Answer:
Let's denote the length of the two equal sides of the isosceles triangle as x cm each and the length of the third side as y cm.
According to the problem we are given that each of the equal sides is twice as large as the third side. This can be expressed as:
x = 2y
The perimeter of a triangle is the sum of the lengths of its sides. In our case the perimeter is given as 30 cm. So we can write the equation:
x + x + y = 30
Substituting the value of x from the first equation we get:
2y + 2y + y = 30
Combining like terms we obtain:
5y = 30
Dividing both sides of the equation by 5 we find:
y = 6
Substituting this value back into the first equation we get:
x = 2y = 2 x 6 = 12
Therefore the length of each equal side of the isosceles triangle is 12 cm and the length of the third side is 6 cm.
[tex]\boxed{\begin{aligned}& \:\sf \: Length\:of\:third\:side = 6 \: cm \qquad \: \\ \\&\:\sf \: Length\:of\:equal\:side = 12 \: cm \end{aligned}} \\ [/tex]
Step-by-step explanation:
Given that, each of the two equal sides of an isosceles triangle is twice as large as the third side.
Let assume that length of third be x cm
So, length of equal side is 2x cm
Further given that, the perimeter of the triangle is 30 cm.
We know, Perimeter of a triangle means sum of its three sides.
[tex]\sf \: x + 2x + 2x = 30 \\ [/tex]
[tex]\sf \: 5x = 30 \\ [/tex]
[tex]\implies \sf \:x = 6 \\ [/tex]
Hence,
[tex]\implies \sf \:\boxed{\begin{aligned}& \:\sf \: Length\:of\:third\:side = x = 6 \: cm \qquad \: \\ \\&\:\sf \: Length\:of\:equal\:side = 2x = 2 \times 6 = 12 \: cm \end{aligned}} \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \large\underline{\sf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}[/tex]
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Answers & Comments
Answer:
Let's denote the length of the two equal sides of the isosceles triangle as x cm each and the length of the third side as y cm.
According to the problem we are given that each of the equal sides is twice as large as the third side. This can be expressed as:
x = 2y
The perimeter of a triangle is the sum of the lengths of its sides. In our case the perimeter is given as 30 cm. So we can write the equation:
x + x + y = 30
Substituting the value of x from the first equation we get:
2y + 2y + y = 30
Combining like terms we obtain:
5y = 30
Dividing both sides of the equation by 5 we find:
y = 6
Substituting this value back into the first equation we get:
x = 2y = 2 x 6 = 12
Therefore the length of each equal side of the isosceles triangle is 12 cm and the length of the third side is 6 cm.
Verified answer
Answer:
[tex]\boxed{\begin{aligned}& \:\sf \: Length\:of\:third\:side = 6 \: cm \qquad \: \\ \\&\:\sf \: Length\:of\:equal\:side = 12 \: cm \end{aligned}} \\ [/tex]
Step-by-step explanation:
Given that, each of the two equal sides of an isosceles triangle is twice as large as the third side.
Let assume that length of third be x cm
So, length of equal side is 2x cm
Further given that, the perimeter of the triangle is 30 cm.
We know, Perimeter of a triangle means sum of its three sides.
[tex]\sf \: x + 2x + 2x = 30 \\ [/tex]
[tex]\sf \: 5x = 30 \\ [/tex]
[tex]\implies \sf \:x = 6 \\ [/tex]
Hence,
[tex]\implies \sf \:\boxed{\begin{aligned}& \:\sf \: Length\:of\:third\:side = x = 6 \: cm \qquad \: \\ \\&\:\sf \: Length\:of\:equal\:side = 2x = 2 \times 6 = 12 \: cm \end{aligned}} \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information
[tex]\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \large\underline{\sf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}[/tex]