Question

The area of the triangle ABC is 72sq. cm and AD= 3cm is the height from A to BC. Find the length of BC​

Answer 1

Answer:

see explanation for answer

Step-by-step explanation:

AD = 3cm

tan 60 = P / B

root 3 = 3 / B

B root 3 = 3

B = 3 / root3

so rationalizing it

B = 3 * root3 / root3* root3

B = 3root3 / 3

cancelling 3 and  3

we get

B = root 3 ( B = DC)

we get ,

BC = BD + DC

BC = root3 + root 3

BC = 2root3

Answer 2

$\Huge\rm\orange{answeR:}$

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

• $\sf{Area~of~∆ABC = 72cm^{2}}$
• $\sf{AD = 3cm}$

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

• $\sf{The~length~BC}$

$\bf{{\underline{Solution}}:}$

$\sf Area \: of \: a \: triangle = \dfrac{1}{2} \times base \times height \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \implies \sf {72cm}^{2} = \frac{1}{2} \times BC \times AD \\ \implies \sf {72cm}^{2} = \frac{1}{2} \times BC \times AD \\ \implies \sf \frac{1}{2} \times BC \times 3 = {72cm}^{2} \: \: \: \: \\ \implies \sf \frac{3}{2} \times BC = 72 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \implies \sf BC = \cancel{72} \: \: ^{24} \times \frac{2}{ \cancel{3} \: ^{1} } \: \: \: \: \: \: \: \: \: \: \\ \implies \sf BC = 24 \times 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \implies \sf BC = \orange{ \underline{ \boxed{ \bf 48cm}}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\bf\therefore{{\underline{Required~answer}}:}$

• $\sf{BC = \orange{ \underline{ \boxed{ \bf 48cm}}} }$