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LET Elementary Mathematics — Plane and Solid GeometryDetailed Explanation

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Detailed explanations for LET Elementary Mathematics — Plane and Solid Geometry. This page treats you like a serious reviewer: we unpack the concepts thoroughly, show worked examples of how Professional Regulation Commission (PRC) frames Plane and Solid Geometry questions, and explain the underlying reasoning that gets you to the right answer every time.

Exam context

The Licensure Examination for Professional Teachers — Elementary is conducted by Professional Regulation Commission (PRC) and is scheduled for Bi-annual. The Mathematics subtest is marked as "Core" in the official pattern, and Plane and Solid Geometry appears in position 4th of 7 in the LET Elementary Mathematics review rotation. Passing mark: Weighted average of 75% with no grade below 50%. Recent LET Elementary 2026 papers have drawn roughly a meaningful share of questions from this subject.

Plane and Solid Geometry - Detailed Explanation

Geometry is one of the most consistently tested areas in the Mathematics component of the Licensure Examination for Teachers (LET) – Elementary Level. As a future elementary school teacher, you will teach geometric concepts to Grades 1–6 pupils under the K–12 Basic Education Curriculum (BEC), so mastering these ideas serves a dual purpose: passing the LET and teaching effectively. This chapter covers plane geometry (flat, two-dimensional figures) and solid geometry (three-dimensional shapes), including angles, triangles, polygons, congruence, similarity, circles, perimeter, area, surface area, volume, and the Pythagorean theorem. The LET tests these topics through formula-application problems set in practical contexts — fencing a school garden, painting a classroom wall, or filling a water drum. The winning strategy is a clean, memorized set of formulas combined with careful unit handling (area in square units, volume in cubic units) and a disciplined, step-by-step problem-solving approach. Work through every worked example in this chapter, and you will be ready for any geometry item the PRC examiners set.

Concepts

Angles and Angle Pairs

An angle is formed when two rays share a common endpoint called the vertex. Angles are measured in degrees (°). Understanding angle classifications and relationships is fundamental because angle-pair rules appear in nearly every geometry item on the LET. CLASSIFICATIONS BY SIZE: • Acute angle: measures between 0° and 90° (less than a right angle) • Right angle: measures exactly 90° (marked with a small square) • Obtuse angle: measures between 90° and 180° • Straight angle: measures exactly 180° (a straight line) • Reflex angle: measures between 180° and 360° KEY ANGLE PAIRS: • Complementary angles: two angles whose measures sum to 90°. Think: 'C' for Corner (right angle). Example: 35° and 55° are complementary because 35 + 55 = 90. • Supplementary angles: two angles whose measures sum to 180°. Think: 'S' for Straight line. Example: 110° and 70° are supplementary because 110 + 70 = 180. • Vertical angles (vertically opposite angles): formed when two straight lines intersect. The angles directly across from each other are EQUAL. Example: if one angle is 65°, the vertical angle is also 65°. • Linear pair: two adjacent angles on a straight line; they are supplementary (sum = 180°). PARALLEL LINES CUT BY A TRANSVERSAL: When a transversal (a line that crosses two or more lines) cuts two PARALLEL lines, special equal and supplementary relationships arise: • Corresponding angles: on the same side, one above and one below the parallel lines — they are EQUAL. • Alternate interior angles: between the parallel lines, on opposite sides of the transversal — they are EQUAL. • Co-interior (same-side interior / consecutive interior) angles: between the parallel lines, on the SAME side of the transversal — they are SUPPLEMENTARY (sum to 180°). These parallel-line relationships are favorite LET items because students often confuse alternate interior angles (equal) with co-interior angles (supplementary).

Examples

This is a classic LET-style algebra-meets-geometry item. Set up two equations — one expressing the angle relationship (complementary = 90°) and one expressing the size relationship (one is 25° more than the other). Substitute and solve.

Scenario

Two angles are complementary. One angle is 25° larger than the other. Find both angles.

Solution

Step 1: Let the smaller angle = x. Then the larger angle = x + 25. Step 2: Since they are complementary: x + (x + 25) = 90 Step 3: Simplify: 2x + 25 = 90 Step 4: Subtract 25 from both sides: 2x = 65 Step 5: Divide by 2: x = 32.5° Step 6: The larger angle = 32.5 + 25 = 57.5° Answer: The two angles are 32.5° and 57.5°. Verification: 32.5 + 57.5 = 90° ✓

Draw the figure mentally: two horizontal parallel lines crossed by a diagonal transversal. The 65° angle and its alternate interior angle are like the letter 'Z' or 'N' — they sit in the Z-shape between the parallel lines. Co-interior angles sit in a 'C' or 'U' shape and must add to 180°.

Scenario

Two parallel lines are cut by a transversal. One of the angles formed measures 65°. Find (a) the alternate interior angle and (b) the co-interior angle on the same side.

Solution

Step 1: Identify the relationship. (a) Alternate interior angles between parallel lines are EQUAL. Alternate interior angle = 65° (b) Co-interior (same-side interior) angles are SUPPLEMENTARY. Co-interior angle = 180° - 65° = 115°

Intersecting lines always create two pairs of vertical (equal) angles. Adjacent angles are supplementary. This pattern allows you to find all four angles from just one measurement.

Scenario

Two lines intersect and form four angles. One angle measures 48°. Find all four angles.

Solution

Step 1: The angle directly opposite (vertical angle) = 48°. Step 2: The adjacent angle forms a linear pair with 48°: 180° - 48° = 132°. Step 3: The fourth angle is vertical to 132°: 132°. Answer: The four angles are 48°, 132°, 48°, 132°.

Applications

Measuring and constructing angles in Grade 4–6 geometry lessons (K–12 BEC Math Curriculum)
Checking if a corner of a classroom or a piece of furniture is a right angle
Understanding how roads intersect and traffic angle signs work
Architecture and carpentry — ensuring walls meet at the correct angles
LET problems about fencing angles and corner measurements of school lots

Misconceptions

WRONG: Co-interior angles are equal. CORRECT: Co-interior angles are supplementary (sum to 180°), not equal. Only alternate interior and corresponding angles are equal.
WRONG: Complementary angles must be adjacent. CORRECT: Two angles are complementary as long as their sum is 90°, regardless of position.
WRONG: Vertical angles are the angles that share a vertex. CORRECT: Vertical angles are the pair of NON-adjacent angles formed by two intersecting lines — they are directly across from each other.
WRONG: All angles formed by a transversal with two lines are equal. CORRECT: The equal and supplementary relationships only apply when the two lines are PARALLEL.

Related Concepts

Triangle angle sum (interior angles of a triangle sum to 180°)
Polygon angle sum formula
Exterior angle theorem for triangles
Coordinate geometry and slopes of perpendicular/parallel lines

Common Exam Questions

Example

Find the supplement of 73°. Answer: 180 - 73 = 107°

Approach

Subtract the given angle from 90° (complement) or 180° (supplement). For 'x is 15° more than its complement,' set up: x + (x - 15) = 90 and solve.

Question Type

Find the complement or supplement of a given angle

Example

If a co-interior angle is 112°, find the other co-interior angle: 180 - 112 = 68°

Approach

Identify which angle pair is involved (corresponding, alternate interior, co-interior). Apply the rule: equal for corresponding and alternate; supplementary for co-interior.

Question Type

Parallel lines and transversal — find a missing angle

Example

Two vertical angles are (3x + 10)° and (5x - 20)°. Then 3x + 10 = 5x - 20, so 30 = 2x, x = 15. Each angle = 3(15)+10 = 55°.

Approach

Set the two vertical angles equal to each other and solve the equation if variables are involved.

Question Type

Vertical angles — find the unknown

Key Points To Remember

Complementary = sum is 90°; Supplementary = sum is 180°
Vertical angles are always EQUAL — this is a theorem, not a coincidence
Alternate interior angles between parallel lines are EQUAL
Co-interior (same-side interior) angles between parallel lines are SUPPLEMENTARY
Corresponding angles between parallel lines are EQUAL
A straight line = 180°; a full rotation = 360°
When solving for an unknown angle, set up an equation using the appropriate relationship

Triangles: Classification, Angle Sum, and the Exterior Angle Theorem

A triangle is a polygon with exactly three sides and three angles. The sum of the interior angles of ANY triangle is always 180°. This is one of the most important and frequently tested geometric facts on the LET. CLASSIFICATION BY SIDES: • Equilateral triangle: all three sides are equal; all three angles are equal (each = 60°). • Isosceles triangle: exactly two sides are equal; the base angles (angles opposite the equal sides) are also equal. • Scalene triangle: all three sides are different; all three angles are different. CLASSIFICATION BY ANGLES: • Acute triangle: all three angles are less than 90°. • Right triangle: exactly one angle is 90° (the side opposite is called the hypotenuse). • Obtuse triangle: exactly one angle is greater than 90°. EXTERIOR ANGLE THEOREM: An exterior angle of a triangle (formed by extending one side of the triangle) equals the SUM of the two NON-adjacent interior angles (called remote interior angles). Formula: Exterior angle = Remote Interior Angle 1 + Remote Interior Angle 2 Example: If the two remote interior angles are 40° and 75°, the exterior angle = 40 + 75 = 115°. TRIANGLE INEQUALITY THEOREM: For three lengths to form a triangle, the sum of ANY two sides must be GREATER THAN the third side. Use this to quickly eliminate impossible triangle options: Test: sides 3, 4, 8 → 3 + 4 = 7, which is NOT greater than 8. These cannot form a triangle. Test: sides 5, 7, 10 → 5+7=12>10 ✓; 5+10=15>7 ✓; 7+10=17>5 ✓. These CAN form a triangle.

Examples

The key insight is recognizing that an isosceles triangle has TWO equal base angles. Once you identify the unique (vertex) angle, subtract it from 180° and divide the remainder equally between the two base angles.

Scenario

An isosceles triangle has a vertex angle of 40°. Find each base angle.

Solution

Step 1: The vertex angle is the unique angle = 40°. Step 2: Let each base angle = x (they are equal in an isosceles triangle). Step 3: Set up equation: 40 + x + x = 180 Step 4: 40 + 2x = 180 Step 5: 2x = 140 Step 6: x = 70° Answer: Each base angle measures 70°. Verification: 40 + 70 + 70 = 180° ✓

This is a right triangle (one angle = 90°). Simply subtract the known angles from 180°. Note: the two non-right angles (53° and 37°) are complementary — they add to 90°. This is always true in a right triangle.

Scenario

In a triangle, one angle measures 90° and another measures 37°. Find the third angle.

Solution

Step 1: Sum of all three angles = 180° Step 2: Third angle = 180° - 90° - 37° Step 3: Third angle = 180° - 127° = 53° Answer: The third angle is 53°.

The exterior angle theorem provides a direct shortcut — no need to find the adjacent interior angle first. Just subtract the known remote angle from the exterior angle.

Scenario

An exterior angle of a triangle measures 120°. One of the remote interior angles is 45°. Find the other remote interior angle.

Solution

Step 1: Exterior angle = Remote angle 1 + Remote angle 2 Step 2: 120 = 45 + Remote angle 2 Step 3: Remote angle 2 = 120 - 45 = 75° Answer: The other remote interior angle is 75°.

Applications

Grade 5–6 lessons on classifying triangles (K–12 BEC Math Competencies)
Architecture: triangular roof trusses use equilateral or isosceles triangles for stability
Navigation and surveying: triangulation uses triangle properties to find distances
LET items about finding missing angles in triangles embedded in larger figures
Creating bulletin board displays shaped as triangular pennants and calculating their areas

Misconceptions

WRONG: An isosceles triangle's equal angles are at the top. CORRECT: The equal angles (base angles) are opposite the equal sides, which may be at any position.
WRONG: The exterior angle is the supplement of the adjacent interior angle only by accident. CORRECT: The exterior angle ALWAYS equals the sum of the two remote interior angles — this is a theorem derived from the 180° angle sum.
WRONG: Equilateral triangles are a separate type from isosceles. CORRECT: An equilateral triangle IS a special isosceles triangle (all three pairs of sides are equal, so all three pairs of base angles are equal).
WRONG: Any three lengths can form a triangle. CORRECT: The triangle inequality must be satisfied — the sum of the two shorter sides must be greater than the longest side.

Related Concepts

Pythagorean theorem (applies specifically to right triangles)
Triangle congruence and similarity
Area of a triangle (½ × base × height)
Polygon angle sum formula (triangles are the building block)

Common Exam Questions

Example

Angles are (x+10)°, (2x)°, and 50°. Then (x+10) + 2x + 50 = 180 → 3x + 60 = 180 → x = 40. Angles: 50°, 80°, 50°.

Approach

Add the two known angles and subtract from 180°. Watch for variables — set up an equation.

Question Type

Find a missing angle in a triangle

Example

A triangle with angles 60°, 60°, 60° is equilateral (and acute).

Approach

For type by sides, compare side lengths. For type by angles, check if any angle is 90° or > 90° or all < 90°.

Question Type

Identify triangle type

Example

Exterior angle = (3x + 5)°; remote angles are 2x° and 35°. So 3x+5 = 2x+35 → x = 30.

Approach

Exterior angle = sum of the two remote interior angles. Set up equation if a variable is involved.

Question Type

Exterior angle theorem application

Key Points To Remember

Sum of interior angles of any triangle = 180°
Equilateral triangle: all sides equal, all angles = 60°
Isosceles triangle: two equal sides means two equal BASE angles
Exterior angle = sum of the two remote (non-adjacent) interior angles
Triangle Inequality: sum of any two sides > third side
In a right triangle, the hypotenuse is ALWAYS the longest side and is opposite the 90° angle
A triangle can have AT MOST one right angle or AT MOST one obtuse angle

The Pythagorean Theorem

The Pythagorean theorem is arguably the single most important formula in elementary geometry and is guaranteed to appear on the LET. It applies ONLY to RIGHT triangles. STATEMENT: In a right triangle, the square of the length of the hypotenuse equals the sum of the squares of the other two sides. FORMULA: a² + b² = c² Where: • a and b are the two LEGS (the sides that form the right angle) • c is the HYPOTENUSE (the side OPPOSITE the right angle; it is always the LONGEST side) COMMON PYTHAGOREAN TRIPLES (memorize these — they appear constantly on the LET): • 3-4-5: 3² + 4² = 9 + 16 = 25 = 5² • 5-12-13: 5² + 12² = 25 + 144 = 169 = 13² • 8-15-17: 8² + 15² = 64 + 225 = 289 = 17² • Multiples: 6-8-10, 9-12-15, 10-24-26, 15-20-25, etc. TESTING IF A TRIANGLE IS A RIGHT TRIANGLE: Given three sides, substitute the LARGEST side as c. If a² + b² = c², it IS a right triangle. SPECIAL RIGHT TRIANGLES: • 45-45-90 triangle: legs are equal; hypotenuse = leg × √2 • 30-60-90 triangle: sides are in ratio 1 : √3 : 2 FINDING A MISSING SIDE: • To find the hypotenuse: c = √(a² + b²) • To find a leg: a = √(c² - b²) Always check: is the answer you found the LONGEST side of the triangle? If not, re-examine which side is which.

Examples

The ladder, wall, and ground form a right triangle. The ladder is the hypotenuse. Recognizing the 6-8-10 pattern (double of 3-4-5) could let you answer instantly without computation.

Scenario

A ladder leans against a classroom wall. Its foot is 6 m from the wall and the top reaches 8 m up the wall. How long is the ladder?

Solution

Step 1: Draw a right triangle. The wall and ground form the right angle. Leg a = 6 m (horizontal distance from wall to ladder foot) Leg b = 8 m (vertical height up the wall) Hypotenuse c = ladder length (unknown) Step 2: Apply the Pythagorean theorem: a² + b² = c² Step 3: 6² + 8² = c² Step 4: 36 + 64 = c² Step 5: 100 = c² Step 6: c = √100 = 10 m Answer: The ladder is 10 m long. Note: This is the 3-4-5 triple scaled by 2: (6-8-10).

Always put the LARGEST side as c when testing for a right triangle. The 9-12-15 pattern is just 3-4-5 multiplied by 3, so recognizing the triple gives an instant answer.

Scenario

A triangle has sides 9 cm, 12 cm, and 15 cm. Is it a right triangle?

Solution

Step 1: The largest side is 15 cm — this would be the hypotenuse if it is a right triangle. Step 2: Check: a² + b² = 9² + 12² = 81 + 144 = 225 Step 3: c² = 15² = 225 Step 4: Since 9² + 12² = 15² (225 = 225), the triangle IS a right triangle. Note: This is the 3-4-5 triple scaled by 3 (9-12-15).

Diagonal of a rectangle problems are extremely common on the LET. The diagonal is always the hypotenuse of the right triangle formed by the length and width. Recognizing 10-24-26 as a Pythagorean triple saves significant computation time.

Scenario

A rectangular school lot measures 24 m by 10 m. What is the length of the diagonal path across the lot?

Solution

Step 1: The diagonal divides the rectangle into two right triangles. Leg a = 24 m, Leg b = 10 m, Hypotenuse c = diagonal (unknown) Step 2: c² = a² + b² = 24² + 10² = 576 + 100 = 676 Step 3: c = √676 = 26 m Answer: The diagonal path is 26 m long. Note: This is the 5-12-13 triple scaled by 2 (10-24-26).

Applications

Finding the diagonal of a rectangular classroom or school lot
Determining if a newly constructed corner is perfectly square (check if 3-4-5 ratio holds)
Navigation: finding the straight-line (as-the-crow-flies) distance when you know horizontal and vertical distances
Grade 6 K–12 BEC lesson: applying the Pythagorean theorem to real-life problems
Distance formula in coordinate geometry — the Pythagorean theorem in disguise

Misconceptions

WRONG: The formula a² + b² = c² can be used for any triangle. CORRECT: It applies ONLY to right triangles. The generalized version for other triangles is the Law of Cosines.
WRONG: Any side can be substituted as c. CORRECT: c MUST be the hypotenuse — the side OPPOSITE the right angle, always the LONGEST side.
WRONG: If c² = a² + b², the triangle is just any special triangle. CORRECT: If and only if c² = a² + b², the triangle is a RIGHT triangle. This is the converse of the theorem.
WRONG: The legs of a right triangle must be equal. CORRECT: Legs are equal only in the special 45-45-90 right triangle. In general, the legs can be any lengths.

Related Concepts

Distance formula in coordinate geometry
Triangle classification (right triangle)
Pythagorean triples
Special right triangles (45-45-90 and 30-60-90)
Diagonal of a rectangle

Common Exam Questions

Example

Legs = 5 m and 12 m. c = √(25+144) = √169 = 13 m (5-12-13 triple)

Approach

Square both legs, add them, take the square root. Check if the numbers form a known triple.

Question Type

Find the hypotenuse given two legs

Example

Hypotenuse = 13, one leg = 5. Missing leg = √(169-25) = √144 = 12

Approach

Rearrange to a = √(c² - b²). Subtract, then take the square root.

Question Type

Find a missing leg given the hypotenuse and one leg

Example

Sides 7, 24, 25: 7²+24²=49+576=625=25². YES, it is a right triangle.

Approach

Put the largest side as c. Compute a² + b² and compare with c². If equal, it is a right triangle.

Question Type

Test whether three given sides form a right triangle

Key Points To Remember

a² + b² = c² applies ONLY to right triangles; c is always the hypotenuse
Memorize the triples: 3-4-5, 5-12-13, 8-15-17 and their multiples
To test for a right triangle: substitute the largest side as c and check if a² + b² = c²
Hypotenuse is ALWAYS the longest side and is OPPOSITE the right angle
When finding a missing LEG: subtract squares (a = √(c² - b²))
The distance formula in coordinate geometry is derived from the Pythagorean theorem
Always check that your computed answer is reasonable (hypotenuse > each leg)

Congruence and Similarity of Triangles

CONGRUENCE means two figures have exactly the SAME shape AND the same SIZE. Every corresponding side and every corresponding angle are equal. The symbol for congruence is ≅. TRIANGLE CONGRUENCE POSTULATES AND THEOREMS: • SSS (Side-Side-Side): All three pairs of corresponding sides are equal. • SAS (Side-Angle-Side): Two pairs of sides and the INCLUDED angle (the angle between those two sides) are equal. • ASA (Angle-Side-Angle): Two pairs of angles and the INCLUDED side (the side between those two angles) are equal. • AAS (Angle-Angle-Side): Two pairs of angles and a NON-included side are equal. • HL (Hypotenuse-Leg): For RIGHT triangles only — the hypotenuse and one leg are equal. Note: SSA (side-side-angle) and AAA (angle-angle-angle) do NOT prove congruence. SIMILARITY means two figures have exactly the SAME SHAPE but may be DIFFERENT SIZES. Corresponding angles are equal; corresponding sides are in proportion. The symbol for similarity is ~. TRIANGLE SIMILARITY POSTULATES: • AA (Angle-Angle): Two pairs of corresponding angles are equal. (The third pair must also be equal since angle sum = 180°, so you only need to verify two.) • SSS (Side-Side-Side Proportionality): All three pairs of corresponding sides are proportional. • SAS (Side-Angle-Side Proportionality): Two pairs of sides are proportional AND the included angle is equal. SCALING IN SIMILAR FIGURES: • If the ratio of corresponding sides (scale factor) is k, then: - Corresponding sides ratio = k : 1 - Corresponding perimeters ratio = k : 1 (same as sides) - Corresponding AREAS ratio = k² : 1 (square of the side ratio) - Corresponding VOLUMES ratio = k³ : 1 (cube of the side ratio) SHADOW PROBLEMS (Classic LET Item): A person and a nearby vertical object (flagpole, tree, post) both cast shadows at the same time. The person, their shadow, and the sun ray form one right triangle; the object, its shadow, and the sun ray form a similar triangle. Their heights and shadow lengths are proportional: person's height / person's shadow = object's height / object's shadow

Examples

Shadow problems are among the most common LET similar-triangle applications. Always write the proportion as (height₁/shadow₁) = (height₂/shadow₂). Make sure both heights are in the SAME units and both shadows are in the SAME units before setting up the ratio.

Scenario

A 1.5 m tall student casts a 2 m shadow at the same time a flagpole casts a 16 m shadow. How tall is the flagpole?

Solution

Step 1: Identify the similar triangles. The student and flagpole are vertical; their shadows are horizontal; the sun's rays create equal angles — AA similarity applies. Step 2: Set up the proportion: Student's height / Student's shadow = Flagpole's height / Flagpole's shadow 1.5 / 2 = h / 16 Step 3: Cross-multiply: 2h = 1.5 × 16 = 24 Step 4: h = 24 / 2 = 12 m Answer: The flagpole is 12 m tall.

This tests whether you know that AREAS scale by the SQUARE of the side ratio, not the side ratio itself. Students who mistakenly multiply 27 × (5/3) = 45 cm² are using the wrong ratio — a classic error.

Scenario

Two similar triangles have corresponding sides in the ratio 3:5. If the area of the smaller triangle is 27 cm², what is the area of the larger triangle?

Solution

Step 1: The side ratio (scale factor) k = 3/5. Step 2: Area ratio = k² = (3/5)² = 9/25. Step 3: Set up proportion: 27 / Area of larger = 9 / 25 Step 4: Cross-multiply: 9 × Area of larger = 27 × 25 = 675 Step 5: Area of larger = 675 / 9 = 75 cm² Answer: The area of the larger triangle is 75 cm².

Applications

Using similar triangles to measure the height of trees, buildings, or flagpoles indirectly
Map scales: a 1:10,000 map uses similarity — every measurement on the map is 10,000 times smaller than reality
Photocopying or enlarging drawings maintains similar proportions
Grade 5–6 K–12 BEC lessons on similar figures and scale drawings
Architectural blueprints are scale drawings using similarity

Misconceptions

WRONG: SAS means any two sides and any angle. CORRECT: SAS requires the angle to be the INCLUDED angle — between the two given sides.
WRONG: Two triangles with equal angles are congruent. CORRECT: Equal angles (AAA) proves only SIMILARITY, not congruence. The triangles could be different sizes.
WRONG: Areas of similar figures have the same ratio as their sides. CORRECT: Areas scale by the SQUARE of the side ratio. If sides are in ratio 2:3, areas are in ratio 4:9.
WRONG: SSA is a valid congruence postulate. CORRECT: SSA (two sides and a non-included angle) does NOT guarantee congruence — it can produce two different triangles (the ambiguous case).

Related Concepts

Angle relationships in triangles
Scale factors and proportional reasoning
Perimeter and area of triangles
Trigonometric ratios (introduced from similar right triangles)

Common Exam Questions

Example

A 2 m post casts a 3 m shadow. A nearby tree casts a 15 m shadow. Tree height = 2/3 × 15 = 10 m.

Approach

Set up proportion: height1/shadow1 = height2/shadow2. Cross-multiply and solve.

Question Type

Shadow/indirect measurement using similar triangles

Example

Two triangles: equal sides of 5 and 7, with the angle between them equal → SAS

Approach

List what is given (sides or angles). Check which postulate matches — especially SAS requires the angle to be BETWEEN the two sides.

Question Type

Identify the congruence postulate used

Example

Triangles with sides 6, 8, 10 and 9, x, 15: ratio = 9/6 = 3/2. x = 8 × 3/2 = 12.

Approach

Identify corresponding sides, set up a proportion, cross-multiply, solve.

Question Type

Find unknown sides in similar triangles

Key Points To Remember

Congruence = same shape AND same size (≅); Similarity = same shape, possibly different size (~)
Congruence shortcuts: SSS, SAS, ASA, AAS, HL — memorize which include/exclude 'angle between sides'
SSA and AAA do NOT prove congruence
Similarity needs only AA (two angle pairs) — most efficient test
Sides of similar triangles are PROPORTIONAL — set up a proportion and cross-multiply
Areas of similar figures scale by the SQUARE of the side ratio
Shadow problems use AA similarity — the sun angle is equal for both objects simultaneously

Polygons and Their Angle Sums

A polygon is a closed plane figure made of straight line segments (sides). 'Poly' means many; 'gon' means angles. NAMING POLYGONS BY NUMBER OF SIDES: • 3 sides: Triangle • 4 sides: Quadrilateral • 5 sides: Pentagon • 6 sides: Hexagon • 7 sides: Heptagon • 8 sides: Octagon • 9 sides: Nonagon • 10 sides: Decagon • 12 sides: Dodecagon A REGULAR polygon has all sides equal AND all angles equal. INTERIOR ANGLE SUM FORMULA: For any n-sided polygon: Sum of interior angles = (n - 2) × 180° Why? Any polygon can be divided into (n - 2) triangles, each contributing 180°. EACH INTERIOR ANGLE OF A REGULAR POLYGON: Each interior angle = (n - 2) × 180° / n EXTERIOR ANGLE SUM: For ANY convex polygon (regardless of number of sides): Sum of exterior angles = 360° This means: Each exterior angle of a regular n-gon = 360° / n QUADRILATERALS (4-sided polygons) — SPECIAL PROPERTIES: • Square: 4 equal sides, 4 right angles. A square is a rectangle AND a rhombus. • Rectangle: opposite sides equal and parallel, 4 right angles. • Rhombus: 4 equal sides, opposite angles equal, opposite sides parallel. • Parallelogram: opposite sides equal and parallel, opposite angles equal. • Trapezoid (Trapezium in British English): exactly ONE pair of parallel sides. • Kite: two pairs of consecutive equal sides. HIERARCHY (Important for LET 'which statement is always true' items): Every square is a rectangle; every square is a rhombus; every rectangle and rhombus is a parallelogram; every parallelogram is a quadrilateral. BUT: not every rectangle is a square (it may not have equal sides); not every rhombus is a square (it may not have right angles).

Examples

The hexagon is important because regular hexagons tile a plane perfectly (like honeycomb cells) — a real-world fact teachers often discuss. The formula (n-2)×180 is derived by drawing all diagonals from one vertex, creating (n-2) triangles.

Scenario

Find the sum of the interior angles of a hexagon, and then find each interior angle of a REGULAR hexagon.

Solution

Step 1: n = 6 for a hexagon. Step 2: Sum = (n - 2) × 180° = (6 - 2) × 180° = 4 × 180° = 720° Step 3: For a regular hexagon, all angles are equal. Each interior angle = 720° ÷ 6 = 120° Answer: Sum = 720°; each angle = 120°.

This is a classic LET item type — given the angle, find the number of sides. The key step is multiplying both sides by n to clear the fraction, then solving the linear equation.

Scenario

A regular polygon has each interior angle measuring 108°. How many sides does it have?

Solution

Step 1: Use the formula: each interior angle = (n - 2) × 180° / n Step 2: Set up equation: 108 = (n - 2) × 180 / n Step 3: Multiply both sides by n: 108n = (n - 2) × 180 Step 4: Expand: 108n = 180n - 360 Step 5: Rearrange: 360 = 180n - 108n = 72n Step 6: n = 360 / 72 = 5 Answer: The polygon has 5 sides (it is a regular pentagon).

Same logic as finding a missing triangle angle, just with a different total (360° instead of 180°). The sum formula is the key.

Scenario

A quadrilateral has three angles measuring 85°, 110°, and 95°. Find the fourth angle.

Solution

Step 1: Sum of interior angles of a quadrilateral = (4 - 2) × 180° = 360° Step 2: Fourth angle = 360° - 85° - 110° - 95° Step 3: = 360° - 290° = 70° Answer: The fourth angle is 70°.

Applications

Grade 4–6 K–12 BEC: classifying polygons, finding unknown angles in quadrilaterals
Floor tiling: knowing which regular polygons tessellate (triangle, square, hexagon) for classroom decoration
School building design: octagonal pavilions, hexagonal garden plots
LET items on quadrilateral properties: 'Which is always true about a parallelogram?'
Philippine peso bills and coins feature geometric polygon shapes in their designs

Misconceptions

WRONG: All polygons have interior angle sums of 180°. CORRECT: Only triangles (n=3) have interior angle sum = 180°. The formula (n-2)×180 gives different values for different n.
WRONG: The sum of exterior angles depends on the number of sides. CORRECT: The sum of exterior angles of any convex polygon is ALWAYS 360°, regardless of the number of sides.
WRONG: A rectangle is always a square. CORRECT: A square is always a rectangle, but a rectangle is a square ONLY when all four sides are equal.
WRONG: A rhombus has right angles. CORRECT: A rhombus has four EQUAL SIDES but right angles only if it is a square.

Related Concepts

Perimeter and area of polygons
Triangle angle sum (the fundamental building block)
Tessellations and tiling patterns
Coordinate geometry and plotting polygon vertices

Common Exam Questions

Example

Pentagon with angles 100°, 110°, 120°, 90°. Missing = 540 - 420 = 120°.

Approach

Compute the angle sum using (n-2)×180°, subtract all known angles.

Question Type

Find missing angle in a polygon

Example

Regular polygon with 150° angles: (n-2)×180 = 150n → 180n-360=150n → 30n=360 → n=12.

Approach

Set (n-2)×180/n = given angle. Solve for n algebraically.

Question Type

Find number of sides given interior angle of regular polygon

Example

True or False: Every rhombus is a square. FALSE — a rhombus has equal sides but not necessarily right angles.

Approach

Know the hierarchy: square ⊂ rectangle ⊂ parallelogram; square ⊂ rhombus ⊂ parallelogram. Read each statement and test against definitions.

Question Type

Properties of special quadrilaterals

Key Points To Remember

Sum of interior angles of an n-gon = (n - 2) × 180°
Each interior angle of a REGULAR n-gon = (n - 2) × 180° ÷ n
Sum of EXTERIOR angles of any convex polygon = 360° (always, regardless of n)
Triangle: 180°; Quadrilateral: 360°; Pentagon: 540°; Hexagon: 720°; Octagon: 1080°
A square is a special rectangle AND a special rhombus
A rhombus has 4 equal sides; a rectangle has 4 right angles; a square has BOTH
Trapezoid has exactly ONE pair of parallel sides; parallelogram has TWO pairs

Perimeter and Area of Plane Figures

PERIMETER is the total distance AROUND the boundary of a figure. It is measured in LINEAR units (m, cm, km). Think: how much fencing to enclose a lot. AREA is the amount of flat SURFACE SPACE inside a figure. It is always measured in SQUARE units (m², cm², km²). Think: how much paint to cover a wall, or how many tiles to cover a floor. KEY FORMULAS TABLE: SQUARE (side = s): • Perimeter = 4s • Area = s² RECTANGLE (length = l, width = w): • Perimeter = 2(l + w) • Area = l × w TRIANGLE (base = b, height = h; sides = a, b, c): • Perimeter = a + b + c (sum of all three sides) • Area = ½ × b × h • Note: h must be the PERPENDICULAR height from base to opposite vertex PARALLELOGRAM (base = b, height = h; adjacent sides = a and b): • Perimeter = 2(a + b) • Area = b × h (NOT the slant side — use the PERPENDICULAR height) TRAPEZOID (parallel sides = a and b; height = h): • Perimeter = sum of all four sides • Area = ½ (a + b) × h CIRCLE (radius = r; diameter = d = 2r): • Circumference (perimeter) = 2πr = πd • Area = πr² • Use π ≈ 3.1416 (or 3.14 for approximation) • Use π ≈ 22/7 when r is a multiple of 7 (gives exact results) COMPOSITE FIGURES: For L-shapes, rectangles with semicircular ends, or any irregular figure: 1. Divide the figure into standard shapes. 2. Compute each area separately. 3. ADD areas together (for union) or SUBTRACT (for cut-outs). CRITICAL: Always convert all measurements to the SAME UNIT before computing. 1 m = 100 cm; 1 m² = 10,000 cm².

Examples

Read the question carefully! 'How much fencing' asks for PERIMETER. 'How much soil/mulch/plants' asks for AREA. This distinction is a very common LET question.

Scenario

A rectangular school garden is 12 m long and 8 m wide. Find (a) the perimeter for fencing and (b) the area for planting.

Solution

Step 1: Identify dimensions: l = 12 m, w = 8 m Step 2: (a) Perimeter = 2(l + w) = 2(12 + 8) = 2(20) = 40 m Step 3: (b) Area = l × w = 12 × 8 = 96 m² Answer: Perimeter = 40 m (need 40 m of fencing); Area = 96 m² (planting surface).

The beauty of using 22/7 when r is a multiple of 7: the 7 in the denominator cancels with the r in the formula, leaving clean integer arithmetic. For r = 7: πr² = (22/7)(49) = 22 × 7 = 154. No decimals needed!

Scenario

A circular table has a radius of 7 cm. Find (a) its area and (b) its circumference. Use π = 22/7.

Solution

Step 1: r = 7 cm. Since r = 7 is a multiple of 7, use π = 22/7. Step 2: (a) Area = πr² = (22/7) × 7² = (22/7) × 49 = 22 × 7 = 154 cm² Step 3: (b) Circumference = 2πr = 2 × (22/7) × 7 = 2 × 22 = 44 cm Answer: Area = 154 cm²; Circumference = 44 cm.

The trapezoid formula averages the two parallel sides and multiplies by the height. Think of it as a rectangle whose width is the average of the two parallel sides: (10+6)/2 = 8 m wide, 4 m tall → 8×4 = 32 m².

Scenario

A trapezoidal school lot has parallel sides of 10 m and 6 m, and a height of 4 m between them. Find the area.

Solution

Step 1: Parallel sides: a = 10 m, b = 6 m; height h = 4 m. Step 2: Area = ½ (a + b) × h Step 3: = ½ (10 + 6) × 4 Step 4: = ½ × 16 × 4 Step 5: = ½ × 64 = 32 m² Answer: The area is 32 m².

For composite figures, always decompose into simpler shapes. Note that the diameter of the semicircle = 6 m = the width of the rectangle, so r = 3 m. Adding the two areas gives the total.

Scenario

A composite figure consists of a rectangle (10 m × 6 m) with a semicircle on one of the shorter ends (radius = 3 m). Find the total area. Use π ≈ 3.14.

Solution

Step 1: Area of rectangle = l × w = 10 × 6 = 60 m² Step 2: Area of semicircle = ½ × πr² = ½ × 3.14 × 3² = ½ × 3.14 × 9 = ½ × 28.26 = 14.13 m² Step 3: Total area = 60 + 14.13 = 74.13 m² Answer: The total area is approximately 74.13 m².

Applications

Estimating the amount of paint needed to cover a classroom wall (area in m²)
Computing how many floor tiles are needed for a room (area ÷ tile size)
Determining how much wire or ribbon is needed to border a bulletin board (perimeter)
Calculating the cost of fencing a school garden lot (perimeter × cost per meter)
Grade 4–6 K–12 BEC competencies: measuring and computing perimeter and area of composite figures

Misconceptions

WRONG: Area and perimeter have the same units. CORRECT: Perimeter is in LINEAR units (m, cm); area is in SQUARE units (m², cm²). The units tell you what formula to use.
WRONG: For a parallelogram, Area = base × slant side. CORRECT: Area = base × PERPENDICULAR height. The slant (oblique) side is used for perimeter, not area.
WRONG: Area of a circle = 2πr. CORRECT: Circumference (perimeter) = 2πr; Area = πr². These are two completely different formulas.
WRONG: Doubling all dimensions of a figure doubles its area. CORRECT: Doubling all dimensions QUADRUPLES the area (since area scales by the square of the linear scale factor: 2² = 4).

Related Concepts

Units of measurement and unit conversion
Composite figures
Similar figures and area ratios
Surface area (the 3D extension of area)

Common Exam Questions

Example

Room: 8m × 5m. Area for tiling = 40 m². Each tile is 0.5m × 0.5m = 0.25 m². Tiles needed = 40/0.25 = 160 tiles.

Approach

A = l × w for rectangle; A = s² for square. Read whether the question asks for area or perimeter.

Question Type

Find area for flooring/painting — rectangle or square

Example

Diameter = 14 cm → r = 7 cm. Area = (22/7)(49) = 154 cm². Circumference = 2(22/7)(7) = 44 cm.

Approach

Identify whether r or d is given. If d is given, r = d/2. Check if r is a multiple of 7 to decide which value of π to use.

Question Type

Find circumference or area of a circle

Example

L-shape: large rectangle 8×6 = 48; subtract small rectangle 3×2 = 6. L-shape area = 42 sq units.

Approach

Split into basic shapes. Identify which areas to ADD and which to SUBTRACT. Ensure units are consistent.

Question Type

Composite figure area

Key Points To Remember

Perimeter = LINEAR distance around (meters, cm) — for fencing, framing, borders
Area = SQUARE surface inside (m², cm²) — for flooring, painting, tiling
Triangle area = ½ × base × PERPENDICULAR height (not the slant side)
Parallelogram area = base × PERPENDICULAR height (not the slant side)
Trapezoid area = ½ × (sum of parallel sides) × height
Circle: use 22/7 when radius is a multiple of 7; otherwise use 3.14
For composite figures: divide into basic shapes, compute each, then add or subtract

Surface Area and Volume of Solids

Solid geometry deals with THREE-DIMENSIONAL figures. We measure two things: • SURFACE AREA: the total area of all the outer faces/surfaces of the solid. Measured in SQUARE units (m², cm²). Think: how much material to wrap a box or paint a tank. • VOLUME: the amount of space INSIDE the solid. Measured in CUBIC units (m³, cm³). Think: how much water a tank holds or how much concrete fills a mold. KEY FORMULAS: CUBE (edge = s): • Volume = s³ • Surface Area = 6s² (6 equal square faces) RECTANGULAR PRISM / BOX (length = l, width = w, height = h): • Volume = l × w × h • Surface Area = 2(lw + lh + wh) CYLINDER (radius = r, height = h): • Volume = πr²h • Surface Area = 2πr² + 2πrh = 2πr(r + h) (factored form) [2πr² for the two circular ends; 2πrh for the curved side] CONE (radius = r, vertical height = h, slant height = l): • Volume = (1/3)πr²h • Surface Area = πr² + πrl = πr(r + l) [πr² for the circular base; πrl for the curved lateral surface] • Slant height: l = √(r² + h²) by the Pythagorean theorem SPHERE (radius = r): • Volume = (4/3)πr³ • Surface Area = 4πr² PYRAMID (base area = B, height = h): • Volume = (1/3) × B × h • Surface Area = base area + sum of all triangular lateral face areas CRITICAL RELATIONSHIP (favorite LET test point): Volume of cone = (1/3) × Volume of cylinder with same base and height Volume of pyramid = (1/3) × Volume of prism with same base and height This means: if you fill a cone and a cylinder of the same radius and height, you need exactly 3 full cones to fill the cylinder.

Examples

For the volume, compute r² first (9), multiply by h (90), then by π. For the surface area, carefully compute each term. The first term (2πr²) is for both circular ends; the second term (2πrh) is for the curved lateral surface.

Scenario

A cylindrical water tank has a radius of 3 m and a height of 10 m. Find (a) its volume and (b) its surface area. Use π ≈ 3.14.

Solution

Step 1: r = 3 m, h = 10 m Step 2: (a) Volume = πr²h = 3.14 × 3² × 10 = 3.14 × 9 × 10 = 3.14 × 90 = 282.6 m³ Step 3: (b) Surface Area = 2πr² + 2πrh = 2(3.14)(3²) + 2(3.14)(3)(10) = 2(3.14)(9) + 2(3.14)(30) = 56.52 + 188.4 = 244.92 m² Answer: Volume = 282.6 m³; Surface Area ≈ 244.92 m².

The three terms lw, lh, and wh represent the three different pairs of faces: bottom/top, front/back, and left/right. Multiplying by 2 accounts for the fact that each face has an opposite equal face.

Scenario

A rectangular storage box measures 8 cm long, 5 cm wide, and 3 cm high. Find (a) its volume and (b) its surface area.

Solution

Step 1: l = 8, w = 5, h = 3 Step 2: (a) Volume = l × w × h = 8 × 5 × 3 = 120 cm³ Step 3: (b) Surface Area = 2(lw + lh + wh) lw = 8 × 5 = 40 lh = 8 × 3 = 24 wh = 5 × 3 = 15 SA = 2(40 + 24 + 15) = 2(79) = 158 cm² Answer: Volume = 120 cm³; Surface Area = 158 cm².

Compute r³ first: 3³ = 27. Then multiply by π: 3.14 × 27 = 84.78. Then multiply by 4/3: either compute (4 × 84.78)/3 = 339.12/3 = 113.04, or multiply by 4 first then divide by 3.

Scenario

Find the volume of a ball (sphere) with radius 3 cm. Use π ≈ 3.14.

Solution

Step 1: r = 3 cm Step 2: Volume = (4/3)πr³ Step 3: = (4/3)(3.14)(3³) Step 4: = (4/3)(3.14)(27) Step 5: = (4/3)(84.78) Step 6: = 4 × 28.26 Step 7: = 113.04 cm³ Answer: Volume ≈ 113.04 cm³.

Applications

Finding how much water a school water drum can hold (volume of cylinder)
Estimating how much paint is needed to coat a spherical water tank (surface area)
Computing how much concrete to pour for a rectangular foundation (volume of rectangular prism)
Grade 5–6 K–12 BEC: measuring and computing volume of 3D shapes
LET problems: 'How many liters of water can a conical container hold?' (1 m³ = 1,000 liters)

Misconceptions

WRONG: Surface area and volume use the same units. CORRECT: Surface area = square units (cm², m²); Volume = cubic units (cm³, m³). Read the question to determine which is asked.
WRONG: Cone volume = πr²h. CORRECT: Cone volume = (1/3)πr²h. Forgetting the one-third is the most common error in solid geometry items.
WRONG: The height of a cone is the slant height. CORRECT: The VERTICAL height (perpendicular from base to apex) is used in the volume formula. The SLANT height (along the surface) is used in the surface area formula.
WRONG: Doubling the radius of a sphere doubles its volume. CORRECT: Volume = (4/3)πr³, so doubling r multiplies volume by 2³ = 8.

Related Concepts

Area of circles (used in cylinder and cone formulas)
Pythagorean theorem (used to find slant height of cone)
Unit conversion (1 m³ = 1,000 L = 1,000,000 cm³)
Composite solids

Common Exam Questions

Example

Cone with r=6, h=7: V = (1/3)(22/7)(36)(7) = (1/3)(22)(36) = (1/3)(792) = 264 cubic units

Approach

Cylinder: V = πr²h. Cone: V = (1/3)πr²h. A cone holds exactly 1/3 the volume of the same-dimension cylinder.

Question Type

Volume of cylinder or cone

Example

Box 4×3×2: SA = 2(12+8+6) = 2(26) = 52 sq units

Approach

SA = 2(lw + lh + wh). Compute each of the three products first, then add, then double.

Question Type

Surface area of a rectangular box

Example

A cylinder and cone have the same base and height. If the cylinder volume is 180 cm³, the cone volume = 180/3 = 60 cm³.

Approach

Recognize that V(cone) = (1/3)V(cylinder) for same base and height. Use this to find one from the other.

Question Type

Comparison of cone/cylinder volumes

Key Points To Remember

Volume is in CUBIC units; Surface area is in SQUARE units — NEVER mix them up
Cone and pyramid volumes are ONE-THIRD of the corresponding cylinder/prism
Sphere volume = (4/3)πr³; Sphere surface area = 4πr²
Cylinder: two circular ends (2πr²) + curved side (2πrh)
Rectangular prism SA = 2(lw + lh + wh) — remember there are 3 pairs of opposite faces
Slant height of cone ≠ vertical height; use Pythagorean theorem: l = √(r² + h²)
Cube SA = 6s²; Cube Volume = s³

Circles: Parts, Properties, and Calculations

A circle is the set of all points in a plane that are at the same distance (the RADIUS) from a fixed point (the CENTER). KEY PARTS OF A CIRCLE: • Center (O): the fixed point equidistant from all points on the circle. • Radius (r): the distance from the center to any point on the circle. • Diameter (d): a chord that passes through the center; d = 2r (the longest chord). • Chord: a line segment connecting any two points on the circle. (The diameter is a special chord.) • Arc: a part of the circumference (curved portion between two points). • Semicircle: half of a circle; an arc formed by a diameter. • Central angle: an angle whose vertex is at the CENTER of the circle. • Inscribed angle: an angle whose vertex is ON the circle, with sides as chords. • Tangent: a line that touches the circle at exactly ONE point (the point of tangency). A tangent is PERPENDICULAR to the radius drawn to the point of tangency. • Secant: a line that intersects the circle at TWO points. FORMULAS: • Circumference C = 2πr = πd • Area A = πr² KNOW-ONE-FIND-ALL: If you know any one measurement, you can find all others: - Given r: d = 2r, C = 2πr, A = πr² - Given d: r = d/2, then apply above - Given C: r = C/(2π), then A = πr² - Given A: r = √(A/π), then C = 2πr ARC LENGTH AND SECTOR AREA (advanced): • Arc length = (central angle / 360°) × 2πr • Sector area = (central angle / 360°) × πr² π APPROXIMATION GUIDELINES: • For most calculations: π ≈ 3.14 • When r is a multiple of 7: π ≈ 22/7 (gives exact integer answers) • When the problem says 'in terms of π': leave π as a symbol in the answer

Examples

Given the circumference, always find the radius first. Since 88 = (44/7)r, multiply both sides by 7/44. Note: r = 14 is a multiple of 7, confirming that π = 22/7 is the right choice for clean arithmetic.

Scenario

The circumference of a circular running track is 88 m. Find (a) the radius and (b) the area of the circle it encloses. Use π = 22/7.

Solution

Step 1: C = 88 m; use C = 2πr Step 2: 88 = 2 × (22/7) × r Step 3: 88 = (44/7) × r Step 4: r = 88 × (7/44) = 88 × 7 / 44 = 616 / 44 = 14 m Step 5: Area = πr² = (22/7) × 14² = (22/7) × 196 = 22 × 28 = 616 m² Answer: Radius = 14 m; Area = 616 m².

Always convert diameter to radius before applying the area formula. This is one of the most common errors: using d instead of r. A = πr² uses the RADIUS squared.

Scenario

A circular garden has a diameter of 10 m. Find the area. Use π ≈ 3.14.

Solution

Step 1: d = 10 m, so r = d/2 = 5 m Step 2: Area = πr² = 3.14 × 5² = 3.14 × 25 = 78.5 m² Answer: The area of the garden is 78.5 m².

Applications

Computing the area of a circular classroom clock face or plate
Finding how much edging (circumference) is needed around a circular garden
Sector areas: computing the area of a pie-chart slice or pizza slice
Grade 5–6 K–12 BEC lessons on circles: parts, circumference, and area
Designing circular bulletin boards and computing material needed

Misconceptions

WRONG: Area = 2πr. CORRECT: Circumference = 2πr; Area = πr². Students confuse these two formulas — remember that area involves r SQUARED.
WRONG: Use diameter directly in A = πr². CORRECT: Always halve the diameter to get the radius before applying A = πr².
WRONG: A chord must pass through the center. CORRECT: A chord connects any two points on the circle. Only the DIAMETER passes through the center.
WRONG: π = 22/7 is exact. CORRECT: 22/7 ≈ 3.142857… is an approximation of π ≈ 3.14159…. Use it when radius is a multiple of 7 for clean computation, but it is not the exact value of π.

Related Concepts

Area and perimeter of plane figures
Sector and arc of a circle
Central and inscribed angles
Cylinder and cone formulas (built on circle area)

Common Exam Questions

Example

d = 14 cm → r = 7 → A = (22/7)(49) = 154 cm²

Approach

Divide diameter by 2 to get radius, then A = πr². Never plug the diameter directly into A = πr².

Question Type

Find area given diameter

Example

C = 44 cm → r = 44/(2 × 22/7) = 44 × 7/44 = 7 cm

Approach

C = 2πr → r = C/(2π). Or C = πd → d = C/π.

Question Type

Find radius or diameter given circumference

Example

Sector with central angle 90° in a circle of r = 6: Area = (90/360) × π(36) = (1/4)(36π) = 9π.

Approach

Multiply the fraction of the full angle (central angle / 360°) by the full area or circumference.

Question Type

Sector area or arc length

Key Points To Remember

Diameter = 2 × Radius; Radius = Diameter / 2
Circumference = 2πr (perimeter of the circle)
Area = πr² — note: it is r SQUARED, not r
Use π = 22/7 when r or d is a multiple of 7
A tangent is perpendicular to the radius at the point of tangency
All radii of the same circle are equal
The diameter is the longest chord and equals twice the radius

Coordinate Geometry: Distance and Midpoint Formulas

Coordinate geometry (also called analytic geometry) places geometric figures on the Cartesian plane — the xy-coordinate system — so that algebra can be used to solve geometric problems. THE DISTANCE FORMULA: The distance between points (x₁, y₁) and (x₂, y₂) is: d = √[(x₂ - x₁)² + (y₂ - y₁)²] This formula is simply the Pythagorean theorem applied to coordinates: • The horizontal difference (x₂ - x₁) is one leg • The vertical difference (y₂ - y₁) is the other leg • The distance d is the hypotenuse STEP-BY-STEP PROCEDURE: 1. Identify (x₁, y₁) and (x₂, y₂) 2. Compute (x₂ - x₁) and (y₂ - y₁) 3. Square both differences 4. Add the squares 5. Take the square root THE MIDPOINT FORMULA: The midpoint M of the segment joining (x₁, y₁) and (x₂, y₂) is: M = ((x₁ + x₂)/2 , (y₁ + y₂)/2) The midpoint is the average of the two x-coordinates and the average of the two y-coordinates. APPLICATIONS: • Finding the length of a side or diagonal of a polygon in the coordinate plane • Finding the center of a circle given the endpoints of a diameter (use midpoint formula) • Determining if a triangle is right-angled (compute sides, check Pythagorean theorem)

Examples

The differences 3 and 4 are the legs of a right triangle, and the distance 5 is the hypotenuse. Recognizing the 3-4-5 triple allows you to skip the square root computation.

Scenario

Find the distance between points A(1, 2) and B(4, 6).

Solution

Step 1: (x₁, y₁) = (1, 2); (x₂, y₂) = (4, 6) Step 2: x₂ - x₁ = 4 - 1 = 3 Step 3: y₂ - y₁ = 6 - 2 = 4 Step 4: d = √(3² + 4²) = √(9 + 16) = √25 = 5 Answer: The distance between A and B is 5 units. Note: Recognize the 3-4-5 Pythagorean triple!

Just average the x-values and average the y-values. The midpoint (5, 5) is equidistant from both P(2,3) and Q(8,7). You can verify: distance from (5,5) to (2,3) = √(9+4) = √13; distance from (5,5) to (8,7) = √(9+4) = √13. ✓

Scenario

Find the midpoint of the segment joining P(2, 3) and Q(8, 7).

Solution

Step 1: x₁ = 2, y₁ = 3, x₂ = 8, y₂ = 7 Step 2: Midpoint x = (x₁ + x₂)/2 = (2 + 8)/2 = 10/2 = 5 Step 3: Midpoint y = (y₁ + y₂)/2 = (3 + 7)/2 = 10/2 = 5 Answer: Midpoint = (5, 5).

Applications

Finding the length of a segment or side of a polygon plotted on graph paper
Verifying geometric properties (is ABCD a rectangle? — check if diagonals bisect each other using midpoint)
Finding the center of a circle (midpoint of the diameter)
Grade 6 K–12 BEC coordinate geometry lessons
Map problems: finding straight-line distances between two coordinate locations

Misconceptions

WRONG: Distance = (x₂-x₁) + (y₂-y₁). CORRECT: Distances are found using the Pythagorean theorem — square each difference, add, then take the square root.
WRONG: Midpoint = (x₂-x₁, y₂-y₁). CORRECT: Midpoint uses ADDITION: ((x₁+x₂)/2, (y₁+y₂)/2). The subtraction form gives displacement, not midpoint.
WRONG: The sign of (x₂-x₁) affects the distance. CORRECT: Since it gets SQUARED, (x₂-x₁)² = (x₁-x₂)². Distance is always positive.

Related Concepts

Pythagorean theorem
Cartesian coordinate system
Slope of a line
Equation of a circle

Common Exam Questions

Example

Points (0,0) and (5,12): d = √(25+144) = √169 = 13 (5-12-13 triple)

Approach

Apply d = √[(x₂-x₁)² + (y₂-y₁)²]. Check if the result matches a Pythagorean triple.

Question Type

Find the distance between two given points

Example

Midpoint of (-3, 4) and (7, -2): ((−3+7)/2, (4+(−2))/2) = (4/2, 2/2) = (2, 1)

Approach

Average the x-coordinates and average the y-coordinates separately.

Question Type

Find the midpoint of a segment

Key Points To Remember

Distance formula = Pythagorean theorem applied to coordinates
d = √[(x₂-x₁)² + (y₂-y₁)²] — SQUARE the differences before adding
Midpoint = average of x-coordinates, average of y-coordinates
Order doesn't matter for distance: (x₂-x₁)² = (x₁-x₂)² because it gets squared
Midpoint formula finds the CENTER of a segment
Use midpoint formula to find the center of a circle when given diameter endpoints

Practice Problems

Supplementary = 180°. Set up the equation using the relationship between the two angles, then solve. Always verify by checking the sum.

Problem

Two angles are supplementary. One angle is three times the other. Find both angles.

Solution

Step 1: Let the smaller angle = x. Then the larger = 3x. Step 2: x + 3x = 180° (supplementary) Step 3: 4x = 180° Step 4: x = 45° Step 5: Larger angle = 3 × 45 = 135° Answer: The angles are 45° and 135°. Verification: 45 + 135 = 180° ✓

Recognizing 5-12-13 as a standard Pythagorean triple saves computation time. Always confirm by checking: 5² + 12² = 25 + 144 = 169 = 13².

Problem

A right triangle has legs of 5 cm and 12 cm. Find the hypotenuse.

Solution

Step 1: Apply a² + b² = c² Step 2: 5² + 12² = c² Step 3: 25 + 144 = c² Step 4: 169 = c² Step 5: c = √169 = 13 cm Answer: The hypotenuse is 13 cm. Note: This is the 5-12-13 Pythagorean triple.

Unit conversion is a frequent LET trap. Always identify the target unit in the question and convert BEFORE or AFTER computing. Here, converting cm to m (÷100 each) then computing, or computing in cm² then dividing by 10,000, both give 0.24 m².

Problem

A triangular bulletin board has a base of 80 cm and a height of 60 cm. Find its area in square meters.

Solution

Step 1: Area of triangle = ½ × b × h = ½ × 80 cm × 60 cm = ½ × 4800 = 2400 cm² Step 2: Convert to m²: 1 m² = 10,000 cm² 2400 cm² ÷ 10,000 = 0.24 m² Answer: The area is 0.24 m².

The formula (n-2)×180° always gives the sum; divide by n for a regular polygon. The octagon appears in many architectural designs (octagonal windows, gazebos in Philippine school grounds).

Problem

A regular octagon has each interior angle measuring how many degrees?

Solution

Step 1: n = 8 for an octagon. Step 2: Sum of interior angles = (n - 2) × 180° = (8 - 2) × 180° = 6 × 180° = 1080° Step 3: Each interior angle of a regular octagon = 1080° ÷ 8 = 135° Answer: Each interior angle measures 135°.

Use π = 22/7 because r = 7 is a multiple of 7, giving clean arithmetic. The unit conversion (cm³ to liters) is a common LET addition. Remember: 1 L = 1000 cm³ = 1000 mL.

Problem

A cylindrical container has a radius of 7 cm and a height of 20 cm. How many liters of water can it hold? Use π = 22/7. (Note: 1 L = 1000 cm³)

Solution

Step 1: V = πr²h = (22/7)(7²)(20) Step 2: = (22/7)(49)(20) Step 3: = 22 × 7 × 20 (the 7 in denominator cancels with 7 in 49) Step 4: = 22 × 140 = 3080 cm³ Step 5: Convert to liters: 3080 cm³ ÷ 1000 = 3.08 liters Answer: The container can hold 3.08 liters.

Working backwards from area to radius to circumference. The key step is isolating r²: divide both sides of πr² = 154 by π. Since r = 7 (a multiple of 7), π = 22/7 was the right choice.

Problem

A circular playground has an area of 154 m². Find the circumference. Use π = 22/7.

Solution

Step 1: A = πr² = 154 Step 2: (22/7)r² = 154 Step 3: r² = 154 × (7/22) = 1078/22 = 49 Step 4: r = √49 = 7 m Step 5: Circumference = 2πr = 2 × (22/7) × 7 = 2 × 22 = 44 m Answer: The circumference is 44 m.

Perimeter ratio equals side ratio for similar figures. Area ratio = (side ratio)² = (4/7)² = 16/49. Setting up the proportion and solving gives 32 cm².

Problem

Two similar triangles have perimeters in the ratio 4:7. If the area of the larger triangle is 98 cm², find the area of the smaller triangle.

Solution

Step 1: Perimeters are in ratio 4:7, so the SIDE ratio (scale factor) k = 4/7. Step 2: Area ratio = k² = (4/7)² = 16/49 Step 3: Area of smaller / Area of larger = 16/49 Step 4: Area of smaller / 98 = 16/49 Step 5: Area of smaller = 98 × (16/49) = (98/49) × 16 = 2 × 16 = 32 cm² Answer: The area of the smaller triangle is 32 cm².

Be careful with negative coordinates: 5 - (-3) = 5 + 3 = 8, not 2. The 6-8-10 triple provides quick recognition of the answer as 10.

Problem

Find the distance between the points (-3, 1) and (5, 7).

Solution

Step 1: x₁ = -3, y₁ = 1, x₂ = 5, y₂ = 7 Step 2: Δx = 5 - (-3) = 8 Step 3: Δy = 7 - 1 = 6 Step 4: d = √(8² + 6²) = √(64 + 36) = √100 = 10 Answer: The distance is 10 units. Note: 6-8-10 is the 3-4-5 triple scaled by 2.

Perimeter uses ALL FOUR sides. Area uses only the two PARALLEL sides and the height. The legs (8 m and 6 m) appear in the perimeter but NOT in the area formula.

Problem

A trapezoid has parallel sides of 15 m and 9 m, and legs of 8 m and 6 m. The height between the parallel sides is 5 m. Find (a) the perimeter and (b) the area.

Solution

Step 1: (a) Perimeter = sum of all four sides = 15 + 9 + 8 + 6 = 38 m Step 2: (b) Area = ½(a + b) × h = ½(15 + 9) × 5 = ½ × 24 × 5 = ½ × 120 = 60 m² Answer: Perimeter = 38 m; Area = 60 m².

This composite problem requires computing wall areas plus ceiling, then subtracting the openings. Note: the floor is NOT included (not being painted). This type of practical problem is common in LET items contextualized in school settings.

Problem

A school buys paint to cover the four walls and ceiling of a rectangular classroom 9 m long, 7 m wide, and 3 m high. There are two doors (each 2 m × 1 m) and three windows (each 1.5 m × 1 m) that will not be painted. Find the paintable area.

Solution

Step 1: Area of four walls: Two long walls: 2 × (9 × 3) = 54 m² Two short walls: 2 × (7 × 3) = 42 m² Total wall area = 54 + 42 = 96 m² Step 2: Area of ceiling: 9 × 7 = 63 m² Step 3: Total surface = 96 + 63 = 159 m² Step 4: Deduct doors: 2 × (2 × 1) = 4 m² Step 5: Deduct windows: 3 × (1.5 × 1) = 4.5 m² Step 6: Paintable area = 159 - 4 - 4.5 = 150.5 m² Answer: The paintable area is 150.5 m².

Exam Preparation Tips

MEMORIZE THE FORMULA TABLE: Write out all formulas (area, perimeter, volume, surface area) on a single sheet and review daily. The LET tests formula APPLICATION, not derivation, so instant formula recall is essential.
MASTER PYTHAGOREAN TRIPLES: Know 3-4-5, 5-12-13, 8-15-17 and their multiples by heart. Recognizing these instantly eliminates lengthy square root calculations in the exam.
READ THE QUESTION TWICE: Determine whether the question asks for PERIMETER/CIRCUMFERENCE (linear units), AREA/SURFACE AREA (square units), or VOLUME (cubic units). Applying the wrong formula is the most common geometry error on the LET.
UNIT CONSISTENCY: Always convert all measurements to the SAME unit BEFORE computing. Mixing meters and centimeters will give answers wrong by a factor of 100 (for lengths) or 10,000 (for areas).
ANGLE SUM SHORTCUTS: Triangle = 180°; Quadrilateral = 360°; Pentagon = 540°; Hexagon = 720°. Memorize these so you don't need to apply the formula (n-2)×180 every time.
π SELECTION STRATEGY: Use π = 22/7 when the radius or diameter is a multiple of 7 (7, 14, 21, 28, etc.). Use π ≈ 3.14 for all other values. This gives cleaner arithmetic and saves time.
CONGRUENCE vs. SIMILARITY: Congruence = same size AND shape (use ≅); Similarity = same shape, possibly different size (use ~). For the LET, know the difference between congruence postulates (SSS, SAS, ASA, AAS, HL) and similarity postulates (AA, SSS-proportional, SAS-proportional).
CONE AND PYRAMID VOLUME: These are ONE-THIRD of the corresponding prism. If a question gives you a cone and a cylinder with the same dimensions, the cone holds exactly 1/3 as much.
SHADOW/SIMILAR TRIANGLE PROBLEMS: Set up as a proportion: height₁/shadow₁ = height₂/shadow₂. Cross-multiply and solve. Make sure the units in the numerators match and the units in the denominators match.
COMPLEMENT vs. SUPPLEMENT: Memory trick: 'C comes before S' in the alphabet, and 90° comes before 180° on the number line. Complementary = 90°; Supplementary = 180°.
COMPOSITE FIGURES: Always identify the component shapes clearly before computing. Draw dotted lines to show where you are dividing the figure. Label each sub-area before adding or subtracting.
ALTITUDE IN TRIANGLES AND PARALLELOGRAMS: The HEIGHT in area formulas is ALWAYS the PERPENDICULAR (vertical) distance, never the slant or oblique side. This is the most frequent conceptual error for these shapes.
EXAM TIME MANAGEMENT: In LET geometry items, if the numbers look messy, check if you may have confused radius and diameter, or misapplied a formula. Clean Pythagorean triples and whole-number answers suggest you are on the right track.
REGULAR POLYGON ANGLES: If given a regular polygon's interior angle and asked for the number of sides, use the reverse formula: n = 360° / exterior angle, where exterior angle = 180° - interior angle.
DISTANCE AND MIDPOINT: The distance formula is the Pythagorean theorem in disguise — the coordinate differences are the legs. The midpoint formula is simply averaging — add coordinates and divide by 2.

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In summary

Plane and Solid Geometry is a high-yield area of the LET Mathematics component that rewards systematic preparation. The key to success lies in three habits: (1) memorizing formulas precisely — especially the (n-2)×180° polygon formula, the Pythagorean theorem, and the one-third factor for cones and pyramids; (2) reading each problem carefully to identify what is being asked (perimeter, area, or volume) and matching it to the correct formula; and (3) maintaining unit consistency throughout every computation. As future elementary school teachers, your mastery of these concepts directly supports quality mathematics instruction in Grades 1–6 under the K–12 Basic Education Curriculum. When you teach Grade 4 pupils how to find the area of a rectangle or guide Grade 6 learners through the Pythagorean theorem, your deep conceptual understanding — built through this kind of rigorous review — will translate into confident, competent teaching. Remember: the PRC and the LET evaluate not just whether you can compute, but whether you have the mathematical content knowledge to serve Filipino children effectively in the classroom. Under RA 7836 (Philippine Teachers Professionalization Act), earning your professional license certifies that competence. Study with that purpose in mind, and approach each geometry problem as both an exam challenge and a future teaching opportunity. Final reminders for exam day: sketch figures for every word problem, label all given information, show all steps, verify units in your final answer, and use Pythagorean triple recognition as your time-saving superpower. You are prepared — trust your preparation, work systematically, and succeed!

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