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Mathematics · Triangles

NCERT Examples

All 8 worked examples from Triangles, with independently verified solutions and direct navigation.

Mathematics · Chapter 6 · NCERT Examples

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NCERT Example

Example 1

basic proportionality theorem · ratios · parallel lines

A line intersects sides AB and AC of △ABC at D and E respectively and is parallel to BC. Prove that AD/AB = AE/AC.

Triangle ABC with D on AB, E on AC, and line DE parallel to BC.
Figure 6-13 from the supplied NCERT chapter.

Solution

Board-exam working:

Given: DE ∥ BC in △ABC.

By the Basic Proportionality Theorem:

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

Taking reciprocals:

DBAD=ECAE\frac{DB}{AD}=\frac{EC}{AE}

Adding 1 to both sides:

DBAD+1=ECAE+1\frac{DB}{AD}+1=\frac{EC}{AE}+1
DB+ADAD=EC+AEAE\frac{DB+AD}{AD}=\frac{EC+AE}{AE}
ABAD=ACAE\frac{AB}{AD}=\frac{AC}{AE}

Taking reciprocals again:

ADAB=AEAC\frac{AD}{AB}=\frac{AE}{AC}

Final answer

Therefore, AD/AB = AE/AC.

NCERT Example

Example 2

basic proportionality theorem · trapezium · parallel lines

ABCD is a trapezium with AB ∥ DC. E and F are points on the non-parallel sides AD and BC respectively such that EF is parallel to AB. Show that AE/ED = BF/FC.

Trapezium ABCD with AB parallel to DC and segment EF parallel to both bases.
Figure 6-14 from the supplied NCERT chapter.

Solution

Board-exam working:

Construction: Join AC and let it intersect EF at G.

The same trapezium ABCD with diagonal AC meeting EF at G.
Figure 6-15 from the supplied NCERT chapter.

Since AB ∥ DC and EF ∥ AB, we also have EF ∥ DC.

In △ADC, EG ∥ DC. By BPT:

AEED=AGGC...(1)\frac{AE}{ED}=\frac{AG}{GC}\qquad ...(1)

In △CAB, GF ∥ AB. By BPT:

CGGA=CFFB\frac{CG}{GA}=\frac{CF}{FB}

Taking reciprocals:

AGGC=BFFC...(2)\frac{AG}{GC}=\frac{BF}{FC}\qquad ...(2)

From (1) and (2):

AEED=BFFC\frac{AE}{ED}=\frac{BF}{FC}

Final answer

Therefore, AE/ED = BF/FC.

NCERT Example

Example 3

converse of BPT · isosceles triangle · parallel lines

In Figure 6.16, PS/SQ = PT/TR and ∠PST = ∠PRQ. Prove that PQR is an isosceles triangle.

Triangle PQR with S on PQ, T on PR, segment ST, and the marked angles PST and PRQ.
Figure 6-16 from the supplied NCERT chapter.

Solution

Board-exam working:

Given:

PSSQ=PTTR\frac{PS}{SQ}=\frac{PT}{TR}

Since S and T divide PQ and PR in the same ratio, ST ∥ QR by the converse of BPT.

Therefore, ∠PST = ∠PQR because they are corresponding angles.

But ∠PST = ∠PRQ is given. Hence ∠PQR = ∠PRQ.

Sides opposite equal angles of a triangle are equal, so PR = PQ.

Final answer

Therefore, △PQR is isosceles with PQ = PR.

NCERT Example

Example 4

AAA similarity · parallel lines

In Figure 6.29, PQ ∥ RS. Prove that △POQ ∼ △SOR.

Lines PS and QR are parallel and diagonals PR and QS intersect at O.
Figure 6-29 from the supplied NCERT chapter.

Solution

Board-exam working:

Given: PQ ∥ RS.

∠OPQ = ∠OSR because they are alternate interior angles.

∠OQP = ∠ORS because they are alternate interior angles.

Also, ∠POQ = ∠SOR because they are vertically opposite angles.

Thus the correspondence is P ↔ S, O ↔ O and Q ↔ R.

Therefore, △POQ ∼ △SOR by the AAA similarity criterion.

Final answer

Therefore, △POQ ∼ △SOR.

NCERT Example

Example 5

SSS similarity · corresponding angles

Observe Figure 6.30 and find ∠P.

Triangles ABC and PQR with side lengths and angles marked for an SSS similarity calculation.
Figure 6-30 from the supplied NCERT chapter.

Solution

Board-exam working:

In △ABC and △RQP:

ABRQ=3.87.6=12\frac{AB}{RQ}=\frac{3.8}{7.6}=\frac12
BCQP=612=12\frac{BC}{QP}=\frac6{12}=\frac12
CAPR=3363=12\frac{CA}{PR}=\frac{3\sqrt3}{6\sqrt3}=\frac12

Thus:

ABRQ=BCQP=CAPR\frac{AB}{RQ}=\frac{BC}{QP}=\frac{CA}{PR}

Therefore, △ABC ∼ △RQP by SSS similarity.

The verified correspondence is A ↔ R, B ↔ Q and C ↔ P. Hence ∠P = ∠C.

By the angle-sum property, ∠C = 180° − 80° − 60° = 40°.

Final answer

∠P = 40°.

NCERT Example

Example 6

SAS similarity · vertically opposite angles

In Figure 6.31, OA × OB = OC × OD. Show that ∠A = ∠C and ∠B = ∠D.

Segments AB and CD intersect through O, forming triangles AOD and COB.
Figure 6-31 from the supplied NCERT chapter.

Solution

Board-exam working:

Given: OA × OB = OC × OD.

Dividing both sides by OC × OB:

OAOC=ODOB...(1)\frac{OA}{OC}=\frac{OD}{OB}\qquad ...(1)

Also, ∠AOD = ∠COB because they are vertically opposite angles. ...(2)

From (1) and (2), the two sides including the equal angle are proportional.

Therefore, △AOD ∼ △COB by SAS similarity, with A ↔ C and D ↔ B.

Corresponding angles of similar triangles are equal, so ∠A = ∠C and ∠D = ∠B.

Final answer

∠A = ∠C and ∠B = ∠D.

NCERT Example

Example 7

AA similarity · shadows · application

A girl of height 90 cm is walking away from the base of a lamp-post at 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.

A lamp-post AB, a girl CD, and the shadow DE along the same ground line.
Figure 6-32 from the supplied NCERT chapter.

Solution

Board-exam working:

Let DE = x m be the girl's shadow after 4 seconds.

Distance walked, BD = 1.2 × 4 = 4.8 m. Therefore, BE = BD + DE = 4.8 + x metres.

AB = 3.6 m and CD = 90 cm = 0.9 m.

In △ABE and △CDE, ∠B = ∠D = 90° and ∠E is common. Therefore, △ABE ∼ △CDE by AA similarity.

The correspondence is A ↔ C, B ↔ D and E ↔ E. Hence:

BEDE=ABCD\frac{BE}{DE}=\frac{AB}{CD}
4.8+xx=3.60.9=4\frac{4.8+x}{x}=\frac{3.6}{0.9}=4
4.8+x=4x4.8+x=4x
4x−x=4.84x-x=4.8
3x=4.83x=4.8
x=1.6x=1.6

Final answer

The girl's shadow is 1.6 m long.

NCERT Example

Example 8

similar triangles · medians · SAS similarity · SSS similarity

CM and RN are respectively the medians of △ABC and △PQR. If △ABC ∼ △PQR, prove that:

(i) △AMC ∼ △PNR

(ii) CM/RN = AB/PQ

(iii) △CMB ∼ △RNQ

Similar triangles ABC and PQR with medians CM and RN respectively.
Figure 6-33 from the supplied NCERT chapter.

Solution

Board-exam working:

Given △ABC ∼ △PQR, the correspondence is A ↔ P, B ↔ Q and C ↔ R.

Therefore, corresponding sides and angles are equal in ratio:

ABPQ=BCQR=CARP...(1)\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}\qquad ...(1)
∠A=∠P,∠B=∠Q,∠C=∠R\angle A=\angle P,\quad \angle B=\angle Q,\quad \angle C=\angle R

(i) Since CM and RN are medians:

AB=2AM,PQ=2PNAB=2AM,\qquad PQ=2PN
AMPN=ABPQ=CARP\frac{AM}{PN}=\frac{AB}{PQ}=\frac{CA}{RP}

Also, ∠MAC = ∠NPR because ∠A = ∠P. Therefore, △AMC ∼ △PNR by SAS similarity.

(ii) From △AMC ∼ △PNR:

CMRN=CARP\frac{CM}{RN}=\frac{CA}{RP}

From (1):

CARP=ABPQ\frac{CA}{RP}=\frac{AB}{PQ}

Therefore:

CMRN=ABPQ\frac{CM}{RN}=\frac{AB}{PQ}

(iii) From (1) and part (ii):

CMRN=BCQR\frac{CM}{RN}=\frac{BC}{QR}

Also:

BM=AB2,QN=PQ2BM=\frac{AB}{2},\qquad QN=\frac{PQ}{2}
BMQN=ABPQ=CMRN\frac{BM}{QN}=\frac{AB}{PQ}=\frac{CM}{RN}

Thus:

CMRN=BCQR=BMQN\frac{CM}{RN}=\frac{BC}{QR}=\frac{BM}{QN}

Therefore, △CMB ∼ △RNQ by SSS similarity.

Final answer

(i) △AMC ∼ △PNR; (ii) CM/RN = AB/PQ; (iii) △CMB ∼ △RNQ.