Opposite vertices.
Two.
Class 9 · Maths · Chapter 8 · बिहार बोर्ड (BSEB)CBSE · NCERT 2026-27
Quadrilaterals
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1. Read — Parts · angle sum · types · Theorems 8.1 to 8.7 · Theorem 8.8 · Exercise 8.1 · mid-point · Exercise 8.2, diagram, worked example, board tip
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चतुर्भुज के भाग · Textbook 8.1 · side · angle · diagonal
Four points make a quadrilateral only when no three lie on one line. There are four sides, four angles and four vertices. The two segments joining opposite vertices are the diagonals.
The book takes only the quadrilateral whose diagonals meet inside the figure. A crossed figure, whose diagonals fall outside, is not a quadrilateral of this chapter. Read the sides in order: AB, BC, CD, DA.
Question: In ABCD the diagonal AC = 10 cm and O is its mid-point. Find AO.
Rule: the mid-point splits the segment into two equal parts.
Substitution: AO = AC / 2 = 10 cm / 2 = 5 cm.
A diagonal is not a side. AC is a side only when it is one of the four boundary sides.
Do not apply the angle sum to a crossed figure by calling it a quadrilateral. The book keeps that figure apart.
Opposite vertices.
Two.
False — no three points may be collinear, and only then is there a closed four-sided figure.
10 / 2.
5 cm.
The diagonals are AC and BD. The sides are AB, BC, CD and DA.
कोणों का योग 360° · Textbook 8.2 · diagonal · two triangles
A diagonal splits a quadrilateral into two triangles. Each triangle gives 180°, so the four angles add to 360°. The check is done by drawing a diagonal. It is not a numbered activity.
In a ratio, first add the parts, then share 360°. In 3 : 5 : 9 : 13 the parts are 3 + 5 + 9 + 13 = 30. One part = 360° / 30 = 12°.
Draw AC in ABCD. The angles of ΔADC and of ΔABC are each 180°. The two sums give 360°. Angle A splits into two pieces on AC, and so does angle C. The pieces join back into the four angles. This is the book’s check.
Question: The angles are in the ratio 3 : 5 : 9 : 13. Find all four measures.
Formula: one part = 360° / (sum of the parts).
Substitution: the sum of the parts = 30. One part = 360° / 30 = 12°. The angles are 36°, 60°, 108° and 156°. The unit is the degree. Check: 36 + 60 + 108 + 156 = 360.
Do not multiply each number of the ratio by 360°. First take 30, then 12°.
Write the check 36 + 60 + 108 + 156 in the answer. If the sum is not 360°, the part is wrong.
Two triangles.
360°.
True — so 180° + 180° = 360°.
360 / 30.
12°.
3 × 12.
36°.
80° + 90° + 110° = 280°. The fourth = 360° − 280° = 80°.
False — the sum is 360°. The angles are 36°, 60°, 108° and 156°.
चतुर्भुज के प्रकार · Textbook 8.3 · trapezium · kite · square
If one pair of opposite sides is parallel, the figure is a trapezium. If both pairs are parallel, it is a parallelogram. A rectangle is a parallelogram with one angle 90°. A rhombus has all four sides equal. A square is both a rectangle and a rhombus.
A kite has two pairs of adjacent sides equal. A kite is not a parallelogram. A rectangle or a rhombus is not a square by itself. In a proof, take a trapezium to mean one pair of parallel sides. A parallelogram needs both pairs.
Question: A parallelogram has all sides equal and one angle 90°. Name it.
Rule: four equal sides make a rhombus. One angle of 90° makes a rectangle. Both together make a square.
Conclusion: the figure is a square. A square is also a parallelogram.
Stopping at “rhombus” for a square is half an answer. Also write that one angle is 90°.
The equal sides of a kite are adjacent. Writing them as opposite becomes a wrong reason for a parallelogram.
Two pairs.
A parallelogram.
False — a kite has equal adjacent sides. Both pairs of opposite sides are not taken as parallel.
Both a rectangle and a rhombus.
A square.
The book’s line.
Both a rectangle and a rhombus.
A trapezium has one pair of opposite sides parallel. A parallelogram has both pairs parallel. A kite is not a parallelogram.
समांतर चतुर्भुज के गुण · Textbook 8.4 · Theorems 8.1 to 8.7
Theorem 8.1: a diagonal divides a parallelogram into two congruent triangles. Theorem 8.2: opposite sides are equal. Theorem 8.4: opposite angles are equal. Theorem 8.6: the diagonals bisect each other.
The converses also make a parallelogram. Theorem 8.3 uses both pairs of opposite sides equal. Theorem 8.5 uses the opposite angles. Theorem 8.7 uses the bisecting diagonals. The diagonals of a rectangle are equal. The diagonals of a rhombus cut at 90°. A square has both facts.
Question: In a parallelogram ∠A = 70°. Find the other three angles.
Rule: opposite angles are equal, Theorem 8.4. Adjacent angles are supplementary, because the opposite sides are parallel and a side is a transversal.
Substitution: ∠C = 70°. ∠B = 180° − 70° = 110°. ∠D = 110°. The unit is the degree.
The opposite of 70° is also 70°. The neighbour is 110°. Do not write all four as 70°.
A rectangle has equal diagonals, and a rhombus has perpendicular diagonals. Write both properties on one figure only when it is a square.
Theorem 8.4.
70°.
True — Theorem 8.6.
180 − 70.
110°.
Opposite sides are equal. Opposite angles are equal. The diagonals bisect each other.
एक जोड़ा बराबर और समांतर — अभ्यास 8.1 · Textbook 8.5 · Theorem 8.8 · Exercise 8.1
Theorem 8.8: if one pair of opposite sides is both equal and parallel, the quadrilateral is a parallelogram. If the sides are only equal, or only parallel, this theorem does not apply. A trapezium has one parallel pair, and equality is not required.
Exercise 8.1 comes at this place. First the ratio of angles, then the tests for a parallelogram. When you choose a test, write the one among 8.3, 8.5, 8.7 and 8.8 that matches the given fact.
The first question is the angle ratio 3 : 5 : 9 : 13. Share 360° into 30 parts. In the later questions, see what is given: both pairs of sides, both pairs of angles, the mid-point of the diagonals, or one pair equal and parallel. Write one test and give the theorem number.
Question: In ABCD, AB = CD and AB ∥ CD. Is ABCD a parallelogram?
Theorem: 8.8. One pair of opposite sides is equal and parallel.
Conclusion: yes, ABCD is a parallelogram. Writing only AB ∥ CD was not enough. Equality is given as well.
Do not apply 8.8 for parallel alone. The answer must contain both words, equal and parallel.
In the first ratio question show 36°, 60°, 108° and 156°. Repeating 3 : 5 : 9 : 13 is not the measure.
10-second revision
Theorem 8.8.
A parallelogram.
False — 8.8 needs both equal and parallel.
13 × 12.
156°.
The converse.
8.7.
8.3 both pairs of opposite sides equal. 8.5 both pairs of opposite angles equal. 8.7 diagonals bisect. 8.8 one pair equal and parallel.
True — in the book 8.1 comes after section 8.5, and the mid-point work is in 8.6.
मध्य-बिंदु प्रमेय · Textbook 8.6 · Theorem 8.9 · Theorem 8.10
Theorem 8.9: the segment joining the mid-points of two sides is parallel to the third side and half of it. The proof draws one more parallel and builds a congruent triangle and a parallelogram.
Theorem 8.10 is the converse. A line through the mid-point of one side, parallel to a second side, bisects the third side. Half means half the length, not half the angle.
Draw a triangle and mark the mid-points E and F of two sides. Join EF. The book says EF will come out parallel to the third side and half of it. This is not a numbered activity. The measurement does not replace Theorem 8.9.
Question: E and F are the mid-points of AB and AC. BC = 10 cm. Find EF.
Formula: EF = BC / 2, and EF ∥ BC. Theorem 8.9.
Substitution: EF = 10 cm / 2 = 5 cm. The unit is cm.
Writing EF = BC is wrong. The mid-point segment is half, not the whole side.
Show 10 cm / 2 = 5 cm. The word “half” without the measure can lose the mark.
10-second revision
Theorem 8.9.
Half and parallel.
True — the parallel line bisects the third side.
Half.
5 cm.
Theorem 8.10.
The mid-point of AC.
The segment joining the mid-points of two sides is parallel to the third and half of it. EF = 18 cm / 2 = 9 cm.
अभ्यास 8.2 — चारों मध्य-बिंदु · Textbook 8.6 · Exercise 8.2
Exercise 8.2 asks you to join the mid-points of the four sides of a quadrilateral. Joined in order, the new quadrilateral is a parallelogram. The same mid-point language runs on a rectangle, a rhombus or a trapezium.
The reason is Theorem 8.9. In each triangle cut by a diagonal, the mid-segment is parallel and half. Two such half segments make one pair of parallel sides. Then Theorem 8.8 or 8.3 gives the parallelogram.
Take P, Q, R and S as the mid-points of the sides and join them in order. Show that PQRS is a parallelogram. If a rhombus or a rectangle is given, find the new angles or sides from the same theorem. In a trapezium with AB ∥ DC and E a mid-point, use 8.10 to catch the other mid-point.
Question: PQRS is formed by the mid-points of ABCD. One diagonal AC = 12 cm. Find the mid-segment that is half of that diagonal.
Formula: mid-segment = third side / 2, when the third side is the diagonal of 12 cm.
Substitution: half = 12 cm / 2 = 6 cm. The unit is cm. PQRS is a parallelogram.
Do not write PQRS as a square unless a 90° angle or equal diagonals are given. The general answer is a parallelogram.
With 12 cm / 2 = 6 cm write theorem number 8.9. Only 6 cm is half a reason.
10-second revision
The tenth line of the summary.
A parallelogram.
False — 8.1 comes first, and 8.2 comes after the mid-point theorem.
Half.
6 cm.
In each triangle cut by a diagonal, Theorem 8.9 makes the mid-segment parallel. Two parallel pairs appear, so PQRS is a parallelogram.
Pick a type. The 34 lesson checks are separate — each lesson has as many as its topic needs. All correct earns mastery ★.
No question is marked as a verified past paper. The BSEB set is a model for practice. CBSE items are CBSE-style, not a copy of any year’s paper.
Two triangles.
360°.
3 × 12.
36°.
Both names.
Both a rectangle and a rhombus.
Theorem 8.4.
Equal.
180 − 70.
110°.
The converse.
8.7.
Theorem 8.8.
A parallelogram.
Theorem 8.9.
Half.
Half.
5 cm.
Exercise 8.2.
A parallelogram.
The summary line.
90°.
Adjacent sides.
Not a parallelogram.
The summary.
Equal.
False — the sum is 360°. 180° belongs to one triangle.
True — and a rectangle as well.
True.
False — the side must also be equal.
False — it is parallel and half.
True — 8.1, before it, asks for angles and the parallelogram tests.
False — in a rectangle the diagonals are equal. Cutting at right angles is a property of a rhombus.
False — the book takes the figure whose diagonals meet inside.
Two triangles.
360.
360 / 30.
12.
180 − 70.
110.
The converse is 8.3.
8.2.
Half.
5.
12 / 2.
6.
A right angle.
90.
One pair.
8.8.
8.2 is the sides, 8.4 the angles, 8.6 the diagonals, and 8.8 equal and parallel.
A parallelogram bisects, a rectangle has equal diagonals, a rhombus cuts at 90°, and the mid-segment is half.
Assertion (A): The angles of a quadrilateral add to 360°.
Reason (R): One diagonal makes two triangles, and each triangle gives 180°.
Both are true and R is the reason.
Assertion (A): The diagonals of a parallelogram bisect each other.
Reason (R): The mid-point segment is half of the third side.
Both are true, but R does not explain the diagonal statement.
Assertion (A): A square is a rectangle.
Reason (R): Every rectangle is a square.
A is true. R is false — the sides of a rectangle need not be equal.
Assertion (A): In Theorem 8.9 the mid-segment equals the third side.
Reason (R): It is parallel to the third side and half of it.
A is false. R is true.
Assertion (A): AB = CD and AB ∥ CD make ABCD a parallelogram.
Reason (R): If one pair of opposite sides is equal and parallel, the quadrilateral is a parallelogram.
Both are true and R is Theorem 8.8.
A square, a rectangle and a rhombus are parallelograms. A kite is not.
Opposite sides are 8.3. Equal and parallel is 8.8. The mid-point theorem is 8.9.
360°.
The diagonals of a parallelogram bisect each other.
36°, 60°, 108° and 156°.
Half, and parallel.
At 90°.
The opposite ∠C = 65°. An adjacent angle = 180° − 65° = 115°. So ∠B = ∠D = 115°.
EF = 14 cm / 2 = 7 cm, and EF ∥ BC. Theorem 8.9.
Theorem 8.8. Parallel alone can also make a trapezium. Equality is the second condition.
The diagonals of a rectangle are equal. The diagonals of a rhombus cut at 90°. In a square the diagonals are equal and they also cut at 90°.
A diagonal makes two triangles, 180° + 180° = 360°. The parts are 30, and one part is 12°. The angles are 36°, 60°, 108° and 156°. The sum is 360°.
Opposite sides are equal. Opposite angles are equal. The diagonals bisect. One pair equal and parallel also gives a parallelogram. The angles are 80°, 100°, 80° and 100°.
8.9: the mid-segment is parallel and half. 8.10: a parallel line bisects the third side. EF = 8 cm. At the four mid-points, Theorem 8.9 makes the parallel pairs.
This model set is for practice. It is not a question from any year’s annual examination. Annual questions will be added only when a source page is available.
180 − 100.
80°.
Theorem 8.9.
Half.
360 − 270.
90.
True.
EF = 11 cm and EF ∥ BC. A measurement makes the figure clear. The reason is Theorem 8.9.
These are competency-based practice questions. They are not copies of a CBSE paper.
The rectangle summary.
Equal, and they bisect each other.
Exercise 8.2.
The quadrilateral of the mid-points is a parallelogram.
Assertion (A): A square is a parallelogram.
Reason (R): Every parallelogram is a square.
A is true. R is false — the sides of a parallelogram need not be equal, nor the angles 90°.
False — the sum is 360°, so the angles are 36°, 60°, 108° and 156°.
The opposite corner is 75°, and the adjacent corners are 105° and 105°. The measure is a check. The reason is Theorem 8.4 and the adjacent sum 180°.
By Theorem 8.9 the length = 24 m / 2 = 12 m. The path is parallel to the third side.
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What you learned
| What | Keep this |
|---|---|
| Angle sum | 360° |
| Opposite sides | equal in a parallelogram |
| Diagonals | bisect each other |
| Rectangle | equal diagonals |
| Rhombus | diagonals at 90° |
| Mid-point | parallel and half |
The notes are original writing. The textbook was used only for activity order and numbers. “Verified” will be used only on a question that has a source page.