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Answers
Q1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : quadrilateral.
Sol: Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.
∵ Sum of all the angles of quadrilateral = 360°
∴ 3x + 5x + 9x + 13x = 360°
⇒ 30x = 360°

∴ 3x = 3 × 12° = 36°
5x = 5 × 12°= 60°
9x = 9 × 12° = 108°
13x = 13 × 12° = 156°
⇒ The required angles of the quadrilateral are 36°, 60°, 108° and 156°.
Q2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Sol: A parallelogram ABCD such that AC = BD In ΔABC and ΔDCB,
AC = DB
[Given]
AB = DC
[Opposite sides of a parallelogram]
BC = CB
[Common]
ΔABC = ΔDCB
[SSS criteria]
∴Their corresponding parts are equal.
⇒∠ABC = ∠DCB
...(1)
∵AB || DC and BC is a transversal.
[∵ ABCD is a parallelogram]
∴∠ABC + ∠DCB = 180°
...(2)
From (1) and (2), we have
∠ABC = ∠DCB = 90°
i.e. ABCD is parallelogram having an angle equal to 90°.
∴ABCD is a rectangle.
Q3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Sol: We have a quadrilateral ABCD such that the diagonals AC and BD bisect each other at right angles at O.
∴ In ΔAOB and ΔAOD, we have
AO = AO
[Common]
OB = OD
[Given that O in the mid-point of BD]
∠AOB = ∠AOD
[Each = 90°]
ΔAOB ≌ ΔAOD
[SAS criteria]
Their corresponding parts are equal.
AB = AD
...(1)
Similarly,
AB = BC
...(2)
BC = CD
...(3)
CD = AD
...(4)
∴ From (1), (2), (3) and (4), we have AB = BC CD = DA
Thus, the quadrilateral ABCD is a rhombus.
Q4. Show that the diagonals of a square are equal and bisect each other at right angles.

Sol: We have a square ABCD such that its diagonals AC and BD intersect at O.
(i) To prove that the diagonals are equal, i.e. AC = BD
In ΔABC and ΔBAD, we have
AB = BA
[Common]
BC = AD
[Opposite sides of the square ABCD]
∠ABC = ∠BAD
[All angles of a square are equal to 90°]
∴ΔABC ≌ ΔBAD
[SAS criteria]
⇒Their corresponding parts are equal.
⇒AC = BD
...(1)
(ii) To prove that 'O' is the mid-point of AC and BD.
∵ AD || BC and AC is a transversal.
[∵ Opposite sides of a square are parallel]
∴∠1 = ∠3
[Interior alternate angles]
Similarly, ∠2 = ∠4
[Interior alternate angles]
Now, in ΔOAD and ΔOCB, we have
AD = CB
[Opposite sides of the square ABCD]
∠1 = ∠3
[Proved]
∠2 = ∠4
[Proved
ΔOAD ≌ ΔOCB
[ASA criteria]
∴Their corresponding parts are equal.
⇒OA = OC and OD = OB
⇒O is the mid-point of AC and BD, i.e. the diagonals AC and BD bisect each other at O.
...(2)
(iii) To prove that AC 1 BD.
In ΔOBA and ΔODA, we have
OB = OD
[Proved]
BA =DA
[Opposite sides of the square]
OA = OA
[Common]
∴
ΔOBA ≌ ΔODA
[SSS criteria]
⇒Their corresponding parts are equal.
⇒ ∠AOB = ∠AOD
But ∠AOB and ∠AOD form a linear pair.
∠AOB + ∠AOD = 180°
∠AOB = ∠AOD = 90°
⇒AC ⊥ BD
...(3)
From (1), (2) and (3), we get AC and BD are equal and bisect each other at right angles.
Q5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Sol: We have a quadrilateral ABCD such that ��O�� is the mid-point of AC and BD. Also AC ⊥ BD.
Now, in ΔAOD and ΔAOB, we have
AO = AO
[Common]
OD = OB
[∵ O is the mid-point of BD]
∠AOD = ∠AOB
[Each = 90°]
∴ ΔAOD ≌ ∠AOB
[SAS criteria]
∴Their corresponding parts are equal.
⇒ AD = AB
...(1)
Similarly, we have
AB = BC
...(2)
BC = CD
...(3)
CD = DA
...(4)
From (1), (2), (3) and (4) we have: AB = BC = CD = DA
∴Quadrilateral ABCD is having all sides equal.
In ΔAOD and ΔCOB, we have
AO = CO
[Given]
OD = OB
[Given]
∠AOD = ∠COB
[Vertically opposite angles]
∴
ΔAOD ≌ ΔCOB
⇒Their corresponding pacts are equal.
⇒ ∠1 = ∠2
But, they form a pair of interior alternate angles.
∴AD || BC
Similarly, AB || DC
∴ ABCD is' a parallelogram.
∵ Parallelogram having all of its sides equal is a rhombus.
∴ ABCD is a rhombus.
Now, in ΔABC and ΔBAD, we have
AC = BD
[Given]
BC = AD
[Proved]
AB = BA
[Common]
ΔABC ≌ ΔBAD
[SSS criteria]
Their corresponding angles are equal.
∠ABC = ∠BAD
Since, AD || BC and AB is a transversal.
∴∠ABC + ∠BAD = 180°
[Interior opposite angles are supplementary]
i.e. The rhombus ABCD is having one angle equal to 90°.
Thus, ABCD is a square.
Q6. Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.

Sol: We have a parallelogram ABCD in which diagonal AC bisects ∠A.
⇒ ∠DAC = ∠BAC
(i) To prove that AC bisects ∠C.

∵ABCD is a parallelogram.
∴AB || DC and AC is a transversal.
∴∠l = ∠3
[Alternate interior angles] ...(1)
Also, BC || AD and AC is a transversal.
∴∠2 = ∠4
[Alternate interior angles] ...(2)
But AC bisects ∠A.
[Given]
∴∠1 = ∠2
...(3)
From (1), (2) and (3), we have
∠3 = ∠4
⇒AC bisects ∠C.
(ii) To prove ABCD is a rhombus.
In ΔABC, we have ∠1 = ∠4
[∵∠1 = ∠2 = ∠4]
⇒
BC = AB
Sides opposite to equal angles are equal] ...(4)
Similarly,
AD = DC
...(5)
But ABCD is a parallelogram
[Given]
AB = DC
[Opposite sides of a parallelogram] ...(6)
From (4), (5) and (6), we have
AB = BC = CD = DA
Thus, ABCD is a rhombus.
In ∆APD and ∆BQC
:D
ADDITIONAL INFORMATION :-
ASA Congruence Rule ( Angle – Side – Angle )
- Two triangles are said to be congruent if two angles and the included side of one triangle are equal to two angles and the included side of another triangle
Side-Side-Side congruence rule
- The SSS rule states that: If three sides of one triangle are equal to three sides of another triangle, then the triangles are congruent.
Side-Angle-Side (SAS) Rule
- If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent.
RHS Congruence Rule
- Two right triangles are congruent if the hypotenuse and one side of one triangle are equal to the corresponding hypotenuse and one side of the other triangle