This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Question 1 (Pythagoras theorem) Step 1: Identify the right triangle PQR right-angled at Q, with PQ = 7 cm and QR = 5 cm. x = PR. By Pythagoras theorem: PR^2 = PQ^2 + QR^2 Step 2: Substitute the values. x^2 = 7^2 + 5^2 x^2 = 49 + 25 Step 3: Simplify. x^2 = 74 x = sqrt(74) sqrt(74) cm (since length is positive). sqrt(74) Question 3 (Tangents to a circle) Given: PA = PB = 10 cm (tangents from P), APB = 60^, O is centre. Find AO = r cm. Step 1: OA PA and OB PB (radius tangent). OP bisects APB, so OPA = OPB = 30^. Step 2: Consider right OAP (right-angled at A). ( OPA) = (OA)/(PA) 30^ = (r)/(10) Step 3: Substitute 30^ = (1)/(sqrt(3)). (1)/(sqrt(3)) = (r)/(10) Step 4: Solve for r. r = (10)/(sqrt(3)) Rationalize denominator: r = (10)/(sqrt(3)) × sqrt(3)sqrt(3) = 10sqrt(3)3 10sqrt(3)3 Question 5 (Quadratic equation) Solve x^2 - 5x + 6 = 0. Step 1: Factorize. Find two numbers that multiply to 6 and add to -5: -2 and -3. x^2 - 5x + 6 = (x - 2)(x - 3) = 0 Step 2: Solve. x - 2 = 0 or x - 3 = 0 x = 2 or x = 3 2,\ 3