Thales Theorem Doc

EXERCISE 1. In Triangle ABC , D and E are points on sides AB and AC respectively and AD 3 DE parallel to BC. If = and AC=15cm, then AE is DB 4 a.

45 7

b.

9 8

c.

15 8

d.

15 7

Sol:

Correct option is (a)

Let AE= x cm. Using Basic Proportionality theorem, we have

AD AE = DB EC

or

3 x = 4 AC − x

or

3 x = 4 15 − x

( Since EC = AC − AE )

By cross multiplication, we get

3(15 − x) = 4 x 45 − 3x = 4 x 7 x = 45 x=

45 7

2. In Triangle ABC as shown in figure 6-12 DE parallel to BC . The value of AB is a. 5 cm b. 7 cm c. 6 cm d.

24 cm 5

Sol:

Correct option is (c)

Let DB = x cm. Then AB =

12 + x cm 5

Using Basic Proportionality theorem (Corollary) We have

12

AD AE = AB AC

5 =2 12 + x 5 5

( Since AC = AE + EC )

By cross-multiplication, we get

(

12 × 5 = 2 12 + x 5 5 12 =

24 + 2x 5

2 x = 12 −

24 5

2x =

60 − 24 5

2x =

36 5

x=

)

36 18 = 10 5

Therefore DB =

18 5

But AB= AD+DB=

12 18 30 + = =6 5 5 5

3. In Triangle PQR , S and T are points on sides PQ and QR respectively and ST parallel to QR . Then side PR of Triangle PQR is a. 8 cm b.14 cm c. 23 cm d. 9 cm

Sol:

Correct option is (b)

Since ST parallel to QR , using basic proportionality theorem, we get PS PQ = SQ TR or

3 3x = 4 3x + 2

By cross-multiplication, we get

3(3 x + 2) = 12 x 9 x + 6 = 12 x 12 x − 9 x = 6

3x = 6 x=

6 =2 3

Therefore PQ = 3 × 2 = 6 cm

and TR= 3 × 2 + 2 = 8 cm Now, PR=PT+TR=6+8=14 cm 4. In Triangle ABC , D and E are points on sides AB and AC respectively such that DE parallel to BC . Then length of side EC is a.

36 cm 5

b. 9 cm c. 5 cm d. 4 cm

Sol:

Correct option is (d)

Since DE parallel to BC , using basic proportionality theorem, we have

AD AE = DB EC

16 2 or = 5 5 EC 2 2 EC =

EC =

16 5 × =8 5 2

8 = 4 cm 2

5. In Triangle ABC , AD=5 cm, DB=3 cm, and AC=16 cm. The length of AE, as shown in figure 6-13 for which DE parallel to BC is a. 6 cm b. 10 cm c. 11 cm d. 15 cm

Sol:

Correct option is (b)

For DE to become parallel to BC it should divide AB and AC in the same ratio. Therefore Using converse of Basic Proportionality Theorem, we have

AD AE = DB EC Let AE= x cm

Therefore EC = 16 − x cm Thus

5 x = 3 16 − x

By cross-multiplication, we get

5(16 − x) = 3 x 80 − 5 x = 3 x −5 x − 3 x = −80 −8 x = −80 x=

−80 −8

x = 10 cm

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