Question

An isosceles triangle has equal sides each 13 cm and a base 24 cm in length.find its height

Answer 1

$\huge\mathbb{\underline{\underline{SOLUTION:-}}}$

$\bold{Given}\begin{cases}\sf{A\: isosceles\:\triangle}\\ \sf{2\:Equal \:sides=13cm}\\ \sf{base=24cm}\end{cases}$

$\huge\mathtt{TO\: FIND:-}$

Height of isosceles triangle

• Now 1st find the area if triangle
• By heron's formula

$\sf\underline{\underline{\red{According\:to\: Question:-}}}$

$\sf{\pink{\sqrt{s(s-a)(s-b)(s-c)}}}$

$\tt s=semi\:perimeter\\ \\ \tt s=\frac{a+b+c}{2}$

$\bold{Given}\begin{cases} \sf{a=13cm}\\ \sf{b=13cm}\\ \sf{c=24cm}\end{cases}$

$\tt semi\:perimeter=\frac{13+13+24}{2}\\ \\ \tt s=\cancel{\frac{50}{2}}=25\\ \\ \tt semi\: perimeter=25cm$

$\sf Area\:of\:_\triangle \sqrt{s(s-a)(s-b)(s-c)}\\ \\ \sf\leadsto Area\:of\:_\triangle=\sqrt{25(25-13)(25-13)(25-24)}\\ \\ \sf\leadsto Area\:of_\triangle =\sqrt{25\times 12\times 12\times 1}\\ \\ \sf\leadsto Area\:of_\triangle=\sqrt{5\times 5\times 4\times 3\times 4\times 3}\\ \\ \sf\leadsto Area\:of_\triangle=5\times 4\times 3\\ \\ \sf Area\:of_\triangle=60cm{}^{2}$

The area of triangle

$\sf \implies 60m{}^{2}$

• Now to find the height of triangle

$\sf\underline{\purple{Area_\triangle=\frac{1}{2}\times base\times height}}$

$\bold{Given}\begin{cases}\sf{Area=60cm{}^{2}}\\ \sf{base=24cm}\end{cases}$

$\sf\ Area\:of\:\triangle=\frac{1}{2}\times base\times height \\ \\ \sf\implies 60=\frac{1}{2}\times 24\times height\\ \\ \sf\implies 60\times 2=24h\\ \\ \sf\implies 120=24h\\ \\ \sf\implies \cancel{\frac{120}{24}}=height\\ \\ \sf\implies height =5$

$\boxed{\pink{\sf{The\: height \:of\: triangle \:is \:5cm}}}$

#BAL

#answerwithquality

Answer 2

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}{24}}=height\\ \\ \sf\implies height =5\end{gathered}

Areaof△=

2

1

×base×height

⟹60=

2

1

×24×height

⟹60×2=24h

⟹120=24h

⟹

24

120

=height

⟹height=5