Question

angle ABC is right angled at B .If AB=5cm and BC =12 cm find its semi perimeter. solve this it's urgent​

Answer 1

Given: Triangle ABC is right angled at B.

AB=5cm and BC =12 cm

To find: The semi perimeter of the triangle

Solution: We can understand from the question that AB is the perpendicular and BC is the base.

Using Pythagoras' theorem,

(perpendicular)² + (base)² = (hypotenuse)²

⇒ AC² = AB² + BC²

⇒ AC² = 5² + 12²

⇒ AC² = 25 + 144

⇒ AC² = 169

⇒ AC = 13 cm

Now, the semi perimeter of a triangle is given by the formula :-

s = (a + b + c)/2  [where a, b and c are the three sides of the triangle]

Hence, the semi perimeter of the triangle ABC

= (5 + 12 + 13)/2

= 30/2

= 15 cm.

Answer: 15 cm

Answer 2

Answer:

Given :

• ABC is right angled at B.
• AB = 5 cm

BC = 12 cm

To Find :

• AC = ?
• Semi perimeter

Solution :

We have the measurement for the sides AB and BC so we will find the measurement of side AC by using the Hypotenuse theorem which is :

$\star{\boxed{ \rm{{Hypotenuse}^{2} = {Perpendicular}^{2} + {Base}^{2}}}}$

Here, we can say that the :

• Perpendicular = AB
• Base = BC
• Hypotenuse = AC

Substituting all the given values in the heron's formula to find AC : (H)² = (P)² + (B)²

$\begin{gathered}\quad{\implies{{(H)}^{2} = {(P)}^{2} + {(B)}^{2}}}\\\\\quad{\implies{{(AC)}^{2} = {(AB)}^{2} + {(BC)}^{2}}}\\\\\quad{\implies{{(AC)}^{2} = {(5)}^{2} + {(12)}^{2}}}\\\\\quad{\implies{{(AC)}^{2} = (5 \times 5) + (12 \times 12)}}\\\\\quad{\implies{{(AC)}^{2} = 25 + 144}}\\\\\quad{\implies{{(AC)}^{2} = 169}}\\\\\quad{\implies{(AC) = \sqrt{169}}}\\\\\quad{\implies{(AC) = \sqrt{13 \times 13} }}\\\\\quad{\implies{(AC) = 13 \: cm}}\end{gathered}$

Hence, the measurement of AC is 13 cm.

$\rule{190}1$

Now, we need to find the semi perimeter. The formula of semi perimeter is :

$\star{\boxed{\rm{ S = \dfrac{a + b + c}{2}}}}$

Substituting all the given values in the formula to find semi perimeter : S = (a+b+c)/2

$\begin{gathered} \quad{\implies{Semi_{(Perimeter)} = \dfrac{a + b + c}{2}}} \\ \\ \quad{\implies{Semi_{(Perimeter)} = \dfrac{5 + 12 + 13}{2}}} \\ \\ \quad{\implies{Semi_{(Perimeter)} = \dfrac{30}{2}}} \\ \\ \quad{\implies{Semi_{(Perimeter)} = \cancel{\dfrac{30}{2}}}} \\ \\ \quad{\implies{Semi_{(Perimeter)} = 15 \: cm}}\end{gathered}$

Hence, the semi perimeter is 15 cm.

$\rule{190}1$

Learn More :

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

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