cosec a - 1+cot a /cot a+1-cosec a = 1 + cos a/ sin a
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Answer:
Given :-
The perimeter of an isosceles triangle is 42 cm.
The ratio of the equal side to its base is 3 : 4.
To Find :-
What is the area of the triangle.
Formula Used :-
\clubsuit♣ Area Of Triangle by Heron's Formula :
\begin{gathered}\footnotesize\mapsto \sf\boxed{\bold{\pink{Area_{(Triangle)} =\: \sqrt{s(s - a)(s - b)(s - c)}}}}\\\end{gathered}
↦
Area
(Triangle)
=
s(s−a)(s−b)(s−c)
where,
s = Semi-Perimeter
a = First Side Of Triangle
b = Second Side Of Triangle
c = Third Side Of Triangle
Solution :-
Let,
\mapsto \bf First\: Side_{(Triangle)} =\: 3a\: cm↦FirstSide
(Triangle)
=3acm
\mapsto \bf Second\: Side_{(Triangle)} =\: 4a\: cm↦SecondSide
(Triangle)
=4acm
\mapsto \bf Third\: Side_{(Triangle)} =\: 3a\: cm↦ThirdSide
(Triangle)
=3acm
As we know that :
\begin{gathered}\footnotesize\bigstar\: \: \sf\boxed{\bold{\pink{Perimeter_{(Triangle)} =\: Sum\: of\: all\: sides}}}\: \: \bigstar\\\end{gathered}
★
Perimeter
(Triangle)
=Sumofallsides
★
Given :
Perimeter of triangle = 42 cm
According to the question by using the formula we get,
\implies \sf 42 =\: 3a + 4a + 3a⟹42=3a+4a+3a
\implies \sf 42 =\: 10a⟹42=10a
\implies \sf \dfrac{42}{10} =\: a⟹
10
42
=a
\implies \sf 4.2 =\: a⟹4.2=a
\implies \sf\bold{\purple{a =\: 4.2\: cm}}⟹a=4.2cm
Hence, the required sides of a triangle are :
❒ First Side Of Triangle :
\implies \sf First\: Side_{(Triangle)} =\: 3a\: cm⟹FirstSide
(Triangle)
=3acm
\implies \sf First\: Side_{(Triangle)} =\: (3 \times 4.2)\: cm⟹FirstSide
(Triangle)
=(3×4.2)cm
\implies \sf\bold{\green{First\: Side_{(Triangle)} =\: 12.6\: cm}}⟹FirstSide
(Triangle)
=12.6cm
❒ Second Side Of Triangle :
\implies \sf Second\: Side_{(Triangle)} =\: 4a\: cm⟹SecondSide
(Triangle)
=4acm
\implies \sf Second\: Side_{(Triangle)} =\: (4 \times 4.2)\: cm⟹SecondSide
(Triangle)
=(4×4.2)cm
\implies \sf\bold{\green{Second\: Side_{(Triangle)} =\: 16.8\: cm}}⟹SecondSide
(Triangle)
=16.8cm
❒ Third Side Of Triangle :
\implies \sf Third\: Side_{(Triangle)} =\: 3a\: cm⟹ThirdSide
(Triangle)
=3acm
\implies \sf Third\: Side_{(Triangle)} =\: (3 \times 4.2)\: cm⟹ThirdSide
(Triangle)
=(3×4.2)cm
\implies \sf\bold{\green{Third\: Side_{(Triangle)} =\: 12.6\: cm}}⟹ThirdSide
(Triangle)
=12.6cm
Now, we have to find the semi-perimeter of a triangle :
As we know that :
\begin{gathered}\footnotesize\bigstar\: \: \sf\boxed{\bold{\pink{Semi-Perimeter =\: \dfrac{Sum\: of\: all\: sides}{2}}}}\: \: \bigstar\\\end{gathered}
★
Semi−Perimeter=
2
Sumofallsides
★
Given :
First Side (a) = 12.6 cm
Second Side (b) = 16.8 cm
Third Side (c) = 12.6 cm
According to the question by using the formula we get
\implies \bf Semi-Perimeter =\: \dfrac{a + b + c}{2}⟹Semi−Perimeter=
2
a+b+c
\implies \sf Semi-Perimeter =\: \dfrac{12.6 + 16.8 + 12.6}{2}⟹Semi−Perimeter=
2
12.6+16.8+12.6
\implies \sf Semi-Perimeter =\: \dfrac{\cancel{42}}{\cancel{2}}⟹Semi−Perimeter=
2
42
\implies \sf\bold{\purple{Semi-Perimeter =\: 21\: cm}}⟹Semi−Perimeter=21cm
Now, we have to find the area of the triangle by using the Heron's Formula :
Given :
Semi-Perimeter (s) = 21 cm
First Side (a) = 12.6 cm
Second Side (b) = 16.8 cm
Third Side (c) = 12.6 cm
According to the question by using the formula we get,
\small\longrightarrow \sf Area_{(Triangle)} =\: \sqrt{21(21 - 12.6)(21 - 16.8)(21 - 12.6)}⟶Area
(Triangle)
=
21(21−12.6)(21−16.8)(21−12.6)
\small\longrightarrow \sf Area_{(Triangle)} =\: \sqrt{21(8.4)(4.2)(8.4)}⟶Area
(Triangle)
=
21(8.4)(4.2)(8.4)
\small\longrightarrow \sf Area_{(Triangle)} =\: \sqrt{21 \times 8.4 \times 4.2 \times 8.4}⟶Area
(Triangle)
=
21×8.4×4.2×8.4
\small\longrightarrow \sf Area_{(Triangle)} =\: \sqrt{6223.392}⟶Area
(Triangle)
=
6223.392
\small\longrightarrow \sf\bold{\red{Area_{(Triangle)} =\: 78.898\: cm^2}}⟶Area
(Triangle)
=78.898cm
2
{\small{\bold{\underline{\therefore\: The\: area\: of\: the\: triangle\: is\: 78.898\: cm^2\: .}}}}
∴Theareaofthetriangleis78.898cm
2
.