Question

Prove
that the
sum oft 360
quadrilateral ABI 23 300
Hint - Drow diagonal BD by
hoints B and D and use single
niny
sean
pusperity of a tuia
trusngle
for
two tressegles?
formed
of
a
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Answer 1

Correct Question:

Prove that the sum of the angles of quadrilateral is 360°.

Solution:

Given: PQRS is quadrilateral.

Construction: Join QS.

To prove: ∠P + ∠Q + ∠R + ∠S = 360°

Proof:

In ΔPQS,

=> ∠SPQ + ∠PQS + ∠PSQ = 180º    ........(1) [Angle sum property]

Similarly in ΔQRS,

=> ∠SQR + ∠QRS + ∠QSR = 180º    ........(2) [Angle sum property]

Now, add Equation (1) and (2), we get

=> ∠SPQ + ∠PQS + ∠PSQ + ∠SQR + ∠QRS + ∠QSR = 180° + 180°

=> ∠P + ∠Q + ∠R + ∠S  = 360º

Hence Proved!!!

Answer 2

$\huge{\mathfrak{\red{\underline{Answer :-}}}}$

Given :-

PQRS is quadrilateral.

Construction :-

★ Join BD

To prove: ∠A + ∠B + ∠C + ∠D = 360°

Solution :-

In ΔPQS,

→ ∠DAB + ∠ABD + ∠ADB = 180°.......(1) [A.S.P]

Now,

In ΔQRS,

→ ∠DBC + ∠BCD + ∠BDC = 180°.......(2) [A.S.P]

From Equation (1) and (2), we get

=> ∠DCB + ∠ABD + ∠ADB + ∠DBC + ∠BCD+ ∠BDC = 180° + 180°

→ ∠A + ∠B + ∠C + ∠D  = 360°

$\huge{\boxed{\bf{\pink{Hence \: Proved}}}}$