# NCERT Solutions for Class 9 Maths Unit 8

### Unit 8 Quadrilaterals Exercise 8.1, 8.2 Solutions

Quadrilateral is a polygon with four sides (or edges) and four vertices or corners. Sometimes, the term quadrangle is used, by analogy with triangle, and sometimes tetragon for consistency with pentagon (5-sided), hexagon (6-sided) and so on.
The origin of the word "quadrilateral" is the two Latin words quadri, a variant of four, and latus, meaning "side".
Quadrilaterals are simple (not self-intersecting) or complex (self-intersecting), also called crossed. Simple quadrilaterals are either convex or concave.
The interior angles of a simple (and planar) quadrilateral ABCD add up to 360 degrees of arc, that is
A + B + C + D = 360 degree.
This is a special case of the n-gon interior angle sum formula (n - 2) × 180°. In a crossed quadrilateral, the four interior angles on either side of the crossing add up to 720°.
All convex quadrilaterals tile the plane by repeated rotation around the midpoints of their edges.

### Exercise 8.1 : Solutions of Questions on Page Number : 146

Q1 :

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

Let the common ratio between the angles be x. Therefore, the angles will be 3x, 5x, 9x, and 13x respectively.

As the sum of all interior angles of a quadrilateral is 360º,

∴ 3x + 5x + 9x + 13x = 360º

30x = 360º

x = 12º

Hence, the angles are

3x = 3 x 12 = 36º

5x = 5 x 12 = 60º

9x = 9 x 12 = 108º

13x = 13 x 12 = 156º

Q2 :

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

Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of its interior angles is 90º.

In ΔABC and ΔDCB,

AB = DC (Opposite sides of a parallelogram are equal)

BC = BC (Common)

AC = DB (Given)

∴ ΔABC ≅ ΔDCB (By SSS Congruence rule)

⇒ ∠ ABC = ∠ DCB

It is known that the sum of the measures of angles on the same side of transversal is 180º.

∠ ABC + ∠ DCB = 180º (AB || CD)

⇒ ∠ ABC + ∠ ABC = 180º

⇒ 2∠ ABC = 180º

⇒ ∠ ABC = 90º

Since ABCD is a parallelogram and one of its interior angles is 90º, ABCD is a rectangle.

Q3 :

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

Q4 :

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

Q5 :

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

Q6 :

Diagonal AC of a parallelogram ABCD bisects ∠ A (see the given figure). Show that

(i) It bisects ∠ C also,

(ii) ABCD is a rhombus.

Q7 :

ABCD is a rhombus. Show that diagonal AC bisects ∠ A as well as ∠ C and diagonal BD bisects ∠ B as well as ∠ D.

Q8 :

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.

Q9 :

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see the given figure). Show that:

(i) ΔAPD ≅ ΔCQB

(ii) AP = CQ

(iii) ΔAQB ≅ ΔCPD

(iv) AQ = CP

(v) APCQ is a parallelogram

Q10 :

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (See the given figure). Show that

(i) ΔAPB ≅ ΔCQD

(ii) AP = CQ

Q11 :

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see the given figure). Show that

(i) Quadrilateral ABED is a parallelogram

(ii) Quadrilateral BEFC is a parallelogram

(iv) Quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) ΔABC ≅ ΔDEF.

Q12 :

ABCD is a trapezium in which AB || CD and AD = BC (see the given figure). Show that

(i) ∠ A = ∠ B

(ii) ∠ C = ∠ D

(iv) diagonal AC = diagonal BD

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

### Exercise 8.2 : Solutions of Questions on Page Number : 150

Q1 :

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.

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.

Q3 :

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.

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 the given figure). Show that F is the mid-point of BC.

Q5 :

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.

Q6 :

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

Q7 :

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)