Question

In ∆AOB and ∆COD, angle B = angle C and O is the midpoint of BC.
Find the values of x and y if AB= 3x units, CD = y+2 units, AO = x+2 units, DO = y units

Answer 1

$\mathcal{\gray{\underline{\underline{\blue{GIVEN:-}}}}}$

• In ∆AOB and ∆COD,

1. ⟨B = ⟨C .
2. And O is the midpoint of BC .

• BO = CO

Let, According to the question

• AB = 3x units

• CD = y + 2 units

• AO = x + 2 units

• DO = y units

$\mathcal{\gray{\underline{\underline{\blue{TO\: FIND:-}}}}}$

• The value of x and y .

$\mathcal{\gray{\underline{\underline{\blue{SOLUTION:-}}}}}$

✍️ See the attachment figure .

$\red\checkmark\:\rm{\pink{\angle{B}\:=\:\angle{C}}}$ [Given]

In the figure,

• AB || CD and BC be line which intersect the two parallel lines .

$\rm{\orange{\implies\:\angle{A}\:=\:\angle{D}}}$ [Alternate angles]

• AD line and BC line intersect each other at O .

$\rm{\green{\implies\:\angle{AOB}\:=\:\angle{COD}}}$ [Opposite angles]

Now, in ∆AOB and ∆COD,

1. ⟨A = ⟨D
2. ⟨AOB = ⟨COD
3. ⟨C = ⟨B

☃️ Hence, $\mathcal\red{\triangle{AOB}\:\cong\:\triangle{COD}\:}$ [AAA criteria]

So,

$\rm{\implies\:AB\:=\:DC}$

$\rm{\implies\:3x\:=\:y\:+\:2\:}$

$\rm\purple{\implies\:3x\:-\:y\:=\:2\:}$ -----(1)

And

$\rm{\implies\:AO\:=\:DO}$

$\rm{\implies\:x\:+\:2\:=\:y\:}$

$\rm\purple{\implies\:x\:-\:y\:=\:-2\:}$ -----(2)

Now, Equation (1) - Equation (2),

=> 3x - y - (x - y) = 2 - (-2)

=> 3x - y - x + y = 2 + 2

=> 2x = 4

=> x = 4/2

$\mathcal\red{\implies\:x\:=\:2\:}$

⚡ Put the value of “ x = 2 ” in equation (2), we get

=> 2 - y = -2.

=> y = 2 + 2

$\mathcal\red{\implies\:y\:=\:4\:}$

$\rm\pink{\therefore}$ The value of “x” is 2 and the value of “y” is 4 .