Welcome to AssamSchool. This article contains 30 Extra Questions from Quadrilaterals from Class 9 Mathematics Chapter 8. These Quadrilaterals Class 9 Extra Questions are helpful to test your understanding and knowledge of the chapter. These additional questions will not only help you to excel in the chapter but also will give you a strong foundation in Mathematics.
We have also provided step-by-step answers to the questions for reference. So, let’s start…
30 Quadrilaterals Class 9 Extra Questions
1. In a parallelogram, if one angle is \(70°\), find the measures of all other angles.
2. True or False: The diagonals of a rhombus are equal in length.
3. What is the length of the line segment joining the mid-points of two sides of a triangle if the third side is \(12 cm\)?
4. Prove that a quadrilateral with diagonals bisecting each other at \(90°\) is a rhombus.
5. In rectangle \(ABCD\), diagonals \(AC\) and \(BD\) intersect at \(O\). If \(∠AOD = 110°\), find \(∠OAD\).
6. If the mid-points of the sides of a quadrilateral are joined in order, prove that the resulting figure is a parallelogram.
7. In a parallelogram \(ABCD\), \(E\) and \(F\) are mid-points of \(AB\) and \(CD\) respectively. Show that \(AF\) and \(EC\) trisect diagonal \(BD\).
8. Prove that the figure formed by joining the mid-points of the sides of a rectangle is a rhombus.
9. In a square, show that the diagonals bisect each other at right angles.
10. In a trapezium \(ABCD\), \(AB || CD\) and \(AD = BC\). Prove that \(∠A = ∠B\).
11. \(ABCD\) is a quadrilateral where diagonals \(AC\) and \(BD\) bisect each other. If \(∠A = 90°\), prove that \(ABCD\) is a rectangle.
12. If two adjacent angles of a parallelogram are in the ratio \(2:3\), find all angles.
13. Prove that the bisectors of two consecutive angles of a parallelogram intersect at a right angle.
14. In \(ΔABC\), \(D\), \(E\), and \(F\) are mid-points of \(AB\), \(BC\), and \(CA\). Show that \(ΔABC\) is divided into four congruent triangles.
15. The diagonals of a parallelogram intersect at point \(O\). If \(AO = 3 cm\), find the length of diagonal \(AC\).
16. Show that if the mid-points of the sides of a rhombus are connected, the resulting figure is a rectangle.
17. Prove that a line segment joining the mid-points of the non-parallel sides of a trapezium is parallel to the bases and equal to half their difference.
18. In a rhombus, one diagonal is \(16 cm\), and the other is \(12 cm\). Find the side length.
19. Prove that in a parallelogram, the angle bisectors of two opposite angles are parallel.
20. \(ABCD\) is a parallelogram. \(P\) and \(Q\) are points on diagonal \(BD\) such that \(DP = PQ = QB\). Prove that \(APCQ\) is a parallelogram.
21. In a rectangle, diagonals are \(10 cm\) each. Find the perimeter if one side is \(6 cm\).
22. In a square \(ABCD\), show that the diagonals are perpendicular bisectors of each other.
23. In \(ΔABC\), \(D\) is the mid-point of \(AB\). A line through \(D\) parallel to \(BC\) meets \(AC\) at \(E\). Prove that \(E\) is the mid-point of \(AC\).
24. In a parallelogram, if \(AB = 5 cm\) and \(AD = 3 cm\), find the lengths of \(BC\) and \(CD\).
25. Prove that the diagonals of an isosceles trapezium are equal.
26. In quadrilateral \(ABCD\), \(AB = CD\) and \(∠A = ∠C = 90°\). Prove that \(ABCD\) is a rectangle.
27. If one diagonal of a rhombus is equal to its side, find its angles.
28. In a parallelogram, show that the sum of squares of all sides equals the sum of squares of the diagonals.
29. \(ABCD\) is a trapezium with \(AB || CD\). \(E\) and \(F\) are mid-points of \(AD\) and \(BC\) respectively. Prove that \(EF = (AB + CD)/2\).
30. In a rhombus, diagonals are \(24 cm\) and \(10 cm\). Find its area.
Answers
- Solution:
In a parallelogram, consecutive angles are supplementary.
Given \(∠A = 70°\), then \(∠B = 180° − 70° = 110°\)
Opposite angles are equal: \(∠C = ∠A = 70°\), \(∠D = ∠B = 110°\) - Solution:
False. The diagonals of a rhombus are unequal but bisect each other at \(90°\) - Solution:
By the Mid-point Theorem, the segment is half the third side: \(12/2 = 6 cm\) - Solution:
Let diagonals \(AC\) and \(BD\) bisect each other at \(90°\)
Since diagonals bisect each other, \(ABCD\) is a parallelogram.
Diagonals intersecting at \(90°\) imply it’s a rhombus (Theorem 8.6 converse). - Solution:
Diagonals in a rectangle are equal and bisect each other.
\(∠AOD = 110°\), so \(∠OAD = (180° − 110°)/2 = 35°\) - Solution:
Let mid-points of quadrilateral \(ABCD\) be (\P, Q, R, S\)
By Mid-point Theorem, \(PQ || AC || SR\) and \(PQ = SR = ½ AC\)
Similarly, \(QR || BD || PS\). Hence, \(PQRS\) is a parallelogram. - Solution:
Let \(BD\) intersect \(AF\) at \(P\) and \(EC\) at \(Q\)
In \(ΔABD\), \(E\) is mid-point of \(AB\), so \(EP || AD\) (Mid-point Theorem)
Similarly, \(FQ || BC\). Using congruency, \(DP = PQ = QB\) - Solution:
Let mid-points of rectangle \(ABCD\) be \(P, Q, R, S\)
\(PQ = QR = RS = SP\) (all sides equal to half diagonals)
Diagonals of rectangle are equal, so \(PQRS\) is a rhombus. - Solution:
In square \(ABCD\), diagonals \(AC\) and \(BD\) intersect at \(O\)
\(AO = OC\), \(BO = OD\) (bisect each other).
\(ΔAOB ≅ ΔBOC\) (SAS), so \(∠AOB = 90°\) - Solution:
Draw \(CE || AD\) meeting \(AB\) at \(E\). \(ADCE\) is a parallelogram.
\(∠A = ∠B\) (since \(AD = BC\) and \(ΔCEB\) is isosceles). - Solution:
Diagonals bisect each other ⇒ \(ABCD\) is a parallelogram.
Given \(∠A = 90°\), so all angles are \(90° ⇒ ABCD\) is a rectangle. - Solution:
Let angles be \(2x\) and \(3x\).(\2x + 3x = 180° ⇒ x = 36°\)
Angles: \(72°, 108°, 72°, 108°\) - Solution:
Let \(∠A\) and \(∠B\) bisectors meet at \(P\)
\(∠PAB + ∠PBA = (∠A/2) + (∠B/2) = (180°)/2 = 90°\)
\(⇒ ∠APB = 90°\) - Solution:
By Mid-point Theorem, \(DE || AC\), \(EF || AB\), \(FD || BC\)
Each small triangle is congruent by SSS. - Solution:
\(AO = 3 cm ⇒ AC = 2 × AO = 6 cm\) - Solution:
Midpoints \(PQRS\) of rhombus \(ABCD\) form a quadrilateral with sides parallel to diagonals.
Diagonals of rhombus are perpendicular ⇒ \(PQRS\) has right angles ⇒ rectangle. - Solution:
Let \(EF\) join midpoints \(E\) and \(F\) of \(AD\) and \(BC\) in trapezium \(ABCD\)
Extend \(AD\) and \(BC\) to meet at \(G\). Use similar triangles to show \(EF = (AB − CD)/2\) - Solution:
Side = \(√[(8)² + (6)²] = √100 = 10 cm\) - Solution:
Let bisectors of \(∠A\) and \(∠C\) meet at \(O\)
Alternate angles formed by bisectors and parallels are equal ⇒ bisectors are parallel. - Solution:
In \(ΔAPD\) and \(ΔCQB\), \(DP = BQ\), \(AD = CB\), \(∠ADP = ∠CBQ\) ⇒ triangles congruent \(⇒ AP = CQ\)
Similarly, \(AQ = CP ⇒ APCQ\) is a parallelogram. - Solution:
Let sides be \(6 cm\) and \(x\). Diagonal \(= 10 cm ⇒ 6² + x² = 10² ⇒ x = 8 cm\)
Perimeter \(= 2(6 + 8) = 28 cm\) - Solution:
Diagonals in a square bisect at \(90°\) and are equal (properties of square). - Solution:
By Mid-point Theorem, \(DE || BC ⇒ E\) is mid-point of \(AC\) (Theorem 8.9) - Solution:
Opposite sides equal: \(BC = AD = 3 cm, CD = AB = 5 cm\) - Solution:
In isosceles trapezium \(ABCD\), \(AD = BC\)
\(ΔABD ≅ ΔBAC (SAS) ⇒ AC = BD\) - Solution:
\(AB = CD\) and \(∠A = ∠C = 90° ⇒ ABCD\) is a parallelogram with right angles ⇒ rectangle. - Solution:
Let diagonal \(AC = side AB\).
Then \(ΔABC\) is equilateral \(⇒ ∠ABC = 60°, ∠BAD = 120°\) - Solution:
In parallelogram, \(AB² + BC² + CD² + DA² = 2(AB² + BC²)\) (since \(AB = CD, BC = AD\))
Diagonals: \(AC² + BD² = 2(AB² + BC²) ⇒\) equality holds. - Solution:
Connect midpoints \(E\) and \(F\). By Mid-point Theorem, \(EF || AB\) and \(EF = (AB + CD)/2\) - Solution:
Area \(= (d1 × d2)/2 = (24 × 10)/2 = 120 cm²\)
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Abdur Rohman is an Electrical Engineer from Charaideo, Assam, who wears multiple hats as a part-time teacher, blogger, entrepreneur, and digital marketer. Passionate about education, he founded The Assam School blog to provide free, comprehensive textbook solutions, MCQs (Multiple Choice Questions), and other academic content for students from Class V to XII.