In ΔABC, ∠135°, CA = $5\sqrt{2}$ cm and AB = 7 cm. E and F are midpoints of sides AC and AB, respectively. The length of EF (in cm) is :

6.5

By using the cosine rule :-

cos 135º = \(\frac{ 7² + (5√2)² - BC²}{2 × 7 ×5√2 }\)

\(\frac{ -1}{√2 }\) = \(\frac{ 49 + 50 - BC²}{2 × 7 ×5√2 }\)

- 70 = 99 - BC²

BC² = 169

= ± 13

So, BC = 13 cm

By using mid-point theorem,

EF = \(\frac{ BC}{2 }\)

= \(\frac{ 13}{2 }\)

= 6.5 cm