Mathematics · Triangles · NCERT Exercises

Exercise 6.2

Complete, independently verified solutions for NCERT Exercise 6.2.

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Mathematics · Chapter 6 · NCERT Exercises

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Verified NCERT questions and solutionsSolutions are checked independently and follow the supplied NCERT chapter edition.

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Question

Question 1

basic proportionality theorem · numerical geometry

In Figure 6.17 (i) and (ii), DE ∥ BC. Find EC in (i) and AD in (ii).

Two triangles for Exercise 6.2 Question 1, each with DE parallel to BC and the required lengths labelled.
Figure 6-17 from the supplied NCERT chapter.

Solution

Given:

In each diagram, DE is parallel to BC and the labelled segment lengths are known.

To find:

EC in part (i) and AD in part (ii).

Two triangles for Exercise 6.2 Question 1, each with DE parallel to BC and the required lengths labelled.
Figure 6-17 from the supplied NCERT chapter.

Solution:

(i) In △ABC, DE ∥ BC. By BPT:

ADDB=AEEC\frac{AD}{DB}=\frac{AE}{EC}

Substituting AD = 1.5 cm, DB = 3 cm and AE = 1 cm:

1.53=1EC\frac{1.5}{3}=\frac{1}{EC}
12=1EC\frac12=\frac{1}{EC}
EC=2 cmEC=2\text{ cm}

(ii) Again, by BPT:

ADDB=AEEC\frac{AD}{DB}=\frac{AE}{EC}

Substituting DB = 7.2 cm, AE = 1.8 cm and EC = 5.4 cm:

AD7.2=1.85.4\frac{AD}{7.2}=\frac{1.8}{5.4}
AD7.2=13\frac{AD}{7.2}=\frac13
AD=7.23=2.4 cmAD=\frac{7.2}{3}=2.4\text{ cm}

Final answer

(i) EC = 2 cm; (ii) AD = 2.4 cm.

Question

Question 2

converse of BPT · parallel lines · ratios

E and F are points on sides PQ and PR respectively of △PQR. State whether EF ∥ QR in each case:

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Solution

Given:

E lies on PQ and F lies on PR in triangle PQR; three different sets of segment lengths are supplied.

To answer:

For each part, determine whether EF is parallel to QR.

Converse of the Basic Proportionality TheoremTriangle PQR with E on PQ, F on PR and segment EF; compare PE to EQ and PF to FR.PQREF
Converse of the Basic Proportionality Theorem — solution sketch (not to scale).

Solution:

By the converse of BPT, EF ∥ QR when:

PEEQ=PFFR\frac{PE}{EQ}=\frac{PF}{FR}

(i)

PEEQ=3.93=1.3\frac{PE}{EQ}=\frac{3.9}{3}=1.3
PFFR=3.62.4=1.5\frac{PF}{FR}=\frac{3.6}{2.4}=1.5

The ratios are unequal. Therefore, EF is not parallel to QR.

(ii)

PEEQ=44.5=89\frac{PE}{EQ}=\frac{4}{4.5}=\frac89
PFFR=89\frac{PF}{FR}=\frac89

The ratios are equal. Therefore, EF ∥ QR.

(iii)

EQ=PQ−PE=1.28−0.18=1.10 cmEQ=PQ-PE=1.28-0.18=1.10\text{ cm}
FR=PR−PF=2.56−0.36=2.20 cmFR=PR-PF=2.56-0.36=2.20\text{ cm}
PEEQ=0.181.10=955\frac{PE}{EQ}=\frac{0.18}{1.10}=\frac9{55}
PFFR=0.362.20=955\frac{PF}{FR}=\frac{0.36}{2.20}=\frac9{55}

The ratios are equal. Therefore, EF ∥ QR.

Final answer

(i) No; (ii) Yes; (iii) Yes.

Question

Question 3

basic proportionality theorem · parallel lines · proof

In Figure 6.18, LM ∥ CB and LN ∥ CD. Prove that AM/AB = AN/AD.

Points M and B on ray AB, N and D on ray AD, and A, L, C collinear, with LM parallel to CB and LN parallel to CD.
Figure 6-18 from the supplied NCERT chapter.

Solution

Given:

LM is parallel to CB and LN is parallel to CD in Figure 6.18.

To prove:

AM/AB = AN/AD.

Points M and B on ray AB, N and D on ray AD, and A, L, C collinear, with LM parallel to CB and LN parallel to CD.
Figure 6-18 from the supplied NCERT chapter.

Proof:

In △ABC, LM ∥ CB. By Example 1 (the BPT result using whole sides):

AMAB=ALAC...(1)\frac{AM}{AB}=\frac{AL}{AC}\qquad ...(1)

In △ADC, LN ∥ CD. By the same result:

ANAD=ALAC...(2)\frac{AN}{AD}=\frac{AL}{AC}\qquad ...(2)

The right-hand sides of (1) and (2) are equal.

Therefore:

AMAB=ANAD\frac{AM}{AB}=\frac{AN}{AD}

Final answer

Therefore, AM/AB = AN/AD.

Question

Question 4

basic proportionality theorem · parallel lines · proof

In Figure 6.19, DE ∥ AC and DF ∥ AE. Prove that BF/FE = BE/EC.

Triangle ABC with D on AB and F and E on BC, where DE is parallel to AC and DF is parallel to AE.
Figure 6-19 from the supplied NCERT chapter.

Solution

Given:

DE is parallel to AC and DF is parallel to AE in Figure 6.19.

To prove:

BF/FE = BE/EC.

Triangle ABC with D on AB and F and E on BC, where DE is parallel to AC and DF is parallel to AE.
Figure 6-19 from the supplied NCERT chapter.

Proof:

In △BAE, DF ∥ AE. By BPT:

BDDA=BFFE...(1)\frac{BD}{DA}=\frac{BF}{FE}\qquad ...(1)

In △BAC, DE ∥ AC. By BPT:

BDDA=BEEC...(2)\frac{BD}{DA}=\frac{BE}{EC}\qquad ...(2)

The left-hand sides of (1) and (2) are equal.

Therefore:

BFFE=BEEC\frac{BF}{FE}=\frac{BE}{EC}

Final answer

Therefore, BF/FE = BE/EC.

Question

Question 5

basic proportionality theorem · converse of BPT · proof

In Figure 6.20, DE ∥ OQ and DF ∥ OR. Show that EF ∥ QR.

Triangle PQR with E on PQ, F on PR, D on PO, and segments DE, DF, EF and OQ, OR shown.
Figure 6-20 from the supplied NCERT chapter.

Solution

Given:

DE is parallel to OQ and DF is parallel to OR in Figure 6.20.

To prove:

EF is parallel to QR.

Triangle PQR with E on PQ, F on PR, D on PO, and segments DE, DF, EF and OQ, OR shown.
Figure 6-20 from the supplied NCERT chapter.

Proof:

In △POQ, DE ∥ OQ. By BPT:

PEEQ=PDDO...(1)\frac{PE}{EQ}=\frac{PD}{DO}\qquad ...(1)

In △POR, DF ∥ OR. By BPT:

PFFR=PDDO...(2)\frac{PF}{FR}=\frac{PD}{DO}\qquad ...(2)

From (1) and (2):

PEEQ=PFFR\frac{PE}{EQ}=\frac{PF}{FR}

Thus E and F divide sides PQ and PR in the same ratio.

Therefore, EF ∥ QR by the converse of BPT.

Final answer

Therefore, EF ∥ QR.

Question

Question 6

basic proportionality theorem · converse of BPT · proof

In Figure 6.21, A, B and C are points on OP, OQ and OR respectively such that AB ∥ PQ and AC ∥ PR. Show that BC ∥ QR.

Triangle PQR with A on OP, B on OQ and C on OR, with AB parallel to PQ and AC parallel to PR.
Figure 6-21 from the supplied NCERT chapter.

Solution

Given:

A, B and C lie on OP, OQ and OR; AB is parallel to PQ and AC is parallel to PR.

To prove:

BC is parallel to QR.

Triangle PQR with A on OP, B on OQ and C on OR, with AB parallel to PQ and AC parallel to PR.
Figure 6-21 from the supplied NCERT chapter.

Proof:

In △OPQ, AB ∥ PQ. By BPT:

OAAP=OBBQ...(1)\frac{OA}{AP}=\frac{OB}{BQ}\qquad ...(1)

In △OPR, AC ∥ PR. By BPT:

OAAP=OCCR...(2)\frac{OA}{AP}=\frac{OC}{CR}\qquad ...(2)

From (1) and (2):

OBBQ=OCCR\frac{OB}{BQ}=\frac{OC}{CR}

Therefore B and C divide OQ and OR in the same ratio.

Hence BC ∥ QR by the converse of BPT.

Final answer

Therefore, BC ∥ QR.

Question

Question 7

basic proportionality theorem · midpoint theorem · proof

Using Theorem 6.1, prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side.

Solution

Given:

D is the midpoint of AB in triangle ABC.

To prove:

A line through D parallel to BC bisects AC.

Construction:

Draw DE parallel to BC, meeting AC at E.

Midpoint theorem constructionTriangle ABC with D at the midpoint of AB and E at the midpoint of AC, joined by DE.ABCDE
Midpoint theorem construction — solution sketch (not to scale).

Proof:

Since D is the midpoint of AB, AD = DB. Therefore:

ADDB=1\frac{AD}{DB}=1

Because DE ∥ BC, BPT gives:

ADDB=AEEC\frac{AD}{DB}=\frac{AE}{EC}

Therefore:

AEEC=1\frac{AE}{EC}=1
AE=ECAE=EC

Hence E is the midpoint of AC, and the parallel line bisects the third side.

Final answer

The line parallel to one side bisects the third side.

Question

Question 8

converse of BPT · midpoint theorem · proof

Using Theorem 6.2, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side.

Solution

Given:

D and E are the midpoints of AB and AC in triangle ABC.

To prove:

DE is parallel to BC.

Construction:

Join the midpoints D and E.

Midpoint theorem constructionTriangle ABC with D at the midpoint of AB and E at the midpoint of AC, joined by DE.ABCDE
Midpoint theorem construction — solution sketch (not to scale).

Proof:

Since D is the midpoint of AB, AD = DB. Therefore:

ADDB=1\frac{AD}{DB}=1

Since E is the midpoint of AC, AE = EC. Therefore:

AEEC=1\frac{AE}{EC}=1

Thus:

ADDB=AEEC\frac{AD}{DB}=\frac{AE}{EC}

Therefore, DE ∥ BC by the converse of BPT.

Final answer

The segment joining the two midpoints is parallel to the third side.

Question

Question 9

AA similarity · trapezium · diagonals · proof

ABCD is a trapezium in which AB ∥ DC, and its diagonals intersect at O. Show that AO/BO = CO/DO.

Solution

Given:

ABCD is a trapezium, AB is parallel to DC, and diagonals AC and BD meet at O.

To prove:

AO/BO = CO/DO.

Trapezium with intersecting diagonalsTrapezium ABCD has AB parallel to DC; diagonals AC and BD intersect at O.ABCDO
Trapezium with intersecting diagonals — solution sketch (not to scale).

Proof:

Consider △AOB and △COD.

∠AOB=∠COD\angle AOB=\angle COD

Vertically opposite angles.

∠ABO=∠CDO\angle ABO=\angle CDO

Alternate interior angles, since AB is parallel to DC and BD is a transversal.

Therefore, △AOB ∼ △COD by AA similarity, with A ↔ C, O ↔ O and B ↔ D.

Corresponding sides are proportional:

AOCO=BODO\frac{AO}{CO}=\frac{BO}{DO}

Rearranging:

AOBO=CODO\frac{AO}{BO}=\frac{CO}{DO}

Final answer

Therefore, AO/BO = CO/DO.

Question

Question 10

SAS similarity · trapezium · converse · proof

The diagonals of quadrilateral ABCD intersect at O such that AO/BO = CO/DO. Show that ABCD is a trapezium.

Solution

Given:

Diagonals AC and BD of quadrilateral ABCD meet at O and AO/BO = CO/DO.

To prove:

AB is parallel to DC, so ABCD is a trapezium.

Trapezium with intersecting diagonalsTrapezium ABCD has AB parallel to DC; diagonals AC and BD intersect at O.ABCDO
Trapezium with intersecting diagonals — solution sketch (not to scale).

Proof:

AOBO=CODO\frac{AO}{BO}=\frac{CO}{DO}

Cross multiplying:

AO×DO=BO×COAO\times DO=BO\times CO

Dividing by CO × DO:

AOCO=BODO...(1)\frac{AO}{CO}=\frac{BO}{DO}\qquad ...(1)

Also, ∠AOB = ∠COD because they are vertically opposite angles. ...(2)

From (1) and (2), △AOB ∼ △COD by SAS similarity, with A ↔ C and B ↔ D.

Therefore, ∠ABO = ∠CDO as corresponding angles.

These are alternate interior angles on transversal BD, so AB ∥ DC.

Hence ABCD has one pair of opposite sides parallel and is a trapezium.

Final answer

Therefore, AB ∥ DC and ABCD is a trapezium.

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