Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
Answer:
According to the Heron's formula, Area (A) of the triangle having sides a,b,c units is
A=
s(s−a)(s−b)(s−c)
where
s=
2
a+b+c
For the given triangle,
a=18 cm
b=24 cm
c=30 cm
s=
2
18+24+30
=36
A=
36(36−18)(36−24)(36−30)
A=
36×18×12×6
A=
216×216
=216 cm
2
Smallest side =18 cm
Area of the triangle =
2
1
×base×altitude=216
2
1
×18× altitude=216
Altitude =
9
216
=24 cm
Step-by-step explanation:
hope this will help
Answer:
[tex]\boxed{\boxed{\pink{\bf \leadsto The \ length \ of \ shortest \ altitude \ is \ 2.4\sqrt{6} \ units . }}}[/tex]
Step-by-step explanation:
Here given measure of sides are 18 , 24 and 30 units. Firstly let's find the area of ∆ using Heron's Formula.
[tex]\boxed{\red{\bf Area_{\triangle} =\sqrt{s(s-a)(s-b)(s-c)}}}[/tex]
Where s is semi Perimeter . And here s will be ( 18 + 24 + 30 ) / 2 = 36 units .
[tex]\bf \implies Area = \sqrt{s(s-a)(s-b)(s-c)} [/tex]
[tex]\bf \implies Area = \sqrt{ 36 ( 36 - 18)(36-24)(36-30)}[/tex]
[tex]\bf \implies Area = \sqrt{ 36 \times 18 \times 12} [/tex]
[tex]\bf \implies Area = \sqrt{ 6^2 \times 6\times 6 \times 2 \times 3} [/tex]
[tex]\bf \implies Area = 6^2\sqrt{6} unit^2 [/tex]
[tex]\bf \boxed{ \implies Area_{triangle} = 36\sqrt{6} units^2} [/tex]
Also we know that ,
[tex]\boxed{\red{\bf Area_{\triangle} = \dfrac{1}{2}\times (base)\times (height)}}[/tex]
Let's find the altitudes now ,
Altitude on side of 18 units :-
[tex]\bf \implies Area = \dfrac{1}{2} \times (base)(height) \\\\\bf \implies 36\sqrt{6} unit^2 = \dfrac{1}{2} \times 18 \times h_1 = 36\sqrt{6} u^2 \\\\\bf\implies h_1 =\dfrac{ 36\sqrt6 \times 2 }{18} \\\\\boxed{\bf\implies h_1 = 4\sqrt6 units }[/tex]
Altitude on side of 24 units :-
[tex]\bf \implies Area = \dfrac{1}{2} \times (base)(height) \\\\\bf \implies 36\sqrt{6} unit^2 = \dfrac{1}{2} \times 24 \times h_1 = 36\sqrt{6} u^2 \\\\\bf\implies h_1 =\dfrac{ 36\sqrt6 \times 2 }{24} \\\\\boxed{\bf\implies h_1 = 3\sqrt6 units }[/tex]
Altitude on side 30 units :-
[tex]\bf \implies Area = \dfrac{1}{2} \times (base)(height) \\\\\bf \implies 36\sqrt{6} unit^2 = \dfrac{1}{2} \times 30 \times h_1 = 36\sqrt{6} u^2 \\\\\bf\implies h_1 =\dfrac{ 36\sqrt6 \times 2 }{30} \\\\\boxed{\bf\implies h_1 = 2.4\sqrt6 units }[/tex]
Hence the lenght of shortest altitude is 2.4√6 units and its on the side of 30 units.
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
