Triangles — Andhra Pradesh (SSC) Class 9 Mathematics Solutions (Free)

SSS Criterion: if all three sides of one triangle are equal in length to the corresponding three sides of another triangle, the two triangles are congruent. Given: AB = 6 cm, BC = 8 cm, CA = 10 cm in triangle ABC, and PQ = 6 cm, QR = 8 cm, RP = 10 cm in triangle PQR. Comparing corresponding sides: AB = PQ = 6 cm BC = QR = 8 cm CA = RP = 10 cm Since all three pairs of corresponding sides are equal, by the SSS congruence criterion: △ABC ≅ △PQR

SAS Criterion: if two sides and the INCLUDED angle (the angle directly between those two sides) of one triangle equal the corresponding two sides and included angle of another triangle, the triangles are congruent. Given: AB = CB, ∠ABD = ∠CBD, BD = BD (common side). In triangles ABD and CBD: - AB = CB (given) - ∠ABD = ∠CBD (given) — and this angle is INCLUDED between sides AB/BC and BD in each triangle, exactly the arrangement SAS requires - BD = BD (common to both triangles) Since two sides and the included angle of triangle ABD equal the corresponding two sides and included angle of triangle CBD, by the SAS congruence criterion: △ABD ≅ △CBD

ASA Criterion: if two angles and the INCLUDED side (the side directly between those two angles) of one triangle equal the corresponding two angles and included side of another triangle, the triangles are congruent. Given: ∠Q = ∠Y, QR = YZ, ∠R = ∠Z. In triangles PQR and XYZ: - ∠Q = ∠Y (given) - QR = YZ (given) — and this side is INCLUDED between angles Q and R (correspondingly Y and Z) in each triangle - ∠R = ∠Z (given) Since two angles and the included side of triangle PQR equal the corresponding two angles and included side of triangle XYZ, by the ASA congruence criterion: △PQR ≅ △XYZ

RHS Criterion: if the hypotenuse and one leg of a right triangle equal the hypotenuse and corresponding leg of another right triangle, the two triangles are congruent. (This is a special case that works with only two sides given, because the right angle itself is already guaranteed to match.) Given: ∠B = ∠E = 90°, AC = DF (hypotenuses), AB = DE (one leg). In right triangles ABC and DEF: - ∠B = ∠E = 90° (both are right triangles, given) - AC = DF (hypotenuses equal, given) - AB = DE (one leg equal, given) Since the hypotenuse and one leg of triangle ABC equal the corresponding hypotenuse and leg of triangle DEF, by the RHS congruence criterion: △ABC ≅ △DEF (Note: the RHS criterion works even though only two pairs of sides are directly stated equal, because the right angle at B and E, both fixed at exactly 90°, is automatically a matching third condition — this is the one exception where AAS-style reasoning isn't needed to justify congruence from just two given measurements.)

Given: triangle ABC with AB = AC (an isosceles triangle). To prove: ∠B = ∠C (the base angles are equal) Proof: Step 1: Draw the bisector of ∠A, meeting BC at point D. Step 2: Consider triangles ABD and ACD: - AB = AC (given) - ∠BAD = ∠CAD (AD bisects ∠A, by construction) - AD = AD (common side) Step 3: By the SAS congruence criterion, △ABD ≅ △ACD. Step 4: Since the triangles are congruent, their corresponding angles are equal: ∠ABD = ∠ACD Which is the same as saying ∠B = ∠C. Hence proved: in an isosceles triangle, the angles opposite the equal sides are themselves equal.

The Angle Sum Property of a triangle states that the three interior angles of any triangle always add up to 180°. ∠A + ∠B + ∠C = 180° 50° + 70° + ∠C = 180° ∠C = 180° − 120° ∠C = 60° Verification: 50° + 70° + 60° = 180° ✓ The sum of all angles in any triangle is always 180°.

In an isosceles triangle, the altitude drawn from the vertex angle (here, P, the angle between the two equal sides) to the base always bisects the base — splitting it into two equal halves — and is also perpendicular to it. Given: PQ = PR = 13 cm, QR = 10 cm. Step 1: Since the altitude bisects QR, each half of the base measures: QR/2 = 10/2 = 5 cm Step 2: The altitude, half the base, and the equal side PQ (or PR) form a right-angled triangle, with PQ as the hypotenuse. Using the Pythagorean theorem: h² + 5² = 13² h² = 169 − 25 h² = 144 h = 12 cm The altitude from P to QR is 12 cm.

The Triangle Inequality: for any triangle with sides a, b, and c, the sum of any two sides must be strictly greater than the third side — if this fails for even one pair, the three lengths cannot form a closed triangle at all. Why this must be true: the shortest path between any two points is a straight line. If you tried to form a triangle where two sides summed to less than (or equal to) the third, the two shorter sides simply would not be able to stretch far enough to meet and close the triangle — they would either fall short or lie flat along the third side without forming a genuine enclosed shape. Checking 4 cm, 5 cm, and 10 cm: 4 + 5 = 9, which must be greater than 10 for a valid triangle. But 9 < 10 — the condition FAILS. Since the sum of the two shorter sides (4 cm and 5 cm) is less than the third side (10 cm), these three lengths CANNOT form a triangle.