Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
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 hope this was helpful. Please come back whenever you need more information or answers to your queries. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
