Question

in the given figure ABCD is a trapezium with DC parallel to AB, angle AOB=126 and angle DCQ=CDP=52,find the values of x and y​

Answer 1

Answer:

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(i) ∠DAB=∠CBA

Construction :- to produce AB to E and draw CE∥DA,MO∥AB

Proof: So AECD is a parallelogram.

In ΔBEC

BC=CE (AD=BC=CE,opposite sides of parallelogram)

∠CBE=∠CEB (opposite to equal sides) ---- (i)

∠ADC=∠BEC (opposite angle of a parallelogram) ----- (ii)

∠DAB+∠ADC=180

∘

(sum of co-interior angles) ------ (iii)

∠ABC=∠EBC=180

∘

(L.P.P.)

∠ABC+∠ADC=180

∘

---- (iv) from (ii)

From (iii) and (iv)

∠DAB=∠CBA

(ii) ∠BCD=∠EBC ---- (v) (alternate interior angles)

From (i),(ii), and (v)

∠ADC=∠BCD proved

(iii) In ΔABD and ΔBAC

∠BAD=∠ABC

AB=AB (common)

AD=BC (Given)

ΔABD≅ΔBAC (by SAS rule)

BD=AC (By C.P.C.T)

(iv) In ΔABD

MO∥AB

MD

AM

=

OD

BO

--- A, (by lemma of B.P.T)

In ΔADC

MO∥DC

MD

AM

=

OC

AO

---- B (by B.P.T.)

From A and B

OD

BO

=

OC

AO

---- C

OD

BO

+1=

OC

AO

+1

OD

BO+DO

=

OC

AO+OC

OD

BD

=

OC

AC

OD=OC (AC=BD)

From C

OB=OA (OD=OC)

Answer 2

Solution :-

Construction :- Produce AP and BQ to meet at R.

In ∆RDC we have,

→ ∠RDC = ∠RCD (given 52°) ---------- Eqn.(1)

so,

→ DR = CR (sides opposite to equal angles are equal in length.)

now,

→ ∠RAB = ∠RDC (given that, ABCD is a trapezium. so, AB || DC , therefore, corresponding angles are equal). -------- Eqn.(2)

Similarly,

→ ∠RBA = ∠RCD (corresponding angles .) -------- Eqn.(3)

from Eqn.(1) , Eqn.(2) and Eqn.(3) we can conclude that,

→ ∠RAB = ∠RBA

or,

→ AR = RB .

then,

→ AD = AR - DR = RB - CR = BC

therefore, we can conclude that, ABCD is an isosceles trapezium.

hence,

→ OA = OB

→ ∠OAB = ∠OBA.

now, in ∆AOB we have,

→ ∠OAB + ∠OBA + ∠AOB = 180° (By angle sum property.)

→ 2∠OAB + 126° = 180°

→ 2∠OAB = 180° - 126°

→ 2∠OAB = 54°

→ ∠OAB = 27°

therefore,

→ ∠x = ∠DAB - ∠OAB

→ ∠x = 52° - 27°

→ ∠x = 25° (Ans.)

also, in ∆ABC, we have,

→ ∠ACB + ∠CAB + ∠ABC = 180° (By angle sum property.)

→ y + 27° + 52° = 180°

→ y = 180° - 27° - 52°

→ y = 180° - 79°

→ y = 101° (Ans.)

Learn more :-

In ABC, AD is angle bisector,

angle BAC = 111 and AB+BD=AC find the value of angle ACB=?

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