EXERCISE SET 12.1

Question 1

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 angle A. List all angles opposite to angle A.
(iii) Define a pair of opposite sides and a pair of opposite angles without using the names of the vertices.

Solution:
(i)

• Adjacent sides to side AB: The sides that share a common vertex with AB are side AD (at vertex A) and side BC (at vertex B).
• Opposite side to side AB: The side that does not share any vertex with AB is side CD.

(ii)

• Adjacent angles to angle A: The angles that share a common side with angle A are angle B (sharing side AB) and angle D (sharing side AD).
• Opposite angle to angle A: The angle that does not share any side with angle A is angle C.

(iii)

• Pair of opposite sides: Two sides of a quadrilateral are said to be opposite if they do not share any common vertex (or endpoint).
• Pair of opposite angles: Two interior angles of a quadrilateral are said to be opposite if they do not share a common side (arm).

Question 2

Precisely define the internal angle of a quadrilateral at a given vertex. Your answer should work for a non-convex quadrilateral too.

Solution:
Let ABCD be a quadrilateral and let V be a chosen vertex (e.g., vertex B, with adjacent vertices A and C). The interior angle at vertex V is defined as the region swept by rotating ray VA into ray VC through the interior points of the quadrilateral.

More precisely, using the opposite vertex D:

1. Draw segment VD.
2. If D lies in the interior of angle ABC (the convex case), the interior angle at V is the union of angle AVD and angle CVD.
3. For a non-convex vertex, the interior angle is reflex (greater than 180 degrees). Formally, the interior angle at vertex V is the angle between the two adjacent sides meeting at V measured through the bounded region enclosed by the four sides of the quadrilateral.

Question 3

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

?
Solution:

1. Case 1: AB || DC. For a simple (non-self-intersecting) quadrilateral, AB || DC implies it is a convex trapezium. The consecutive interior angles sum to 180 degrees, forcing all angles to be less than 180 degrees.
2. Case 2: AB = CD. Yes, ABCD can be non-convex. We can construct a “dart” shape where vertex B is pushed inward towards the interior of the shape while keeping the lengths of AB and CD equal.
3. Case 3: angle A = angle C. Yes, ABCD can be non-convex. For instance, in a symmetric dart-shaped quadrilateral, one of the remaining angles (B or D) can be reflex (> 180 degrees) while angles A and C remain equal acute or obtuse angles.

Question 4

Classification of ABCD based on the location of D relative to triangle ABC.

Solution:
The lines AB, BC, and CA divide the plane into 7 regions. The classification is:

• D inside triangle ABC: ABCD is non-convex (a dart/arrowhead shape) with a reflex angle at D.
• D in regions vertically opposite to vertices A, B, or C: The ordering A-B-C-D produces a convex quadrilateral.
• D in regions between parallel strip extensions of sides: The edges will cross, making ABCD self-intersecting.

Question 5

Can a quadrilateral be both self-intersecting and non-planar?

Solution:
Yes. A quadrilateral is non-planar (skew) when its four vertices do not lie in a single plane. If we choose four non-coplanar points in 3D space, the resulting segments AB, BC, CD, and DA form a skew quadrilateral which, when projected onto a 2D plane, often appears self-intersecting.

EXERCISE SET 12.2

Question 1

True or false?
(i) A parallelogram with a right angle is a rectangle.
(ii) A rhombus with perpendicular diagonals is a square.
(iii) If the diagonals of a parallelogram are equal, then it is a rectangle.

Solution:
(i) True. In a parallelogram, adjacent angles are supplementary. If one is 90°, the others must also be 90°.
(ii) False. Every rhombus has perpendicular diagonals by definition. To be a square, it must also have equal diagonals or right angles.
(iii) True. If diagonals are equal, triangles ABC and DCB are congruent (SSS), making angle ABC = angle DCB = 90°.

Question 2

The diagonal AC of a parallelogram ABCD bisects angle A. Show that it also bisects angle C and that ABCD is a rhombus.

Solution:

1. AC bisects angle C: Since AB || CD, angle BAC = angle DCA (alternate interior). Since AD || BC, angle DAC = angle BCA (alternate interior). Given angle BAC = angle DAC, it follows that angle DCA = angle BCA.
2. ABCD is a rhombus: In triangle ABC, angle BAC = angle BCA, so AB = BC. Since opposite sides of a parallelogram are equal (AB = CD, BC = AD), all four sides are equal.

Question 3

Answer with Yes/No. If No, what extra condition is needed?
(i) If diagonals bisect angles, is it a rhombus?
(ii) If diagonals bisect each other at right angles, is it a rhombus?
(iii) If diagonals are of equal length, is it a rectangle?

Solution:
(i) Yes.
(ii) Yes.
(iii) No. Extra condition: ABCD must be a parallelogram. (An isosceles trapezium also has equal diagonals).

Question 4

Show that the intersection points of the four angle bisectors of a parallelogram form a rectangle. Why assume AB != BC?

Solution:
In parallelogram ABCD, angle A + angle D = 180°. Their halves sum to 90°. In triangle ASD, the third angle at S must be 90°. Applying this to all four triangles formed by the bisectors results in a quadrilateral with four 90° angles, which is a rectangle.

• Assumption AB != BC: If AB = BC (a rhombus), the angle bisectors are the diagonals, which all meet at a single point. The rectangle would collapse into a point.

EXERCISE SET 12.3

Question 1

(i) Congruence of midsegment triangles. (ii) Reconstruction.

Solution:
(i) By the Midpoint Theorem, each small triangle formed by the midpoints (PQR, QPA, PBR, RQC) has side lengths equal to half the sides of ABC. By SSS, all four are congruent.
(ii) Yes. To reconstruct ABC, draw lines through P, Q, and R parallel to the opposite sides of triangle PQR.

Question 2

In triangle ABC, let M and N be midpoints of AB and AC. Show that MN bisects AD.

Solution:
By the Midpoint Theorem, MN || BC. In triangle ABD, M is the midpoint of AB and the segment MK (part of MN) is parallel to BD. By the Converse of the Midpoint Theorem, K must be the midpoint of AD.

Question 3

In a trapezium ABCD (AB || DC), let G and H be midpoints of diagonals AC and BD. Prove GH || AB.

Solution:
Let M be the midpoint of AD. In triangle ADC, MG || CD and MG = 1/2 CD. In triangle ABD, MH || AB and MH = 1/2 AB. Since CD || AB, both MG and MH are parallel to AB and share point M, meaning M, G, and H are collinear. Thus, GH || AB.

• AB != CD: If AB = CD, it is a parallelogram where diagonals bisect each other, making G and H the same point (GH = 0).

Question 4

Midpoints P, Q, R, S of sides AB, BC, CD, DA.
(i) Show PR and QS bisect each other.
(ii) Show if AC = BD, then PR is perpendicular to QS.

Solution:
(i) PQRS is a Varignon parallelogram because PQ || SR || AC and PQ = SR = 1/2 AC. Diagonals of a parallelogram bisect each other.
(ii) If AC = BD, then PQ = 1/2 AC = 1/2 BD = PS. A parallelogram with equal adjacent sides is a rhombus, and its diagonals (PR, QS) are perpendicular.

EXERCISE SET 12.4

Question 1

Justify why the plane cannot be tiled with a regular pentagon.

Solution:
The interior angle of a regular pentagon is 108°. For a shape to tile the plane, the interior angles meeting at a vertex must sum to 360°. Since 360 is not an integer multiple of 108 (360/108 = 3.33), regular pentagons leave gaps or overlap.

Question 2

Tile the plane with a non-convex DART.

Solution:
Method: Pair two congruent DARTs to form a parallelogram. Since any parallelogram can tile the plane via translation, the DART tiles the plane.

END-OF-CHAPTER EXERCISES

Question 3

Self-intersecting ABCD where AB and CD intersect at E. Show sum of angles < 360°.

Solution:
In triangle AED, angle A + angle D = 180° – angle AED. In triangle BEC, angle B + angle C = 180° – angle BEC.
Since angle AED = angle BEC (vertically opposite), the total sum is 360° – 2(angle AED). Since angle AED > 0, the sum is less than 360°.

Question 9

Trapezium ABCD (AD=3, BC=5). Find ratio of areas of AEFD and EBCF (E, F are midpoints).

Solution:
EF = (3+5)/2 = 4. Let height of each part be h.
Area(AEFD) = 1/2 (3+4)h = 3.5h.
Area(EBCF) = 1/2 (4+5)h = 4.5h.
Ratio = 3.5 : 4.5 = 7 : 9.

Question 20

Counting diagonals of an n-gon.

Polygon	Vertices (n)	Diagonals
Triangle	3	0
Quadrilateral	4	2
Pentagon	5	5
Hexagon	6	9
Octagon	8	20

Formula: (n * (n – 3)) / 2.

Question 24

What fraction of the square is shaded?

Solution:
In the figure where vertices are connected to midpoints of opposite sides, the central square is formed by the intersection of four lines. Using similar triangles or area dissection, the central square’s area is exactly 1/5 (20%) of the total area.