From the figure we have FG―=EF― and ∠GHF=90° ⇒∠EHF=90° also GH―=93 units Since △GHF is right angle triangle, hence GH―2+HF―2=GF―2 932+HF―2=GF―2 GF―2−HF―2=932-(1) Also since △FHE is right angle triangle, hence FH―2+HE―2=EF―2 HE―2=EF―2−FH―2 =GF―2−HF―2 =932, using (1) ∴HF―=93 Therefore GE―=GH―+HE― =93+93 =186 ∴GE―=186

"What is EG?" solved and explained

Sheelmgal1p

Answered question

2021-11-26

What is EG?

Answer & Explanation

Ida Perry

Beginner2021-11-27Added 20 answers

From the figure we have

$\overset{―}{F G} = \overset{―}{E F}$
and $∠ G H F = 90 °$
$\Rightarrow ∠ E H F = 90 °$
also $\overset{―}{G H} = 93$ units
Since $△ G H F$ is right angle triangle, hence
${\overset{―}{G H}}^{2} + {\overset{―}{H F}}^{2} = {\overset{―}{G F}}^{2}$
$93^{2} + {\overset{―}{H F}}^{2} = {\overset{―}{G F}}^{2}$
${\overset{―}{G F}}^{2} - {\overset{―}{H F}}^{2} = 93^{2}$ -(1)
Also since $△ F H E$ is right angle triangle, hence
${\overset{―}{F H}}^{2} + {\overset{―}{H E}}^{2} = {\overset{―}{E F}}^{2}$
${\overset{―}{H E}}^{2} = {\overset{―}{E F}}^{2} - {\overset{―}{F H}}^{2}$
$= {\overset{―}{G F}}^{2} - {\overset{―}{H F}}^{2}$
$= 93^{2}$ , using (1)
$∴ \overset{―}{H F} = 93$
Therefore $\overset{―}{G E} = \overset{―}{G H} + \overset{―}{H E}$
$= 93 + 93$
$= 186$
$∴ \overset{―}{G E} = 186$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in High school geometry

Didn't find what you were looking for?