# Quadrilaterals (Exercise 8.2) NCERT Solutions - Mathematics (Maths) Class 9

Q1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig). AC is a diagonal.
Show that:
(i) SR || AC and SR = (1/2) AC
(ii) PQ = SR
(iii) PQRS is a parallelogram. Ans: We have P as the mid-point of AB, Q as the mid-point of BC, R as the mid-point of CD, S as the mid-point of DA, and AC as the diagonal of quadrilateral ABCD.
(i) To prove that SR =(1/2) AC and SR || AC.
In ΔACD, we have
S as the mid-point of AD, R as the mid-point of CD.
∵ The line segment joining the mid-point of any two sides of a triangle is parallel to the third side and half of it.
∴ SR = (1/2)AC and SR || AC
(ii) To prove that PQ = SR.
In ΔABC, we have P is the mid-point of AB, Q is the mid-point of BC.
∴ PQ = (1/2) AC            …(1)
Also, SR = (1/2) AC            [Proved]             …(2)
From (1) and (2), PQ = SR
(iii) To prove that PQRS is a parallelogram.
In ΔABC, P and Q are the mid-points of AB and BC.
∴ PQ = (1/2)AC and PQ || AC             ...(3)
In ΔACD, S and R are the mid-points of DA and CD.
∴ SR = (1/2)AC and SR || AC             …(4)
From (3) and (4), we get
PQ =(1/2)AC = SR and PQ || AC || SR
⇒ PQ = SR and PQ || SR
i.e. One pair of opposite sides in quadrilateral PQRS is equal and parallel.
∴ PQRS is a parallelogram.

Q2. 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.
Ans: We have P as the mid-point of AB, Q as the midpoint of BC, R as the mid-point of CD, S as the mid-point of DS. We have to prove that PQRS is a rectangle.
Let us join AC.
∵ In ΔABC, P and Q are the mid-points of AB and BC.
∴ PQ =(1/2)AC and PQ || AC             …(1)
Also in ΔADC, R and S are the mid-points of CD and DA.
∴ SR = (1/2)AC and SR || AC
From (1) and (2), we get
PQ =(1/2) AC = SR and PQ || AC || SR
⇒ PQ = SR and PQ || SR
i.e. One pair of opposite sides of quadrilateral PQRS is equal and parallel.
∴ PQRS is a parallelogram.
Now, in ΔERC and ΔEQC, ∠1= ∠2            [∵ The diagonal of a rhombus bisects the opposite angles]
CR = CQ            [Each is equal to(1/2) of a side of rhombus]CE = CE [Common]
∴ ΔERC ≌ ΔEQC            [SAS criteria]
⇒ ∠3= ∠4[c.p.c.t.]
But ∠3 + ∠4 = 180º [Linear pair] ⇒ ∠3= ∠4 = 90°
But ∠5= ∠3            [Vertically opposite angles]
∴ ∠5 = 90º PQ || AC
⇒ PQ || EF
∴ PQEF is a quadrilateral having a pair of opposite sides parallel and one of the angles is 90º.
∴ PQEF is a rectangle.
⇒ ∠RQP = 90º
∴ One angle of parallelogram PQRS is 90º.
Thus, PQRS is a rectangle.

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.
Ans: Given in the question,
ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively.
Construction,
Join AC and BD.
To Prove,
PQRS is a rhombus.
Proof:
In ΔABC
P and Q are the mid-points of AB and BC, respectively
, PQ || AC and PQ = ½ AC (Midpoint theorem) — (i)
SR || AC and SR = ½ AC (Midpoint theorem) — (ii)
So, PQ || SR and PQ = SR
As in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, so, it is a parallelogram.
, PS || QR and PS = QR (Opposite sides of parallelogram) — (iii)
Now,
In ΔBCD,
Q and R are mid points of side BC and CD, respectively.
, QR || BD and QR = ½ BD (Midpoint theorem) — (iv)
AC = BD (Diagonals of a rectangle are equal) — (v)
From equations (i), (ii), (iii), (iv) and (v),
PQ = QR = SR = PS
So, PQRS is a rhombus.
Hence Proved

Q4. 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 Fig). Show that F is the mid-point of BC. Ans: Given that,
ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD.
To prove,
F is the mid-point of BC.
Proof,
BD intersected EF at G.
E is the mid point of AD and also EG || AB.
Thus, G is the mid point of BD (Converse of mid point theorem)
Now,
In ΔBDC,
G is the mid point of BD and also GF || AB || DC.
Thus, F is the mid point of BC (Converse of mid point theorem)

Q5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig). Show that the line segments AF and EC trisect the diagonal BD.
Ans: We have ABCD is a parallelogram such that: E is the mid-point of AB and F is the mid-point of CD. Let us join the opposite vertices B and D.
Since, the opposite sides of a parallelogram are parallel and equal.
∴ AB || DC ⇒ AE || FC            …(1)
Also AB = DC
or (1/2) AB =(1/2)DC ⇒ AE = FC            …(2)
From (1) and (2),
we can say that AECF is quadrilateral having a pair of the opposite sides as parallel and equal.
∴ AEFC is a parallelogram.
⇒ AE || CF
Now, in DABC, F is the mid-point of DC             [Given]
and FP || CQ             [∵ AF || CE]
⇒ P is the mid-point of DQ             [Converse of mid-point theorem]
⇒ DP = PQ             …(3)
Similarly, in DBAP, BQ = PQ             …(4)
∴ From (3) and (4), we have
DP = PQ = BQ
⇒ The line segments AF and EC trisect the diagonal BD.

Q6. 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) CM = MA = 1/2 AB
Ans: We have a triangle ABC, such that ∠C = 90º M is the mid-point of AB and MD || BC (i) To prove that D is the mid-point of AC.
In ΔACB, we have M as the mid-point of AB.             [Given]
MD || BC            [Given]
∴ Using the converse of mid-point theorem, D is the mid-point of AC.
(ii) To prove that MD ⊥ AC.
Since, MD || BC             [Given] and AC is a transversal.
∴ ∠ MDA = ∠BCA            [Corresponding angles]
But ∠BCA = 90º            [Given]
∴ ∠MDA = 90º ⇒ MD ⊥ AC.
(iii) To prove that CM = MA =(1/2) AB
In ΔADM and ΔCDM, we have
∠ADM = ∠CDM            [Each = 90º]
MD = MD            [Common]
AD = CD            [∵ M is the mid-point of AC (Proved)]
∴ ΔADM ≌ ΔCOM            [SAS criteria]
⇒ MA = MC [c.p.c.t.]             …(1)
∵ M is the mid-point AB.            [Given]
∴ MA =(1/2)AB              …(2)
From (1) and (2), we have
CM = MA = (1/2) AB

The document Quadrilaterals (Exercise 8.2) NCERT Solutions | Mathematics (Maths) Class 9 is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

## FAQs on Quadrilaterals (Exercise 8.2) NCERT Solutions - Mathematics (Maths) Class 9

 1. What are the properties of a quadrilateral? Ans. A quadrilateral is a polygon with four sides. The properties of a quadrilateral include: - The sum of its interior angles is always equal to 360 degrees. - Opposite angles are equal. - The opposite sides are parallel. - Diagonals bisect each other.
 2. How many diagonals does a quadrilateral have? Ans. A quadrilateral has two diagonals. Diagonals are line segments that connect two non-adjacent vertices of a polygon. In a quadrilateral, there are two pairs of non-adjacent vertices, and each pair can be connected by a diagonal.
 3. What are the types of quadrilaterals? Ans. There are several types of quadrilaterals, including: - Parallelogram: A quadrilateral with opposite sides parallel. - Rectangle: A parallelogram with all angles equal to 90 degrees. - Square: A rectangle with all sides equal in length. - Rhombus: A parallelogram with all sides equal in length. - Trapezium: A quadrilateral with one pair of parallel sides. - Kite: A quadrilateral with two pairs of adjacent sides equal in length.
 4. How can we determine if a quadrilateral is a parallelogram? Ans. To determine if a quadrilateral is a parallelogram, we need to check if its opposite sides are parallel. This can be done by comparing the slopes of the opposite sides. If the slopes are equal, then the sides are parallel, and the quadrilateral is a parallelogram.
 5. What is the formula to find the area of a quadrilateral? Ans. The formula to find the area of a quadrilateral depends on its type. Some common formulas include: - For a parallelogram: Area = base x height. - For a rectangle: Area = length x width. - For a square: Area = side x side. - For a rhombus: Area = (diagonal1 x diagonal2) / 2. - For a trapezium: Area = (sum of parallel sides) x height / 2.

## Mathematics (Maths) Class 9

62 videos|426 docs|102 tests

## Mathematics (Maths) Class 9

62 videos|426 docs|102 tests
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