SPECIAL TYPES OF QUADRILATERALS - Q&A
EXERCISE 17
6. ABCD is a rectangle. If ∠BPC = 124°, calculate:
(i) ∠BAP
(ii) ∠ADP
Solution:
In a rectangle, diagonals are equal and bisect each other. Thus, in ΔBPC, PB = PC.
So, ΔBPC is an isosceles triangle.
∠PBC = ∠PCB = (180° - 124°)/2 = 56°/2 = 28°.
(i) Each angle of a rectangle is 90°.
∠ABC = 90°.
∠ABP = 28°.
∠BAP is not part of ΔBPC directly. However, ΔAPB is isosceles (PA=PB).
∠APB + 124° = 180° (Linear pair) ⇒ ∠APB = 56°.
In ΔAPB: ∠BAP = ∠ABP = (180° - 56°)/2 = 62°.
(ii) ∠ADP:
Diagonals bisect each other, so PD = PA. ΔAPD is isosceles.
∠APD = ∠BPC = 124° (Vertically opposite angles).
∠ADP = ∠PAD = (180° - 124°)/2 = 28°.
Answer: (i) 62° (ii) 28°
7. ABCD is a rhombus. If ∠BAC = 38° find:
(i) ∠ACB
(ii) ∠DAC
(iii) ∠ADC
Solution:
(i) In a rhombus, sides are equal (AB = BC).
ΔABC is isosceles ⇒ ∠ACB = ∠BAC.
∠ACB = 38°.
(ii) Diagonals bisect interior angles in a rhombus.
∠DAC = ∠BAC = 38°.
(iii) ∠DAB = ∠DAC + ∠BAC = 38° + 38° = 76°.
Consecutive angles are supplementary: ∠ADC + ∠DAB = 180°.
∠ADC = 180° - 76° = 104°.
Answer: (i) 38° (ii) 38° (iii) 104°
8. ABCD is a rhombus. If ∠BCA = 35° find ∠ADC.
Solution:
Diagonals bisect angles: ∠DCA = ∠BCA = 35°.
Total ∠BCD = 70°.
Opposite angles are equal: ∠BAD = 70°.
Consecutive angles are supplementary: ∠ADC = 180° - 70° = 110°.
Answer: 110°
9. PQRS is a parallelogram whose diagonals intersect at M. If ∠PMS = 54°, ∠QSR = 25° and ∠SQR = 30° find:
(i) ∠RPS
(ii) ∠PRS
(iii) ∠PSR
Solution:
(i) PS || QR ⇒ ∠PSQ = ∠SQR = 30°.
In ΔPMS: ∠MPS + ∠MSP + ∠PMS = 180°.
∠MPS + 30° + 54° = 180° ⇒ ∠MPS = 96°.
Since M lies on PR, ∠RPS = 96°.
(ii) In ΔMQR: ∠RMQ = ∠PMS = 54° (Vertically opposite).
∠MQR = 30°.
∠MRQ = 180° - (54° + 30°) = 96°.
∠PRS = 96°.
(iii) ∠PSR = ∠PSQ + ∠QSR = 30° + 25° = 55°.
Answer: (i) 96° (ii) 96° (iii) 55°
10. Given Parallelogram ABCD in which diagonals AC and BD intersect at M. Prove: M is the mid-point of LN.
Solution:
Consider ΔLMD and ΔNMB (assuming line L-M-N cuts AD and BC).
1. ∠LDM = ∠NBM (Alternate interior angles).
2. DM = BM (Diagonals bisect each other).
3. ∠DML = ∠BMN (Vertically opposite).
ΔLMD ≅ ΔNMB (ASA).
LM = NM (CPCTC).
Thus, M is the mid-point of LN.
11. In an isosceles-trapezium, show that the opposite angles are supplementary.
Solution:
In isosceles trapezium ABCD (AB || DC, AD=BC):
Base angles are equal: ∠A = ∠B and ∠C = ∠D.
Consecutive interior angles (between parallel lines) sum to 180°: ∠A + ∠D = 180°.
Substitute ∠A with ∠B: ∠B + ∠D = 180°.
Thus, opposite angles are supplementary.
12. ABCD is a parallelogram. What kind of quadrilateral is it if:
(i) AC = BD and AC is perpendicular to BD?
(ii) AC is perpendicular to BD but is not equal to it?
(iii) AC = BD but AC is not perpendicular to BD?
Solution:
(i) Equal and perpendicular diagonals ⇒ Square.
(ii) Perpendicular diagonals only ⇒ Rhombus.
(iii) Equal diagonals only ⇒ Rectangle.
13. Prove that the diagonals of a parallelogram bisect each other.
Solution:
In ΔAOB and ΔCOD (where diagonals intersect at O):
AB = CD (Opposite sides).
∠OAB = ∠OCD (Alternate angles).
∠OBA = ∠ODC (Alternate angles).
ΔAOB ≅ ΔCOD (ASA).
OA = OC and OB = OD.
14. If the diagonals of a parallelogram are of equal lengths, the parallelogram is a rectangle. Prove it.
Solution:
Consider ΔABC and ΔBAD.
AB = AB (Common).
BC = AD (Opposite sides).
AC = BD (Given).
ΔABC ≅ ΔBAD (SSS).
∠ABC = ∠BAD. Since AD || BC, sum is 180°.
2∠ABC = 180° ⇒ ∠ABC = 90°.
Hence, it is a rectangle.
15. In parallelogram ABCD, E is the mid-point of AD and F is the mid-point of BC. Prove that BFDE is a parallelogram.
Solution:
AD || BC and AD = BC.
ED = 1/2 AD and BF = 1/2 BC.
So ED = BF and ED || BF.
One pair of opposite sides is equal and parallel.
Therefore, BFDE is a parallelogram.
Test yourself
1. Multiple Choice Type:
(i) The given figure shows a parallelogram. The value of x for which it will be a rhombus is:
(a) 35° (b) 25° (c) 15° (d) 45°
Solution:
Diagonals of a rhombus intersect at 90°.
3x + 15 = 90 ⇒ 3x = 75 ⇒ x = 25.
Answer: (b) 25°
(ii) A rhombus will be a square, if:
(a) its diagonals bisect each other
(b) its diagonals are perpendicular to each other
(c) one of its angles is 60°
(d) none of these
Solution:
A rhombus becomes a square if diagonals are equal or one angle is 90°. None of the options a, b, c describe this condition.
Answer: (d) none of these
(iii) The diagonals of a quadrilateral, bisect each other at right-angle. The quadrilateral is a:
(a) trapezium (b) square (c) rhombus (d) parallelogram
Solution:
This is the defining property of a rhombus.
Answer: (c) rhombus
(iv) In a parallelogram ABCD; ∠A = (3x + 2)° and ∠D = (2x + 3)°; the value of x is:
(a) 37 (b) 35 (c) 30 (d) 40
Solution:
(3x + 2) + (2x + 3) = 180 ⇒ 5x + 5 = 180 ⇒ 5x = 175 ⇒ x = 35.
Answer: (b) 35
(v) In a trapezium ABCD, AB//DC and AD = BC. If ∠A = (5x + 8)° and ∠D = (4x + 10)° the measure of angle B is
(a) 98° (b) 82° (c) 96° (d) none of these
Solution:
(5x + 8) + (4x + 10) = 180 ⇒ 9x = 162 ⇒ x = 18.
Angle A = 98°. Since isosceles trapezium, Angle B = Angle A = 98°.
Answer: (a) 98°
(vi) Statement 1: In order to prove that a given parallelogram is a rectangle, we must prove that (a) any angle of it is 90° or (b) its diagonals are equal.
Statement 2: A kite is an arrowhead in which two pairs of adjacent sides are equal.
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and statement 2 is false.
(d) Statement 1 is false, and statement 2 is true.
Solution:
Both statements describe correct geometric definitions/properties.
Answer: (a) Both the statements are true.
(vii) Assertion (A): In a parallelogram, the bisectors of any two pair of adjacent angles meet at a right angle.
Reason (R): In a parallelogram, opposite angles are equal.
(a) (1) (b) (2) (c) (3) (d) (4)
Solution:
A is true. R is true. R does not explain A (A depends on adjacent angles being supplementary).
Answer: (b) (2)
(viii) Assertion (A): One of the diagonals of a rhombus is equal to one of its sides. The angles of a rhombus are 60°, 120°, 60°, 120°.
Reason (R): All sides of a rhombus are equal.
(a) (1) (b) (2) (c) (3) (d) (4)
Solution:
A is true (equilateral triangles formed). R is true and explains the formation of equilateral triangles.
Answer: (a) (1)
(ix) Assertion (A): If one angle of a parallelogram measures 90°, the parallelogram is a rectangle.
Reason (R): If each interior angle of a parallelogram is 90° then all sides of it are equal.
(a) (1) (b) (2) (c) (3) (d) (4)
Solution:
A is true. R is false (could be a rectangle with unequal adjacent sides).
Answer: (c) (3)
(x) Assertion (A): The adjacent angles of a parallelogram are in the ratio 2:3. The remaining angles are 78° and 102°.
Reason (R): Opposite angles of a parallelogram are equal.
(a) (1) (b) (2) (c) (3) (d) (4)
Solution:
2x + 3x = 180 ⇒ x=36. Angles: 72°, 108°. Assertion values are wrong.
Answer: (d) (4)
2. The adjacent sides of a parallelogram are in the ratio 5: 4. If the perimeter of the parallelogram is 108 cm, find the length of its sides.
Solution:
2(5x + 4x) = 108
18x = 108 ⇒ x = 6.
Sides: 30 cm and 24 cm.
3. In the given figure, ABCD is a parallelogram. If OA = 6 cm and AC - BD = 2 cm; find the length of BD.
Solution:
AC = 2 * OA = 12.
12 - BD = 2 ⇒ BD = 10 cm.
4. The diagonals of a parallelogram ABCD intersect each other at point O. If OA = x + y, OC = 20, OD = x + 3 and OB = 18; find the values of x and y.
Solution:
OD = OB ⇒ x + 3 = 18 ⇒ x = 15.
OA = OC ⇒ x + y = 20 ⇒ 15 + y = 20 ⇒ y = 5.
Answer: x = 15, y = 5.
5. One of the diagonals of a rhombus and its sides are equal. Find the angles of the rhombus.
Solution:
Triangle formed by two sides and one diagonal is equilateral (all sides equal).
Angles of triangle = 60°.
Rhombus angles: 60° and (60+60)=120°.
Answer: 60°, 120°, 60°, 120°.
6. In a parallelogram ABCD, E is the mid-point of side AB and CE bisects angle BCD. Prove that:
(i) AE = AD
(ii) DE bisects ∠ADC and
(iii) Angle DEC is a right angle.
Solution:
(i) CE bisects C ⇒ ∠BCE = ∠DCE. AB || DC ⇒ ∠CEB = ∠DCE.
So ∠BCE = ∠CEB ⇒ BE = BC.
AE = BE (Mid-point) ⇒ AE = BC.
Since AD = BC, AE = AD.
(ii) In ΔADE, AE = AD ⇒ ∠ADE = ∠AED.
∠AED = ∠CDE (Alt. int). ⇒ ∠ADE = ∠CDE.
(iii) 2∠CDE + 2∠DCE = 180° ⇒ ∠CDE + ∠DCE = 90°.
∠DEC = 180 - 90 = 90°.
7. In the diagram given below, the bisectors of interior angles of the parallelogram PQRS enclose a quadrilateral ABCD. Show that:
(i) ∠PSB + ∠SPB = 90°
(ii) ∠PBS = 90°
(iii) ∠ABC = 90°
(iv) ∠ADC = 90°
(v) ∠A = 90°
(vi) ABCD is a rectangle
Solution:
(i) Sum of consecutive angles P and S = 180°. Halves sum to 90°.
(ii) In ΔPBS, remaining angle = 180 - 90 = 90°.
(iii) Vertically opposite or same angle at vertex B = 90°.
(iv, v) Similarly for other vertices.
(vi) All angles 90° ⇒ Rectangle.
8. In a parallelogram ABCD, X and Y are mid-points of opposite sides AB and DC respectively. Prove that:
(i) AX = YC
(ii) AX is parallel to YC
(iii) AXCY is a parallelogram.
Solution:
(i) AB = DC ⇒ 1/2 AB = 1/2 DC ⇒ AX = YC.
(ii) AB || DC ⇒ AX || YC.
(iii) AX = YC and AX || YC ⇒ Parallelogram.
9. The given figure shows a parallelogram ABCD. Points M and N lie in diagonal BD such that DM = BN. Prove that:
(i) ΔDMC ≅ ΔBNA and so CM = AN
(ii) ΔAMD ≅ ΔCNB and so AM = CN.
(iii) ANCM is a parallelogram.
Solution:
(i) CD = AB, ∠CDM = ∠ABN, DM = BN. SAS Congruence.
(ii) AD = CB, ∠ADM = ∠CBN, DM = BN. SAS Congruence.
(iii) Opposite sides equal (CM = AN, AM = CN) ⇒ Parallelogram.
10. The given figure shows a rhombus ABCD in which angle BCD = 80°. Find angles x and y.
Solution:
x is angle at diagonal intersection = 90°.
∠BCD = 80 ⇒ ∠BCA = 40.
In ΔBMC, y + 40 + 90 = 180 ⇒ y = 50°.
(Assuming y corresponds to ∠CBD or similar).
Answer: x = 90°, y = 50°.
11. Use the information given in the following diagram to find the values of x, y and z.
Solution:
Opposite sides equal: 3x + 14 = 2x + 25 ⇒ x = 11.
Adjacent angles supplementary: (3y + 5) + (y + 9) = 180 ⇒ 4y = 166 ⇒ y = 41.5.
Alternate angles: z = 24.
Answer: x = 11, y = 41.5, z = 24.
12. The following figure is a rectangle in which x : y = 3 : 7; find the values of x and y.
Solution:
x + y = 90.
3k + 7k = 90 ⇒ 10k = 90 ⇒ k = 9.
x = 27°, y = 63°.
13. In the given figure, AB // EC, AB = AC and AE bisects ∠DAC. Prove that:
(i) ∠EAC = ∠ACB
(ii) ABCE is a parallelogram.
Solution:
(i) Isosceles ΔABC: ∠B = ∠C. Ext ∠DAC = 2∠C.
AE bisects ∠DAC ⇒ ∠DAE = ∠EAC = ∠C.
(ii) ∠EAC = ∠C (Alt int) ⇒ AE || BC.
Given AB || EC.
Two pairs parallel ⇒ Parallelogram.