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Q1. The angles of a quadrilateral are in the ratio 3: 5: 9: 13. Find all angles of the quadrilateral.
Answer
Let x be the common ratio between the angles.
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°
Q2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Answer
Given,
AC = BD
To show,
To show ABCD is a rectangle we have to prove that one of
its interior angle is rightangled.
Proof,
In ΔABC and ΔBCD,
BC = BC (Common)
AB = DC (Opposite sides of a parallelogram are equal)
AC = BD (Given)
Therefore, ΔABC ≅ ΔBCD by SSS congruence condition.
∠B = ∠C (by CPCT)
also,
∠A + ∠B = 180° (Sum of the angles on the same side of the
transversal)
⇒ 2∠A = 180°
⇒ ∠A = 90° = ∠B
Thus ABCD is a rectangle.
Q3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Answer 
Let ABCD be a quadrilateral whose diagonals bisect each
other at right angles.
Given,
OA = OC, OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA =
90°
To show,
ABCD is parallelogram and AB = BC = CD = AD
Proof,
In ΔAOB and ΔCOB,
OA = OC (Given)
∠AOB = ∠COB (Opposite sides of a parallelogram are
equal)
OB = OB (Common)
Therefore, ΔAOB ≅ ΔCOB by SAS congruence condition.
Thus, AB = BC (by CPCT)
Similarly we can prove,
AB = BC = CD = AD
Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.
Thus, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.
Q4. Show that the diagonals of a square are equal and bisect each other at right angles.
Answer 
Let ABCD be a square and its diagonals AC and BD
intersect each other at O.
To show,
AC = BD, AO = OC and ∠AOB = 90°
Proof,
In ΔABC and ΔBAD,
BC = BA (Common)
∠ABC = ∠BAD = 90°
AC = AD (Given)
Therefore, ΔABC ≅ ΔBAD by SAS congruence condition.
Thus, AC = BD by CPCT. Therefore, diagonals are equal.
Now,
In ΔAOB and ΔCOD,
∠BAO = ∠DCO (Alternate interior angles)
∠AOB = ∠COD (Vertically opposite)
AB = CD (Given)
Therefore, ΔAOB ≅ ΔCOD by AAS congruence condition.
Thus, AO = CO by CPCT. (Diagonal bisect each other.)
Now,
In ΔAOB and ΔCOB,
OB = OB (Given)
AO = CO (diagonals are bisected)
AB = CB (Sides of the square)
Therefore, ΔAOB ≅ ΔCOB by SSS congruence condition.
also, ∠AOB = ∠COB
∠AOB + ∠COB = 180° (Linear pair)
Thus, ∠AOB = ∠COB = 90° (Diagonals bisect each other at
Q5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Answer 
Given,
Let ABCD be a quadrilateral in which diagonals AC and BD
bisect each other at a right angle at O.
To prove,
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)
Therefore, ΔAOB ≅ ΔCOD by SAS congruence condition.
Thus, AB = CD by 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)
Therefore, ΔAOD ≅ ΔCOD by SAS congruence condition.
Thus, AD = CD by CPCT.  (ii)
also,
AD = BC and AD = CD
⇒ AD = BC = CD = AB  (ii)
also, ∠ADC = ∠BCD by CPCT.
and ∠ADC + ∠BCD = 180° (cointerior angles)
⇒ 2∠ADC = 180°
⇒ ∠ADC = 90°  (iii)
One of the interior ang is right angle.
Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a
square.
Q6. Diagonal AC of a parallelogram ABCD bisects ∠A (seeFig.). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.
Answer 
(i) In ΔADC and ΔCBA,
AD = CB (Opposite sides of a gm)
DC = BA (Opposite sides of a gm)
AC = CA (Common)
Therefore, ΔADC ≅ ΔCBA by SSS congruence condition.
Thus,
∠ACD = ∠CAB by CPCT
and ∠CAB = ∠CAD (Given)
⇒ ∠ACD = ∠BCA
Thus, AC bisects ∠C also.
(ii) ∠ACD = ∠CAD (Proved)
⇒ AD = CD (Opposite sides of equal angles of a triangle are
equal)
Also, AB = BC = CD = DA (Opposite sides of a gm)
Thus, ABCD is a rhombus.
Q7. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
Answer  Let ABCD is a rhombus and AC and BD are its diagonals.
Proof,
AD = CD (Sides of a rhombus)
∠DAC = ∠DCA (Angles opposite of equal sides of a triangle
are equal.)
also, AB  CD
⇒ ∠DAC = ∠BCA (Alternate interior angles)
⇒ ∠DCA = ∠BCA
Therefore, AC bisects ∠C.
Similarly, we can prove that diagonal AC bisects ∠A.
Also, by preceding above method we can prove that
diagonal BD bisects ∠B as well as ∠D.
Q8. ABCD is a rectangle 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 
(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)
Therefore, 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 ∠D
Q9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig.). Show that:
(i) ΔAPD ≅ΔCQB
(ii) AP = CQ
(iii) ΔAQB ≅ ΔCPD
(iv) AQ = CP
(v) APCQ is a parallelogram
Answer 
(i) In ΔAPD and ΔCQB,
DP = BQ (Given)
∠ADP = ∠CBQ (Alternate interior angles)
AD = BC (Opposite sides of a gm)
Thus, ΔAPD ≅ ΔCQB by SAS congruence condition.
(ii) AP = CQ by CPCT as ΔAPD ≅ ΔCQB.
(iii) In ΔAQB and ΔCPD,
BQ = DP (Given)
∠ABQ = ∠CDP (Alternate interior angles)
AB = BCCD (Opposite sides of a gm)
Thus, ΔAQB ≅ ΔCPD by SAS congruence condition.
(iv) AQ = CP by CPCT as ΔAQB ≅ ΔCPD.
(v) From (ii) and (iv), it is clear that APCQ has equal
opposite sides also it has equal opposite angles. Thus,
APCQ is a gm.
Q10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig.). Show that
(i) ΔAPB ≅ ΔCQD
(ii) AP = CQ
Answer
(i) In ΔAPB and ΔCQD,
∠ABP = ∠CDQ (Alternate interior angles)
∠APB = ∠CQD (equal to right angles as AP and CQ are
perpendiculars)
AB = CD (ABCD is a parallelogram)
Thus, ΔAPB ≅ ΔCQD by AAS congruence condition.
(ii) AP = CQ by CPCT as ΔAPB ≅ ΔCQD.
Q11. In ΔABC and ΔDEF,AB = DE, AB  DE, BC = EF and BC  EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig). Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(iii) AD  CF and AD = CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC = DF
(vi) ΔABC ≅ ΔDEF
Answer:
(i) AB = DE and AB  DE (Given)
Thus, quadrilateral ABED is a parallelogram because two
opposite sides of a quadrilateral are equal and parallel to
each other.
(ii) Again BC = EF and BC  EF.
Thus, quadrilateral BEFC is a parallelogram.
(iii) Since ABED and BEFC are parallelograms.
⇒ AD = BE and BE = CF (Opposite sides of a parallelogram
are equal)
Thus, AD = CF.
Also, AD  BE and BE  CF (Opposite sides of a
parallelogram are parallel)
Thus, AD  CF
(iv) AD and CF are opposite sides of quadrilateral ACFD
which are equal and parallel to each other. Thus, it is a
parallelogram.
(v) AC  DF and AC = DF because ACFD is a parallelogram.
(vi) In ΔABC and ΔDEF,
AB = DE (Given)
BC = EF (Given)
AC = DF (Opposite sides of a parallelogram)
Thus, ΔABC ≅ ΔDEF by SSS congruence condition.
Q12. ABCD is a trapezium in which AB  CD and AD = BC (see Fig). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ΔABC ≅ ΔBAD
(iv) diagonal AC =diagonal BD
[Hint : Extend AB anddraw a line through C parallel to DA intersecting AB produced at E.]
Answer 
Construction: Draw a line through C parallel to DA intersecting AB produced at E.
(i) CE = AD (Opposite sides of a parallelogram)
AD = BC (Given)
Therefor, BC = CE
⇒ ∠CBE = ∠CEB
also,
∠A + ∠CBE = 180° (Angles on the same side of transversal
and ∠CBE = ∠CEB)
∠B + ∠CBE = 180° (Linear pair)
⇒ ∠A = ∠B
(ii) ∠A + ∠D = ∠B + ∠C = 180° (Angles on the same side of
transversal)
⇒ ∠A + ∠D = ∠A + ∠C (∠A = ∠B)
⇒ ∠D = ∠C
(iii) In ΔABC and ΔBAD,
AB = AB (Common)
∠DBA = ∠CBA
AD = BC (Given)
Thus, ΔABC ≅ ΔBAD by SAS congruence condition.
(iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA
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76 videos397 docs109 tests
