Module 03 of 06 · Full Solution Edition
90°.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:
Adjacent angles inside a right angle must sum to 90°.
x + 44 = 90 ⇒ x = 46°.
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:
A straight angle = 180°. Two adjacent angles on the line sum to 180°.
x + 130 = 180 ⇒ x = 50°.
"Straight angle" means a line, not a right angle. Use 180°, not 90°.
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.
Three concurrent lines create 6 angles in 3 vertical-angle pairs around the point. Around a point, all angles sum to 360°.
Use the two straight lines in the diagram, then apply the given equality.
∠2 + ∠3 = 180°, so ∠3 = 180° − 140° = 40°.∠5 + ∠4 + ∠3 = 180°, so ∠5 + ∠4 = 140°.∠1 + ∠5 + ∠4 = 180°. Since ∠5 + ∠4 = 140°, ∠1 = 40°.∠1 + ∠5 = ∠3 + ∠4. Since ∠1 = ∠3 = 40°, we get ∠5 = ∠4.∠5 = ∠4 = 140° ÷ 2 = 70°.B — 70°
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?
The 4 interior angles of any quadrilateral sum to 360°.
Sum all 5 numbers. The "extra" one = total − 360.
62 + 85 + 99 + 108 + 114 = 468.468 − 360 = 108.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?
Use angle sums in the small right-side triangle, then use the isosceles-triangle base-angle theorem.
First find the angle at R in the large triangle. Since PQ = PR, the two base angles at Q and R are equal.
180° − 95° = 85°.R is 180° − 60° − 85° = 35°.∠PRQ of the large triangle. Since PQ = PR, ∠PQR = ∠PRQ = 35°.x = ∠QPR = 180° − 35° − 35° = 110°.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.
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?
Exterior Angle Theorem: an exterior angle of a triangle equals the sum of the two non-adjacent interior angles.
75° = ∠BAC + ∠ABC = 35° + ∠ABC.∠ABC = 75° − 35° = 40°.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.
B — 40°
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:
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:
Let width = w, then length = 2w. Apply P = 2(l + w).
P = 2(2w + w) = 6w.6w = 120 ⇒ w = 20.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:
Union of two overlapping rectangles: Area(A ∪ B) = Area(A) + Area(B) − Area(A ∩ B).
11 × 8 = 88 cm². Combined = 176 cm².176 − 64 = 112 cm².Don't just add the two areas. The overlap region was counted twice and needs to be subtracted once.
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:
Find the pattern: how does the perimeter grow as n increases? Set up a linear formula, then solve for n.
60 cm: removing that corner removes 10 + 5 cm of outer edge and adds the same 10 + 5 cm as a step.2 × 5 = 10 cm each time.P(n) = 60 + 10(n − 1) = 10n + 50.10n + 50 = 710, so 10n = 660 and n = 66.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.
C — n = 66
The perimeter of the L-shaped figure shown is — with side labels 3, 2, 5, 3, 2.
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.
2 × (5 + 5) = 20.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.
D — 20
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:
3 × PQ + 16 = 40 ⇒ 3 × PQ = 24 ⇒ PQ = 8.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?
Place the right angle at the origin. Use coordinates to compute the area directly.
B = (0, 0), A = (0, 24), C = (24, 0).(0, 12).(12, 0).½ × 12 × 12 = 72 cm².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:
All 60° angles + parallel sides hint at equilateral triangle sub-structures.
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.
a.QR = PQ + TS − a − a + ... . Carefully tracking: QR = 16, PT = SR = 5.PQ + QR + RS + ST + TP = 10 + 16 + 5 + 6 + 5 = 42 cm.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.
A — 42 cm
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:
Use sum = mean × count with areas.
30π × 3 = 90π.π·1² + π·5² + π·x² = 90π.1 + 25 + x² = 90 ⇒ x² = 64 ⇒ x = 8.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:
Area scales as the square of the radius. Triple the radius ⇒ multiply area by 9.
Original / New = (π × 2²) / (π × 6²) = 4 / 36 = 1/9.
If radius is halved: area becomes (1/2)² = 1/4 of original. If radius is k times: area is k².
C — 1/9
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?
A vertical line of symmetry means folding the shape along a vertical axis produces a perfect match between the two halves.
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.
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?
"Viewed from both sides" = horizontal mirror image. A letter looks the same iff it has a vertical line of symmetry.
Letters with vertical symmetry: A, H, I, M, O, T, U, V, W, X, Y.
Check each letter in "PUG FOR SALE":
Total: U, O, A → 3 letters.
E has horizontal symmetry but not vertical — so it looks different when mirrored left-right. Don't confuse the two types.
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?
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.
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?
A translation slides the shape without rotating or flipping. The result has the same orientation as the original.
Check which triangle has the same shape and the same direction the right-angle points.
Only triangle D has identical orientation to the shaded triangle. The others are either rotated or reflected.
"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.
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?
Two reflections across intersecting lines compose to a rotation about the intersection point, by twice the angle between the lines.
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.
Two reflections across parallel lines = a translation by twice the distance between the lines.
E
In the diagram, a right angle and a reflex angle of measure k° share a vertex. The value of k is:
All angles around a single point sum to 360°. A reflex angle is one greater than 180°.
k = 360° − 90° = 270°.
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:
Angles on one side of a straight line sum to 180°: 90 + 20 + x = 180 ⇒ x = 70.
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:
An isosceles triangle has two equal angles. The given 50° angle could either be one of the equal pair or the unique angle.
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°). ✓
Option E (60°, 70°) sums to 180° but makes all three angles different, so the triangle is NOT isosceles.
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:
The diagonal of a square bisects the 90° corner angle, creating two 45° angles.
Since the diagonal of a square makes a 45° angle with each side: x = 45°.
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:
Compute x using isosceles-triangle base angles. y is the corner of a rectangle.
∠PQR = ∠PRQ = (180° − 70°) / 2 = 55°. So x = 55.x + y = 55 + 90 = 145.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?
Use vertical angles at E, then the angle sum in the right triangle.
∠BEC = 60°.∠AED = 60°.∠EAD = 90°, ∠AED = 60°. Angle sum: x = 180° − 90° − 60° = 30°.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?
The longest chord of a circle is its diameter, which passes through the centre. Diameter = 2 × radius.
Diameter = 2 × 4 = 8 cm.
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?
14 + 12 + 12 = 38 cm.2(8 + x) = 16 + 2x.16 + 2x = 38 ⇒ 2x = 22 ⇒ x = 11.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:
The shaded regions are the parts of the square that lie outside the overlapping rectangle. Compute square area minus the overlap.
8 × 8 = 64 cm².8 × 2 = 16 cm².64 − 16 = 48 cm².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?
A single reflection reverses orientation (chirality). For each candidate, check whether you can find a line such that T reflects exactly onto it.
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.
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.
E — triangle E