Question

in a trapezium AB || DC with centre O and OD = x-2,OC = x+3,OA = x+5,OB = x-1.
Find x value

Answer 1

Question:

In figure, if AB || DC, then find the value of x.

{OD= 3 cm, OC = (x - 5) cm

OA = (3x – 19) cm, OB = (x – 3) cm}

Answer:

x = 8 or 9

Note:

• If two lines intersect at a point, then the vertically opposite angles are equal.

• A transversal is a line which cuts two lines at two distinct points in the same plane.

• If a transversal cuts two parallel lines, then the alternative interior angles are equal.

• If the crossponding sides of two triangles are proportional then both the triangles are said to be similar triangles.

• If two triangles are similar then their crossponding angles are equal.

Solution:

Given:

AB || DC

OD= 3 cm

OC = (x - 5) cm

OA = (3x - 19) cm

OB = (x- 3) cm

Since,

AB || DC and DB is transversal ,

Thus ,

Angle CDO = Angle OBA

{Alternate interior angles are equal}

Again,

AB || DC and CA is transversal ,

Thus ,

Angle DCO = Angle OAB

{Alternate interior angles are equal}

Also,

Angle COD = Angle AOB

{Vertically opposite angles are equal}

Now,

In ∆COD and ∆AOB

Angle CDO = Angle OBA

Angle DCO = Angle OAB

Angle COD = Angle AOB

Thus,

∆COD and ∆AOB are similar triangles.

{By A.A.A. similarity property of triangles}

Since ,

∆COD and ∆AOB are similar triangles.

Thus,

The crossponding sides of ∆COD and ∆AOB must be proportional.

ie; CO/OA = DO/OB = CD/AB

Thus;

=> CO/OA = DO/OB

=> (x-5)/(3x-19) = 3/(x-3)

=> (x-5)•(x-3) = 3•(3x-19)

=> x^2 - 3x - 5x + 15 = 9x - 57

=> x^2 - 8x + 15 = 9x - 57

=> x^2 - 8x - 9x + 15 + 57 = 0

=> x^2 - 8x - 9x + 72 = 0

=> x(x - 8) - 9(x - 8) = 0

=> (x - 8)(x - 9) = 0

=> x = 8 , 9

Hence,

The required values of x are 8 and 9 .