Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The perimeter of the triangle is
Answers
if the equal sides of an isosceles triangle is 13 then
13cm+13cm+24cm=perimeter of the triangle
13+13+24=?
43+24
67cm
67cm is your answer
Step-by-step explanation:
Height of isosceles triangle
Now 1st find the area if triangle
By heron's formula
\sf\underline{\underline{\red{According\:to\: Question:-}}}
AccordingtoQuestion:−
\sf{\pink{\sqrt{s(s-a)(s-b)(s-c)}}}
s(s−a)(s−b)(s−c)
\begin{gathered}\tt s=semi\:perimeter\\ \\ \tt s=\frac{a+b+c}{2}\end{gathered}
s=semiperimeter
s=
2
a+b+c
\begin{gathered}\bold{Given}\begin{cases} \sf{a=13cm}\\ \sf{b=13cm}\\ \sf{c=24cm}\end{cases}\end{gathered}
Given
⎩
⎪
⎪
⎨
⎪
⎪
⎧
a=13cm
b=13cm
c=24cm
\begin{gathered}\tt semi\:perimeter=\frac{13+13+24}{2}\\ \\ \tt s=\cancel{\frac{50}{2}}=25\\ \\ \tt semi\: perimeter=25cm\end{gathered}
semiperimeter=
2
13+13+24
s=
2
50
=25
semiperimeter=25cm
\begin{gathered}\sf Area\:of\:_\triangle \sqrt{s(s-a)(s-b)(s-c)}\\ \\ \sf\leadsto Area\:of\:_\triangle=\sqrt{25(25-13)(25-13)(25-24)}\\ \\ \sf\leadsto Area\:of_\triangle =\sqrt{25\times 12\times 12\times 1}\\ \\ \sf\leadsto Area\:of_\triangle=\sqrt{5\times 5\times 4\times 3\times 4\times 3}\\ \\ \sf\leadsto Area\:of_\triangle=5\times 4\times 3\\ \\ \sf Area\:of_\triangle=60cm{}^{2}\end{gathered}
Areaof
△
s(s−a)(s−b)(s−c)
⇝Areaof
△
=
25(25−13)(25−13)(25−24)
⇝Areaof
△
=
25×12×12×1
⇝Areaof
△
=
5×5×4×3×4×3
⇝Areaof
△
=5×4×3
Areaof
△
=60cm
2
The area of triangle
\sf \implies 60m{}^{2}⟹60m
2
Now to find the height of triangle
\sf\underline{\purple{Area_\triangle=\frac{1}{2}\times base\times height}}
Area
△
=
2
1
×base×height
\begin{gathered}\bold{Given}\begin{cases}\sf{Area=60cm{}^{2}}\\ \sf{base=24cm}\end{cases}\end{gathered}
Given{
Area=60cm
2
base=24cm
\begin{gathered}\sf\ Area\:of\:\triangle=\frac{1}{2}\times base\times height \\ \\ \sf\implies 60=\frac{1}{2}\times 24\times height\\ \\ \sf\implies 60\times 2=24h\\ \\ \sf\implies 120=24h\\ \\ \sf\implies \cancel{\frac{120}{24height =5
Areaof△=
2
1
×base×height
⟹60=
2
1
×24×height
⟹60×2=24h
⟹120=24h
⟹
24
120
=height
⟹height=5