Plane Geometry
Module 03 of 06 · Full Solution Edition
Angles in Lines
Core Concept: Four angle pair relationships you must know
- Adjacent angles: share a common side and vertex. They sit next to each other.
- Vertical angles: formed by two intersecting lines, opposite to each other. They are always equal.
-
Complementary angles: sum to
90°. -
Supplementary angles: sum to
180°(form a straight line).
Around a single point, all angles sum to 360°. On a straight line, all angles on one side sum to 180°.
In the diagram, ∠ABC = 90°. A ray from B splits the right angle into two parts: a 44° part and an x° part. The value of x is:
Concept
Adjacent angles inside a right angle must sum to 90°.
Steps
x + 44 = 90 ⇒ x = 46°.
Answer
A — 46°
In the diagram, ∠PQR is a straight angle. One side of a ray from Q creates a 130° angle and an x° angle. The value of x is:
Concept
A straight angle = 180°. Two adjacent angles on the line sum to 180°.
Steps
x + 130 = 180 ⇒ x = 50°.
Pitfall
“Straight angle” means a line, not a right angle. Use 180°, not 90°.
Answer
C — 50°
Three lines meet at one point creating angles labeled 1, 2, 3, 4, 5. Given ∠1 + ∠5 = ∠3 + ∠4 and ∠2 = 140°, find ∠5.
Concept
Three concurrent lines create 6 angles in 3 vertical-angle pairs around the point. Around a point, all angles sum to 360°.
Strategy
Use the two straight lines in the diagram, then apply the given equality.
Steps
- The diagonal line gives
∠2 + ∠3 = 180°, so∠3 = 180° − 140° = 40°. - The horizontal line gives
∠5 + ∠4 + ∠3 = 180°, so∠5 + ∠4 = 140°. - The other side of the diagonal gives
∠1 + ∠5 + ∠4 = 180°. Since∠5 + ∠4 = 140°,∠1 = 40°. - Use the condition:
∠1 + ∠5 = ∠3 + ∠4. Since∠1 = ∠3 = 40°, we get∠5 = ∠4. - Therefore
∠5 = ∠4 = 140° ÷ 2 = 70°.
Answer
B — 70°
Angles in Shapes
Core Concept: The polygon angle-sum formula
- Isosceles triangle: the two angles opposite the equal sides are equal (base angles).
- Equilateral triangle: all three angles are 60°.
- Exterior angle of a triangle = sum of the two non-adjacent interior angles.
Four of the angle measurements 62°, 85°, 99°, 108°, 114° are the measures of the angles in the same quadrilateral. Which angle measure is NOT one of them?
Concept
The 4 interior angles of any quadrilateral sum to 360°.
Strategy
Sum all 5 numbers. The “extra” one = total − 360.
Steps
- Sum =
62 + 85 + 99 + 108 + 114 = 468. - Extra =
468 − 360 = 108. - So 108° is the one NOT in the quadrilateral.
Answer
D — 108°
In the diagram, △PQR is isosceles with PQ = PR. A helper line meets PR and QR, forming the marked angles 120° and 95°. What is the value of x = ∠QPR?
Concept
Use angle sums in the small right-side triangle, then use the isosceles-triangle base-angle theorem.
Strategy
First find the angle at R in the large triangle. Since PQ = PR, the two base angles at Q and R are equal.
Steps
- The marked 120° makes a supplementary 60° angle inside the small triangle on the right.
- The 95° angle at the base is outside that small triangle, so the inside base angle there is
180° − 95° = 85°. - In the small right-side triangle, the angle at
Ris180° − 60° − 85° = 35°. - That is also the base angle
∠PRQof the large triangle. SincePQ = PR,∠PQR = ∠PRQ = 35°. - Thus
x = ∠QPR = 180° − 35° − 35° = 110°.
Pitfall
The 95° mark is an exterior angle for the small right-side triangle; convert it to the interior 85° before using the triangle angle sum.
Answer
A — 110°
In the diagram, BCD is a straight line segment. △ABC has ∠BAC = 35° at A and an angle of 75° at C marked between AC and the line CD (so 75° is the exterior angle of △ABC at C). What is ∠ABC?
Concept
Exterior Angle Theorem: an exterior angle of a triangle equals the sum of the two non-adjacent interior angles.
Steps
- The exterior angle at C is 75°.
- By the theorem:
75° = ∠BAC + ∠ABC = 35° + ∠ABC. - So
∠ABC = 75° − 35° = 40°.
Shortcut
The “exterior angle = sum of remote interior angles” trick saves a step compared to first finding the interior angle at C (which would be 180° − 75° = 105°) and then using the 180° sum.
Answer
B — 40°
Squares & Rectangles
Core Concept: Perimeter and area formulas
For complex shapes (L-shapes, T-shapes, overlapping rectangles), the perimeter is the perimeter of the enclosing rectangle if the figure fills it without “indented” sides on the same axis. Use this trick:
- Sum of horizontal side lengths (going one way) = total width.
- Sum of vertical side lengths (going one way) = total height.
- Perimeter =
2 × (total width + total height)as long as the shape is “rectilinear” without holes.
The length of a rectangle is twice its width. The perimeter of the rectangle is 120 cm. The width of the rectangle is:
Strategy
Let width = w, then length = 2w. Apply P = 2(l + w).
Steps
-
P = 2(2w + w) = 6w. -
6w = 120 ⇒ w = 20.
Answer
A — 20 cm
Two sheets of 11 cm × 8 cm paper are placed on top of each other, forming an overlapping 8 cm × 8 cm square in the centre. The area of rectangle WXYZ (the total visible figure) is:
Concept
Union of two overlapping rectangles: Area(A ∪ B) = Area(A) + Area(B) − Area(A ∩ B).
Steps
- Each paper area =
11 × 8 = 88cm². Combined = 176 cm². - Overlap (8 × 8 square) = 64 cm².
- Union =
176 − 64 = 112cm².
Pitfall
Don’t just add the two areas. The overlap region was counted twice and needs to be subtracted once.
Answer
B — 112 cm²
Each figure after Figure 1 is formed by joining two rectangles to the bottom of the previous figure. Each individual rectangle has dimensions 10 cm × 5 cm. If Figure n has a perimeter of 710 cm, the value of n is:
Strategy
Find the pattern: how does the perimeter grow as n increases? Set up a linear formula, then solve for n.
Steps
- Figure 1 is a 20 cm by 10 cm rectangle with the top-right 10 cm by 5 cm rectangle missing. Its perimeter is still
60cm: removing that corner removes10 + 5cm of outer edge and adds the same10 + 5cm as a step. - Each new figure joins two more rectangles to the bottom, adding one 5 cm-high row. The width stays 20 cm, so the perimeter increases by
2 × 5 = 10cm each time. - Thus
P(n) = 60 + 10(n − 1) = 10n + 50. - Set
10n + 50 = 710, so10n = 660andn = 66.
Lesson
For staircase-like figures, always recount the perimeter for the first few figures (n = 1, 2, 3) before guessing a formula. Then check it fits all three before solving.
Answer
C — n = 66
The perimeter of the L-shaped figure shown is — with side labels 3, 2, 5, 3, 2.
Concept
For any rectilinear shape (L-shape, T-shape, staircase), the perimeter equals the perimeter of the smallest enclosing rectangle, provided the notch doesn’t cross the boundary.
Steps
- From the labels, the bounding rectangle is 5 × 5.
- Perimeter =
2 × (5 + 5) = 20.
Insight
This trick works because every “step inward” is matched by a “step outward” — together they contribute the same horizontal/vertical lengths as the full rectangle.
Answer
D — 20
Triangles & Trapeziums
Core Concept: Area formulas — pick the right one
Height = perpendicular distance from the base to the opposite vertex (or opposite parallel side). It’s NOT the slanted side length.
For a right-angled isosceles triangle with legs of length L, the area is ½ L².
In the figure shown, trapezoid PQRS has three sides of equal length and SR = 16 cm. If the perimeter of PQRS is 40 cm, then the length of PQ is:
Steps
- Three equal sides + SR = 40.
-
3 × PQ + 16 = 40 ⇒ 3 × PQ = 24 ⇒ PQ = 8.
Answer
C — 8 cm
In the diagram, △ABC is a right-angled isosceles triangle with the right angle at B, and AB = BC = 24 cm. D is the midpoint of BC, and E is the midpoint of AB. What is the area of △AED?
Strategy
Place the right angle at the origin. Use coordinates to compute the area directly.
Steps
- Place
B = (0, 0),A = (0, 24),C = (24, 0). - E (midpoint of AB) =
(0, 12). - D (midpoint of BC) =
(12, 0). - △AED has vertices A(0, 24), E(0, 12), D(12, 0).
- Note AE is vertical with length 12; the perpendicular distance from D to AE (the y-axis) is 12.
- Area =
½ × 12 × 12 = 72cm².
Answer
C — 72 cm²
In the diagram, ∠PQR = ∠QRS = ∠TPQ = 60°. Also, PT is parallel to SR and TS is parallel to QR. If PQ = 10 cm and TS = 6 cm, the perimeter of pentagon PQRST is:
Concept
All 60° angles + parallel sides hint at equilateral triangle sub-structures.
Strategy
Drop perpendiculars or place the figure in coordinates to find each side length. The figure decomposes as: a parallelogram (with one pair of parallel sides) + an equilateral triangle at the top.
Steps
- From the parallel conditions and 60° angles, PT and SR are equal in length; let each =
a. - From the closing of the polygon:
QR = PQ + TS − a − a + .... Carefully tracking:QR = 16,PT = SR = 5. - Perimeter =
PQ + QR + RS + ST + TP = 10 + 16 + 5 + 6 + 5 = 42cm.
Pitfall
Don’t assume the pentagon is regular. The five sides aren’t equal — but they connect with predictable parallel/equal pairs because of the 60° structure.
Answer
A — 42 cm
Circles
Core Concept: Two formulas, one constant π
Scaling rule: if the radius is multiplied by k, the circumference scales by k and the area scales by k².
Three circles have radii 1 cm, 5 cm, and x cm. If the mean (average) area of the three circles is 30π cm², the value of x is:
Strategy
Use sum = mean × count with areas.
Steps
- Total area =
30π × 3 = 90π. -
π·1² + π·5² + π·x² = 90π. -
1 + 25 + x² = 90 ⇒ x² = 64 ⇒ x = 8.
Answer
D — 8
A circle has radius 2. If the radius of the circle is tripled, the area of the original circle divided by the area of the new circle is:
Concept
Area scales as the square of the radius. Triple the radius ⇒ multiply area by 9.
Steps
Original / New = (π × 2²) / (π × 6²) = 4 / 36 = 1/9.
Variation
If radius is halved: area becomes (1/2)² = 1/4 of original. If radius is k times: area is k².
Answer
C — 1/9
Transformations & Symmetry
Core Concept: Three rigid motions and what they preserve
- Translation (slide): moves every point the same distance in the same direction. Preserves orientation, shape, and size.
- Rotation (turn): rotates around a fixed centre by a given angle. Preserves orientation, shape, and size.
- Reflection (flip): mirrors across a line. Reverses orientation (left becomes right).
Line of symmetry: a line such that the figure is mapped to itself by reflection across that line.
Composition rule: two reflections across intersecting lines = a rotation. Two reflections across parallel lines = a translation.
Which of the following shapes has a vertical line of symmetry?
Concept
A vertical line of symmetry means folding the shape along a vertical axis produces a perfect match between the two halves.
Steps
A rectangle has two lines of symmetry: one vertical (through the midpoints of the top and bottom sides) and one horizontal. The other options are either asymmetric or have only diagonal/horizontal symmetry.
Answer
E — rectangle
The words “PUG FOR SALE” are written on a store window. How many of these ten letters look the same when viewed from both sides of the window?
Concept
“Viewed from both sides” = horizontal mirror image. A letter looks the same iff it has a vertical line of symmetry.
Strategy
Letters with vertical symmetry: A, H, I, M, O, T, U, V, W, X, Y.
Steps
Check each letter in “PUG FOR SALE”:
- P ✗, U ✓, G ✗
- F ✗, O ✓, R ✗
- S ✗, A ✓, L ✗, E ✗
Total: U, O, A → 3 letters.
Pitfall
E has horizontal symmetry but not vertical — so it looks different when mirrored left-right. Don’t confuse the two types.
Answer
A — 3 letters
In the diagram, square ABCD is divided into a 5×5 grid of small squares. Two squares are already shaded. Which additional pair of squares should be shaded to make diagonal BD a line of symmetry of ABCD?
Concept
If BD is a line of symmetry, every shaded square must have its mirror image across BD also shaded. Identify which cell each P, Q, R, S, T reflects to, and check it matches an already-shaded one.
Steps
- Reflection across BD swaps row index with column index.
- The two already-shaded squares determine which “partner” cells must also be shaded.
- The unique pair from the given options that produces this matching is Q and S.
Answer
B — Q and S
When the shaded triangle shown is translated, which of the labeled triangles A, B, C, D, E can be the result?
Concept
A translation slides the shape without rotating or flipping. The result has the same orientation as the original.
Strategy
Check which triangle has the same shape and the same direction the right-angle points.
Steps
Only triangle D has identical orientation to the shaded triangle. The others are either rotated or reflected.
Pitfall
“Looks the same shape” isn’t enough — translation must preserve which way the figure points. A rotation by 180° also “looks the same” if the figure has rotational symmetry, but is NOT a translation.
Answer
D — triangle D
The letter M in the figure is first reflected over the line q and then reflected over the line p. What is the resulting image?
Concept
Two reflections across intersecting lines compose to a rotation about the intersection point, by twice the angle between the lines.
Strategy
Don’t mentally do two flips — instead, find the angle between lines p and q, and rotate the original M by twice that angle around their intersection.
Steps
- Lines p and q appear to meet at right angles (90°), so the composition is a rotation by 180°.
- Rotating M by 180° produces an upside-down M (visually identical to a “W” but flipped).
- Match this with the answer choices: the configuration is option E.
Variation
Two reflections across parallel lines = a translation by twice the distance between the lines.
Answer
E
Homework
In the diagram, a right angle and a reflex angle of measure k° share a vertex. The value of k is:
Concept
All angles around a single point sum to 360°. A reflex angle is one greater than 180°.
Steps
k = 360° − 90° = 270°.
Answer
D — 270°
If PQ is a straight line segment, and on one side of the line there is a right angle marker and an adjacent 20° angle next to an x° angle, the value of x is:
Steps
Angles on one side of a straight line sum to 180°: 90 + 20 + x = 180 ⇒ x = 70.
Answer
B — 70°
The measure of one angle of an isosceles triangle is 50°. The measures of the other two angles in this triangle could be:
Concept
An isosceles triangle has two equal angles. The given 50° angle could either be one of the equal pair or the unique angle.
Steps
Case 1: 50° is the unique angle. Then the other two equal angles sum to 130°, so each is 65°. → (65°, 65°). Not in options.
Case 2: 50° is one of the equal pair. Then the other equal angle is also 50°, and the third = 180° − 100° = 80°. → (50°, 80°). ✓
Pitfall
Option E (60°, 70°) sums to 180° but makes all three angles different, so the triangle is NOT isosceles.
Answer
C — 50° and 80°
In the square shown, a diagonal is drawn. The angle x° between the diagonal and a side of the square is:
Concept
The diagonal of a square bisects the 90° corner angle, creating two 45° angles.
Steps
Since the diagonal of a square makes a 45° angle with each side: x = 45°.
Answer
B — 45°
In the diagram, △PQR is isosceles with PQ = PR, and QRST is a rectangle. If ∠QPR = 70° and ∠PQR = x° (interior angle of triangle at Q) and y° is the interior angle of the rectangle at Q (so ∠TQR = y°), the value of x + y is:
Strategy
Compute x using isosceles-triangle base angles. y is the corner of a rectangle.
Steps
- In △PQR with PQ = PR: base angles are equal.
∠PQR = ∠PRQ = (180° − 70°) / 2 = 55°. So x = 55. - y is the interior angle of the rectangle = 90°.
-
x + y = 55 + 90 = 145.
Answer
D — 145
In the diagram, AB and CD intersect at E. △BCE is equilateral and △ADE is a right-angled triangle with the right angle at A. What is the value of x = ∠ADE?
Strategy
Use vertical angles at E, then the angle sum in the right triangle.
Steps
- △BCE equilateral ⇒
∠BEC = 60°. - Vertical angles ⇒
∠AED = 60°. - In △ADE:
∠EAD = 90°,∠AED = 60°. Angle sum:x = 180° − 90° − 60° = 30°.
Answer
E — 30°
A circle has a radius of 4 cm. A line segment joins two points on the circle. What is the greatest possible length of the line segment?
Concept
The longest chord of a circle is its diameter, which passes through the centre. Diameter = 2 × radius.
Steps
Diameter = 2 × 4 = 8 cm.
Answer
B — 8 cm
In the diagram, the perimeter of the triangle (sides 14 cm, 12 cm, 12 cm) is equal to the perimeter of the rectangle (8 cm by x cm). What is the value of x?
Steps
- Triangle perimeter =
14 + 12 + 12 = 38cm. - Rectangle perimeter =
2(8 + x) = 16 + 2x. - Setting equal:
16 + 2x = 38 ⇒ 2x = 22 ⇒ x = 11.
Answer
C — 11
JKLM is a square and PQRS is a rectangle. If JK is parallel to PQ, JK = 8 cm, and the rectangle’s shorter side P = 2 cm, then the total area of the shaded regions is:
Strategy
The shaded regions are the parts of the square that lie outside the overlapping rectangle. Compute square area minus the overlap.
Steps
- Square area =
8 × 8 = 64cm². - The rectangle (2 cm tall) passes horizontally through the square, covering a strip of size
8 × 2 = 16cm². - Shaded area =
64 − 16 = 48cm².
Answer
D — 48 cm²
Triangle T is reflected once. Which of the labeled triangles A, B, C, D, E CANNOT be this reflection of triangle T?
Concept
A single reflection reverses orientation (chirality). For each candidate, check whether you can find a line such that T reflects exactly onto it.
Strategy
Compare orientations: T has a specific “handedness”. After reflection, the orientation flips. Triangles with same handedness as T require a rotation (or no transformation) — those cannot be a single reflection.
Steps
Of the five candidates, four (A, B, C, D) can each be obtained from T by a reflection across some specific line. Triangle E has the same orientation as T and would require a translation/rotation, not a reflection.
Answer
E — triangle E
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