Question

prove that p=1/3+√7+1/√7+√5+1/√5+√3+1/√3+1 =1
Plz Answer fast..
I will mark u as brainliest

Answer 1

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

$\tt A\: 10 \:m\: long\: ladder\: is\: placed\: against \:a \:wall.\: The\\\tt foot\: of \:the\: ladder\: is\: 6\: m\: from\: the\: wall. \:How\: far\\\tt up \:the\: wall \:does\: the \:top\: of\: the \:ladder\: reach?$

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

$\sf \bf {\boxed {\mathbb {GIVEN}}}$

$\sf \bf Length \:of \:the \:ladder =10\:m$

$\sf \bf Foot \:of \:the \:ladder =6\:m$

$\sf \bf {\boxed {\mathbb {TO\:FIND}}}$

$\sf \bf How\: much\: far\: does\: the\: end\: of\: the\: ladder\: reaches\: the\: wall$

$\sf \bf {\boxed {\mathbb {SOLUTION}}}$

$\sf \bf In\: \triangle ABC, \angle B ={90}^{\circ}$

${\color {gold}\underline {\sf By pythagoras theorem}}$

${\boxed {\boxed{\boxed {\color {blue} {\sf \bf {AC}^{2}={AB}^{2}+{BC}^{2}}}}}}$

${\underbrace {\overbrace {\color {orange} {\bf Substitute \:the \:values}}}}$

$\sf \bf \implies {\Big(10\Big)}^{2}={\Big(AB\Big)}^{2}+{\Big(6\Big)}^{2}$

$\sf \bf \implies 100={AB}^{2}-36$

$\sf \bf \implies {AB}^{2}=100-36$

$\sf \bf \implies {AB}^{2}=64$

$\sf \bf \implies AB=\sqrt{64}$

$\implies{\boxed {\boxed {\color {aqua} {\sf \bf AB=8 \:m}}}}$

${\underbrace {\color {red} {\underline{\overline {\color {red} {\sf \therefore The\:end\:of \:the \:ladder \:reached \:8\:m\:of \:the \:wall}}}}}}$

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

__________________________________________

$\sf \bf \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

${\underbrace {\color {brown} {\sf Pythagoras \:theorem \:statement}}}$

$\tt \bf In\: right \:angled \:triangle\: square\: on\: the\: hypotenuse\\\tt \bf is\: equal\: to\: the\: sum\: of\: the\: squares\: of\:other \\\tt\bf two\: sides\: of \:the\: triangle$

Answer 2

}}}

SOLUTION

\sf \bf In\: \triangle ABC, \angle B ={90}^{\circ}In△ABC,∠B=90

∘

{\color {gold}\underline {\sf By pythagoras theorem}}

Bypythagorastheorem

{\boxed {\boxed{\boxed {\color {blue} {\sf \bf {AC}^{2}={AB}^{2}+{BC}^{2}}}}}}

AC

2

=AB

2

+BC

2

{\underbrace {\overbrace {\color {orange} {\bf Substitute \:the \:values}}}}

Substitutethevalues

⟹(10)

2

=(AB)

2

+(6)

2

100={AB}^{2}-36⟹100=AB

2

−36

{AB}^{2}=100-36⟹AB

2

=100−36

=64⟹AB

2

=64

⟹AB=

64

AB=8m

∴Theendoftheladderreached8mofthewall