Question

InABC, AD perpendicular to BC and BE perpendicular to AC. If BC = 16 cm, AD = 6 cm and BE = 8 cm, then find AC.​

Answer 1

Answer:

10cm

Step-by-step explanation:

Using pythagoras theorem:

16cm can be divided as 8cm and 8 cm

And hence AC would be:

$8^{2}$+$6^{2}$ = AC

64 + 36 = 100

Hence, AC = $\sqrt{100}$= 10 cm

Answer 2

Answer:

$\large{ \sf{ \underline{ \underline \color{blue}{☆ Question}}}}$

In ∆ ABC, AD

$\perp$ BC and BE $\perp$ AC. If BC = 16 cm, AD = 6 cm and BE = 8 cm, then find AC.

$\large{ \sf{ \underline{ \underline \pink{☆Answer}}}}$

$\large{ \bf{ \underline{To \: find :}}}$

Value of side AC.

$\large{ \bf{ \underline{Given :}}}$

$\bf \: AD \perp BC$

$\bf \: BE \perp AC$

$\bf \: BC = 16 cm$

$\bf \: AD = 6 cm$

$\bf \: BE = 8 cm$

$\large{ \bf{ \underline{Solⁿ :}}}$

${ \boxed{ \sf{ \dag{ \: Area \: of \: triangle = \frac{1}{2} \times b \times h}}}}$

Area of this triangle

$\sf \frac{1}{2} \times AD \times BC = \frac{1}{2} \times BE \times AC$

$\sf \frac{1}{2} \times 6cm \times 16cm = \frac{1}{2} \times 8cm \times AC$

$\sf 48cm {}^{2} = \frac{1}{2} \times 8cm \times AC$

$\sf \: AC = \frac{48cm {}^{2} }{4cm}$

$\sf ∴ AC = 12cm$

$\sf \: Hope \: it \: helps \: you...$

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