EXERCISE 8.2
Q 1, Ex 8.2, Page No 150, Quadrilaterals, NCERT Class 9th Maths
Page No 150:
Question 1:
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the given figure). AC is a diagonal. Show that:
(i) SR || AC and SR =
AC(ii) PQ = SR
(iii) PQRS is a parallelogram.
Answer:
(i) In ΔADC, S and R are the mid-points of sides AD and CD respectively.In a triangle, the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it.
∴ SR || AC and SR = $\frac{1}{2}$AC … (1)
(ii) In ΔABC, P and Q are mid-points of sides AB and BC respectively. Therefore, by using mid-point theorem,
PQ || AC and PQ = $\frac{1}{2}$AC … (2)
Using equations (1) and (2), we obtain
PQ || SR and PQ = SR … (3)
⇒ PQ = SR
(iii) From equation (3), we obtained
PQ || SR and PQ = SR
Clearly, one pair of opposite sides of quadrilateral PQRS is parallel and equal.
Hence, PQRS is a parallelogram.
Q 2, Ex 8.2, Page No 150, Quadrilaterals, CBSE Class 9th Maths
Question 2:
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.Answer:
In ΔABC, P and Q are the mid-points of sides AB and BC respectively.
∴ PQ || AC and PQ = $\frac{1}{2}$AC (Using mid-point theorem) … (1)
In ΔADC,
R and S are the mid-points of CD and AD respectively.
∴ RS || AC and RS = $\frac{1}{2}$AC (Using mid-point theorem) … (2)
From equations (1) and (2), we obtain
PQ || RS and PQ = RS
Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to
each other, it is a parallelogram.
Let the diagonals of rhombus ABCD intersect each other at point O.
In quadrilateral OMQN,
MQ || ON (
PQ || AC)QN || OM (
QR || BD)Therefore, OMQN is a parallelogram.
⇒ ∠MQN = ∠NOM
⇒ ∠PQR = ∠NOM
However, ∠NOM = 90° (Diagonals of a rhombus are perpendicular to each other)
∴ ∠PQR = 90°
Clearly, PQRS is a parallelogram having one of its interior angles as 90º.
Hence, PQRS is a rectangle.
Q 3, Ex 8.2, Page No 150, Quadrilaterals, Class 9th
Question 3:
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.Answer:
Let us join AC and BD.
In ΔABC,
P and Q are the mid-points of AB and BC respectively.
∴ PQ || AC and PQ = $\frac{1}{2}$AC (Mid-point theorem) … (1)
Similarly in ΔADC,
SR || AC and SR = $\frac{1}{2}$ AC (Mid-point theorem) … (2)
Clearly, PQ || SR and PQ = SR
Since in quadrilateral PQRS, one pair of opposite sides is equal and parallel to
each other, it is a parallelogram.
∴ PS || QR and PS = QR (Opposite sides of parallelogram)… (3)
In ΔBCD, Q and R are the mid-points of side BC and CD respectively.
∴ QR || BD and QR =$\frac{1}{2}$BD (Mid-point theorem) … (4)
However, the diagonals of a rectangle are equal.
∴ AC = BD …(5)
By using equation (1), (2), (3), (4), and (5), we obtain
PQ = QR = SR = PS
Therefore, PQRS is a rhombus.
Q 4, Ex 8.2, Page No 150, Quadrilaterals, Class 9th Maths
Question 4:
ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid – point of AD. A line is drawn through E parallel to AB intersecting BC at F (see the given figure). Show that F is the mid-point of BC.
Answer:
Let EF intersect DB at G.
By converse of mid-point theorem, we know that a line drawn through the mid-point of any side of a triangle and parallel to another side, bisects the third side.
In ΔABD,
EF || AB and E is the mid-point of AD.
Therefore, G will be the mid-point of DB.
As EF || AB and AB || CD,
∴ EF || CD (Two lines parallel to the same line are parallel to each other)
In ΔBCD, GF || CD and G is the mid-point of line BD. Therefore, by using converse of mid-point theorem, F is the mid-point of BC.
Q 5, Ex 8.2, Page No 151, Quadrilaterals, Class 9th Maths
Page No 151:
Question 5:
In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). Show that the line segments AF and EC trisect the diagonal BD.
Answer:
ABCD is a parallelogram.∴AB || CD
And hence, AE || FC
Again, AB = CD (Opposite sides of parallelogram ABCD)
$\frac{1}{2}$AB =$\frac{1}{2}$CD
AE = FC (E and F are mid-points of side AB and CD)
In quadrilateral AECF, one pair of opposite sides (AE and CF) is parallel and equal to each other. Therefore, AECF is a parallelogram.
⇒ AF || EC (Opposite sides of a parallelogram)
In ΔDQC, F is the mid-point of side DC and FP || CQ (as AF || EC). Therefore, by using the converse of mid-point theorem, it can be said that P is the mid-point of DQ.
⇒ DP = PQ … (1)
Similarly, in ΔAPB, E is the mid-point of side AB and EQ || AP (as AF || EC). Therefore, by using the converse of mid-point theorem, it can be said that
Q is the mid-point of PB.
⇒ PQ = QB … (2)
From equations (1) and (2),
DP = PQ = BQ
Hence, the line segments AF and EC trisect the diagonal BD
Q 6, Ex. 8.2, Page No 151, Quadrilaterals - NCERT Class 9th Maths
Question 6:
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.Answer:
Let ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively. Join PQ, QR, RS, SP, and BD.
In ΔABD, S and P are the mid-points of AD and AB respectively. Therefore, by using mid-point theorem, it can be said that
SP || BD and SP = $\frac{1}{2}$ BD … (1)
Similarly in ΔBCD,
QR || BD and QR = $\frac{1}{2}$BD … (2)
From equations (1) and (2), we obtain
SP || QR and SP = QR
In quadrilateral SPQR, one pair of opposite sides is equal and parallel to
each other. Therefore, SPQR is a parallelogram.
We know that diagonals of a parallelogram bisect each other.
Hence, PR and QS bisect each other.
Q 7, Ex 8.2, Page No 151, Quadrilaterals, Class 9th Mathematics
Question 7:
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that(i) D is the mid-point of AC
(ii) MD ⊥ AC
(iii) $\mathrm{CM}=\mathrm{MA}=\frac{1}{2} \mathrm{AB}$
Answer:
(i) In ΔABC,
It is given that M is the mid-point of AB and MD || BC.
Therefore, D is the mid-point of AC. (Converse of mid-point theorem)
(ii) As DM || CB and AC is a transversal line for them, therefore,
∠MDC + ∠DCB = 180º (Co-interior angles)
∠MDC + 90º = 180º
∠MDC = 90º
∴ MD ⊥ AC
(iii) Join MC.
In ΔAMD and ΔCMD,
AD = CD (D is the mid-point of side AC)
∠ADM = ∠CDM (Each 90º)
DM = DM (Common)
∴ΔAMD ≅ ΔCMD (By SAS congruence rule)
Therefore, AM = CM (By CPCT)
However, AM = $\frac{1}{2}$AB (M is the mid-point of AB)
Therefore, it can be said that
CM = AM = $\frac{1}{2}$AB
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