Grade 10 Mathematics Term 3 Geometry Revision Guide

Course Overview and Attribution

• School: Salamah Bint Butti School (مءرسة سلامة بنت بطي)
• Ministry: Ministry of Education, UAE
• Subject: Mathematics 10 General (Reveal)
• Term: End of Term 3 Exam Revision
• Academic Year (AY): 2526
• Instructor: Teacher Rhoda (Ms. Rhoda)
• Instructional Message: "ان شاء الله كل مجهوداتك ستؤتي ثماره" (By God's will, all your efforts will bear fruit).
• Additional Formulas Noted on Title Page:
  • Circumference/Geometry fragment: $+2\pi r$
  • Algebraic fragments: $2x$ and $x + b$
  • Distance Formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
  • Area fragment: $A = 1$

M14L1: The Triangle-Sum Theorem

• Key Concept: The interior angles of any triangle always sum to exactly $180^\circ$.
• Application Problems:
  • 1. Given a figure, identify $m\angle 1$ and $m\angle 2$. For instance, if the angles are complementary in a right triangle context, values often result in $30^\circ$ and $60^\circ$.
  • 2. Find the measures of $\angle 1$, $\angle 2$, and $\angle 3$ in complex multi-triangle diagrams. Based on the answer key, the specifics for this problem are $m\angle 1 = 109^\circ$, $m\angle 2 = 29^\circ$, and $m\angle 3 = 71^\circ$.
  • 3. In another triangle set, find the measures resulting in $m\angle 1 = 56^\circ$, $m\angle 2 = 56^\circ$, and $m\angle 3 = 74^\circ$.
  • 4. Isosceles Context: In a figure where two base angles are congruent and the vertex angle is given as $146^\circ$, find the measure of the base angle $\angle 1$.
  • Calculation: $\frac{180^\circ - 146^\circ}{2} = \frac{34^\circ}{2} = 17^\circ$.

M14L1: The Exterior Angle Theorem

• Definition: The measure of an exterior angle of a triangle is equal to the sum of the measures of its two remote interior angles.
• Examples:
  • Example 3: Calculating $m\angle DAB$. Based on the guide, the target measure is $115^\circ$.
  • Problem 15-17:
  • To find the value of $x$ in an algebraic triangle, the answer is $40^\circ$.
  • To find the measure of an angle labeled $2x^\circ$: $2 \times 40 = 80^\circ$ (Note: Key for question 16 suggests $60^\circ$, which implies $x = 30^\circ$ in that specific diagram setup).
  • The value of the exterior angle in this set is evaluated as $120^\circ$.

M14L2: Name and Use Corresponding Parts of Congruent Triangles

• CPCTC Principle: Corresponding Parts of Congruent Triangles are Congruent.
• Congruence Statements: The order of vertices in a congruence statement must match the corresponding parts exactly.
  • Example: $\triangle ABC \cong \triangle DBC$ implies $A$ corresponds to $D$, $B$ to $B$, and $C$ to $C$.
  • Another example: $\triangle XYZ \cong \triangle ABC$ or $\triangle XYZ \cong \triangle BAC$; the correct statement depends on identifying congruent angles (e.g., $\angle X \cong \angle A$, $\angle Y \cong \angle B$, etc.).
• Algebraic Proving:
  • If $\triangle ABC \cong \triangle FDE$, corresponding sides like $BC$ and $DE$ or angles like $\angle A$ and $\angle F$ are equal.
  • Example values: In one problem find $x = 36^\circ$ and $y = 48^\circ$ based on triangle congruence properties.
• Real-World Applications:
  • Cell Phone Case Design: If $\angle ABC = 35^\circ$, $\angle BAC = 29^\circ$, and $\angle ACB \cong \angle DEB$, then $m\angle DEB = 180 - (35 + 29) = 116^\circ$.
  • Dining Table Construction: If $\angle PRQ \cong \angle TVU$ and $m\angle RPQ = 49^\circ$, identifying the corresponding angle $\angle TVU$ yields $41^\circ$ or $49^\circ$ depending on orientation (Key says $41^\circ$).

M14L3: Triangle Congruence Postulates (SSS and SAS)

• Postulate 14.1: Side-Side-Side (SSS) Congruence: If three sides of one triangle are congruent to three sides of a second triangle, then the triangles are congruent.
• Postulate 14.2: Side-Angle-Side (SAS) Congruence: If two sides and the included angle of one triangle are congruent to two sides and the included angle of a second triangle, then the triangles are congruent.
• Logic and Definitions:
  • Midpoint: If $B$ is the midpoint of $AD$, then $\overline{AB} \cong \overline{BD}$.
  • Segment Bisector: If line segment $\overline{AC}$ bisects $\overline{BD}$, it divides $\overline{BD}$ into two congruent segments ($\overline{BC} \cong \overline{CD}$).
  • Identifying Missing Information: To prove congruence by SSS, if two pairs of sides are known, the third pair must be stated (e.g., $\overline{RT} \cong \overline{RT}$ via the Reflexive Property).

M14L4: Triangle Congruence Postulates (ASA and AAS)

• ASA (Angle-Side-Angle): Requires two angles and the included side.
• AAS (Angle-Angle-Side): Requires two angles and a non-included side.
• Key Definitions in Proofs:
  • Angle Bisector: If $\overline{BD}$ bisects $\angle ABC$, then $\angle ABD \cong \angle CBD$.
  • Midpoint Application: Used to establish that segments are congruent (e.g., $\overline{UV} \cong \overline{VX}$ if $V$ is the midpoint of $\overline{UX}$).
• Calculating Measures from Congruent Triangles:
  • In a set of congruent triangles, side lengths like $AC$, $AB$, and $BC$ can be identified. Examples from the text yield measures of $41$, $21$, and $40$.

M14L6: Properties of Equilateral Triangles

• Equilateral Triangle: A triangle where all three sides are congruent.
• Equiangular Property: All equilateral triangles are equiangular, meaning each interior angle measures exactly $60^\circ$.
• Problems:
  • If a triangle is equilateral with side lengths given as expressions (e.g., $x + 5$ and $15$), then $x = 10$ because all sides are equal.
  • Finding $m\angle Y$ and side $WY$: If the triangle is equilateral, $m\angle Y = 60^\circ$ and $WY$ will equal the other given side lengths (e.g., $7\text{ in.}$).

M1L1: Perpendicular Bisectors

• Theorem 6.1: Perpendicular Bisector Theorem: If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
• Mathematical Applications:
  • Find $FM$ if given a point on the bisector: Resulting in $2.8$.
  • Find $XW$ given distances to endpoints are equal ($2x-5 = x+5$ or similar logic): Resulting in $7.5$.
  • Solve for $x$: Expressions like $3x - 10 = 2x + 5$ are typically used to find the variable.

M1L2: Angle Bisectors and the Incenter Theorem

• Incenter Theorem: Mentioned in relation to point $P$ in $\triangle XYZ$.
• Calculation Example: If $P$ is the incenter, it is the meeting point of the triangle's angle bisectors. Finding $m\angle LZP$ yields $40.3^\circ$.

M1L3: The Centroid Theorem

• Centroid Property: The centroid of a triangle is located two-thirds of the distance from each vertex to the midpoint of the opposite side.
• Fundamental Relationship: $BP = \frac{2}{3}(BL)$ and $PL = \frac{1}{3}(BL)$, creating the ratio $BP = 2(PL)$.
• Calculations in $\triangle CDE$ (U is centroid, $UK=12$, $EM=21$, $UD=9$):
  • Find $CU$: $18$
  • Find $MU$: $7$
  • Find $EU$: $14$
  • Find $JD$: $13.5$
• Coordinate Geometry of Centroid:
  • To find the centroid of $\triangle ABC$ with vertices $A(-3, 3)$, $B(2, 5)$, and $C(4, -3)$, use the formula:     $Centroid = (\frac{x_1 + x_2 + x_3}{3}, \frac{y1 + y2 + y3}{3})$$Centroid = (\frac{-3+2+4}{3}, \frac{3+5-3}{3}) = (\frac{3}{3}, \frac{5}{3}) = (1, \frac{5}{3})$

M1L3 & M1L6: Triangle Inequalities

• Exterior Angle Inequality Theorem: The measure of an exterior angle of a triangle is greater than the measure of either of its remote interior angles.
• Angle-Side Inequalities:
  • If one side of a triangle is longer than another side, the angle opposite the longer side is greater than the angle opposite the shorter side.
  • Example: Listing angles of $\triangle RST$ from smallest to largest based on side lengths gives $\angle T < \angle R < \angle S$.
  • Example: Listing sides of $\triangle RST$ from shortest to longest based on angle measures gives $\overline{ST} < \overline{RT} < \overline{RS}$.
• Triangle Inequality Theorem: The sum of any two side lengths of a triangle must be greater than the third side length. For a third side $x$ with known sides $a$ and $b$:   $|a - b| < x < a + b$
  • Range of Third Side Examples:
  • 1. Sides $13$ and $6$: $7 < x < 19$ (Note: Key says B, which is $13 < x < 25$ for a different set of numbers like 6 and 19).
  • 2. Sides $7$ and $29$: Range is $22 < x < 36$.
  • 3. Sides $13$ and $27$: Range is $14 < x < 40$.
  • 4. Sides $18$ and $23$: Range is $5 < x < 41$.

M2L2 & M2L3: Parallelograms

• Basic Properties:
  • Opposite sides are parallel and congruent.
  • Opposite angles are congruent (e.g., $\angle R \cong \angle P$).
  • Consecutive angles are supplementary ($m\angle R + m\angle S = 180^\circ$).
• Diagonal Properties: The diagonals of a parallelogram bisect each other.
• Problems:
  • If $m\angle P = 52^\circ$, find $m\angle R$: Answer is $52^\circ$.
  • Find $x$ and $y$ involving diagonals: If diagonal segments are $3x$ and $15$, then $3x = 15 \implies x = 5$.
  • Tests for Parallelograms: Determine if a quad is a parallelogram by checking if opposite sides are equal ($2x+3 = x+7$) or if diagonals bisect each other accurately.

M2L4: Rectangles

• Definition: A parallelogram with four right angles.
• Properties: All properties of parallelograms plus diagonals are congruent ($AC \cong BD$).
• Calculations: If $AE = 44.5\,ft$ (where $E$ is the intersection of diagonals), then the full diagonal $AC = 89\,ft$. Also, segments like $DG$ would be equal to other half-diagonals (e.g., $39\,ft$).
• Proving a Rectangle: Quadrilateral $ABCD$ is a rectangle if opposite sides are congruent and diagonals are congruent.

M2L5: Rhombi and Squares

• Properties of a Rhombus:
  • All four sides are congruent.
  • Diagonals are perpendicular ($m\angle CEB = 90^\circ$).
  • Diagonals bisect opposite angles.
• Properties of a Square:
  • All properties of a rectangle (right angles, congruent diagonals) and a rhombus (all sides congruent, perpendicular diagonals).
• Problems with Rhombi:
  • If $AE = 8$, then $AC = 16$.
  • If $AB = 26$ and $BD = 20$, use the Pythagorean Theorem in the triangle formed by diagonals to find $AE = 24$.
• Problems with Squares:
  • In square $BCDF$ with $FD = 55$, the side $BC = 55$.
  • Diagnostic values: For square diagonals, half-diagonal $GD \approx 38.9$ and full diagonal $BD \approx 77.8$.

M2L6: Trapezoids and Kites

• Trapezoid Midsegment: The segment connecting the midpoints of the legs. Its length is the average of the two bases:   $Midsegment = \frac{base1 + base2}{2}$
  • Example: If base $LK = 19$ and base $TS = 15$, midsegment $HJ = \frac{19 + 15}{2} = 17$. (Note cross-check with answer key shows variability based on labels).
  • Example: If $LK = 27.4$ and $TS = 24$, midsegment $HJ = 25.7$ (Key says $20.6$, suggesting different positioning of labels).
• Kite Properties:
  • Exactly two pairs of consecutive congruent sides.
  • Diagonals are perpendicular.
  • Exactly one pair of opposite angles is congruent (the angles between non-congruent sides).
  • Calculations: Find perimeter by using the Pythagorean Theorem to find side lengths from diagonals. For instance, finding $BC = \sqrt{side^2}$ to get simplified radicals like $4\sqrt{34}$.