1. Find the area of an equilateral triangle whose perimeter is 24 cm.
2. The length of hypotenuse of a right angled triangle is 13 and the length of another side is 5 cm. Find its area.
Share
1. Find the area of an equilateral triangle whose perimeter is 24 cm.
2. The length of hypotenuse of a right angled triangle is 13 and the length of another side is 5 cm. Find its area.
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Answer:
>>> 1) 16√3 cm²
>>> 2) 30 cm²
Step-by-step explanation:
#Question 1 :-
Given that ,
We have to find ,
Solution :
Finding Length of equilateral triangle using its Perimeter.
[tex] \: \: \: \boxed{\sf{Perimeter_{(Equilateral \: Triangle)} = 3 \times a }}[/tex]
Where ,
a refers to length/side of triangle
[tex] \longmapsto \: \: \: \sf{3 \times a = 24}[/tex]
Dividing both sides with 3 :
[tex]\longmapsto \: \: \: \sf{ \dfrac{( \cancel{3} \times a )}{ \cancel{3}}= \cancel{\dfrac{ 24}{3}}} [/tex]
We get ,
[tex]\longmapsto \: \: \: \sf{ \bold{a = 8 \: cm}}[/tex]
Now , Finding area of equilateral triangle :
[tex] \: \: \: \: \: \: \: \boxed{\sf{Area_{(Equilateral \: Triangle)} = \dfrac{ \sqrt{3} \: a {}^{2} }{4} }}[/tex]
Where ,
[tex] \dashrightarrow \: \: \: \sf{ \dfrac{ \sqrt{3} \: \times {8}^{2} }{4}}[/tex]
Simplifying :
[tex]\dashrightarrow \: \: \: \sf{ \dfrac{ \sqrt{3} \: \times \cancel{64}}{ \cancel{4}}}[/tex]
[tex]\dashrightarrow \: \: \: \underline{ \boxed{ \sf{ \bold{16 \sqrt{3} \: cm {}^{2} }}}} \: \: \: \bigstar[/tex]
#Question 2 :-
Given Right Angle Triangle ,
We have to find ,
Solution :
Let us say that ,
For finding area of triangle we need to find the length of its third side using Pythagoras Theorem :
[tex] \longmapsto \: \: \: \sf{x {}^{2} + {5}^{2} = 13 {}^{2} }[/tex]
[tex]\longmapsto \: \: \: \sf{x {}^{2} + 25 = 169 }[/tex]
Transposing 25 to RHS :
[tex]\longmapsto \: \: \: \sf{x {}^{2} = 169 - 25 }[/tex]
[tex]\longmapsto \: \: \: \sf{x {}^{2} = 144 }[/tex]
[tex]\longmapsto \: \: \: \sf{x = \sqrt{144} }[/tex]
[tex]\longmapsto \: \: \: \sf{ \bold{x = 12 \: cm }}[/tex]
Now, finding area of right angled triangle :
[tex] \: \: \boxed{\sf{Area_{(Right \: Angled \: Triangle)} = \dfrac{1}{2} \times (ab)}}[/tex]
Where ,
[tex] \dashrightarrow \: \: \: \sf{ \dfrac{1}{ \cancel{2}} \times ( \cancel{12 }\times 5) }[/tex]
[tex] \dashrightarrow \: \: \: \sf{6 \times 5}[/tex]
We get ,
[tex]\dashrightarrow \: \: \: \underline{ \boxed{ \sf{ \bold{30 \: cm {}^{2} }}}} \: \: \: \bigstar[/tex]