MathsICSE Class 8 · Understanding Quadrilaterals

Angle Sum Property of a Quadrilateral: Ratios, Algebra and Bisectors (ICSE Class 8)

Updated 17 October 2026 · Fact-checked

The interior angles of any quadrilateral add up to $360^\circ$, because a diagonal splits it into two triangles of $180^\circ$ each. In a parallelogram, adjacent angles add up to $180^\circ$ and opposite angles are equal.

Notice the diagonal AC cuts the quadrilateral into two triangles, and their six angles together make the four angles of ABCD, so the sum is 360°.

Illustration drawn for QuestionBuddy.

Understand Angle Sum Property of a Quadrilateral

The CISCE curriculum lists the angle sum property among the properties of quadrilaterals.

**Why $36 0^{\circ}$ .** Draw a diagonal from one vertex of a quadrilateral $A B C D$ . It makes two triangles, $A B C$ and $A C D$ , each with angle sum $18 0^{\circ}$ . The angles of the two triangles together fill the four angles of the quadrilateral, so $∠ A + ∠ B + ∠ C + ∠ D = 36 0^{\circ}$ . This works for convex quadrilaterals (all interior angles less than $18 0^{\circ}$ ).

Using the property.

Missing angle: subtract the three known angles from $36 0^{\circ}$ .
Ratio form: if the angles are in the ratio $a : b : c : d$ , one part is $\frac{36 0 ^{\circ}}{a + b + c + d}$ .
Algebraic form: write each angle as an expression in $x$ and set the sum equal to $36 0^{\circ}$ .
Exterior angle: an exterior angle plus its interior angle on the same side form a linear pair, $18 0^{\circ}$ .

Parallelogram. Opposite sides are parallel, so each pair of adjacent angles (co-interior angles) add to $18 0^{\circ}$ , and opposite angles are equal. If one angle is known, all four are known.

Bisector result (for practice). If the bisectors of $∠ A$ and $∠ B$ of a quadrilateral $A B C D$ meet at $O$ , then $∠ A O B = \frac{1}{2} (∠ C + ∠ D)$ , because in triangle $A O B$ the angle at $O$ is $18 0^{\circ} - \frac{1}{2} (A + B)$ and $A + B = 36 0^{\circ} - (C + D)$ .

Checking. After finding all angles, add them: the total must be $36 0^{\circ}$ .

Angle Sum Property of a Quadrilateral: formulas to remember

Angle sum: $∠ A + ∠ B + ∠ C + ∠ D = 36 0^{\circ}$; Any convex quadrilateral.
Parallelogram angles: $∠ A + ∠ B = 18 0^{\circ}, ∠ A = ∠ C$; Adjacent supplementary, opposite equal.
Bisector angle: $∠ A O B = \frac{1}{2} (∠ C + ∠ D)$; Bisectors of angles A and B meeting at O.

How to solve Angle Sum Property of a Quadrilateral questions

Write each unknown angle in terms of one letter and equate the sum to 360 degrees.

1Sketch the quadrilateral and mark the given angles.
2Express every unknown angle in one variable.
3Equate the sum of all four to $36 0^{\circ}$ .
4Solve and find each angle.
5Check that the four angles add up to $36 0^{\circ}$ .

Angle Sum Property of a Quadrilateral in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Four corners of a plot

Suppose a quadrilateral-shaped plot has three corner angles measured as 85 degrees, 105 degrees and 70 degrees.

Find the fourth corner angle.

Try it first, then show the solutionSolution

Sum of three $= 85 + 105 + 70 = 260$ .
Fourth $= 360 - 260 = 100$ degrees.

Answer: $10 0^{\circ}$ .

Memory hook: Four corners always total a full turn.

Tricks to remember Angle Sum Property of a Quadrilateral

Short memory aids. Say each one aloud once, then use it on a question the same day.

Two triangles
A quadrilateral is two triangles stuck together: 2 x 180 = 360.
Use it when: Remembering the sum.
One part
In a ratio problem, 360 divided by the total number of parts gives the value of one part.
Use it when: Ratio angle questions.

Angle Sum Property of a Quadrilateral: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Missing angle · 1 mark · MCQBoard level

Three angles of a quadrilateral are $8 5^{\circ}$ , $10 5^{\circ}$ and $7 0^{\circ}$ . The fourth angle is:
(a) $9 0^{\circ}$ (b) $10 0^{\circ}$ (c) $11 0^{\circ}$ (d) $12 0^{\circ}$

Try it first, then show the solutionSolution

$360 - (85 + 105 + 70) = 360 - 260 = 100$ .

Answer: (b)

Example 2: Angles in a ratio · 3 marks · Short answerBoard level

The angles of a quadrilateral are in the ratio $2 : 3 : 4 : 6$ . Find each angle.

Try it first, then show the solutionSolution

Parts $= 2 + 3 + 4 + 6 = 15$ ; one part $= \frac{360}{15} = 2 4^{\circ}$ .
Angles: $4 8^{\circ}, 7 2^{\circ}, 9 6^{\circ}, 14 4^{\circ}$ .
Check: $48 + 72 + 96 + 144 = 360$ .

Answer: $4 8^{\circ}, 7 2^{\circ}, 9 6^{\circ}, 14 4^{\circ}$

Example 3: Algebraic angles · 3 marks · Short answerHarder board level

The angles of a quadrilateral are $x^{\circ}$ , $(x + 10)^{\circ}$ , $2 x^{\circ}$ and $(2 x + 20)^{\circ}$ . Find $x$ and each angle.

Try it first, then show the solutionSolution

$x + (x + 10) + 2 x + (2 x + 20) = 360 \Rightarrow 6 x + 30 = 360$ .
$6 x = 330 \Rightarrow x = 55$ .
Angles: $5 5^{\circ}, 6 5^{\circ}, 11 0^{\circ}, 13 0^{\circ}$ (sum $36 0^{\circ}$ ).

Answer: $x = 55$ ; angles $5 5^{\circ}, 6 5^{\circ}, 11 0^{\circ}, 13 0^{\circ}$

Example 4: Angles of a parallelogram · 4 marks · Short answerHarder board level

In parallelogram $A B C D$ , $∠ A$ is $3 0^{\circ}$ more than twice $∠ B$ . Find all four angles.

Try it first, then show the solutionSolution

Adjacent angles: $∠ A + ∠ B = 18 0^{\circ}$ . Let $∠ B = x$ , $∠ A = 2 x + 30$ .
$3 x + 30 = 180 \Rightarrow x = 50$ ; $∠ A = 13 0^{\circ}$ .
Opposite angles equal: $∠ C = 13 0^{\circ}$ , $∠ D = 5 0^{\circ}$ .
Check: $130 + 50 + 130 + 50 = 360$ .

Answer: $∠ A = ∠ C = 13 0^{\circ}$ , $∠ B = ∠ D = 5 0^{\circ}$

Example 5: Bisectors of two angles · 4 marks · Long answerHOTS / competency

In quadrilateral $A B C D$ , $∠ C = 10 0^{\circ}$ and $∠ D = 7 0^{\circ}$ . The bisectors of $∠ A$ and $∠ B$ meet at $O$ . Find $∠ A O B$ and justify your answer.

Try it first, then show the solutionSolution

$∠ A + ∠ B = 360 - (100 + 70) = 19 0^{\circ}$ .
In $△ A O B$ : $∠ O A B + ∠ O B A = \frac{1}{2} (190) = 9 5^{\circ}$ .
$∠ A O B = 180 - 95 = 8 5^{\circ}$ .
Check with the result $\frac{1}{2} (∠ C + ∠ D) = \frac{1}{2} (170) = 8 5^{\circ}$ .

Answer: $∠ A O B = 8 5^{\circ}$

Where students lose marks in Angle Sum Property of a Quadrilateral

The slips examiners see most often, and what to write instead.

Mistake: Using 180 degrees for a quadrilateral
That is the triangle sum.
Wrong working
Fourth angle $= 180 - 260$ .
Right working
$360 - 260 = 10 0^{\circ}$ .
Do this: Say "four angles, 360" aloud.
Usually costs: Answer mark.
Mistake: Applying the parallelogram rule to any quadrilateral
Adjacent angles are supplementary only when the sides are parallel.
Wrong working
In quadrilateral $A B C D$ , $∠ A = 10 0^{\circ}$ so $∠ B = 8 0^{\circ}$ .
Right working
Only the sum of all four is fixed at $36 0^{\circ}$ for a general quadrilateral.
Do this: Check whether it is stated to be a parallelogram.
Usually costs: Method marks.

Exam tips

Show the sum equation explicitly and state it equals 360 degrees.
Finish with a check line that adds all four angles.
For ratio questions, write "one part".

Practice questions on Angle Sum Property of a Quadrilateral

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The sum of the interior angles of any quadrilateral is:
(a) $18 0^{\circ}$ (b) $27 0^{\circ}$ (c) $36 0^{\circ}$ (d) $54 0^{\circ}$
Hint
Two triangles.
Answer
(c)
Q2 · 2 marks · Short answerBoard level
Three angles of a quadrilateral are $9 0^{\circ}$ , $7 5^{\circ}$ and $11 0^{\circ}$ . Find the fourth.
Hint
Subtract from 360.
Answer
$360 - 275 = 8 5^{\circ}$ .
Q3 · 3 marks · Short answerHarder board level
The angles of a quadrilateral are in the ratio $3 : 5 : 7 : 9$ . Find the angles and state whether any two of them are supplementary.
Hint
One part is 360/24.
Answer
One part $= 1 5^{\circ}$ ; angles $4 5^{\circ}, 7 5^{\circ}, 10 5^{\circ}, 13 5^{\circ}$ . $45 + 135 = 180$ and $75 + 105 = 180$ : two supplementary pairs.
Q4 · 3 marks · Short answerHarder board level
An exterior angle at vertex $A$ of quadrilateral $A B C D$ is $7 0^{\circ}$ , and $∠ B = 10 0^{\circ}$ , $∠ C = 8 5^{\circ}$ . Find $∠ D$ .
Hint
Interior A + exterior A = 180.
Answer
$∠ A = 180 - 70 = 11 0^{\circ}$ . $∠ D = 360 - (110 + 100 + 85) = 6 5^{\circ}$ .
Q5 · 4 marks · Long answerHOTS / competency
The four angles of a quadrilateral are such that each is $2 0^{\circ}$ more than the previous one. Find them. Can such a quadrilateral have two right angles?
Hint
Let the smallest be x.
Answer
$x + (x + 20) + (x + 40) + (x + 60) = 360 \Rightarrow 4 x + 120 = 360 \Rightarrow x = 60$ : angles $6 0^{\circ}, 8 0^{\circ}, 10 0^{\circ}, 12 0^{\circ}$ . None is $9 0^{\circ}$ and all differ, so it cannot have two right angles.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Angle sum of a quadrilateral?Show answer
360 degrees.
2.In a parallelogram, adjacent angles add to?Show answer
180 degrees.
3.Opposite angles of a parallelogram?Show answer
Equal.
4.Angles in ratio 1:2:3:4: smallest?Show answer
36 degrees.

Angle Sum Property of a Quadrilateral: frequently asked questions

Does the 360 sum hold for a concave quadrilateral?

Yes, if the reflex angle is measured as the interior angle.

Why do two triangles give the sum?

Because the angles of the two triangles together make up all four angles of the quadrilateral.