Question

graingle ABC is right angled at A. P &Q are points on AB such that AP=PQ=QB. If 3CB^2+5CP^2=kCQ^2

Answer 1

Answer:

$\huge \mathfrak \blue{answer}$

We have,

CB² = CA² + BA²

= CA² + 3(AP)²

= CA² + 9AP²

CP² = CA² + AP²

Thus, 3CB² + 5CP²

= 3(CA² + 9AP²) + 5(CA² + AP²)

= 8CA²= 32AP²

= 8(CA² + 4AP²)

8(CA² + (2AP)²)

= 8(CA² + AQ²)

= 8CQ²

S

$\huge \mathfrak \purple{k = 8}$