Ch 6: Triangles (Part 1)
Based on the latest CBSE rationalized syllabus. Master the Basic Proportionality Theorem (BPT) and Similarity Criteria (AAA, SSS, SAS).
Exercise 6.1 Solutions
Q1. Fill in the blanks using the correct word given in brackets:
- (i) All circles are similar. (congruent, similar)
- (ii) All squares are similar. (similar, congruent)
- (iii) All equilateral triangles are similar. (isosceles, equilateral)
- (iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.
Q2. Give two different examples of pair of: (i) similar figures (ii) non-similar figures.
Solution:
(i) Similar figures: Two equilateral triangles of sides 2cm and 4cm. Two squares of sides 3cm and 5cm.
(ii) Non-similar figures: A triangle and a square. A rhombus and a rectangle.
Q3. State whether the following quadrilaterals are similar or not:
Solution:
Given figures are a Rhombus (sides 1.5 cm) and a Square (sides 3 cm).
The ratio of corresponding sides is equal (1.5/3 = 1/2). However, their corresponding angles are not equal (Square has 90° angles, Rhombus does not).
Therefore, the quadrilaterals are NOT similar.
Exercise 6.2 (BPT / Thales Theorem)
💡 Thales Theorem (BPT): If DE || BC in ∆ABC, then AD/DB = AE/EC.
Q1. In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Solution for (i):
Given: AD = 1.5 cm, DB = 3 cm, AE = 1 cm. DE || BC.
By Basic Proportionality Theorem (BPT):
AD / DB = AE / EC
1.5 / 3 = 1 / EC
1/2 = 1 / EC ⇒ EC = 2 cm
Solution for (ii):
Given: DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm. DE || BC.
By BPT:
AD / 7.2 = 1.8 / 5.4
AD / 7.2 = 1 / 3
AD = 7.2 / 3 ⇒ AD = 2.4 cm
Q2. E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR:
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
PE/EQ = 3.9 / 3 = 1.3
PF/FR = 3.6 / 2.4 = 1.5
Since PE/EQ ≠ PF/FR, EF is not parallel to QR.
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
PE/QE = 4 / 4.5 = 8/9
PF/RF = 8 / 9
Since PE/QE = PF/RF, by the converse of BPT, EF || QR.
Q6. In Fig. 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.
Proof:
In ∆OPQ, AB || PQ (Given).
By BPT: OA/AP = OB/BQ --- (1)
In ∆OPR, AC || PR (Given).
By BPT: OA/AP = OC/CR --- (2)
From (1) and (2), we get:
OB/BQ = OC/CR
Now, in ∆OQR, since OB/BQ = OC/CR, by the Converse of BPT, we have:
BC || QR. (Hence Proved)
Q9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.
Proof:
Construction: Draw a line EO through O parallel to DC (and hence parallel to AB), meeting AD at E.
In ∆ADC, EO || DC.
By BPT: AE/ED = AO/OC --- (1)
In ∆ABD, EO || AB.
By BPT: DE/EA = DO/OB ⇒ AE/ED = BO/OD --- (2)
From (1) and (2):
AO/OC = BO/OD
Rearranging the terms:
AO/BO = CO/DO. (Hence Proved)
Exercise 6.3 (Similarity Criteria)
Q2. In Fig. 6.35, ∆ODC ~ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
Solution:
DOB is a straight line. Therefore, angles on a straight line add up to 180°.
∠DOC + ∠BOC = 180°
∠DOC + 125° = 180° ⇒ ∠DOC = 55°
In ∆DOC, sum of angles is 180°.
∠DCO + ∠CDO + ∠DOC = 180°
∠DCO + 70° + 55° = 180°
∠DCO + 125° = 180° ⇒ ∠DCO = 55°
Given ∆ODC ~ ∆OBA. Corresponding angles are equal.
Therefore, ∠OAB = ∠OCD (or ∠DCO).
∠OAB = 55°
Q4. In Fig. 6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR.
Proof:
In ∆PQR, ∠1 = ∠2 (Given).
Since sides opposite to equal angles are equal, PR = PQ.
Given: QR/QS = QT/PR.
Substitute PR with PQ: QR/QS = QT/PQ --- (1)
Now, in ∆PQS and ∆TQR:
From (1): QR/QS = QT/PQ (Sides are proportional)
∠PQS = ∠TQR (∠1 is common to both triangles)
By SAS Similarity Criterion:
∆PQS ~ ∆TQR. (Hence 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.
Solution:
Let AB be the pole (6m) and BC be its shadow (4m). Let PQ be the tower (h) and QR be its shadow (28m).
At the same time of day, the sun's elevation is the same. Therefore:
In ∆ABC and ∆PQR:
∠B = ∠Q = 90° (Both are vertical to the ground)
∠C = ∠R (Angle of elevation of the sun)
By AA Similarity Criterion, ∆ABC ~ ∆PQR.
Since triangles are similar, their corresponding sides are proportional:
AB/PQ = BC/QR
6/h = 4/28
6/h = 1/7
h = 6 × 7 = 42 m.
The height of the tower is 42 m.
Q16. If AD and PM are medians of triangles ABC and PQR, respectively where ∆ABC ~ ∆PQR, prove that AB/PQ = AD/PM.
Proof:
Given ∆ABC ~ ∆PQR. Therefore:
AB/PQ = BC/QR = AC/PR --- (1)
and ∠B = ∠Q --- (2)
Since AD and PM are medians, BD = BC/2 and QM = QR/2.
From (1): AB/PQ = (2BD)/(2QM) = BD/QM --- (3)
Now, consider ∆ABD and ∆PQM:
∠B = ∠Q (from 2)
AB/PQ = BD/QM (from 3)
By SAS Similarity Criterion, ∆ABD ~ ∆PQM.
Since these triangles are similar, their corresponding sides are proportional:
AB/PQ = AD/PM. (Hence Proved)