Question

Altitude on the hypotenuse of a right angled triangle divides it in two parts of Length 16 cm and 9 cm Find length of the altitude​

Answer 1

$\large\orange \bigstar \: \red{ \underline{ \underline{ \green{ \rm given : }}}}$

$\: \: \: \: \: \: \: \: \: \: \: \rightarrow \: \purple{1 \rm side \: of \: (h) = 16cm}$

$\: \: \: \: \: \: \: \: \: \: \: \rightarrow \: \purple{ \rm 2side \: of(h) = 9cm}$

$\large\blue \bigstar \: \orange{ \underline{ \underline{ \pink{ \rm solution : }}}}$

$\bigstar$ ABC right angle triangle is drawn in such a way that ∠ABC = 90° .

• D is the point on side AC where BD ⊥ AC .

$\rightarrow$ Also D divides the side length AC in two parts AD and DC of lengths 9cm and 16 cm respectively.

$\large \: \green{ \underline{ \large{ \red{ \rm Now, }}}}$ from ∆ABD and ∆BDC

$\rightarrow$∠ADB = ∠BDC = 90°

$\rightarrow$∠BAD = ∠DBC ( see figure , if we assume ∠DBC = x°

$\large \: \orange{ \underline{ \large{ \purple{ \rm then, }}}}$ DBA = 90 - x° then from ∆ABD , ∠BAD = x° )

$\bullet \: \: \green{ \boxed{ \red{from \: \: A - A \: \: similarity \: \: rule , </p><p>}}}$

∆DAB ~ ∆DBC

so,

$\rightarrow$BD/DC = AD/BD

⇒BD² = DC × AD

= 16cm × 9 cm

• taking square root both sides,

$\implies \: \sqrt{16} \: \times \: \sqrt{9}$

$\implies \: 4 \times 3$

$\therefore \rm \large \boxed{BD = 12cm}$

Hence ,

$\large \: \bigstar \: \red{ \boxed{ \pink{ \boxed{ \blue{ \underline{ \green{altitude = 12cm }}}}}}}$

Answer 2

Answer:

BD = 12 cm

Step-by-step explanation:

altitude = 12 cm