Understanding Quadrilaterals - 8th math

Angle Sum Property of a Polygon

Angle Sum Property of a Triangle

The sum of all the three angles of a triangle is equal to 180⁰.

Example (i):

Here, angle A + angle B + angle C = 180⁰

If angle A = 60⁰ and angle B = 45⁰, then angle C = ?

We know that, sum of all the three angles of a triangle = 180⁰

Therefore, angle A + angle B + angle C = 180⁰

⇒ 60⁰ + 45⁰ + angle C = 180⁰

⇒ 105⁰ + angle C = 180⁰

⇒ angle C = 180⁰ – 105⁰

⇒ angle C = 75⁰ Answer

Example (ii) : In the given triangle if two angles are 45⁰ and 75⁰, then find other angles.

Here, angle c = 45⁰ and angle y = 75⁰

And, since angle c and angle n are vertically opposite angles hence they are equal.

Thus, angle n = angle c = 45⁰

Now, we know that, angle c and angle m are supplementary

Thus, angle c + angle m = 180⁰

⇒ 45⁰ + angle m = 180⁰

⇒ angle m = 180⁰ – 45⁰

⇒ angle m = 135⁰

And, since angle m and angle g are vertically opposite angles hence they are equal.

Thus, angle m = angle g = 135⁰

And, since angle c and angle n are vertically opposite angles hence they are equal.

Thus, angle n = angle c = 45⁰

And, since angle y and angle b are vertically opposite angles hence they are equal.

Thus, angle b = angle y = 75⁰

Now, angle y and angle b are supplementary as they together form a straight line.

Thus, angle y + angle a = 180⁰

⇒ 75⁰ + angle a = 180⁰

⇒ angle a = 180⁰ – 75⁰

⇒ angle a = 105⁰

And, since angle a and angle d are vertically opposite angles hence they are equal.

Thus, angle a = angle d = 105⁰

Now, in the given triangle,

Angle a = 105⁰, angle c = 45⁰

And we know that, sum of all the three angles of a triangle = 180⁰

Therefore, angle a + angle c + angle h = 180⁰

⇒ 105⁰ + 45⁰ + angle h = 180⁰

⇒ 150⁰ + angle h = 180⁰

⇒ angle h = 180⁰ – 150⁰

⇒ angle h = 30⁰

Now, since angle h and angle q are vertically opposite angle and hence are equal

Thus, angle q = angle h = 30⁰

Now, angle h and angle r together form a straight line and hence are supplementary

Thus, angle h + angle r = 180⁰

⇒ 30⁰ + angle r = 180⁰

⇒ angle r = 180⁰ – 30⁰

⇒ angle r = 150⁰

Now, angle r and angle p are vertically opposite angles, thus they are equal

Thus, angle p = angle r = 150⁰

Angle sum property of a quadrilateral

Sum of all the four angles of a quadrilateral is equal to 360⁰

This means, in a quadrilateral

Angle a + angle b + angle a + angle b = 360⁰

Example (i): If three angles of a quadrilateral are 65⁰, 95⁰ and 80⁰ respectively find the fourth angle of the given quadrilateral.

Solution

We know that, sum of all the four angles of a quadrilateral = 360⁰

Therefore, 65⁰ + 95⁰ + 80⁰ + angle D = 360⁰

⇒ 240⁰ + angle D = 360⁰

⇒ angle D = 360⁰ – 240⁰

⇒ angle D = 120⁰ Answer

Example (ii): If four angles of a pentagon are 65⁰, 75⁰, 95⁰ and 70⁰ respectively, then find the fifth unknown angle.

Solution

We know that, Sum of all the angles of a polygon = 360⁰

Therefore, in the given figure of pentagon

65⁰ + 75⁰ + 95⁰ + 70⁰ + a = 360⁰

⇒ 305⁰ + a = 360⁰

⇒ a = 360⁰ – 305⁰

⇒ a = 55⁰ Answer

Exterior Angle Sum Property of a Polygon

The sum of all the exterior angles of a polygon is equal to 360⁰.

In the given figure of quadrilateral,

`/_a+/_b+/_c+/_d =360^o`

Example (i) If the four exterior angles of a pentagon are 50⁰, 65⁰, 60⁰ and 75⁰ respectively, then find the fifth exteriror angle of that pentagon.

Solution

Let the figure of pentagon is given below.

And as given in the question, angle a = 50⁰

Angle b = 65⁰

Angle c = 60⁰

And angle d = 75⁰

Then angle e = ?

We know that, sum of all the exterior angles of a pentagon = 360⁰

Therefore, in the given pentagon,

`50^o\+65^o\+60^o\+75^o\+/_e=360^o`

`:. /_e+250^0=360^o`

`=>/_e=360^0-250^o`

Thus, angle a = 110⁰ Answer

Example (ii) Find all the unknown angles in the given pentagon.

In ∠DAE and ∠ PAE

∠DAE and ∠ PAE both form a straight line and hence are supplementary angles.

Thus, ∠DAE + 75⁰ = 180⁰

⇒ ∠DAE = 180⁰ – 75⁰

⇒ ∠DAE = 105⁰

Now, since ∠ PAE and ∠ DAO are vertically opposite angles, and hence are equal

Thus, ᩐ DAO = ∠ PAE = 75⁰

And since ∠ PAO and ∠ DAO are vertically opposite angles, and hence are equal

Thus, ∠ PAO = ∠ DAO = 105⁰

Now, in between angle AEF and angle GEB

∠ AEF and ∠GEB are vertically opposite angles and hence are equal

Thus, ∠ AEF = ∠ GEB = 65⁰

Now, in between ∠GEB and ∠AEB

Both angles GEB and AEB together form a straight line and hence are supplementary.

Thus, ∠ GEB + ∠ AEB = 180⁰

⇒ 65⁰ + ∠ AEB = 180⁰

⇒ ∠ AEB = 180⁰ – 65⁰

⇒ ∠ AEF = 115⁰

Now, between angles AEB and angle FEG

Both the angles AEB and FEG are vertically opposite angles and hence are equal

Thus, ∠ AEF = ∠ FEG = 115⁰

Now, between angles EBC and angle HBJ

∠ EBC and ∠ HBJ are vertically opposite angles and hence are equal

Thus, ∠ EBC = ∠ HBJ = 110⁰

Now, between angles EBH and EBC

Both of the angles EBH and EBC form a straight line and hence are supplementary

Thus, ∠ EBH + ∠ EBC = 180⁰

⇒ ∠ EBH + 110⁰ = 180⁰

⇒ ∠EBH = 180⁰ – 110⁰

⇒ ∠ EBH = 70⁰

Now between angles EBH and CBJ

Both the angles EBH and CBJ are vertically opposite angles.

Thus, ∠ EBH = ∠ EBJ = 70⁰

Now, between angles DCB and LCK

Both the angles DCB and LCK are vertically opposite angles

Thus, ∠ECB = ∠ LCK = 115⁰

Between angles DCB and BCK

Angles DCB and BCK together form a straight line, and hence are supplementary angles.

Thus, ∠ DCB + ∠ BCK = 180⁰

⇒ 115⁰ + ∠ BCK = 180⁰

⇒ ∠ BCK = 180⁰ – 115⁰

⇒ ∠ BCK = 65⁰

Between angles BCK and DCL

Angles BCK and DCL are vertically opposite angles, and hence are equal

Thus, ∠ BCK = ∠ DCL = 65⁰

Between angles ADC and NDM

Angles ADC and NDM are vertically opposite angles, and hence are equal

Thus, ∠ ADC = ∠ NDM = 120⁰

Between angles ADC and CDM

Angles ADC and CDM together form a straight line, and hence are supplementary angles.

Thus, ∠ ADC + ∠ CDM = 180⁰

⇒ 120⁰ + ∠ CDM = 180⁰

⇒ ∠ CDM = 180⁰ – 120⁰

⇒ ∠ CDM = 60⁰

Now, between angles CDM and ADN

Since, angles CDM and ADN are vertically opposite angles and hence are equal.

Thus, ∠ CDM = ∠ ADN = 60⁰

Thus,

∠ DAE = 105⁰

∠ PAO = 105⁰

∠ OAD = 75⁰

∠ FEA = 65⁰

∠ AEB = 115⁰

∠ FEG = 115⁰

∠ EBH = 70⁰

∠ HBJ = 110⁰

∠ CBJ = 70⁰

∠ BCK = 65⁰

∠ KCL = 115⁰

∠ DCL = 65⁰

∠ CDM = 60⁰

∠ NDM = 120⁰

And, ∠ ADN = 60⁰

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