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# Class 9 NCERT Solutions- Chapter 8 Quadrilaterals – Exercise 8.1

• Difficulty Level : Basic
• Last Updated : 03 Jan, 2021

### Question 1. The angles of quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.

Solution:

As we know that, the sum of the angles of a quadrilateral is 360Â° (Angle sum property of quadrilateral)

As they are in ratio 3 : 5 : 9 : 13, so we can assume angles be as 3x, 5x, 9x and 13x.

So,

3x + 5x + 9x + 13x = 360Â°

30x = 360Â°

x = 360/30 = 12Â°

So the angles will be as follows:

3x = 3Ã—12 = 36Â°

5x = 5Ã—12 = 60Â°

9x = 9Ã—12 = 108Â°

13x = 13Ã—12 = 156Â°

Hence, the angles of quadrilateral are 36Â°, 60Â°, 108Â° and 156Â°.

### Question 2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Solution:

As mentioned parallelogram, so let PQRS be a parallelogram

where, given

PR = QS

In âˆ†PQR and âˆ†QRS,

PR = QS      …………………[Given]

PQ = RS      …………………[Opposite sides of a parallelogram]

QR = RQ     …………………[Common side]

âˆ´ âˆ†PQR â‰… âˆ†QRS           [By SSS congruency]

so, âˆ PQR = âˆ QRS [By C.P.C.T.] â€¦……………………………………….(1)

Now, PQ || RS and QR is a transversal. …………………….[PQRS is a parallelogram]

âˆ´ âˆ PQR + âˆ QRS = 180Â°  [Co-interior angles of parallelogram]…………………………………â€¦ (2)

From (1) and (2), we have

âˆ PQR = âˆ QRS = 90Â°

i.e., PQRS is a parallelogram having an angle equal to 90Â°.

Hence, PQRS is a rectangle. (having all angles equal to 90Â° and opposite sides are equal)

### Question 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Solution:

where, given

PO = RO

SO = QO

In âˆ†POQ and âˆ†POS, we have

PO = PO [Common]

OQ = OS [O is the mid-point of QS]

âˆ POQ= âˆ POS [Each 90Â°]

âˆ´ âˆ†POQ â‰… âˆ†POS [By, SAS congruency]

âˆ´ PQ = PS [By C.P.C.T.] â€¦â€¦..    (1)

Similarly, PQ = QR …………………..(2)

QR = RS â€¦…………………………………..(3)

RS = SP â€¦â€¦…………………………………(4)

âˆ´ From (1), (2), (3) and (4), we have

PQ = QR = RS = SP

Thus, the quadrilateral PQRS is a rhombus.

Alternative Solution:

So as it is given that diagonals of a quadrilateral PQRS bisect each other

According to Theorem 8.7 NCERT it is a parallelogram

PQRS can be proved first a parallelogram then proving one pair of adjacent sides equal will result in rhombus.

In âˆ†POQ and âˆ†POS, we have

PO = PO [Common]

OQ = OS [O is the mid-point of QS]

âˆ POQ= âˆ POS [Each 90Â°]

âˆ´ âˆ†POQ â‰… âˆ†POS [By,SAS congruency]

âˆ´ PQ = PS [By C.P.C.T.] â€¦â€¦..    (1)

Similarly, PQ = QR …………………..(2)

QR = RS â€¦…………………………………..(3)

RS = SP â€¦â€¦…………………………………(4)

From (1), (2), (3) and (4), we have

PQ = QR = RS = SP

Hence, as a parallelogram has all sides equal then it is called a rhombus.

### Question 4. Show that the diagonals of a square are equal and bisect each other at right angles.

Solution:

As it is mentioned that it is a square so,

all sides are equal. (PQ = QR = RS = SP)

all angles at four corners = 90Â°

So, to prove that the diagonals are equal, we need to prove PR = QS.

In âˆ†PQR and âˆ†QPS, we have

PQ = QP ……………….[Common]

PS = PQ ………………..[Sides of a square PQRS]

âˆ PQR = âˆ QPS …….[Each angle is 90Â°]

âˆ†PQR â‰… âˆ†QPS [By SAS congruency]

AC = BD [By C.P.C.T.] â€¦(1)

Now as we know square is also a parallelogram

In âˆ†POQ and âˆ†ROS, we have

âˆ POQ = âˆ ROS ……………….[Opposite angles of two intersecting lines]

PQ = RS ………………..[Sides of a square PQRS]

âˆ PQO = âˆ RSO …….[Alternate interior angles are equal]

âˆ†PQR â‰… âˆ†QPS [By ASA congruency]

OP = OQ = OR = OS (Hence the diagonals are equal and bisect each other)

Now , In âˆ†OQP and âˆ†OSP, we have

OQ = OS [Proved]

QP = SP [Sides of a square PQRS]

OP = OP [Common]

âˆ†OQP â‰… âˆ†OSP [By SSS congruency]

âˆ POQ = âˆ POS [By C.P.C.T.] â€¦(3)

âˆ POQ + âˆ POS = 180Â° (âˆµ âˆ POQ and âˆ POS form a linear pair)

âˆ POQ = âˆ POS = 90Â° [By(3)]

PR âŠ¥ QS â€¦(4)

From (1), (2) and (4), we get PR and QS are equal and bisect each other at right angles (90Â°).

### 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:

where,

PR = QS

OP = OR = OQ = OS

âˆ POQ = âˆ QOR = âˆ ROS = âˆ SOP = 90Â°

Now, in âˆ†POS and âˆ†POQ, We have

âˆ POS = âˆ POQ [Each 90Â°]

PO = PO [Common]

OS= OQ [ âˆµ O is the midpoint of BD]

âˆ†POS â‰… âˆ†POQ [By SAS congruency]

PS = PQ [By C.P.C.T.] â€¦(1)

Similarly, we have

PQ = QR â€¦ (2)

QR = RS â€¦(3)

RS = SPâ€¦(4)

From (1), (2), (3) and (4), we have

PQ = QR = RS = SP

Hence, Quadrilateral PQRS have all sides equal.

Now, In âˆ†POS and âˆ†ROQ, we have

PO = RO [Given]

OS = OQ [Given]

âˆ POS = âˆ ROQ [Vertically opposite angles]

So, âˆ†POS â‰… âˆ†ROQ [By SAS congruency]

âˆ PSO = âˆ RQO [By C.P.C.T.]

So as they form a pair of alternate interior angles.

PS || QR

Similarly, PQ || RS

PQRS is a parallelogram.

Parallelogram having all its sides equal is a rhombus.

PQRS is a rhombus.

Now, in âˆ†PQR and âˆ†QPS, we have

PR = QS [Given]

QR = SP [Proved]

PQ = QP [Common]

âˆ†PQR â‰… âˆ†QPS [By SSS congruency]

âˆ PQR = âˆ QPS [By C.P.C.T.] â€¦â€¦(5)

Since, PS || QR and PQ is a transversal.

âˆ PQR + âˆ QPS = 180Â° .. .(6) [ Co â€“ interior angles]

âˆ PQR = âˆ QPS = 90Â° [By(5) & (6)]

So, rhombus PQRS is having corners angle equal to 90Â°.

Hence, PQRS is a square.

### Question 6. Diagonal AC of a parallelogram ABCD bisects âˆ A (see Fig. 8.19). Show that

(i) it bisects âˆ C also,

(ii) ABCD is a rhombus.

Solution:

(i) As, ABCD is a parallelogram.

âˆ BAC = âˆ DCA â€¦…………………………(1) [Alternate interior angles are equal]

âˆ CAD = âˆ BCA â€¦……………………….(2) [Alternate interior angles are equal]

Also, âˆ CAD = âˆ CAB â€¦………………….(3) [ (Given) as AC bisects âˆ A]

From (1), (2) and (3), we have

âˆ DCA = âˆ BCA

Hence, AC bisects âˆ C.

(ii) In âˆ†ABC,

âˆ BAC = âˆ DCA â€¦……………………….. [Alternate interior angles are equal]

BC = AB â€¦………………………….(4) [ Sides opposite to equal angles of a âˆ† are equal]

But, ABCD is a parallelogram. [Given]

AB = DC â€¦……………….(6) (opposite sides of parallelogram)

From (4), (5) and (6), we have

AB = BC = CD = DA

As, ABCD is a parallelogram having all sides equal then it is a rhombus.

### Question 7. ABCD is a rhombus. Show that diagonal AC bisects âˆ  A as well as âˆ  C and diagonal BD bisects âˆ  B as well as âˆ  D.

Solution:

As, ABCD is a rhombus, so

AB = BC = CD = DA

âˆ ADB= âˆ ABDâ€¦â€¦………………….(1) [Angles opposite to equal sides of a triangle are equal]

âˆ ADB = âˆ CBD â€¦………………..(2) [ âˆµ Alternate interior angles are equal] (Rhombus is a parallelogram)

From (1) and (2), we have

âˆ CBD = âˆ ABD â€¦………………..(3)

âˆ ABD= âˆ CDB â€¦…………………..(4)      [ âˆµ Alternate interior angles are equal]

From (1) and (4),

we have âˆ ADB = âˆ CDB

Hence, BD bisects âˆ B as well as âˆ D.

Similarly, we can prove that AC bisects âˆ C as well as âˆ A..

### Question 8. ABCD is a rectangle in which diagonal AC bisects âˆ  A as well as âˆ  C. Show that:

(i) ABCD is a square

(ii) diagonal BD bisects âˆ  B as well as âˆ  D.

Solution:

There is rectangle ABCD such that AC bisects âˆ A as well as âˆ C, so

âˆ BAC = âˆ DAC and,

âˆ DCA = âˆ BCA â€¦â€¦…………………………(1)

(i) As we know that every rectangle is a parallelogram.

ABCD is a parallelogram.

âˆ BCA = âˆ DAC â€¦………………….(2) [ Alternate interior angles are equal]

From (1) and (2), we have

âˆ DCA= âˆ DAC……………………….(3)

In âˆ†ABC, âˆ DCA= âˆ DAC then,

CD = DA [Sides opposite to equal angles of a âˆ† are equal]

Similarly, AB = BC

So, ABCD is a rectangle having adjacent sides equal.

ABCD is a square.

(ii) Since, ABCD is a square

AB = BC = CD = DA

so, In âˆ†ABD, as AB = AD

âˆ ABD = âˆ ADB [Angles opposite to equal sides of a âˆ† are equal]……………………..(1)

Similarly, âˆ CBD = âˆ CDB…………………..(2)

âˆ CBD = âˆ ADB               [Alternate interior angles are equal]………………(3)

From (1) and (3)

âˆ CBD = âˆ ABD

From (2) and (3)

So, BD bisects âˆ B as well as âˆ D.

### Question 9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that :

(i) âˆ†APD â‰… âˆ†CQB

(ii) AP = CQ

(iii) âˆ†AQB â‰… âˆ†CPD

(iv) AQ = CP

(v) APCQ is a parallelogram

Solution:

ABCD is a parallelogram

DP = BQ

(i) As ABCD is a parallelogram

âˆ ADB = âˆ CBD [Alternate interior angles are equal]……………….(1)

âˆ ABD = âˆ CDB   [Alternate interior angles are equal]…………………(2)

Now, in âˆ†APD and âˆ†CQB, we have

AD = CB [Opposite sides of a parallelogram ABCD are equal]

PD = QB [Given]

âˆ ADP = âˆ CBQ [Proved]

Hence, âˆ†APD â‰… âˆ†CQB [By SAS congruency]

(ii) As, âˆ†APD â‰… âˆ†CQB [Proved]

AP = CQ [By C.P.C.T.]…………………(3)

(iii) Now, in âˆ†AQB and âˆ†CPD, we have

QB = PD [Given]

âˆ ABQ = âˆ CDP [Proved]

AB = CD [ Opposite sides of a parallelogram ABCD are equal]

Hence, âˆ†AQB â‰… âˆ†CPD [By SAS congruency]

(iv) As, âˆ†AQB â‰… âˆ†CPD [Proved]

AQ = CP [By C.P.C.T.] …………………………..(4)

Opposite sides are equal. [From (3) and (4)]

Hence, APCQ is a parallelogram. (NCERT Theorem 8.3)

### Question 10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that

(i) âˆ†APB â‰… âˆ†CQD

(ii) AP = CQ

Solution:

ABCD is a parallelogram

DP = BQ

(i) In âˆ†APB and âˆ†CQD, we have

âˆ APB = âˆ CQD [Each 90Â°]

AB = CD [ Opposite sides of a parallelogram ABCD are equal]

âˆ ABP = âˆ CDQ [Alternate angles are equal as AB || CD and BD is a transversal]

Hence, âˆ†APB  â‰…  âˆ†CQD [By AAS congruency]

(ii) As, âˆ†APB â‰… âˆ†CQD [Proved]

AP = CQ [By C.P.C.T.]

### Question 11. In âˆ† ABC and âˆ† DEF, 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. 8.22). Show that

(i) quadrilateral ABED is a parallelogram

(ii) quadrilateral BEFC is a parallelogram

(iv) quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) âˆ†ABC â‰… âˆ†DEF.

Solution:

AB = DE, and AB || DE,

BC = EF and BC || EF

(i) We have AB = DE and AB || DE [Given]

so here., ABED is a quadrilateral in which a pair of opposite sides (AB and DE) are parallel and of equal length.

Hence, ABED is a parallelogram.

(ii) BC = EF and BC || EF [Given]

i.e. BEFC is a quadrilateral in which a pair of opposite sides (BC and EF) are parallel and of equal length.

Hence, BEFC is a parallelogram.

(iii) as, ABED is a parallelogram [Proved]

âˆ´ AD || BE and AD = BE â€¦(1) [Opposite sides of a parallelogram are equal and parallel]

Also, BEFC is a parallelogram. [Proved]

BE || CF and BE = CF â€¦(2) [Opposite sides of a parallelogram are equal and parallel]

From (1) and (2), we have

i.e., In quadrilateral ACFD, one pair of opposite sides (AD and CF) are parallel and of equal length.

Hence, Quadrilateral ACFD is a parallelogram.

(v) Since, ACFD is a parallelogram. [Proved]

So, AC =DF [Opposite sides of a parallelogram are equal]

(vi) In âˆ†ABC and âˆ†DEF, we have

AB = DE [Given]

BC = EF [Given]

AC = DE [Proved in (v) part]

âˆ†ABC â‰… âˆ†DEF [By SSS congruency]

### Question 12. ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that

(i) âˆ A = âˆ B

(ii) âˆ C = âˆ D

(iv) diagonal AC = diagonal BD

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Solution:

so basically here is a trapezium ABCD in which AB || CD and AD = BC.

Extended AB and draw a line through C parallel to DA intersecting AB produced at E

(i) AB || DC , AE || DC Also AD || CE

then, AECD is a parallelogram.

AD = CE â€¦………………………(1) [Opposite sides of the parallelogram are equal]

But AD = BC â€¦(2) [Given]

From (1) and (2),

BC = CE

Now, in âˆ†BCF, we have BC = CF

âˆ CEB = âˆ CBE â€¦(3)  [Angles opposite to equal sides of a triangle are equal]

Also, âˆ ABC + âˆ CBE = 180Â° â€¦ (4) [Linear pair]

and âˆ A + âˆ CEB = 180Â° â€¦(5) [Co-interior angles of a parallelogram ADCE]

From (4) and (5), we get

âˆ ABC + âˆ CBE = âˆ A + âˆ CEB

âˆ ABC = âˆ A [From (3)]

âˆ B = âˆ A â€¦(6)

(ii) AB || CD and AD is a transversal.

âˆ A + âˆ D = 180Â° â€¦(7) [Co-interior angles in parallelogram]

Similarly, âˆ B + âˆ C = 180Â° â€¦ (8)

From (7) and (8), we get

âˆ A + âˆ D = âˆ B + âˆ C

âˆ C = âˆ D [From (6)]

(iii) In âˆ†ABC and âˆ†BAD, we have

AB = BA [Common]

âˆ ABC = âˆ BAD [Proved]

Hence, âˆ†ABC â‰… âˆ†BAD [By SAS congruency]

(iv) Since, âˆ†ABC â‰… âˆ†BAD [Proved]

AC = BD [By C.P.C.T.]

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