Exercise 6.3 – Stepwise Solutions
1. State which pairs of triangles in Fig. 6.34 are similar. Write the criterion used.
(i)
Angles given in both triangles: 60°, 80°, 40°
All corresponding angles are equal.
Criterion: AAA
ΔABC ∼ ΔPQR
(ii)
Small triangle sides: 2, 2.5, 3
Large triangle sides: 4, 5, 6
Ratios: 2/4 = 2.5/5 = 3/6 = 1/2
Criterion: SSS
ΔABC ∼ ΔPQR
(iii)
First triangle sides: 2, 2.7, 3
Second triangle sides: 4, 5, 6
Ratios not equal.
Triangles are NOT similar.
(iv)
Given angle = 70° in both triangles.
Sides around angle proportional:
5/10 = 1/2
Corresponding sides in small triangle also proportional.
Criterion: SAS
Triangles are similar.
(v)
One triangle has angle 80°.
Other triangle also has angle 80°.
Two sides proportional.
Criterion: SAS
Triangles are similar.
(vi)
Angles given: 70°, 80° in first triangle.
Angles given: 80°, 30° in second triangle.
Angles are not matching.
Triangles are NOT similar.
2. In Fig. 6.35, ΔODC ∼ ΔOBA
Given: ∠BOC = 125°, ∠CDO = 70°
Step 1: Linear pair at O
∠DOC = 180° − 125°
∠DOC = 55°
Step 2: In ΔODC
∠DCO = 180° − (70° + 55°)
∠DCO = 55°
Step 3: Since ΔODC ∼ ΔOBA
Corresponding angles equal
∠OAB = 55°
3. Prove OA/OC = OB/OD
Given: AB ∥ DC
In triangles OAB and OCD:
∠OAB = ∠OCD (Alternate interior angles)
∠OBA = ∠ODC (Alternate interior angles)
∠AOB = ∠COD (Vertical angles)
Therefore, triangles OAB and OCD are similar (AAA).
Hence, corresponding sides are proportional:
OA / OC = OB / OD
Hence proved.
Final Answers Summary
1. (i) Similar (AAA)
1. (ii) Similar (SSS)
1. (iii) Not Similar
1. (iv) Similar (SAS)
1. (v) Similar (SAS)
1. (vi) Not Similar
2. ∠DOC = 55°, ∠DCO = 55°, ∠OAB = 55°
3. OA/OC = OB/OD proved
Exercise 6.3 – Stepwise Solutions (Q4 to Q10)
4. In Fig. 6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ∼ ΔTQR.
Given: QR / QS = QT / PR and ∠PQS = ∠TRQ (∠1 = ∠2)
Since the sides including the equal angles are proportional,
By SAS similarity criterion:
ΔPQS ∼ ΔTQR
5. Show that ΔRPQ ∼ ΔRTS
Given: ∠P = ∠RTS
Also, ∠RPQ = ∠RST (common angle structure)
Two angles equal.
By AA similarity criterion:
ΔRPQ ∼ ΔRTS
6. In Fig. 6.37, if ΔABE ≅ ΔACD, show that ΔADE ∼ ΔABC.
Given ΔABE ≅ ΔACD
So, AB = AC and AE = AD
Angles at A are common.
Thus, corresponding angles of ΔADE and ΔABC are equal.
By AA similarity:
ΔADE ∼ ΔABC
7. In Fig. 6.38, altitudes AD and CE intersect at P.
(i) Prove ΔAEP ∼ ΔCDP
∠AEP = ∠CDP = 90°
Vertical opposite angles equal at P
By AA: ΔAEP ∼ ΔCDP
(ii) Prove ΔABD ∼ ΔCBE
∠ADB = ∠CEB = 90°
Another pair of angles equal
By AA: ΔABD ∼ ΔCBE
(iii) Prove ΔAEP ∼ ΔADB
Right angles and common angles equal
By AA: ΔAEP ∼ ΔADB
(iv) Prove ΔPDC ∼ ΔBEC
Right angles equal
Another corresponding angle equal
By AA: ΔPDC ∼ ΔBEC
8. Show that ΔABE ∼ ΔCFB
ABCD is a parallelogram.
AB ∥ CD and AD ∥ BC
Alternate interior angles equal.
Two pairs of angles equal.
By AA similarity:
ΔABE ∼ ΔCFB
9. Prove:
(i) ΔABC ∼ ΔAMP
Both triangles right-angled at B and M respectively.
Common acute angle at A.
By AA similarity: ΔABC ∼ ΔAMP
(ii) CA/PA = BC/MP
From similarity:
Corresponding sides proportional
CA/PA = BC/MP
10. If ΔABC ∼ ΔFEG, show that:
(i) CD/GH = AC/FG
Since triangles similar, corresponding sides proportional.
CD/GH = AC/FG
(ii) ΔDCB ∼ ΔHGE
Angles correspond and sides proportional.
By AA: ΔDCB ∼ ΔHGE
(iii) ΔDCA ∼ ΔHGF
Corresponding angles equal.
By AA: ΔDCA ∼ ΔHGF
Final Summary
4) ΔPQS ∼ ΔTQR (SAS)
5) ΔRPQ ∼ ΔRTS (AA)
6) ΔADE ∼ ΔABC (AA)
7) All four parts proved by AA
8) ΔABE ∼ ΔCFB (AA)
9) (i) ΔABC ∼ ΔAMP (ii) CA/PA = BC/MP
10) All results follow from similarity
Exercise 6.3 – Stepwise Solutions (Q11 to Q16)
11. Prove that ΔABD ∼ ΔECF
Given: ABC is isosceles, AB = AC
AD ⟂ BC and EF ⟂ AC
∠ADB = 90° and ∠EFC = 90°
Since AB = AC (isosceles triangle), base angles are equal:
∠ABD = ∠ECF
Thus, two angles are equal:
By AA similarity:
ΔABD ∼ ΔECF
12. Show that ΔABC ∼ ΔPQR
Given:
AB / PQ = BC / QR = AD / PM
Since two sides and corresponding medians are proportional,
the included angles are equal.
By SAS similarity:
ΔABC ∼ ΔPQR
13. Prove that CA² = CB × CD
Given: ∠ADC = ∠BAC
In triangles ADC and BAC:
∠ADC = ∠BAC
∠ACD = ∠BCA (common angle)
Thus, ΔADC ∼ ΔBAC (AA similarity)
From similarity:
AC / BC = CD / AC
Cross multiplying:
AC² = BC × CD
14. Show that ΔABC ∼ ΔPQR
Given:
AB / PQ = AC / PR = AD / PM
Two sides and medians proportional.
By SAS similarity:
ΔABC ∼ ΔPQR
15. Find the height of the tower
Height of pole = 6 m
Shadow of pole = 4 m
Shadow of tower = 28 m
Since sun rays form similar triangles:
Height / Shadow = constant
6 / 4 = h / 28
Cross multiply:
6 × 28 = 4h
168 = 4h
h = 42 m
Height of tower = 42 m
16. Prove that AB/PQ = AD/PM
Given: ΔABC ∼ ΔPQR
Since triangles are similar, corresponding sides are proportional:
AB / PQ = AC / PR
Medians of similar triangles are also proportional.
Therefore:
AB / PQ = AD / PM