB and ∟EFG such that D and H lie on sides AB and FE of ∆ABC and ∆EFG such that D and H lie on side AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ∆FEG show that:

CD/GH = AC/FG
∆DCB ∆HGE
∆DCA ∆HGF

Solution:-

Given, 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.

(i) From the given condition,

ΔABC ~ ΔFEG.

∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE

Since, ∠ACB = ∠FGE

∴ ∠ACD = ∠FGH (Angle bisector)

And, ∠DCB = ∠HGE (Angle bisector)

In ΔACD and ΔFGH,

∠A = ∠F

∠ACD = ∠FGH

∴ ΔACD ~ ΔFGH (AA similarity criterion)

⇒CD/GH = AC/FG

(ii) In ΔDCB and ΔHGE,

∠DCB = ∠HGE (Already proved)

∠B = ∠E (Already proved)

∴ ΔDCB ~ ΔHGE (AA similarity criterion)

(iii) In ΔDCA and ΔHGF,

∠ACD = ∠FGH (Already proved)

∠A = ∠F (Already proved)

∴ ΔDCA ~ ΔHGF (AA similarity criterion)

In the following figure, 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.

Ncert solutions class 10 chapter 6-21

Solution:

Given, ABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ABD = ∠ECF

In ΔABD and ΔECF,

∠ADB = ∠EFC (Each 90°)

∠BAD = ∠CEF (Already proved)

∴ ΔABD ~ ΔECF (using AA similarity criterion)

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.

Ncert solutions class 10 chapter 6-22

Solution:

Given, ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR

i.e. AB/PQ = BC/QR = AD/PM

We have to prove: ΔABC ~ ΔPQR

As we know here,

AB/PQ = BC/QR = AD/PM

⇒AB/PQ = BC/QR = AD/PM (D is the midpoint of BC. M is the midpoint of QR)

⇒ ΔABD ~ ΔPQM [SSS similarity criterion]

∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]

⇒ ∠ABC = ∠PQR

In ΔABC and ΔPQR

AB/PQ = BC/QR ………………………….(i)

∠ABC = ∠PQR ……………………………(ii)

From equation (i) and (ii), we get,

ΔABC ~ ΔPQR [SAS similarity criterion]

D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA2= CB.CD

Solution:

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

In ΔADC and ΔBAC,

∠ADC = ∠BAC (Already given)

∠ACD = ∠BCA (Common angles)

∴ ΔADC ~ ΔBAC (AA similarity criterion)

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

∴ CA/CB = CD/CA

⇒ CA2 = CB.CD.

Hence, proved.

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.

Solution:

Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that;

AB/PQ = AC/PR = AD/PM

We have to prove, ΔABC ~ ΔPQR

Let us construct first: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.

In ΔABD and ΔCDE, we have

AD = DE [By Construction.]

BD = DC [Since, AP is the median]

and, ∠ADB = ∠CDE [Vertically opposite angles]

∴ ΔABD ≅ ΔCDE [SAS criterion of congruence]

⇒ AB = CE [By CPCT] …………………………..(i)

Also, in ΔPQM and ΔMNR,

PM = MN [By Construction.]

QM = MR [Since, PM is the median]

and, ∠PMQ = ∠NMR [Vertically opposite angles]

∴ ΔPQM = ΔMNR [SAS criterion of congruence]

⇒ PQ = RN [CPCT] ………………………………(ii)

Now, AB/PQ = AC/PR = AD/PM

From equation (i) and (ii),

⇒CE/RN = AC/PR = AD/PM

⇒ CE/RN = AC/PR = 2AD/2PM

⇒ CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]

∴ ΔACE ~ ΔPRN [SSS similarity criterion]

Therefore, ∠2 = ∠4

Similarly, ∠1 = ∠3

∴ ∠1 + ∠2 = ∠3 + ∠4

⇒ ∠A = ∠P …………………………………………….(iii)

Now, in ΔABC and ΔPQR, we have

AB/PQ = AC/PR (Already given)

From equation (iii),

∠A = ∠P

∴ ΔABC ~ ΔPQR [ SAS similarity criterion]

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.

Solution:

Given, Length of the vertical pole = 6m

Shadow of the pole = 4 m

Let Height of tower = h m

Length of shadow of the tower = 28 m

In ΔABC and ΔDEF,

∠C = ∠E (angular elevation of sum)

∠B = ∠F = 90°

∴ ΔABC ~ ΔDEF (AA similarity criterion)

∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional)

∴ 6/h = 4/28

⇒h = (6×28)/4

⇒ h = 6 × 7

⇒ h = 42 m

Hence, the height of the tower is 42 m.

If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM.

Solution:

Given, ΔABC ~ ΔPQR

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

∴AB/PQ = AC/PR = BC/QR……………………………(i)

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ………….…..(ii)

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

∴ BD = BC/2 and QM = QR/2 ……………..………….(iii)

From equations (i) and (iii), we get

AB/PQ = BD/QM ……………………….(iv)

In ΔABD and ΔPQM,

From equation (ii), we have

∠B = ∠Q

From equation (iv), we have,

AB/PQ = BD/QM

∴ ΔABD ~ ΔPQM (SAS similarity criterion)

⇒AB/PQ = BD/QM = AD/PM