Question

For a right triangle if a = 11in , c = 15 in find b Quickkk please help

Answer 1

The answer to this problem is 2√26

15^2-11^2 = 104
sq root 104

Answer 2

Answer:

$\huge\boxed{ \boxed{ \sf{2 \sqrt{26} }}}$

Step-by-step explanation:

to understand this

you need to know about:

• Pythagoras theorem
• PEMDAS

tips and formulas:

• $\sf Pythagoras\: theorem:\\ \sf a²+b²=c²$

given:

• a=11 in
• c=15 in

let's solve:

$\sf step - 1 : make \: b \: the \: subject \: of \: the \: equation$

1. $\sf \: shift \: {a}^{2} \: to \: the\: left \: hand \: side \: and \: change \: its \: sign : \\ \tt {b}^{2} = {c}^{2} - {a}^{2}$
2. $\sf squre \: root \: both \: sides : \\ \tt\sqrt{ {b}^{2} } = \sqrt{ {c}^{2} - {a}^{2} } \\ \tt \: b = \sqrt{ {c}^{2} - {a}^{2} }$

$\sf step - 2 : sustitute \:the \: value \: of \: c \: and \: a \: then \: simplify$

1. $\sf substitute \: the \: given \: value \: of \: c \: and \: b : \\ \tt b = \sqrt{ {15}^{2} - {11}^{2} }$
2. $\sf simplify \: squres : \\ \tt b = \sqrt{225 - 121}$
3. $\sf substract : \\ \tt b = \sqrt{104}$
4. $\sf simplify \: redica l : \\ \tt b = \sqrt{ {2}^{2} \times 26 } \\ \tt \therefore b = 2 \sqrt{26}$