Total 608. Let one side be a. The other is 343-a. So asquareplus 343-a the whole square. So you get a quadratic equation involving a. So it will have two roots. One of them is right.

One of the sides =16×6=96 Third side=256-9=247 Perimeter =265+247+96=608 If hypotenuse is 245=5×53 Other sides can be 5×28,5×45 140,225 But perimeter=140+225+265=630. As some body said the sides can be 264,23 Here 265+264=529=23^2 But perimeter is 552

A + B + C = 608 C = 265 A + B = 608 - 265 A + B = 343 ; A = 343 - B or B = 343 - A A^2 + B^2 = 265^2 Solutions : A = 96 ; B = 247 A = 247 ; B = 96 Checking Solutions : P = 608 = 96 + 247 + 265 ; 608 = 608 : True! 96^2 + 247^2 = 265^2 9.216 + 61.009 = 70.225 70.225 = 70.225 : True Sides equal to 96 and 247.

c = 265 c² = a² + b² 265² = a² + b² P = 608 = a + b + c 608 - c = a + b 608 - 265 = a + b 343 = a + b ============= (a+b)² = 343² a²+b²+2ab = 343² 2ab = 343² - (a²+b²) 2ab = 343² - 265² 2ab = 608*78 ab = 304*78 ============= Term : c > a > b c > a < b 304 = 2⁴*19 78 = 2*3*13 Let check : ab = (2⁴*19)(2*3*13) = 304*78 --> 304 > 265 (Invalid) ab = (2⁴*13)(2*3*19) = 208*114 --> 208 < 265 (Valid) ab = (2⁴*2*13)(3*19) = 416*57 --> 416 > 265 (Invalid) ab = (2⁴*2*3)(13*19) = 96*247 --> 247 < 265 (Valid) Other combination useless. The valid values : ab = 208*114 ab = 96*247 =============== Known : 265² = a² + b² 343 = a + b So let's validated : ab = 208*114 208² + 114² < 265² 208 + 114 < 343 (Unmatch) ab = 96*247 96² + 247² = 265² 96 + 247 = 343 (Match) So the true value of (a,b) is (96,247) or (247,96) 8:03

when you know a+b=m & ab=n ,you can get the value of (a,b) as the only two roots of equation x^2-mx+n=0. For my understanding, a+b=m & ab=n is another form of quadratic equation that is more often seen in practical math problems.

Yes, that's right.

a^2 + (608-265-a)^2 = 265^2 a^2 + (343-a)^2 = 265^2 a^2 + 117649 - 686a + a^2 = 70225 2a^2 - 686a + 47425 = 0 solving the quadratic gives a=96 and a=247 which turn out to be the two unknown sides

EXCELLENTONIO PROFESSOR!

65=343, b=343-a, a^2+b^2=265^2, a^2+(343-a)^2==70225, a^2+117649-686a+a^2-70225=0, 2a^2-686a+47424=0, a^2-343a+23712=0, a1=(343+\/22801)/2= (343+151)/2=96б, а2=(343-151)/2=96, а2=b=96.

The Hypotenuse is 365.perimeter is 608. So both sides put together is 343. If one is a then the other is 343-a। So the sum of the squares of 343आ एंड स्क्वायर of a which is a square is 365 squared। Hence you can get a then bis 343आ। सो थे आंसर

608H/A/ABSino° =2.178 H/A/ABSino°2^1 10^10 2^39 1^1.2^52^5 2^39^1 1^1^1^1 2^1^1 2^1 (H/A/Sino°AB ➖ 2H/A/Sino°AB+1)

φ = 30° → sin⁡(3φ) = 1; ∆ ABC → AB = c = 265; AC = a; BC = b; sin⁡(BCA) = 1 a + b + c = 608 = 265 + a + b → a + b = 343 → b = 343 - a → a^2 + b^2 = c^2 → a = 96 → b = 247 = pyth. triple (96 - 247 - 265)

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264^2+23^2=265^2.so perimeter=265+264+23=552. Check it. A different answer to

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Pythagorean triplets are represented by 2n,n^2-1,n^2+1herel last one is hypotenuse.put n=23 we get triplet 530,528,46.dividing by 2 we get triplet 265,264,23 which can be sides of right angled triangle.so perimeter will be 552. which is another answer.

3,4,5. Multiple side 265 53 times 5 so 3x53 and 4 x 53. Could not understand one word guy said

Geometric?no,no,no,it's algebraic.

For me a=151 and b= 192 151+192+265 = 608 . The problem has different solution!!!!

Again, truly I tell you, if two of you on earth agree about any matter that you pray for, it will be done for you by my Father in heaven.

a^2+b^2=603--265