Thales Theorem

EXERCISE 1. In figure 6-37, PQ parallel to AB, AC=18cm, PC=8cm, BQ=6cm. The length of side BC is a.

28 cm 5

b.

24 cm 5

c. 4 cm d. 5 cm Sol:

Correct option is (b)

Here AC= 18 cm, But AC = AP+PC or AC = AP + 8 or AP =18-8 = 10 cm Since PQ parallel to AB, , In Triangle ABC ,

CP CQ = AP QB or

(Basic proportionality theorem)

8 CQ = 10 6

or CQ =

8 × 6 48 24 = = cm 10 10 5

2. In figure 6-38, AB = 6 cm, BD= 3 cm. If DF=12 cm, then EF is a. 4 cm b. 9 cm c. 6 cm d.

15 cm 4

Sol:

In

Triangle ADF , EG parallel to AF Therefore Triangle DGE Similar to DAF

Therefore

DA DE = GA EF

(Basic proportionality theorem)

… (1)

In Triangle ABC , E is the mid-point of BC and EG parallel to CA

Therefore G is the mid point of AB Therefore AG = BG =

1 1 AB = × 6 = 3 cm 2 2

Now, AG=BG=BD=3 cm From (1), we have

or

AB DE = AG EF

6 DE = 3 EF

or 2 =

DE EF

or 2EF = DE or DE = 2 EF Now, DF=DE+EF or DF=2EF+EF (Q DE=2EF) or DF=3EF or 12 = 3EF or

12 = EF 3

or EF =

12 = 4 cm 3

3. In figure 6-39, DE parallel to BC . The value of x is a. 7 cm b. 5 cm c. 12 cm d. none of the above Sol:

Since

DE parallel to BC Therefore Triangle ADE Similar To Triangle ABC

Thus

AD AE = DB EC

or

6 8 = 9 x

or 6 x = 72 or x =

72 = 12 cm 6

4. In figure 6-40, DE parallel to BC , If AD= x , DB= x − 2 , AE= x + 2 and EC= x − 1 , then x is a. 5 b. 7 c. 6 d. 4 Sol: Since

Correct option is (d)

DE parallel to BC Therefore Triangle ADE Similar to ABC

Therefore

or

AE AD = EC DB

x+2 x = x −1 x − 2

By cross multiplication, we get

( x + 2)( x − 2) = x( x − 1) or x 2 − 4 = x 2 − x or −4 = − x or 4 = x ⇒ x = 4 5. In a trapezium ABCD, AB parallel to DC and AB=7cm DC=14 cm, DE is drawn parallel to AB cuts AD in F and AC in E. BE=3 cm and EC = 4cm. DB intersects EF at G. Then length of FE is a. 10 cm b. 7 cm c. 12 cm d. 9 cm Sol:

Correct option is (a)

In Triangle DFG and Triangle DAB

∠ADB = ∠ADB

(Common to both triangles)

∠DFG = ∠DAB

(Corresponding angles)

Therefore criterion) Thus,

Triangle DFG similar to Triangle DAB

DF FG = DA AB

(Basic proportionality theorem)

But in trapezium, ABCD EF parallel to DC

AF BE = DF EC or

AF 3 = DF 4

(Since BE=3 cm and EC= 4 cm)

Adding 1 to both sides, we get AF 3 +1 = +1 DF 4 AF + DF 3 + 4 = 4 DF AD 7 = DF 4

(AA

similarity

… (1)

or

DF 4 = AD 7

… (2)

Since GE parallel to DC (AAA criterion of similarity)

Triangle BEG Similar to Triangle BCD So,

BE EG = BC CD

or

3 EG = 3 + 4 CD

or

3 EG = 7 CD

or

EG 3 = CD 7

( BE= 3 cm and EC = 4 cm)

3 3 or EG = × CD = × 14 = 6 cm 7 7

… (4)

Now FE=FG+EG = 4+6=10 cm 6. In figure 6-42, AB parallel to EF parallel to CD . If EG=5cm, GC=10cm, DC=18cm. Then EF is a. 25 cm b. 15 cm c. 9 cm d. 16 cm Sol:

Correct option is (c)

Since

EF parallel to DC Therefore Triangle EFG Similar toTriangle CDG

So,

EF CD FG = = EG CG DG

or,

EF 18 = 5 10

or EF =

(AAA similarity)

(basic proportionality theorem)

18 × 5 = 9 cm 10

7. In figure 6-43, Triangle ABC Similar to Triangle AQP . The length of CA is a. 6 cm b. 8 cm c. 10 cm d. 9 cm Sol:

Correct option is (c)

Since Triangle ABC Similar to Triangle AQP Therefore

BC CA = PQ PA

(Basic Proportionality Theorem)

or

12 CA = 8 6

or

12 × 6 = CA 8

or CA =

12 × 6 = 9 cm 8

8. In figure 6-44, the value of x for which ED parallel to AB is a. 5 b. 2 c. 3 d.

7 9

Sol:

Correct option is (b)

In figure 6-42, ED parallel to AB if CD CE = BD EQ

(Converse of Basic Proportionality Theorem)

or

x x+3 = 9x − 8 9x + 7

By cross multiplication, we get x(9 x + 7) = (9 x − 8)( x + 3)

or 9 x 2 + 7 x = 9 x 2 − 8 x + 27 x − 24 or −12 x = −24 or 12 x = 24 or x =

24 = 2 cm 12

9. In figure 6-45, ∠BAD = ∠CAD . If AC=4 cm, DC=8 cm and BC=26 cm, then AB is a. 12 cm b. 8 cm c. 13 cm d. 9 cm Sol:

Correct option is (d)

Here BD=BC-DC or BD=26-8 or BD=18 cm In Triangle BAC , since ∠BAD = ∠CAD

Therefore AD is the bisector of ∠A . So,

or

AB BD = AC DC AB 18 = 4 8

or AB =

18 × 4 = 9 cm 8

10. In figure 6-46 from the given values, one can say that a. Triangle ADE ≅ Triangle ABC b. Triangle ADE Similar to Triangle ABC c. Triangle ADE and Triangle ABC have no relation d. none of the above

Sol:

Correct option is (b)

In Triangle ADE and Triangle ABC , we have AD 9 = DB 4

and

or

… 91)

AE 27 = EC 12

AE 9 = EC 4

… (2)

From (1) and (2), we have AD AE = DB EC

(Equals of equals are equal)

Therefore By converse of basic proportionality theorem, we can say that Triangle ADE Similar to Triangle ABC . 11. In figure 6-47, AD bisects ∠A and intersects BC in D. If BC=20 cm, AC=20 cm and AB=5 cm then BD is: a. 4 cm b. 6 cm c. 8 cm

d. 2 cm Sol:

Correct option is (a)

Since AD is the bisector of ∠A Therefore

DC AC = BD AB

or

DC 20 = BD 5

or

DC =4 BD

(Angle bisector theorem)

Now, adding 1 to both the sides we get DC +1 = 4 +1 BD

or

DC + BD =5 BD

or

BC =5 BD

or

20 =5 BD

… (1)

or BD =

20 = 4 cm 5

12. In figure 8-48, if ∠A = ∠CED , then (i) a. Triangle CAB Similar to Triangle CED b. Triangle CAB ≅ Triangle CED c. Triangle CAB and Triangle CED have no relation d. Triangle CAB Similar to Triangle ECD (ii) Find x

Sol:

Correct option is (a)

(i) In Triangle CAB and Triangle CED , we have

∠1 = ∠2

(Given)

∠ACB = ∠DEC

(Common angle in both the triangle)

Therefore Triangle CAB Similar to Triangle CED similarity)

(AA

criterion

of

(ii) Since Triangle CAB Similar to Triangle CED (proved above)

Therefore

BC AB AC = = DC ED EC

or

BC AB = CD DE

or

10 + 2 9 = 8 x

or

12 9 = 8 x

By cross multiplying we get 12 x = 9 × 6

or x =

9× 6 =6 12

13. In figure 6-49, a vertical stick 10 cm long casting a shadow 6 cm long. The height of the tower casting a shadow of length 30 m at the same time is: a.

75 m 2

b. 40 m c. 50 m d. 38 m Sol:

Correct option is (c)

As the stick and the tower are casting shadows on the ground at the same time.

Therefore ∠ACB = ∠PRQ ∠ABC = ∠PRQ

(each = 90o )

Triangle ABC Similar to Triangle PQR similarity) So,

AB BC = PQ QR

or

10 6 = h 3000

or

h 3000 = 10 6

or h =

14. In figure 6-50 the length of PQ is

b. 23 cm c. 5 cm d. 12 cm Sol:

Correct option is (d)

In right triangle PSR, we have

criterion

(Basic proportionality theorem)

3000 × 10 = 5000cm = 50 m 6

a. 10 cm

(A

of

PS 2 + SR 2 = PR 2

(Pythagoras Theorem)

or 32 + 42 = PR 2 or 9 + 16 = PR 2 or 25 = PR 2 or PR = 25 = 5 cm

Now in Triangle PQR PQ 2 + PR 2 = QR 2

(Pythagoras Theorem)

or PQ 2 + 52 = 132 or PQ 2 + 25 = 169 or PQ 2 = 169 − 25 = 144 or PQ 2 = 144 = 12 cm 15. In figure 6-51 AE=DC=13 cm, BE=5 cm ∠ABC = 90o and AD=EC= x cm. Then x is a. 9 cm b. 7 cm

c. 11 cm d. 8 cm

Sol:

Correct option is (b)

In right Triangle ABE

AB 2 + BE 2 = AE 2

(Pythagoras theorem)

or AB 2 = AE 2 − BE 2 or AB 2 = 132 − 52 = 169 − 25 = 144 or AB = 144 = 12 cm Now, in right Triangle BCD

CD 2 = BD 2 + BC 2 or 132 = ( AB − AD) 2 + ( BE + EC ) 2 or 132 = (12 − x) 2 + (5 + x) 2 or 169 = 144 + x 2 − 24 x + 25 + x 2 + 10 x or 169 = 169 + 2 x 2 − 14 x

or 2 x 2 − 14 x = 0 or 2 x( x − 7) = 0 or x = 0

or x − 7 = 0 ⇒ x = 7

16. In figure 6-52 is of Triangle ABC in which BD ⊥ AC and AC 2 = AB 2 + BC 2 . If BD= 15 cm and CD=9 cm, then AD is a. 15 cm b. 24 cm c. 8 cm d. 25 cm Sol:

Correct option is (d)

In Triangle ABC , it is given that

AC 2 = AB 2 + BC 2 By converse of Pythagoras theorem, we find

∠ABC = 90o Since, BD ⊥ AC , Therefore ∠CDB = 90o

Now, in Triangle ABC and Triangle CDB , we have ∠C = ∠C

(common to both the triangles)

∠ABC = ∠CDB

(Each= 90o )

Therefore Triangle ABC Similar to Triangle CDB similarity) … (1)

(AA-criterion

of

(AA-criterion

of

Now in Triangle BDA and Triangle ABC , we have

∠A = ∠A

(common to both the triangles)

∠BDA = ∠ABC

(Each= 90o )

Therefore Triangle BDA Similar to Triangle ABC similarity) .... (2) From (1) and (2), we have

Triangle CDB Similar to Triangle BDA Therefore

(Transitive property)

BD DC = DA BD

Cross-multiplying, we have

BD 2 = DC × DA or BD 2 = CD × AD Substituting the given values of BD and CD in the above equation, we have

(15) 2 = 9 × AD or 225 = 9 × AD or AD =

225 = 25 cm 9 *******************************************