Question

AB||DC find the value of x
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Answer 1

Answer:

Hi there... here's the answer for your question..

Explanation:

REF.Image.

Here ΔAOB is similar

to ΔCOD

because

∠CDO=∠OBA

∠DCO=∠OAB

Alternate interior angle an equal in ∥ lines

∠DOC=∠AOB (opposite angles are equal)

So its similar by (AAA)

(Angle Angle Angle ) criteria.

So ΔAOB∼ΔCOD

⇒

CO

AO

=

DO

BO

from similatrily properly

x+3

x+5

=

x−2

x−1

(x+5)(x−2)=(x−1)(x+3)

x

2

−2x+5x−10=x

2

−x+3x−3

3x−10=2x−3

x=7

Hope it helps you..

If any doubt ask me..

Thankyou...

Answer 2

Answer:

Given AB is parallel to DC in trapezium ABCD.

Explanation:

In triangle ABE and CDE

Angle AEB =DEC(vertically opposite angles are equal)

angle A=C

angle B=D(alternative angles)

By AAA similarly

∆ABE~∆CDE

Then by criteria of similarly their corresponding angles are equal and their corresponding sides are in proportion.

Therefore AE/EC=BE/DE

7x-9/2x-1=9x-8/3x

21x²-27x=18x²-16x-9x+8

3x²-2x -8=0

3x²-(6x-4x)-8=0

-3x(x+2)-4(x+2)=0

(x+2)(-3x-4)=0

x=-2or x=4/3