NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1.
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 9 |
| Subject | Maths |
| Chapter | Chapter 8 |
| Chapter Name | Quadrilaterals |
| Exercise | Ex 8.1 |
| Number of Questions Solved | 12 |
| Category | NCERT Solutions |
NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals Ex 8.1
Question 1.
The angles of quadrilateral are in the ratio 3: 5: 9:13. Find all the angles of the quadrilateral.
Solution.
Given, the ratio of the angles of quadrilateral are 3:5:9:13.
Let the angles of the quadrilateral are 3x, 5x, 9x and 13x.
We know that, sum of angles of a quadrilateral = 360°
∴ 3x + 5x + 9x +13x = 360°
∴ 30x = 360° ⇒ 
∴ Angles of the quadrilateral are
3x = 3 x 12 = 36°
5x = 5 x 12 = 60°
9x = 9 x 12 = 108°
and 13x =13 x 12=156°
Question 2.
If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Solution.
We have a parallelogram ABCD such that AC = BD.
In Δ ABC and Δ DCB
AC = DB [given]
AB = DC (opposite sides of a parallogram)
BC=CB (Common)
Δ ABC ≅ Δ DCB [by SAS congrunce rule]
∠ABC ≅ ∠DCB ………(i)
∵ AB || DC and BC is a transversal. [ ∵ ABCD is a parallelogram]
∴ ∠ABC + ∠DCB =180°
(interior opposite angles are supplementry]
From eqs. (i) and (ii), we have
∠ABC = ∠DCB =90°
i.e., ABCD is a parallelogram having an angle equal to 90°.
∴ ABCD is a rectangle.
Hence proved.
Question 3.
Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Solution.
We have a quadrilateral ABCD such that the diagonals AC and BD bisect each other at right angles at O.
∴ In Δ AOB, and Δ AOD, we have
AO = AO common]
OB = OD [∵ O is the mid-point of BD]
∠AOB = ∠AOD [each = 90 °]
Δ AOB ≅ Δ AOD [by SAS congruence rule]
AB = AD [by CPCT]… (i)
Similarly, AB = BC …(ii)
BC = CD …(iii)
CD = AD …(iv)
∴ From eqs.(i), (ii), (iii) and (iv), we have AB = BC = CD = DA Hence, the quadrilateral ABCD is a rhombus.
Question 4.
Show that the diagonals of a square are equal and bisect each other at right angles.
Solution.
Given: A square PQRS in which diagonals PR and QS intersect each other at 0.
To prove : PR = SQ, PO = OR, QO = OS and PR ⊥ SQ.
Proof: In Δ POQ and Δ ROS
PQ = RS [sides of a square]
∠PQO=∠RSO [alternative interior angles]
∠QPO = ∠SRO [alternative interior angles]
⇒ Δ POQ ≅ Δ ROS [by SAS congruence rule]
⇒ PO = OR [by CPCT]
and QO = OS [by CPCT]
Thus, PR and SQ bisect each other at O.
Also, in Δ POQ and Δ ROQ,
PQ = RQ
PO = OR
OQ = OQ
⇒ ΔPOQ ≅ ΔROQ by SAS congruence rule]
⇒ ∠POQ = ∠ROQ
Also.,∠POQ + ∠ROQ = 180°
⇒ ∠POQ + ∠POQ = 180°
⇒2 ∠POQ = 180°
⇒ ∠POQ =
Thus, diagonals PR and SQ are perpendicular to each other.
Again, in Δ PQS and Δ QPR
PQ = PQ
PS = QR
∠QPS – ∠PQR
⇒ ΔPQS = ΔQPR
⇒ SQ = PR
Hence, diagonals of a square are equal and bisect each other at right angles.
Question 5.
Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Solution.
Given: A quadrilateral ABCD in which AC = BD and AC⊥ BD such that OA = OC and OB = OD. So, ABCD is a parallelogram.
To prove: ABCD is a square.
Proof: Let AC and BD intersect at a point O.
In Δ ABO and Δ ADO, we have
BO = OD (Given)
AO = OA (Common)
∴ ∠AOB = ∠AOD = 90° (Given)
∴ ΔABO ≅ ΔADO (By SAS)
AB = AD (By CPCT)
Also,AB = DC
and AD = BC (Opposite sides of parallelogram)
∴ AB = BC = DC = AD .(i)
Again, in Δ ABC and Δ BAD, we have AB – BA AC = BD BC – AD
∴ Δ ABC = Δ BAD
∴ ∠ABC = ∠BAD…..(ii)
But ∠ABC + ∠BAD = 180°(Sum of interior angles of a parallelogram)
∴ ∠ABC = ∠BAD = 90° [From Eq. (ii)]
Thus, AB = BC = CD = DA and ∠A = 90°
∴ ABCD is a square.
Question 6.
Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.
Solution.
(i) Given, diagonal AC of a parallelogram ABCD bisects ∠A.
i.e., ∠DAC =∠BAC = ∠BAD … (i)
Here, AB || DC and AC is a transversal.
∴ ∠DCA = ∠BAC …(ii)[pair of alternate angles]
Similarly, BC || AD and BD is a transversal.
∴ ∠BCA = ∠DAC … (iii) [pair of alternate angles]
From eqs. (i), (ii) and (iii), we get
∠DAC = ∠BCA = ∠BAC = ∠DCA .. (iv)
Now, ∠BCD = ∠BCA + ∠DCA
= ∠DAC + ∠BAC = ∠BAD
Hence, diagonal AC also bisects ∠C.
(ii) From eq. (iv), ∠DAC = ∠DCA
⇒ CD = DA [∵ angles opposite to equal sides are equal]
But AB = DC and AD =BC [∵ ABCD is a parallelogram]
∴ AB = BC = CD = DA
Hence, ABCD is a rhombus.
Question 7.
ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
Solution.
Given : ABCD is a rhombus in which AB = BC =CD = AD.
To prove :
(i) Diagonal AC bisects ∠A as well as ∠C.
(ii) Diagonal BD bisects ∠B as well as
Construction: Join diagonals AC and BD.
Proof : ABCD is a rhombus.
∴ AB = BC = CD = AD
Also, AB || CD and AD || BC
Now,AD = CD ⇒ ∠1=∠2 …(i) [angles opposite to equal sides are equal]
Also, CD || AB [opposite sides of the parallelogram]
and AC is transversal. ∠1 = ∠3 … (ii) [alternate interior angles]
From eqs. (i) and (ii), we have
∠2 = ∠3 and ∠1 = ∠4 ⇒ AC bisects ∠C as well as ∠A.
Similarly, we prove that BD bisects ∠B as well as ∠D.
Question 8.
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 ZB as well as
Solution.
We have a rectangle ABCD such that AC bisects ∠A as well as ∠C.
i.e. ∠1 = 4 and ∠2 = ∠3 …(i)
(i) Since rectangle is a parallelogram.
∴ ABCD is a parallelogram.
⇒ AS || CD and AC is a transversal.
∴ ∠2=∠4 … (ii) [alternate interior angles]
From eqs. (i) and (ii), we have
∠3 = ∠4 ⇒ AB = BC [∵ sides opposite to equal angles in Δ ABC are equal]
∴ AB = BC = CD = AD
⇒ ABCD is a rectangle having all of its sides equal.
∴ ABCD is a square.
(ii) Since ABCD is a square, and diagonals of a square bisect the opposite
∴ BD bisects ZB as well as ZD.
Question 9.
In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that
(i) Δ APD = Δ CQB
(ii) AP = CQ
(iii) Δ AQB ≅ Δ CPD
(iv) AQ = CP
(v) APCQis a parallelogram.
Solution.
Now, these equal angles form a pair of alternate angle when line segment AP and QC are intersected by a transversal PQ.
∴AP || CQ and AQ || CP
Now, both pairs of opposite sides of quadrilateral Δ PCQ are parallel. Hence, ΔPCQ is a parallelogram. Hence proved.
Question 10.
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that
(i) Δ APB ≅ Δ CQD
(ii) AP = CQ
Solution.
(i) In Δ APB and Δ CQD, we have
∠ APB = ∠CQD [each = 90 °]
AB = CD [opposite sides of parallelogram ABCD]
∠ABP= ∠CDQ
Δ APB = Δ CQD
(ii) Since, Δ APB = Δ CQD
∴ Their corresponding parts are equal.
AP = CQ
Question 11.
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 the figure). 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
Solution.
Given, in Δ ABC and Δ DEF,
AB = DE,AB || DE, BC = EF and BC || EF
(i) Now, in quadrilateral ABED,
AB = DE and AB || DE [given]
So, ABED is a parallelogram. [∵ a pair of opposite side is equal and parallel]
(ii) In quadrilateral BEFC,
BC = EF and BC || EF [given]
So, BEFC is a parallelogram. [∵ a pair of opposite sides is equal and parallel]
(iii) Since, ABED is a parallelogram.
∴ AD || BE and AD = BE …(i)
Also, BEFC is a parallelogram.
∴ CF || BE and CF = BE …(ii)
From eqs. (i) and (ii), we get
AD || CF and AD = CF
(iv) In quadrilateral ACFD
∴ AD || CF and AD = CF [from part (iii)]
So, ACFD is a parallelogram.
Since, a pair of opposite sides is equal and parallel.
(v) Since, ACFD is a parallelogram.
∴ AC = DF and AC 11 DF
(vi) Now, in Δ ABC and Δ DEF,
AB = DE [given]
BC = EF [given]
and AC = DF [from part (v)]
∴ Δ ABC ≅ Δ DEF [by SSS congruence rule]
Hence proved.
Question 12.
ABCD is a trapezium in which AB || CD and AD = BC (see the figure). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) Δ ABC = Δ BAD
(iv) Diagonal AC = Diagonal
[Hint : Extend AB and draw a line through C parallel to DA . intersecting AB produced at E.]
Solution.
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