Answer:
answer for Q1
step by step explaination:
12x+17x+25x=540
54x=540
x=540/54
x=10
12(10)=120
17(10)=170
25(10)=250
side=
a=120
b=170
c=250
s=(a+b+c)/2
s=540/2
s=270
now by using herons formula
A=√s(s - a)(s - b)(s - c)
A=√270(270-120)(270-170)(270-250)
A=√270 x 150 x 100 x 20
A=9000cm^2
Q2
a=18
b=10
c=?
perimeter=42
18+10+x=42
28+x=42
x=14
c=14
s=42/2
s=21
A=√21(21-18)(21-10)(21-14)
A= √21 x 3 x 11 x 7
A=21√11cm^2
thank you
[tex]\large\underline{\sf{Solution-1}}[/tex]
Let assume that sides of a triangle as: a = 12x, b = 17x, c = 25x
Further given that, Perimeter of a triangle = 540 cm
[tex]\sf\: 12x + 17x + 25x = 540 \\ [/tex]
[tex]\sf\: 54x = 540 \\ [/tex]
[tex]\implies\sf\:x = 10 \\ [/tex]
Thus, we have sides of a triangle as
[tex]\sf\: a = 12 \times 10 = 120 \\ [/tex]
[tex]\sf\: b = 17 \times 10 = 170 \\ [/tex]
[tex]\sf\: c = 25 \times 10 = 250 \\ [/tex]
Now,
[tex]\sf\: Semi-perimeter, s = \dfrac{Perimeter}{2} = \dfrac{540}{2} = 270 \: cm \\ [/tex]
[tex]\sf\: Area\:of\:triangle \\ [/tex]
[tex]\sf\: = \: \sqrt{s(s - a)(s - b)(s - c)} \\ [/tex]
[tex]\sf\: = \: \sqrt{270(270 - 120)(270 - 170)(270 - 250)} \\ [/tex]
[tex]\sf\: = \: \sqrt{270(150)(100)(20)} \\ [/tex]
[tex]\sf\: = \: 100\sqrt{(27)(15)(10)(2)} \\ [/tex]
[tex]\sf\: = \: 100\sqrt{(3 \times 3 \times 3)(5 \times 3)(2 \times 5)(2)} \\ [/tex]
[tex]\sf\: = \: 100 \times 3 \times 3 \times 5 \times 2 \\ [/tex]
[tex]\sf\: = \: 9000 \: {cm}^{2} \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\:Area\:of\:triangle \: = \: 9000 \: {cm}^{2} \: }\\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex]\large\underline{\sf{Solution-2}}[/tex]
Given that, Perimeter of a triangle is 42 cm and two sides of a triangle are 18 cm and 10 cm
Let assume that three sides of a triangle be represented as a, b and c such that a = 18 cm and b = 10 cm
[tex]\sf\: Perimeter = a + b + c \\ [/tex]
[tex]\sf\: 42 = 18 + 10 + c \\ [/tex]
[tex]\sf\: 42 = 28 + c \\ [/tex]
[tex]\implies\sf\:c = 14 \: cm \\ [/tex]
[tex]\sf\: Semi-perimeter, s = \dfrac{Perimeter}{2} = \dfrac{42}{2} = 21 \: cm \\ [/tex]
[tex]\sf\: = \: \sqrt{21(21 - 18)(21 - 10)(21 - 14)} \\ [/tex]
[tex]\sf\: = \: \sqrt{21(3)(11)(7)} \\ [/tex]
[tex]\sf\: = \: 21 \sqrt{11} \: {cm}^{2} \\ [/tex]
[tex]\implies\sf\:\boxed{\bf\:Area\:of\:triangle \: = \: 21 \sqrt{11} \: {cm}^{2} \: }\\ [/tex]
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Answers & Comments
Answer:
answer for Q1
step by step explaination:
12x+17x+25x=540
54x=540
x=540/54
x=10
12(10)=120
17(10)=170
25(10)=250
side=
a=120
b=170
c=250
s=(a+b+c)/2
s=540/2
s=270
now by using herons formula
A=√s(s - a)(s - b)(s - c)
A=√270(270-120)(270-170)(270-250)
A=√270 x 150 x 100 x 20
A=9000cm^2
Q2
a=18
b=10
c=?
perimeter=42
18+10+x=42
28+x=42
x=14
a=18
b=10
c=14
s=(a+b+c)/2
s=42/2
s=21
A=√s(s - a)(s - b)(s - c)
A=√21(21-18)(21-10)(21-14)
A= √21 x 3 x 11 x 7
A=21√11cm^2
thank you
Verified answer
[tex]\large\underline{\sf{Solution-1}}[/tex]
Let assume that sides of a triangle as: a = 12x, b = 17x, c = 25x
Further given that, Perimeter of a triangle = 540 cm
[tex]\sf\: 12x + 17x + 25x = 540 \\ [/tex]
[tex]\sf\: 54x = 540 \\ [/tex]
[tex]\implies\sf\:x = 10 \\ [/tex]
Thus, we have sides of a triangle as
[tex]\sf\: a = 12 \times 10 = 120 \\ [/tex]
[tex]\sf\: b = 17 \times 10 = 170 \\ [/tex]
[tex]\sf\: c = 25 \times 10 = 250 \\ [/tex]
Now,
[tex]\sf\: Semi-perimeter, s = \dfrac{Perimeter}{2} = \dfrac{540}{2} = 270 \: cm \\ [/tex]
Now,
[tex]\sf\: Area\:of\:triangle \\ [/tex]
[tex]\sf\: = \: \sqrt{s(s - a)(s - b)(s - c)} \\ [/tex]
[tex]\sf\: = \: \sqrt{270(270 - 120)(270 - 170)(270 - 250)} \\ [/tex]
[tex]\sf\: = \: \sqrt{270(150)(100)(20)} \\ [/tex]
[tex]\sf\: = \: 100\sqrt{(27)(15)(10)(2)} \\ [/tex]
[tex]\sf\: = \: 100\sqrt{(3 \times 3 \times 3)(5 \times 3)(2 \times 5)(2)} \\ [/tex]
[tex]\sf\: = \: 100 \times 3 \times 3 \times 5 \times 2 \\ [/tex]
[tex]\sf\: = \: 9000 \: {cm}^{2} \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\:Area\:of\:triangle \: = \: 9000 \: {cm}^{2} \: }\\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
[tex]\large\underline{\sf{Solution-2}}[/tex]
Given that, Perimeter of a triangle is 42 cm and two sides of a triangle are 18 cm and 10 cm
Let assume that three sides of a triangle be represented as a, b and c such that a = 18 cm and b = 10 cm
[tex]\sf\: Perimeter = a + b + c \\ [/tex]
[tex]\sf\: 42 = 18 + 10 + c \\ [/tex]
[tex]\sf\: 42 = 28 + c \\ [/tex]
[tex]\implies\sf\:c = 14 \: cm \\ [/tex]
Now,
[tex]\sf\: Semi-perimeter, s = \dfrac{Perimeter}{2} = \dfrac{42}{2} = 21 \: cm \\ [/tex]
Now,
[tex]\sf\: Area\:of\:triangle \\ [/tex]
[tex]\sf\: = \: \sqrt{s(s - a)(s - b)(s - c)} \\ [/tex]
[tex]\sf\: = \: \sqrt{21(21 - 18)(21 - 10)(21 - 14)} \\ [/tex]
[tex]\sf\: = \: \sqrt{21(3)(11)(7)} \\ [/tex]
[tex]\sf\: = \: 21 \sqrt{11} \: {cm}^{2} \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\:Area\:of\:triangle \: = \: 21 \sqrt{11} \: {cm}^{2} \: }\\ [/tex]