Class 9 Mathematics Chapter 7 Triangles

Key Concept: Basic Definitions

c) Three sides and three angles

[Solution Description]

A triangle is defined by three non-collinear points (vertices) connected by three line segments (sides). These sides form three angles. Hence, the minimum elements required are three sides and three angles.

Key Concept: SSS Congruence Rule

a) SSS

[Solution Description] According to Theorem 7.4 (SSS congruence rule), when three sides of one triangle exactly match three sides of another triangle, they are congruent. This does not require any angle measurements.

Key Concept: AAS Congruence Rule

b) AAS Congruence

[Solution Description]

Given: Two angles $\angle L = \angle X$ and $\angle M = \angle Y$, with non-included side $LN = XZ$. Using AAS congruence rule, when two angles and any side of one triangle equal those of another, the triangles are congruent. Since angles sum to $180^\circ$, the third angles $\angle N$ and $\angle Z$ are equal. Hence, $\Delta LMN \cong \Delta XYZ$ by AAS.

Key Concept: Real-world application, SSS congruence

a) Yes, by SSS congruence

[Solution Description] Despite differing order in side lengths, the sets {30, 40, 50} and {40, 50, 30} are identical. By rearranging the correspondence (e.g., AB $\leftrightarrow$ RP, BC $\leftrightarrow$ PQ, CA $\leftrightarrow$ QR), all three sides match. Thus, SSS congruence holds.

Key Concept: SAS Congruence in Isosceles Triangle

c) $AD = AE$

[Solution Description]

Since $BE = CD$, subtract $DE$ from both: $BD = CE$. In $\triangle ABD$ and $\triangle ACE$, $AB = AC$ (given). $\angle B = \angle C$ (base angles of isosceles triangle). Also, $BD = CE$ (proved). Thus, $\triangle ABD \cong \triangle ACE$ by SAS. Hence, $AD = AE$ by CPCT.

Key Concept: Real-life SSS Congruence

d) $\text{SSS (Side-Side-Side)}$

[Solution Description]

Since all three corresponding sides of the triangles are equal ($5 cm, 7 cm, 9 cm$), the $\text{SSS (Side-Side-Side)}$ congruence rule applies. This rule states that triangles with all sides equal are congruent, ensuring identical structural strength.

Key Concept: Combined Properties

c) All angles measure $60^\circ$

[Solution Description]

Side equality makes it equilateral. Syllabus point 10 states each angle is $60^\circ$. By Theorem 7.2, equal sides imply equal opposite angles. Hence all angles equal $60^\circ$.

Key Concept: SSS Congruence Rule

d) All of P, Q, and R

[Solution Description]

All three triangles have sides 5 cm, 6 cm, and 7 cm but arranged in different orders. SSS congruence states that if all three sides of one triangle are equal to another (regardless of order), they are congruent. Hence, all pairs (P-Q, Q-R, P-R) are congruent by SSS.

Key Concept: Obtuse Angle Recognition

c) $150^\circ$

[Solution Description]

Obtuse angles are greater than $90^\circ$ but less than $180^\circ$. $150^\circ$ lies between these limits.

Key Concept: SAS Congruence Rule Application

c) SAS congruence

[Solution Description] Here, two sides ($LM=XY$, $MN=YZ$) and their included angle ($\angle M = \angle Y$) are equal. This matches SAS rule conditions.

Key Concept: RHS, SSS, Hypotenuse Leg Theorem

d) No, legs do not satisfy RHS/SSS

[Solution Description] Check using RHS: Both are right-angled, but their hypotenuses are equal (13 cm). However, the legs differ (5 cm vs. 9 cm). For RHS congruence, one leg and hypotenuse must match. Since neither leg matches between the triangles, they are not congruent. Similarly, SSS fails as the sides differ (5,12,13 vs. 8,9,13).

Key Concept: SSS Congruence Rule

d) SSS

[Solution Description] According to Theorem 7.4 (SSS congruence rule), if $AB = DE$, $BC = EF$, and $AC = DF$ in two triangles $\triangle ABC$ and $\triangle DEF$, they are congruent. This matches the condition described in the question.

Key Concept: Perpendicular bisector property

c) $PB = PC$

[Solution Description] The key property states any point on the perpendicular bisector of a line segment is equidistant from its endpoints. Thus, $PB = PC$.

Key Concept: SAS Congruence Rule Error

b) Two sides and a non-included angle of both triangles are equal

[Solution Description]

For SAS congruence, the angle must be between the two sides. If two sides and a non-included angle are given (e.g., $AB = PQ$, $\angle C = \angle R$, and $BC = QR$), it violates the SAS rule. Only two sides and their INCLUDED angle ensure congruence. Options listing non-included angles (like $\angle C$) lead to error.

Key Concept: CPCT Application

a) $EF$

[Solution Description]

RHS congruence requires hypotenuse and one leg. Here, $AC = DF$ (hypotenuse) and $AB = DE$ (one leg). By CPCT, remaining corresponding sides are equal.

Corresponding to $BC$ in $\triangle ABC$, the matching side in $\triangle DEF$ is $EF$.

Key Concept: Hypotenuse Identification

a) PR = XZ \& PQ = XY

[Solution Description]

For RHS rule: Need equal hypotenuses (PR = XZ) and one pair of equal corresponding legs (either PQ = XY or QR = YZ). Right angles already present.

Key Concept: RHS Congruence

c) RHS congruence

[Solution Description]

Both triangles are right-angled. Hypotenuses (10 m) and one corresponding side (ground distance = 6 m) are equal. By RHS congruence rule, they are congruent.

Key Concept: AAA Insufficiency

c) Not necessarily congruent

[Solution Description]

AAA (three equal angles) ensures similarity but not congruence. From the syllabus: “equality of three angles (AAA) is not sufficient for congruence.” Without equal sides, triangles can have different sizes.

Step-by-Step:

1. Sum of angles is same (180°) → Similar.

2. Missing side-length equality → Not congruent.

3. Correct answer: not congruent.

Key Concept: RHS Congruence Rule

c) RHS congruence

[Solution Description] Since both triangles are right-angled ($∠B = ∠Q = 90°$), their hypotenuse $AC = PR$, and one leg $AB = PQ$ are equal. By Theorem 7.5 (RHS rule), $\Delta ABC ≅ \Delta PQR$.

Key Concept: Vertex Correspondence, CPCT

b) $\angle T = 40^\circ$

[Solution Description]

Correspondence for $\triangle PQR \cong \triangle STU$ is $P \leftrightarrow S$, $Q \leftrightarrow T$, $R \leftrightarrow U$. CPCT gives $\angle Q = \angle T = 40^\circ$ and $PQ = ST = 5$ cm. However, $PQ$ corresponds to $ST$, but option b) refers to $QR$, which equals $TU$, not given. Correct answer is $\angle T = 40^\circ$.

Key Concept: Definition of Congruent Figures

d) Same shape and size

[Solution Description] Congruent figures have exactly the same shape and size. This means their corresponding sides and angles are equal.

Key Concept: Altitude Properties

c) Bisects both $BC$ and $\angle A$

[Solution Description]

In isosceles $\triangle ABC$, $AB = AC$. By property, the altitude from vertex $A$ (which is $AD$) also acts as median and angle bisector. Step 1: Since $AD$ is altitude, $\angle ADB = \angle ADC = 90^\circ$. Step 2: Hypotenuse $AB = AC$ (given). Step 3: Common side $AD = AD$. Hence, $\triangle ABD \cong \triangle ACD$ by RHS, implying $BD = DC$ (bisects $BC$) and $\angle BAD = \angle CAD$.

Key Concept: SSS Congruence Rule

c) SSS congruence rule

[Solution Description]

Both triangles have all three corresponding sides equal (5 cm = 5 cm, 6 cm = 6 cm, 7 cm = 7 cm). By the SSS congruence rule, triangles with equal corresponding sides are congruent.

$\text{SSS rule is satisfied} \rightarrow \text{Triangles are congruent}$

Key Concept: SAS Congruence Rule

a) SAS

[Solution Description]

Step 1: Identify corresponding parts.

– Side $AB = DE$ (given).

– Side $BC = EF$ (given).

– Included angle $\angle B = \angle E$ (given).

Step 2: Apply SAS Congruence Rule. Two sides and the included angle are equal, so $\Delta ABC \cong \Delta DEF$ by SAS.

Key Concept: Sides Identification

b) $AB$, $BC$, $CA$

[Solution Description] In triangle $ABC$, the sides are named using pairs of its vertices. By definition, the sides of $\Delta ABC$ are $AB$, $BC$, and $CA$.

Key Concept: Bangles and Coins

d) Neither the bangles nor the coins are congruent

[Solution Description]

Bangles require identical dimensions. Thickness difference (2 mm vs. 4 mm) means they are not congruent. Coins minted in different years are not explicitly stated in the syllabus as congruent, even if size matches. The syllabus only confirms same-year coins as congruent. Thus, neither pair is congruent.

Key Concept: SSS Congruence Application

d) SSS rule

[Solution Description] Since all three corresponding sides of $\triangle ABC$ and $\triangle PQR$ are equal ($AB = PQ = 5$ cm, $BC = QR = 7$ cm, $AC = PR = 6$ cm), they satisfy the SSS congruence rule. No angles are required to be equal for this rule.

Key Concept: Angle Measurement

c) $60^\circ$

[Solution Description]

In an equilateral triangle, all sides are equal. By Theorem 7.2, all angles are equal. Using the angle sum property of triangles:

$3 \times \text{(Each angle)} = 180^\circ$

Dividing both sides by 3:

$\text{Each angle} = \frac{180^\circ}{3} = 60^\circ$

Key Concept: Theorem Identification

c) Angles opposite to equal sides of an isosceles triangle are equal

[Solution Description]

According to Theorem 7.2 in the syllabus, in an isosceles triangle, the angles opposite to the equal sides are equal. This theorem directly addresses the relationship between equal sides and their opposite angles in such triangles.

Key Concept: Alternate Interior Angles and Parallel Lines

c) Alternate interior angles are equal

[Solution Description] Since triangles are congruent by SAS, $\angle QPR = \angle UTS$. These angles are alternate interior angles formed by transversal PT intersecting PQ and ST. Hence, PQ is parallel to ST because alternate interior angles are equal.

Key Concept: Angle Sum Property \& Equilateral Triangle

c) $60^\circ$

[Solution Description]

Since XY = XZ, triangle XYZ is isosceles with $\angle Y = \angle Z$. Let each be $y$. Using angle sum property:

$60^\circ + y + y = 180^\circ$

$2y = 120^\circ$

$y = 60^\circ$.

Thus, all angles are $60^\circ$, making it equilateral.

Key Concept: SAS Congruence Rule

b) SAS

[Solution Description]

By Axiom 7.1 (SAS Congruence Rule), two triangles are congruent if two sides and the included angle of one triangle equal the corresponding parts of the other. Here, sides XY = LM, YZ = MN, and included angle $\angle Y = \angle M$ are equal. Hence, SAS applies.

Key Concept: SSA Invalid Criterion

c) Side-Side-Angle (SSA)

[Solution Description]

\underline{Step 1}: Identify valid congruence criteria from the syllabus.

Valid criteria are SAS (with angle included), ASA, SSS, RHS.

\underline{Step 2}: Recognize that SSA/ASS lacks an included angle between the given sides, making it invalid.

Key Concept: Midpoint Theorem – AAS Congruence

d) AAS

[Solution Description]

Since $AB \parallel CD$, alternate angles $\angle ABO = \angle DCO$. Vertical angles $\angle AOB = \angle DOC$ are equal. Given $OA = OD$ (midpoint), triangles are congruent by AAS rule.

Key Concept: Components of Triangle

c) 3

[Solution Description]

By definition, a triangle has exactly three angles formed at its three vertices. Regardless of naming (PQR), the count remains consistent. Each vertex forms one angle, totaling three angles in any triangle.