# Chapter 1 - Section 1.1 - Angles, Degrees, and Special Triangles - 1.1 Problem Set: 43

AB = $\sqrt 41 \approx 6.40$

#### Work Step by Step

In given figure 23, we can see two right triangles BCD and BCA, both right angled at C. Given- BC = 4, BD = 5 , AD = 2 In right triangle BCD, applying Pythagoras theorem- $BD^{2}$ = $BC^{2}$ + $DC^{2}$ Therefore- $DC^{2}$ = $BD^{2}$- $BC^{2}$ $DC^{2}$ = $5^{2}$- $4^{2}$ = 25-16=9 $DC = \sqrt 9$ = 3 Now we can calculate AC as- AC = AD + DC = 2 + 3 = 5 Now applying Pythagoras theorem in right triangle BCD- $AB^{2}$ = $BC^{2}$ + $AC^{2}$ $AB^{2}$ = $4^{2}$ + $5^{2}$ $AB^{2}$ = 16 + 25 = 41 Therefore AB = $\sqrt 41 \approx 6.40$

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