Quadrilateral Angle Sum Property: Theorem, Types & Example

Ans. The sum of all the interior angles of a quadrilateral is 360 degrees

Let the unknown angle be x

We have, 

90^o + 45^o + 60^o + x = 360^o

195^o + x = 360^o

x = 165^o

Ans. Let us assume the fourth angle as x since the sum of four interior angles of a quadrilateral is 360^o

Thus , x + 240^o = 360^o

x = 120^o

Ans. Let the angles be 6x, 3x, 4x and 5x

According to the angle sum property of the quadrilateral

6x + 3x + 4x + 5x = 360^o

18x = 360^o

x = 20^o

Thus the four angles are 120^o, 60^o, 80^o and 100^o.

Ans. Let us say the measure of equal angles is x

Since the sum of interior angles of a quadrilateral is 360^o

Thus x +x + 200^o = 360^o 

→ 2x + 200^o = 360^o

x = 80^o

Thus the measure of each angle is 80^o

Ans. The angle sum property of a quadrilateral states that the sum of all four interior angles of a quadrilateral is 360^o

Ans. We can use this concept in other geometric proofs, such as the sum of all the interior angles of a quadrilateral.

Ans. The formula of the sum of the interior angle = (n -2) * 180^o

Ans. We know that the sum of all the exterior angles of a polygon is 360^o. A quadrilateral is a polygon with four sides. Therefore, the sum of the exterior angles of a quadrilateral is 360^o.

Ans. Let sides AB and CD of the quadrilateral ABCD be equal and also AD=BC. Draw diagonal AC

Δ ABC ≅ Δ CDA

so, ∠ BAC = ∠ DCA

and ∠ BCA = ∠ DAC

ABCD is a parallelogram, where each pair of opposite sides is equal and conversely if each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Ans. Here both its diagonals are intersecting at point O . Measure the length of OA, OB, OC and OD

you will observe that OA = OC and OB = OD

Thus O is the midpoint of both the diagonals. Repeat this activity with some more parallelograms, each time you will find that O is the midpoint of both the diagonals.

Ans. Let ABCD is a parallelogram such that AC = BD.

In ΔABC and ΔDCB,

AC = DB [Given]

AB = DC [Opposite sides of a parallelogram]

BC = CB [Common]

∴ ΔABC ≅ ΔDCB [By SSS congruency]

⇒ ∠ABC = ∠DCB [By C.P.C.T.] …(1)

Now, AB || DC and BC is a transversal. [ Δ ABCD is a parallelogram]

∴ ∠ABC + ∠DCB = 180° … (2) [Co-interior angles]

From (1) and (2), we have

∠ABC = ∠DCB = 90°

i.e., ABCD is a parallelogram having an angle equal to 90°.

∴ ABCD is a rectangle.

Ans.  Let ABCD be a quadrilateral such that the diagonals AC and BD bisect each other at right angles at O.

∴ In ΔAOB and ΔAOD, we have

AO = AO [Common]

OB = OD [O is the mid-point of BD]

∠AOB = ∠AOD [Each 90]

∴ ΔAQB ≅ ΔAOD [By,SAS congruency

∴ AB = AD [By C.P.C.T.] ……..(1)

Similarly, AB = BC .. .(2)

BC = CD …..(3)

CD = DA ……(4)

∴ From (1), (2), (3) and (4), we have

AB = BC = CD = DA

Thus, the quadrilateral ABCD is a rhombus.

Ans. In a parallelogram ABCD, 

Let ∠A be x and ∠B be \(\frac{2}{3}\)x

∴ ∠A + ∠B = 180^o

⇒ x + \(\frac{2}{3}\)x = 180^o

⇒ \(\frac{5}{3}\)x = 180^o

⇒ x^o = 180^o * \(\frac{3}{5}\)  = 108^o

∠A = 108^o, ∠A = \(\frac{2}{3}\) * 108^o = 72^o

Ans. Here, PQ = SR = 12 cm

Let PS = x and PS =QR

Therefore,

x + 12 + x +12 = perimeter

2x + 24 = 40

2x = 16

x = 8

Hence the length of each side of the parallelogram is 12 cm, 8cm, 12 cm and 8 cm.

Ans.  

(i) In ΔAPB and ΔCQD, we have

∠APB = ∠CQD [Each 90°]

AB = CD [ Δ Opposite sides of a parallelogram ABCD are equal]

∠ABP = ∠CDQ

[ Δ Alternate angles are equal as AB || CD and BD is a transversal]

∴ ΔAPB = ΔCQD [By AAS congruency]

(ii) Since, ΔAPB ≅ ΔCQD [Proved]

⇒ AP = CQ [By C.P.C.T.]