Triangles — Karnataka (SSLC) Class 9 Mathematics Solutions (Free)

Step 1: SSS congruence criterion states that if all three sides of one triangle are equal to the corresponding three sides of another triangle, then the two triangles are congruent. Step 2: In triangle ABC: AB = 3 cm BC = 4 cm CA = 5 cm Step 3: In triangle PQR: PQ = 3 cm QR = 4 cm RP = 5 cm Step 4: Comparing corresponding sides: AB = PQ = 3 cm BC = QR = 4 cm CA = RP = 5 cm Step 5: Since all three sides of triangle ABC are equal to corresponding sides of triangle PQR, by SSS criterion, triangle ABC ≅ triangle PQR. Answer: Triangle ABC is congruent to triangle PQR (triangle ABC ≅ triangle PQR).

Step 1: SAS congruence criterion states that if two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of another triangle, then the triangles are congruent. Step 2: In triangle ABD: AB = given BD = common side angle ABD = given Step 3: In triangle CBD: CB = given BD = common side angle CBD = given Step 4: Comparing the triangles: AB = CB (given) angle ABD = angle CBD (given) BD = BD (common side) Step 5: The included angle between sides AB and BD in triangle ABD is angle ABD. The included angle between sides CB and BD in triangle CBD is angle CBD. Step 6: Since two sides and the included angle of triangle ABD are equal to corresponding two sides and included angle of triangle CBD, by SAS criterion, triangle ABD ≅ triangle CBD. Answer: Triangle ABD is congruent to triangle CBD (triangle ABD ≅ triangle CBD).

Step 1: ASA congruence criterion states that if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of another triangle, then the triangles are congruent. Step 2: In triangle PQR: angle Q = given QR = given (included side between angles Q and R) angle R = given Step 3: In triangle XYZ: angle Y = given YZ = given (included side between angles Y and Z) angle Z = given Step 4: Comparing the triangles: angle Q = angle Y (given) QR = YZ (given) angle R = angle Z (given) Step 5: QR is the side included between angles Q and R in triangle PQR. YZ is the side included between angles Y and Z in triangle XYZ. Step 6: Since two angles and the included side of triangle PQR are equal to the corresponding two angles and included side of triangle XYZ, by ASA criterion, triangle PQR ≅ triangle XYZ. Answer: Triangle PQR is congruent to triangle XYZ (triangle PQR ≅ triangle XYZ).

Step 1: RHS congruence criterion states that if the hypotenuse and one side of a right triangle are equal to the hypotenuse and corresponding side of another right triangle, then the triangles are congruent. Step 2: In triangle ABC: Angle B = 90° (given) AC = hypotenuse AB = one side Step 3: In triangle DEF: Angle E = 90° (given) DF = hypotenuse DE = one side Step 4: Comparing the triangles: Angle B = Angle E = 90° (both right angles) AC = DF (given hypotenuses are equal) AB = DE (given, corresponding sides are equal) Step 5: By RHS criterion, when the hypotenuse and one side of a right triangle equal the hypotenuse and corresponding side of another right triangle, the triangles are congruent. Step 6: Therefore, triangle ABC ≅ triangle DEF. Answer: Triangle ABC is congruent to triangle DEF (triangle ABC ≅ triangle DEF).

Step 1: Given: AB = AC in triangle ABC. To prove: angle C = angle B (angles opposite equal sides are equal). Step 2: Draw the angle bisector of angle A, and let it meet BC at point D. Step 3: In triangle ABD and triangle ACD: AB = AC (given) AD = AD (common side) angle BAD = angle CAD (AD is angle bisector of angle A) Step 4: By SAS congruence criterion, triangle ABD ≅ triangle ACD. Step 5: From the congruence, corresponding angles are equal: angle ABD = angle ACD angle ADB = angle ADC BD = CD Step 6: angle ABD is the angle at B in triangle ABC, which is angle B. angle ACD is the angle at C in triangle ABC, which is angle C. Step 7: Therefore, angle B = angle C. Answer: In a triangle, angles opposite to equal sides are equal (angle B = angle C).

Step 1: By the angle sum property of a triangle, the sum of all three angles equals 180°. Step 2: Given: angle A = 60° angle B = 80° angle C = ? Step 3: Using angle sum property: angle A + angle B + angle C = 180° Step 4: Substitute known values: 60° + 80° + angle C = 180° 140° + angle C = 180° Step 5: Solve for angle C: angle C = 180° - 140° angle C = 40° Step 6: Verify: 60° + 80° + 40° = 180° ✓ Answer: angle C = 40°; Sum of all angles in the triangle = 180°.

Step 1: Given isosceles triangle PQR with PQ = PR = 5 cm and QR = 6 cm. Step 2: The altitude from P to QR bisects the base QR (property of isosceles triangles). Step 3: Let the altitude from P meet QR at point M. Then QM = MR = QR/2 = 6/2 = 3 cm Step 4: In right triangle PQM: angle PMQ = 90° (altitude is perpendicular) PQ = 5 cm (hypotenuse) QM = 3 cm (base) PM = h (altitude/height) Step 5: By Pythagoras theorem: PQ² = PM² + QM² 5² = h² + 3² 25 = h² + 9 h² = 25 - 9 h² = 16 h = 4 cm Step 6: Verify with triangle PMR: PR² = PM² + MR² 5² = 4² + 3² 25 = 16 + 9 25 = 25 ✓ Answer: h = 4 cm.

Step 1: Triangle Inequality Theorem: The sum of any two sides of a triangle must be greater than the third side. Step 2: For a valid triangle with sides a, b, c: a + b > c b + c > a a + c > b Step 3: Check if sides 3 cm, 4 cm, and 8 cm form a triangle: Let a = 3 cm, b = 4 cm, c = 8 cm Step 4: Check condition 1: a + b > c 3 + 4 > 8 7 > 8 (FALSE) Step 5: Since the first condition itself fails, these sides cannot form a triangle. Step 6: Verification with other conditions: Condition 2: b + c > a → 4 + 8 > 3 → 12 > 3 (TRUE) Condition 3: a + c > b → 3 + 8 > 4 → 11 > 4 (TRUE) Step 7: For a valid triangle, ALL three conditions must be satisfied. Since 3 + 4 = 7 is not greater than 8, these three measurements cannot form the sides of a triangle. Answer: Sides 3 cm, 4 cm, and 8 cm CANNOT form a triangle because 3 + 4 is not greater than 8.