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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.
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