8. Quadrilaterals

Class 9 Mathematics Chapter 8. Quadrilaterals

Chapter 8. Quadrilaterals

Important Note :

1. Sum of the angles of a quadrilateral is 360°.
2. A diagonal of a parallelogram divides it into two congruent triangles.
3. In a parallelogram , opposite sides are equal .

4. In a parallelogram, opposite angles are equal
5. In a parallelogram, diagonals bisect each other .
6. A quadrilateral is a parallelogram, if opposite sides are equal .

7. A quadrilateral is a parallelogram, if opposite angles are equal .

8. A quadrilateral is a parallelogram, if diagonals bisect each other .
9. A quadrilateral is a parallelogram, if a pair of opposite sides is equal and parallel .
10. Diagonals of a rectangle bisect each other and are equal and vice-versa.
11. Diagonals of a rhombus bisect each other at right angles and vice-versa.
12. Diagonals of a square bisect each other at right angles and are equal, and vice-versa.
13. The line-segment joining the mid-points of any two sides of a triangle is parallel to the third side and is half of it.
14. A line through the mid-point of a side of a triangle parallel to another side bisects the third side.
15. The quadrilateral formed by joining the mid-points of the sides of a quadrilateral, in order, is a parallelogram.

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EXERCISE 8.1

1. The angle of quadrilateral are in the ratio . Find all the angles of the quadrilateral .

Solution: let, and are the angle of quadrilateral respectively .

We know that , the sum of the angles of a quadrilateral is 360° .

And

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

Solution: let ABCD is a parallelogram and AC = BD . Then we show that ABCD is a rectangle .

Proof : Since, ABCD is a parallelogram , then

AB = CD and AD = BC

In and, we have

AB=AB [ Common side ]

AD=BC [Given]

BD=AC [Given]

[SSS]

[CPCT]

Again, and is a transversal .

So, ABCD is a parallelogram in which one angle is 90° .

Therefore, ABCD is a rectangle .

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

Solution: let ABCD is a quadrilateral such that ,

and . AC and BD are the diagonals . Then we show that quadrilateral ABCD is rhombus .

Proof: In figure :

In and,we have

[SAS]

[CPCT]

Similarly, , and

So,

Therefore, the quadrilateral ABCD is rhombus .

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

Solution: let ABCD is a square and the diagonals AC = BD . Then we show that

Proof : Since, ABCD is a square .

OA = OC and OB=OD

AB = BC = CD = AD

In and ,we have

[SSS]

[CPCT]

Now,

So,

and .

Proved .

5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angle , then it is a square .

Solution: let ABCD is a quadrilateral and AC = BD and .

Then we show that ABCD is a square .

Proof: Since , the diagonals AC and BD bisect , then

OA = OC and OB = OD

In and,we have

[given]

[common side]

[SAS]

[CPCT]

Similarly , , and

So, AB = BC = CD = AD

In and ,we have

[given]

[common side]

[SAS]

[CPCT]

Again,

So,

Similarly ,

and

Therefore, ABCD is a square .

6. Diagonal AC of a parallelogram ABCD bisects (see Fig. 8.19) . Show that (i) it bisects also, (ii) ABCD is a rhombus .

Solution: Given, the diagonal AC of a parallelogram ABCD bisects . Then we show that :

(i) it bisects also,

(ii) ABCD is a rhombus .

Proof: Since, ABCD is a parallelogram .

So, AB = CD and AD = BC

And AC is the bisector of .

(i) In and , we have

[common side]

[given]

[SAS]

So, is the bisector of .

Therefore. AC bisect . Proved.

(ii) In and , we have

[common side]

[given]

[SAS]

[CPCT]

Therefore , ABCD is a rhombus .

7. ABCD is a rhombus . Show that diagonal AC bisects as well as and diagonal BD bisects and .

Solution: Given, ABCD is a rhombus . Then we show that diagonal AC bisects as well as and diagonal BD bisects and .

Proof : Since, ABCD is a rhombus .

So, AB = BC = CD = AD

In and , we have

[common side]

[given]

[SSS]

[CPCT]

So, AC bisect

and

So, AC bisect .

Therefore, the diagonal AC bisects as well as .

Second part :

In and , we have

[common side]

[given]

[SSS]

[CPCT]

So, BD bisect

and

So, BD bisect .

Therefore, the diagonal BD bisects as well as .

8. ABCD is a rectangle in which diagonal AC bisects as well as . Show that : (i) ABCD is a square (ii) diagonal BD bisects as well as .

Solution: Given, ABCD is a rectangle in which diagonal AC bisects as well as .Then we show that : (i) ABCD is a square (ii) diagonal BD bisects as well as .

Proof: Since, ABCD is rectangle .

So, AB = CD and AD = BC

(i) We have, AC bisects .

and AC bisects

In and , we have

[common side]

[given]

[SSS]

[CPCT]

So, AB = BC = CD = AD

and

Therefore, ABCD is a square . Proved .

(ii) In and , we have

[common side]

[given]

[SSS]

[CPCT]

So, BD bisect

and

So, BD bisect .

Therefore, the diagonal BD bisects as well as .

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

(i) (ii) (iii) (iv) (v) is a parallelogram .

Solution: Given, In parallelogram ABCD , two points P and Q are taken on diagonal BD such that . Then we show that : (i) (ii) (iii) (iv) (v) is a parallelogram .

Proof: Since, ABCD is a parallelogram .

So, and

and

(i) In and , we have

[Given]

[Alternative interior angle]

[Given]

[SAS]

Proved.

(ii) In and , we have

[Given]

[Alternative interior angle]

[Given]

[SAS]

[CPCT]

Proved.

(iii) In and , we have

[Given]

[Alternative interior angle]

[Given]

[SAS]

Proved.

(iv) In and , we have

[Given]

[Alternative interior angle]

[Given]

[SAS]

Proved.

(v) Proof: Since, [SAS]

And [SAS]

We know that a quadrilateral is a parallelogram, then a pair of opposite sides is equal and parallel .

So, and

Again, and

Thus, APCQ is a parallelogram .

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

(i) (ii)

Solution: Given, ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD . Then we show that : (i) (ii)

Proof: (i) In and , we have

[ ABCD is a parallelogram]

[Alternative interior angle ]

[ASA]

(ii) In and , we have

[ ABCD is a parallelogram]

[Alternative interior angle ]

[ASA]

[CPCT]

11. In and , and . 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

(iii) and

(iv) quadrilateral ACFD is a parallelogram

(v)

(vi)

Solution: Given, and , and . Vertices A , B and C are joined to vertices D , E and F respectively .

Then we show that :

(i) quadrilateral ABED is a parallelogram

(ii) quadrilateral BEFC is a parallelogram

(iii) and

(iv) quadrilateral ACFD is a parallelogram

(v)

(vi) .

Proof: In ABED quadrilateral .

We know that, a quadrilateral is a parallelogram , when a pair of opposite angle is equal and parallel .

So, and

Therefore, the quadrilateral ABED is a parallelogram .

(ii) In BEFC quadrilateral .

We know that, a quadrilateral is a parallelogram , when a pair of opposite angle is equal and parallel .

So, and

Therefore, the quadrilateral BEFC is a parallelogram .

(iii) Since, the quadrilateral ABED is a parallelogram .

So, and

Again, the quadrilateral BEFC is a parallelogram .

So, and

From and we get

and Proved.

(iv) In ACFD quadrilateral .

We know that, a quadrilateral is a parallelogram , when a pair of opposite angle is equal and parallel .

So, and

Therefore, the quadrilateral ACFD is a parallelogram .

Proved.

(v) Since, the quadrilateral ACFD is a parallelogram .

So, and

Therefore, AC = DF

Proved.

(vi) In and , we have

AB=DE [given]

BC=EF [given]

AC=DF [given]

[SSS] Proved.

12. ABCD is a trapezium in which and (see Fig. 8.23) . Show that :

(i)

(ii)

(iii)

(iv) diagonal diagonal .

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

Solution: Given, ABCD is a trapezium in which and .Then we show that

(i)

(ii)

(iii)

(iv) diagonal diagonal .

Construction: Extend AB and we draw a line through C parallel to DA intersecting AB produced at E .

Proof: Since,

And AD = BC , then AD = BC = CE .

In , we have

So, BCE is an isosceles triangle.

Again, and is transversal .

and [Linear pair of angles]

From and , we get

Proved.

(ii) Since, and is transversal .

Again, and is transversal .

From (1) and (2) , we get

[ ]

Proved.

(iii) In and ,we have

[common side]

[given]

[SAS]

(iv) In and ,we have

[common side]

[given]

[SAS]

[CPCT]

Therefore, Diagonal AC = Diagonal BD . Proved.

EXERCISE 8.2

1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that : (i) and (ii) PQ = SR (iii) PQRS is a parallelogram.

Solution: Given, ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA . AC is a diagonal.

Then we show that : (i) and (ii) PQ = SR (iii) PQRS is a parallelogram.

Proof: (i) Since, S and R are the mid-point of the sides AD and CD of a respectively .

and Proved.

(ii) Since, S and R are the mid-point of the sides AD and CD of a respectively .

and

Again, Since, P and Q are the mid-point of the sides AB and BC of a respectively .

and

From(i) and (ii) , we get Proved.

(iii) Since, S and R are the mid-point of the sides AD and CD of a respectively .

and

Again, Since, P and Q are the mid-point of the sides AB and BC of a respectively .

and

From(i) and (ii) , we get

and

We know that , a quadrilateral is a parallelogram then a pair of opposite sides is equal and parallel .

Thus, PQRS is a parallelogram . Proved.

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.

Solution: Given, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Then we show that the quadrilateral PQRS is a rectangle.

Proof: Since, P and Q are the mid-point of the sides AB and BC of a respectively .

and

Again, S and R are the mid-point of the sides AD and CD of a respectively .

and

From(i) and (ii) , we get

Similarly,

So, PQRS is a parallelogram .

Since, ABCD is rhombus .

AB = BC = CD = AD

And AP = BP = BQ = CQ = CR = DR = AS = DS

In ,We have

So, APS is an isosceles triangle .

In and , we have

BP = DR [Given ]

[Given]

BQ = DS [Given]

[SAS]

[CPCT]

We have,

[Linear pair of angles]

and

From (i) and (ii) , we get

[ and ]

Since, and is a transversal .

So, PQRS is a parallelogram in which one angle is 90° .

Therefore, PQRS is a rectangle . Proved

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.

Solution: Given, ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Then we show that the quadrilateral PQRS is a rhombus.

Proof: Since, P and Q are the mid-point of the sides AB and BC of a respectively .

and

Again, S and R are the mid-point of the sides AD and CD of a respectively .

and

From(i) and (ii) , we get

So, PQRS is a parallelogram .

Since, ABCD is rectangle .

So, AP = BP [P is the mid-point of AB]

And AS = BQ [S and Q is the mid-point of AD and BC]

In and , we have

[Given]

[SAS]

[CPCT]

Hence , PQRS is a rhombus . Proved.

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 Fig. 8.30). Show that F is the mid-point of BC.

Solution: Given, ABCD is a trapezium in which , 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 . Then we show that F is the mid-point of BC.

Proof: Since, ABCD is a trapezium and .

Let BD and EF intersecting at M .

By the mid-point theorem, we get

and E is the mid-point of AD .

So, and M is the mid-point of BD .

Also

In and

So, F is the mid-point of BC . Proved.

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD.

Solution: Given, a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively . Then we show that the line segments AF and EC trisect the diagonal BD.

Proof: Since, E and F are the mid-points of sides AB and CD respectively .

So, and

ABCD is a parallelogram .

So, and

and AE∥CE

Hence, AECF is a parallelogram .

So,

By the mid-point theorem ,

In ,we have

Since, E is a mid-point of AB and ,then Q is the mid-point of BP .

In ,we have

Since, F is a mid-point of CD and ,then P is the mid-point of DQ .

From (i) and (ii) , we get

Hence, the line segments AF and EC trisect the diagonal BD. Proved.

6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Solution: Given, PQRS is a quadrilateral and A, B, C and D are mid-points of the sides PQ, QR, RS and PS respectively .Then we show that the diagonals AC and BD bisect each other .

Proof: Since, A and B are the mid-point of the sides PQ and QR of a respectively .

and

Again, Since, C and D are the mid-point of the sides RS and PS of a respectively .

and

From(i) and (ii) , we get

and

Similarly, and

and

From (iii) and (iv), we get

and

Thus, ABCD is a parallelogram .

We know that the diagonals of a parallelogram bisect each other .

Hence, the diagonals AC and BD bisect each other .

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) (iii) .

Solution: Given, 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.

Then we show that (i) D is the mid-point of AC (ii) (iii)

Proof: In figure:

(i) In , we have

Since, M is the mid-point of hypotenuse AB and .

By the mid-point theorem, we get

D is the mid-point of the side AC . Proved

(ii) Given,

Since,

So, [ Corresponding angle]

And also

Thus, Proved.

(iii) In and , we have

[D is a mid-point of AC]

[Common side]

[RHS]

[CPCT]

So,

Proved .