RowQ
The Vault
RowQ
The Vault
CBSE Class 9 Maths · 11 questions · 26 marks
A quadrilateral is simply a closed figure with four sides, but the family it belongs to — parallelogram, rhombus, rectangle, square, trapezium or kite — decides which shortcuts you are allowed to use. This chapter builds every property from the congruence rules you already know, so nothing has to be memorised blindly. The Mid-point Theorem at the end is the quiet star: it links triangles and quadrilaterals in one line.
The angles of a quadrilateral are in the ratio 3 : 4 : 5 : 6. The largest angle is:
Answer
The largest angle is 120°. Let the angles be 3x, 4x, 5x and 6x. The angle sum of a quadrilateral is 360°, so 3x + 4x + 5x + 6x = 360°. 18x = 360°, giving x = 20°. The angles are 60°, 80°, 100° and 120°, so the largest is 6x = 120°.
In parallelogram ABCD, ∠A = 108°. The measure of ∠B is:
Answer
∠B = 72°. ∠A and ∠B are adjacent (co-interior) angles of a parallelogram, and adjacent angles of a parallelogram are supplementary because AD is parallel to BC with AB as a transversal. So ∠B = 180° - 108° = 72°.
In triangle PQR, M and N are the mid-points of PQ and PR respectively. If QR = 13.4 cm, then MN equals:
Answer
MN = 6.7 cm. By the Mid-point Theorem, the segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half its length. So MN = (1/2)QR = (1/2)(13.4) = 6.7 cm.
Which quadrilateral has diagonals that are equal in length AND bisect each other at right angles?
Answer
The square. A rhombus has diagonals that bisect each other at right angles, but they are unequal. A rectangle has equal diagonals that bisect each other, but not at right angles. A kite has perpendicular diagonals, but only one diagonal is bisected. A square is both a rhombus and a rectangle, so it inherits both properties: equal diagonals that bisect each other perpendicularly.
Assertion (A): Every rhombus is a square. Reason (R): In a rhombus all four sides are equal in length.
Answer
A is false but R is true. Reason R is true — equal sides on all four edges is the defining property of a rhombus. Assertion A is false because a square needs equal sides AND four right angles. A rhombus with angles 50°, 130°, 50°, 130° has four equal sides but is not a square. The correct statement is the reverse: every square is a rhombus.
In parallelogram ABCD, ∠A = (3x + 10)° and ∠C = (5x - 30)°. Find the value of x and the measure of ∠B.
Answer
∠A and ∠C are opposite angles of a parallelogram, so they are equal. 3x + 10 = 5x - 30 40 = 2x, so x = 20. Therefore ∠A = 3(20) + 10 = 70°. ∠B is adjacent to ∠A and adjacent angles are supplementary, so ∠B = 180° - 70° = 110°.
Prove that a diagonal of a parallelogram divides it into two congruent triangles.
Answer
Let ABCD be a parallelogram with diagonal AC drawn. Since AB is parallel to DC and AC is a transversal, ∠BAC = ∠DCA (alternate interior angles). Since AD is parallel to BC and AC is a transversal, ∠DAC = ∠BCA (alternate interior angles). Also AC = CA (common side). In triangles ABC and CDA, we have two pairs of equal angles with the included side AC equal, so by the ASA criterion, triangle ABC ≅ triangle CDA. Hence the diagonal AC splits the parallelogram into two congruent triangles.
The sides of a triangle measure 10 cm, 14 cm and 18 cm. Find the perimeter of the triangle formed by joining the mid-points of its sides.
Answer
By the Mid-point Theorem, each side of the mid-point triangle equals half of one side of the original triangle. The three sides are (1/2)(10) = 5 cm, (1/2)(14) = 7 cm and (1/2)(18) = 9 cm. Perimeter = 5 + 7 + 9 = 21 cm, which is exactly half the perimeter of the original triangle (42 cm).
(a) Prove that the diagonals of a parallelogram bisect each other. (b) In parallelogram PQRS the diagonals meet at O with PO = (2x + 1) cm, OR = (x + 7) cm, QO = (3y - 4) cm and OS = (y + 6) cm. Find x, y and the lengths of both diagonals.
Answer
(a) Let PQRS be a parallelogram whose diagonals PR and QS meet at O. Since PQ is parallel to SR and PR is a transversal, ∠QPO = ∠SRO (alternate interior angles). Similarly, with QS as the transversal, ∠PQO = ∠RSO. Also PQ = SR (opposite sides of a parallelogram). In triangles POQ and ROS, two angles and the included side are equal, so by ASA, triangle POQ ≅ triangle ROS. By CPCT, PO = OR and QO = OS, which means O is the mid-point of both diagonals. Hence the diagonals bisect each other. (b) Using PO = OR: 2x + 1 = x + 7, so x = 6. Then PO = OR = 13 cm and diagonal PR = 13 + 13 = 26 cm. Using QO = OS: 3y - 4 = y + 6, so 2y = 10 and y = 5. Then QO = OS = 11 cm and diagonal QS = 11 + 11 = 22 cm.
ABCD is a quadrilateral and P, Q, R, S are the mid-points of AB, BC, CD and DA respectively. (a) Prove that PQRS is a parallelogram. (b) If the diagonals AC and BD are both 12 cm long, find the perimeter of PQRS and name the special quadrilateral it becomes, with a reason.
Answer
(a) Join the diagonal AC. In triangle ABC, P and Q are the mid-points of AB and BC, so by the Mid-point Theorem PQ is parallel to AC and PQ = (1/2)AC. In triangle ADC, S and R are the mid-points of AD and DC, so SR is parallel to AC and SR = (1/2)AC. Therefore PQ is parallel to SR and PQ = SR. A quadrilateral in which one pair of opposite sides is both equal and parallel is a parallelogram, so PQRS is a parallelogram. (b) Joining diagonal BD in the same way gives QR = SP = (1/2)BD. With AC = BD = 12 cm, every side is (1/2)(12) = 6 cm. Perimeter = 4 × 6 = 24 cm. Since PQRS is a parallelogram with all four sides equal, it is a rhombus.
Read the following and answer the questions that follow: A craft club stitches kite-shaped banners. Each banner is a quadrilateral KLMN in which KL = KN and ML = MN, with a bamboo rod along the diagonal KM. For one design the club sets ∠K = 84° and ∠M = 116°. (a) Prove that triangle KLM ≅ triangle KNM. (b) Using part (a), explain why ∠L = ∠N. (c) Find the measures of ∠L and ∠N. (d) Does the rod KM bisect ∠K? Justify your answer.
Answer
(a) In triangles KLM and KNM: KL = KN (given) ML = MN (given) KM = KM (common side) All three pairs of sides are equal, so by the SSS criterion, triangle KLM ≅ triangle KNM. (b) Corresponding parts of congruent triangles are equal (CPCT). ∠L in triangle KLM corresponds to ∠N in triangle KNM, so ∠L = ∠N. (c) The angle sum of a quadrilateral is 360°, so ∠K + ∠L + ∠M + ∠N = 360°. 84° + ∠L + 116° + ∠N = 360°, giving ∠L + ∠N = 160°. Since ∠L = ∠N from part (b), each is 160°/2 = 80°. So ∠L = ∠N = 80°. (d) Yes. By CPCT applied to the same congruent triangles, ∠LKM = ∠NKM, so the rod KM divides ∠K into two equal parts and therefore bisects it. Each half measures 84°/2 = 42°.
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