Ch 12 Quadrilaterals Solutions Class 9 -

Table of Contents

Exercise Set 12.1

Q1

Let ABCD be a quadrilateral.
(i) List all sides of ABCD adjacent to side AB. List all sides
opposite to AB.
(ii) List all angles of ABCD adjacent to ∠A. List all angles opposite
to ∠A.
(iii) Define a pair of opposite sides and a pair of opposite angles
without using the names of the vertices.

Solution:

Given,

ABCD is a quadrilateral.

(i)

Adjacent to AB: BC and AD

Opposite to AB: CD

(ii)

Adjacent to ∠A: ∠B and ∠D

Opposite to ∠A: ∠C

(iii)

Opposite sides: two sides with no common endpoint.

Opposite angles: two angles with no common side.

Q2

You have used internal angles of quadrilaterals, but they too require an exact definition, just like how we gave one for a quadrilateral. Precisely define the internal angle of a quadrilateral
at a given vertex. Your answer should work for a non-convex quadrilateral too. (Hint: use the opposite vertex as well.)

Solution:

Given,

ABCD is a quadrilateral. It can be non-convex.

A is the vertex. B and D are adjacent vertices. C is the opposite vertex.

Non-convex ABCD. A is inside triangle BCD.

Definition:

Rays AB and AD make two angles at A.

One is ∠BAD (less than 180°). The other is 360° − ∠BAD (more than 180°).

If A is outside △BCD, then the internal angle is ∠BAD.

If A is inside △BCD, then the internal angle is 360° − ∠BAD.

Q3

In a quadrilateral ABCD, suppose AB ‖ DC. Can ABCD be non-convex? What if we instead assume AB = CD? What if we instead assume ∠A = ∠C?

Solution:

Given,

ABCD is a quadrilateral.

(i) AB ∥ DC

Lines AB and DC are parallel.

C and D lie on one side of line AB.

A and B lie on one side of line DC.

If A and D were on opposite sides of line BC, side AD would cross side BC.

This is not allowed. So A and D lie on one side of BC.

In the same way, B and C lie on one side of AD.

If AB ∥ DC, then ABCD is convex.

(ii) AB = CD

ABCD can be non-convex.

Take A(0, 0), B(4, 0), C(2, 6), D(2, 2).

AB = 4 and CD = 4.

D is inside △ABC. So ∠D is more than 180°.

(iii) ∠A = ∠C

ABCD can be non-convex.

Take A(−3, 0), B(0, 6), C(3, 0), D(0, 2).

The y-axis is a mirror line. It swaps A and C.

So ∠A = ∠C.

D is inside △ABC. So ∠D is more than 180°.

∴

(i) AB ∥ DC: non-convex is No

(ii) AB = CD: non-convex is Yes

(iii) ∠A = ∠C: non-convex is Yes

} Ans.

Q4

Consider three non-collinear points A, B, C and draw the lines AB, BC, CA. For every possible location of point D in the plane outside these lines, decide if ABCD is self-intersecting,
non-convex, or convex. (Hint: the three lines divide the plane into 7 regions.)

Solution:

Given,

A, B, C are non-collinear points.

Lines AB, BC, CA divide the plane into 7 regions.

D is a point outside the three lines.

Result

D in region	Reason	ABCD
1 (inside △ABC)	D is a dent. ∠D > 180°.	Non-convex
2 (beyond AC)	BD cuts AC. Diagonals meet.	Convex
3 (beyond AB)	CD cuts AB.	Self-intersecting
4 (beyond BC)	AD cuts BC.	Self-intersecting
5 (vertex A zone)	A is inside △BCD. ∠A > 180°.	Non-convex
6 (vertex B zone)	B is inside △ACD. ∠B > 180°.	Non-convex
7 (vertex C zone)	C is inside △ABD. ∠C > 180°.	Non-convex