Question

∆ ABC is a right-angled triangle at B. If AB = 3cm and AC = 5cm, calculate BC.​

Answer 1

Answer:

4cm

Step-by-step explanation:

by pythagoras theorem

AC² = BC²+AB²

5² = BC²+ 3²

25-9 = BC²

√16 = BC

BC = 4

Answer 2

Answer:

Given :

∆ABC is a right-angled triangle.

AB = 3cm and AC = 5cm

To Find :

BC in right angled triangle

Solution :

Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”. 

$\longrightarrow{\sf{{(AC)}^{2} = {(AB)}^{2} + {(BC)}^{2}}}$

In this question, we have provided that :

• ➛ AB = 3 cm
• ➛ AC = 5 cm
• ➛ BC = ?

Substituting all the given values in the formula to find the value of BC in right angled triangle :

$\begin{gathered}\qquad\longrightarrow{\sf{{(AC)}^{2} = {(AB)}^{2} + {(BC)}^{2}}} \\ \\ \qquad\longrightarrow{\sf{{(5)}^{2} = {(3)}^{2} + {(BC)}^{2}}} \\ \\ \qquad\longrightarrow{\sf{(5 \times 5) = (3 \times 3) + {(BC)^{2} }}} \\ \\ \qquad\longrightarrow{\sf{25 = 9 + {(BC)^{2}}}} \\ \\ \qquad\longrightarrow{\sf{{(BC)^{2} = 25 - 9}}} \\ \\ \qquad\longrightarrow{\sf{{(BC)^{2} = 16}}} \\ \\ \qquad\longrightarrow{\sf{{(BC) = \sqrt{16}}}} \\ \\ \qquad\longrightarrow{\sf{{(BC) = \sqrt{2 \times 2 \times 2 \times 2}}}} \\ \\ \qquad\longrightarrow{\sf{{(BC) = 2 \times 2}}} \\ \\ \qquad\longrightarrow{\sf{\underline{\underline{\red{{(BC) =4 \: cm}}}}}}\end{gathered}$

Hence, the value of BC in right angled triangle is 4 cm.

$\rule{300}{1.5}$

Learn More :

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Triangle:-}\\ \\ \star\sf Triangle \:area = \dfrac{1}{2}\times b \times h\\ \\ \star\sf Triangle \: perimeter=a+b+c\\\\ \star\sf Scalene\:\triangle=\sqrt{s (s-a)(s-b)(s-c)}\\\\\star\sf Equilateral\: \triangle\:area = \dfrac{\sqrt{3}}{4}\times{side}^{2}\\\\\star\sf Equilateral \:\triangle\:perimeter = 3 \times side\\\\\star\sf Isosceles\: \triangle\:area= \dfrac{3}{4}\sqrt{{4b}^{2}-{a}^{2}}\\\\\star\sf Isosceles\:\triangle\:perimeter=a+2b\end{minipage}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

$\rule{220pt}{4pt}$