NCERT Solutions Class 9 – Chapter 8 Quadrilaterals – Exercise 8.1
Question 1. 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 ∆PQR and ∆QRS,
PR = QS …………………[Given]
PQ = RS …………………[Opposite sides of a parallelogram]
QR = RQ …………………[Common side]
∴ ∆PQR ≅ ∆QRS [By SSS congruency]
so, ∠PQR = ∠QRS [By C.P.C.T.] ………………………………………….(1)
Now, PQ || RS and QR is a transversal. …………………….[PQRS is a parallelogram]
∴ ∠PQR + ∠QRS = 180° [Co-interior angles of parallelogram]…………………………………… (2)
From (1) and (2), we have
∠PQR = ∠QRS = 90°
i.e., PQRS is a parallelogram having an angle equal to 90°.
Hence, PQRS is a rectangle. (having all angles equal to 90° and opposite sides are equal)
Question 2. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Solution:
As mentioned quadrilateral, so let PQRS be a quadrilateral
where, given
PO = RO
SO = QO
In ∆POQ and ∆POS, we have
PO = PO [Common]
OQ = OS [O is the mid-point of QS]
∠POQ= ∠POS [Each 90°]
∴ ∆POQ ≅ ∆POS [By, SAS congruency]
∴ PQ = PS [By C.P.C.T.] …….. (1)
Similarly, PQ = QR …………………..(2)
QR = RS ……………………………………..(3)
RS = SP ………………………………………(4)
∴ 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 ∆POQ and ∆POS, we have
PO = PO [Common]
OQ = OS [O is the mid-point of QS]
∠POQ= ∠POS [Each 90°]
∴ ∆POQ ≅ ∆POS [By,SAS congruency]
∴ PQ = PS [By C.P.C.T.] …….. (1)
Similarly, PQ = QR …………………..(2)
QR = RS ……………………………………..(3)
RS = SP ………………………………………(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 3. Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.
Solution:
(i) As, ABCD is a parallelogram.
∠BAC = ∠DCA ……………………………(1) [Alternate interior angles are equal]
∠CAD = ∠BCA ………………………….(2) [Alternate interior angles are equal]
Also, ∠CAD = ∠CAB …………………….(3) [ (Given) as AC bisects ∠A]
From (1), (2) and (3), we have
∠DCA = ∠BCA
Hence, AC bisects ∠C.
(ii) In ∆ABC,
∠BAC = ∠DCA ………………………….. [Alternate interior angles are equal]
BC = AB …………………………….(4) [ Sides opposite to equal angles of a ∆ are equal]
Similarly, AD = DC ……..(5)
But, ABCD is a parallelogram. [Given]
AB = DC ………………….(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 4. 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.
Solution:
There is rectangle ABCD such that AC bisects ∠A as well as ∠C, so
∠BAC = ∠DAC and,
∠DCA = ∠BCA ………………………………(1)
(i) As we know that every rectangle is a parallelogram.
ABCD is a parallelogram.
∠BCA = ∠DAC …………………….(2) [ Alternate interior angles are equal]
From (1) and (2), we have
∠DCA= ∠DAC……………………….(3)
In ∆ABC, ∠DCA= ∠DAC then,
CD = DA [Sides opposite to equal angles of a ∆ 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 ∆ABD, as AB = AD
∠ABD = ∠ADB [Angles opposite to equal sides of a ∆ are equal]……………………..(1)
Similarly, ∠CBD = ∠CDB…………………..(2)
∠CBD = ∠ADB [Alternate interior angles are equal]………………(3)
From (1) and (3)
∠CBD = ∠ABD
From (2) and (3)
∠ADB = ∠CBD
So, BD bisects ∠B as well as ∠D.
Question 5. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that :
(i) ∆APD ≅ ∆CQB
(ii) AP = CQ
(iii) ∆AQB ≅ ∆CPD
(iv) AQ = CP
(v) APCQ is a parallelogram
Solution:
ABCD is a parallelogram
DP = BQ
(i) As ABCD is a parallelogram
∠ADB = ∠CBD [Alternate interior angles are equal]……………….(1)
∠ABD = ∠CDB [Alternate interior angles are equal]…………………(2)
Now, in ∆APD and ∆CQB, we have
AD = CB [Opposite sides of a parallelogram ABCD are equal]
PD = QB [Given]
∠ADP = ∠CBQ [Proved]
Hence, ∆APD ≅ ∆CQB [By SAS congruency]
(ii) As, ∆APD ≅ ∆CQB [Proved]
AP = CQ [By C.P.C.T.]…………………(3)
(iii) Now, in ∆AQB and ∆CPD, we have
QB = PD [Given]
∠ABQ = ∠CDP [Proved]
AB = CD [ Opposite sides of a parallelogram ABCD are equal]
Hence, ∆AQB ≅ ∆CPD [By SAS congruency]
(iv) As, ∆AQB ≅ ∆CPD [Proved]
AQ = CP [By C.P.C.T.] …………………………..(4)
(v) In a quadrilateral APCQ,
Opposite sides are equal. [From (3) and (4)]
Hence, APCQ is a parallelogram. (NCERT Theorem 8.3)
Question 6. 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) ∆APB ≅ ∆CQD
(ii) AP = CQ
Solution:
ABCD is a parallelogram
DP = BQ
(i) In ∆APB and ∆CQD, we have
∠APB = ∠CQD [Each 90°]
AB = CD [ Opposite sides of a parallelogram ABCD are equal]
∠ABP = ∠CDQ [Alternate angles are equal as AB || CD and BD is a transversal]
Hence, ∆APB ≅ ∆CQD [By AAS congruency]
(ii) As, ∆APB ≅ ∆CQD [Proved]
AP = CQ [By C.P.C.T.]
Question 7. ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC ≅ ∆BAD
(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 …………………………(1) [Opposite sides of the parallelogram are equal]
But AD = BC …(2) [Given]
From (1) and (2),
BC = CE
Now, in ∆BCF, we have BC = CF
∠CEB = ∠CBE …(3) [Angles opposite to equal sides of a triangle are equal]
Also, ∠ABC + ∠CBE = 180° … (4) [Linear pair]
and ∠A + ∠CEB = 180° …(5) [Co-interior angles of a parallelogram ADCE]
From (4) and (5), we get
∠ABC + ∠CBE = ∠A + ∠CEB
∠ABC = ∠A [From (3)]
∠B = ∠A …(6)
(ii) AB || CD and AD is a transversal.
∠A + ∠D = 180° …(7) [Co-interior angles in parallelogram]
Similarly, ∠B + ∠C = 180° … (8)
From (7) and (8), we get
∠A + ∠D = ∠B + ∠C
∠C = ∠D [From (6)]
(iii) In ∆ABC and ∆BAD, we have
AB = BA [Common]
BC = AD [Given]
∠ABC = ∠BAD [Proved]
Hence, ∆ABC ≅ ∆BAD [By SAS congruency]
(iv) Since, ∆ABC ≅ ∆BAD [Proved]
AC = BD [By C.P.C.T.]
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