Theorems

SOLVED EXAMPLES 1. In figure 6-25, DE parallel to BC . Find the value of CE, if AD=8 cm, AB=12 cm and AE=12 cm. Sol:

In Triangle ABC , since DE parallel to BC .

Therefore

AD AE = DB EC

Let EC= x cm

(Basic Proportionality Theorem) since AD=8 cm Therefore DB = 12 − 8 = 4 cm

Substituting the values in the above equation, we have

8 12 = 4 x Cross multiplying , we have

8 x = 12 × 4 8 x = 48

x = 6 cm

2. In figure 6-26, DE parallel to BC . Find the value of x ? Sol:

Since DE parallel to BC , applying Basic proportionality theorem

AD AE = DB EC Using the values given in the figure 6-26, we get

x x+2 = x − 2 x −1 By cross multiplication, we get

x( x − 1) = ( x + 2)( x − 2) x2 − x = x2 − 4 [ (a + b)(a − b) = a 2 − b 2

x 2 − x − x 2 = −4 − x = −4 x = 4 cm

3. In figure 6-27, AC=8cm, BF=6cm and BC=12 cm E and F are points respectively on AC and BC. State whether the line joining EF will be parallel to AB or not. Sol:

From the given figure 6-27, it is clear that AE+EC=AC or 4+EC=8 or EC=8-4=4

and BF+FC=BC or 6+FC=12 or FC=12-6=6 Now,

and

CF 6 = =1 FB 6

… (1)

CE 4 = =1 EA 4

… (2)

From (1) and (2), we have

CF CE = FB EA

(Equals of equals are equal)

By converse of Basic proportionality Theorem, we find that EF parallel to AB . 4. In figure 6-28, AC= a , AP= b , PB= c and PQ= x . Express x in terms of a, b, c . Sol:

Since ∠A = ∠P = 62o

Therefore PQ parallel to AC . Now, ∠B is common to both the triangles, Triangle BQP and Triangle BCA

Triangle BQP Similar to Triangle BCA BQ QP PB = = BC CA AB

So,

(AA corollary)

(Corresponding parts of similar triangles are proportional)

or

PQ PB = AC AB

or

x c = a AP + PB

or

x c = a b+c

or x =

( Since AB = AP + PB )

ac b+c

5. In figure 6-29, AD is the bisector of ∠A that meets BC at D. If AB=9cm, BC=10cm and AC=6cm. Find BD. Sol:

Let BD= x cm. Then DC= 10 − x cm Since AD is the bisector of ∠A

Therefore

or

AB BD = AC CD

9 x = 6 10 − x

Cross-multiplying, we get 9(10 − x) = 6 x 90 − 9 x = 6 x

−9 x − 6 x = −90 −15 x = −90 15 x = 90

x=

90 =6 15

Therefore BD = 6 cm 6. In figure 6-30, DC parallel to AB . Find x Sol:

Construction: Draw OP parallel to DC

Therefore OP parallel to AB

(parallels of parallel are parallel to each other)

Now, in Triangle BCD , since OP parallel to DC , using Basic Proportionality Theorem, we have

BP BO = PC OD

… (1)

Also, OP parallel to AB, Therefore

CP CO = PB OA or,

in Triangle ABC

(basic proportionality theorem)

PB OA = CP CO

… (2)

From (1) and (2), we have

BO OA = OD CO Therefore

(Equals of equals are equal)

2 x + 1 3x − 1 = 6x − 5 9x − 3

By cross-multiplication, we get

(2 x + 1)(9 x − 3) = (3x − 1)(6 x − 5) 18 x 2 − 6 x + 9 x − 3 = 18 x 2 − 15 x − 6 x + 5 18 x 2 − 6 x + 9 x − 18 x 2 + 15 x + 6 x = 5 + 3 24 x = 8

x=

8 24

or x =

1 3

7. A ladder 5 m long reaches a window 4 m high on one side of a street. The ladder is then turned over to the opposite side of the street and is found to reach another window 3m high. Find the width of the street. Sol:

Let AB and AC be the two positions of the ladder when it touches the windows at B and C respectively. BE represents the first window and CD represents the second window. As the wall is always at right angles to the ground.

Triangle ABE and Triangle ACD are right triangles right angled at E and D respectively. Here DE is the width of the streets. Now in right Triangle ABE , using Pythagoras theorem

AE 2 + BE 2 = AB 2 or AE 2 + 42 = 52 or AE 2 + 16 = 25 or AE 2 = 25 − 16 = 9 or AE = 9 = 3 m

… (1)

Now, in Triangle ACD, using Pythagoras theorem

AD 2 + CD 2 = AC 2 or AD 2 + 32 = 52 or AD 2 + 9 = 25 or AD 2 = 25 − 9 = 16 or AD = 16 = 4 m

… (2)

Now DE=AD+AE or DE= 4m + 3m = 7 m

Therefore Width of the street is 7 m 8. A man goes 8 m due west and then 16 m due north. Find the distance between the starting point and the terminal point. Sol: In right triangle OAB, OA = 8 m (Due west)

AB= 15 m (Due north)

By Pythagoras theorem, we have

OA2 + AB 2 = OB 2 or 82 + 152 = OB 2 or 64 + 225 = OB 2 or 289 = OB 2 or

289 = OB

or 17 = OB or OB = 17 m

Therefore The man is 17 m away from the starting point.

9. A tree 16 m high broke at a point but did not separate. Its top touched the ground at a distance of 8 m from the base. Find the height of the point from the ground, at which the tree broke? Sol: Let the tree broke at a height of x m from the base. In figure 5-33, BD= x m AD= 16 m, AB=AD-DB=16- x

Therefore BC=16- x (Since it is the same part AB of the tree)

Now, in right triangle BCD, we have

CD 2 + BD 2 = BC 2 or 82 + x 2 = (16 − x) 2 or 64 + x 2 = 256 + x 2 − 32 x

(By Pythagoras theorem)

or x 2 − x 2 + 32 x = 256 − 64 or 32 x = 192 or x =

192 = 6m 32

Therefore The tree broke at a height of 6 m from the ground. 10. Two poles of heights 9 m and 21 m stand on a playground. If the distance between their feet is 35 m, find the distance between their tops. Sol: Let PQ=9 m Construct QS parallel to PR

Now ∠QST = ∠PRS = 90o

(Corresponding angles)

Therefore The figure PQSR is a rectangle. Since the opposite sides of a rectangle are equal

Therefore SR=9 m and QS= 35 m

Now TS = TR-SR = 21-9 = 12 m In Triangle QST QS= 35 m, TS = 12 m By Pythagoras theorem, we have

QS 2 + ST 2 = QT 2 or QT 2 = QS 2 + ST 2 or QT 2 = 352 + 122 = 1125 + 144 = 1369 or QT = 1369 = 37 m

Therefore The distance between the tops of the poles is 37 m

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