Chapter 8 Quadrilaterals Ex 8.1 Solutions
Question - 1 : - The angles ofquadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of thequadrilateral.
Answer - 1 : -
Let the common ratio between the angles be = x.
We know that the sum of the interior angles of the quadrilateral= 360°
Now,
3x+5x+9x+13x = 360°
⇒ 30x =360°
⇒ x =12°
, Angles of the quadrilateral are:
3x = 3×12° = 36°
5x = 5×12° = 60°
9x = 9×12° = 108°
13x = 13×12° = 156°
Question - 2 : - If the diagonals ofa parallelogram are equal, then show that it is a rectangle.
Answer - 2 : - 
Given that,
AC = BD
To show that, ABCD is a rectangle if the diagonals of aparallelogram are equal
To show ABCD is a rectangle we have to prove that one of itsinterior angles is right angled.
Proof,
In ΔABC and ΔBAD,
BC = BA (Common)
AC = AD (Opposite sides of a parallelogram are equal)
AC = BD (Given)
Therefore, ΔABC ≅ ΔBAD[SSS congruency]
∠A = ∠B [Corresponding parts of CongruentTriangles]
also,
∠A+∠B = 180° (Sum of the angles on the sameside of the transversal)
⇒ 2∠A = 180°
⇒ ∠A = 90° = ∠B
, ABCD is a rectangle.
Hence Proved.
Answer - 3 : - 
Let ABCD be a quadrilateral whose diagonals bisect each other atright angles.
Given that,
OA = OC
OB = OD
and ∠AOB = ∠BOC = ∠OCD = ∠ODA =90°
To show that,
if the diagonals of a quadrilateral bisect each other at rightangles, then it is a rhombus.
i.e., we have to prove that ABCD is parallelogram and AB = BC =CD = AD
Proof,
In ΔAOB and ΔCOB,
OA = OC (Given)
∠AOB = ∠COB (Opposite sides of a parallelogram areequal)
OB = OB (Common)
Therefore, ΔAOB ≅ ΔCOB[SAS congruency]
Thus, AB = BC [CPCT]
Similarly we can prove,
BC = CD
CD = AD
AD = AB
, AB = BC = CD = AD
Opposites sides of a quadrilateral are equal hence ABCD is aparallelogram.
, ABCD is rhombus as it is a parallelogram whose diagonalsintersect at right angle.
HenceProved.Show that thediagonals of a square are equal and bisect each other at right angles.
Answer - 4 : - 
Let ABCD be a square and its diagonals AC and BD intersect eachother at O.
To show that,
AC = BD
AO = OC
and ∠AOB =90°
Proof,
In ΔABC and ΔBAD,
AB = BA (Common)
∠ABC = ∠BAD = 90°
BC = AD (Given)
ΔABC ≅ ΔBAD[SAS congruency]
Thus,
AC = BD [CPCT]
diagonals are equal.
Now,
In ΔAOB and ΔCOD,
∠BAO = ∠DCO (Alternate interior angles)
∠AOB = ∠COD (Vertically opposite)
AB = CD (Given)
, ΔAOB ≅ ΔCOD[AAS congruency]
Thus,
AO = CO [CPCT].
, Diagonal bisect each other.
Now,
In ΔAOB and ΔCOB,
OB = OB (Given)
AO = CO (diagonals are bisected)
AB = CB (Sides of the square)
, ΔAOB ≅ ΔCOB[SSS congruency]
also, ∠AOB = ∠COB
∠AOB+∠COB = 180° (Linear pair)
Thus, ∠AOB = ∠COB = 90°
,Diagonals bisect each other at right anglesShow that if thediagonals of a quadrilateral are equal and bisect each other at right angles,then it is a square.
Answer - 5 : - 
Given that,
Let ABCD be a quadrilateral and its diagonals AC and BD bisecteach other at right angle at O.
To prove that,
The Quadrilateral ABCD is a square.
Proof,
In ΔAOB and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOB = ∠COD (Vertically opposite)
OB = OD (Diagonals bisect each other)
, ΔAOB ≅ ΔCOD[SAS congruency]
Thus,
AB = CD [CPCT] — (i)
also,
∠OAB = ∠OCD (Alternate interior angles)
⇒ AB ||CD
Now,
In ΔAOD and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOD = ∠COD (Vertically opposite)
OD = OD (Common)
, ΔAOD ≅ ΔCOD[SAS congruency]
Thus,
AD = CD [CPCT] — (ii)
also,
AD = BC and AD = CD
⇒ AD =BC = CD = AB — (ii)
also, ∠ADC = ∠BCD [CPCT]
and ∠ADC+∠BCD = 180° (co-interior angles)
⇒ 2∠ADC = 180°
⇒∠ADC =90° — (iii)
One of the interior angles is right angle.
Thus, from (i), (ii) and (iii) given quadrilateral ABCD is asquare.
Hence Proved.
Diagonal AC of aparallelogram ABCD bisects ∠A (see Fig. 8.19). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.
Answer - 6 : -
(i) In ΔADC and ΔCBA,
AD = CB (Opposite sides of a parallelogram)
DC = BA (Opposite sides of a parallelogram)
AC = CA (Common Side)
, ΔADC ≅ ΔCBA[SSS congruency]
Thus,
∠ACD = ∠CAB by CPCT
and ∠CAB = ∠CAD (Given)
⇒ ∠ACD = ∠BCA
Thus,
AC bisects ∠C also.
(ii) ∠ACD = ∠CAD (Proved above)
⇒ AD =CD (Opposite sides of equal angles of a triangle are equal)
Also, AB = BC = CD = DA (Opposite sides of a parallelogram)
Thus,
ABCDis a rhombus.
Question - 7 : - ABCD is arhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
Answer - 7 : -
Given that,
ABCD is a rhombus.
AC and BD are its diagonals.
Proof,
AD = CD (Sides of a rhombus)
∠DAC = ∠DCA (Angles opposite of equal sides of atriangle are equal.)
also, AB || CD
⇒∠DAC = ∠BCA (Alternate interior angles)
⇒∠DCA = ∠BCA
, AC bisects ∠C.
Similarly,
We can prove that diagonal AC bisects ∠A.
Following the same method,
Wecan prove that the diagonal BD bisects ∠Band ∠D.ABCD is arectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
(i) ABCD is a square
(ii) Diagonal BD bisects ∠B as well as ∠D.
Answer - 8 : - 
(i) ∠DAC = ∠DCA (AC bisects ∠A as well as ∠C)
⇒ AD =CD (Sides opposite to equal angles of a triangle are equal)
also, CD = AB (Opposite sides of a rectangle)
,AB = BC = CD = AD
Thus, ABCD is a square.
(ii) In ΔBCD,
BC = CD
⇒ ∠CDB = ∠CBD (Angles opposite to equal sides are equal)
also, ∠CDB = ∠ABD (Alternate interior angles)
⇒ ∠CBD = ∠ABD
Thus, BD bisects ∠B
Now,
∠CBD = ∠ADB
⇒ ∠CDB = ∠ADB
Thus,BD bisects ∠DIn parallelogramABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig.8.20). Show that:
(i) ΔAPD ≅ ΔCQB
(ii) AP = CQ
(iii) ΔAQB ≅ ΔCPD
(iv) AQ = CP
(v) APCQ is a parallelogram
Answer - 9 : -
(i) In ΔAPD and ΔCQB,
DP = BQ (Given)
∠ADP = ∠CBQ (Alternate interior angles)
AD = BC (Opposite sides of a parallelogram)
Thus, ΔAPD ≅ ΔCQB[SAS congruency]
(ii) AP = CQ by CPCT as ΔAPD ≅ ΔCQB.
(iii) In ΔAQB and ΔCPD,
BQ = DP (Given)
∠ABQ = ∠CDP (Alternate interior angles)
AB = CD (Opposite sides of a parallelogram)
Thus, ΔAQB ≅ ΔCPD[SAS congruency]
(iv) As ΔAQB ≅ ΔCPD
AQ = CP [CPCT]
(v) From the questions (ii) and (iv), it is clear that APCQ hasequal opposite sides and also has equal and opposite angles. , APCQ is aparallelogram.
Question - 10 : - ABCD is aparallelogram and AP and CQ are perpendiculars from vertices A and C ondiagonal BD (see Fig. 8.21). Show that
(i) ΔAPB ≅ ΔCQD
(ii) AP = CQ
Answer - 10 : -
(i) In ΔAPB and ΔCQD,
∠ABP = ∠CDQ (Alternate interior angles)
∠APB = ∠CQD (= 90o asAP and CQ are perpendiculars)
AB = CD (ABCD is a parallelogram)
, ΔAPB ≅ ΔCQD[AAS congruency]
(ii) As ΔAPB ≅ ΔCQD.
,AP = CQ [CPCT]