Verified answer

Answer:

Given :-

An isosceles triangle has a base of 20 cm and each of its equal sides is 26 cm long.

To Find :-

What is the area of an isosceles triangle.

Solution :-

Let us consider,

\begin{gathered}\leadsto \bf \triangle ABC\: is\: an\: isosceles\: triangle\: .\\\end{gathered}

⇝△ABCisanisoscelestriangle.

Given :

\mapsto \sf AB =\: 26\: cm↦AB=26cm

And,

\mapsto \sf AC =\: 26\: cm↦AC=26cm

So, we can write as,

\begin{gathered}\mapsto \bf AB =\: AC =\: 26\: cm\\\end{gathered}

↦AB=AC=26cm

\begin{gathered}\mapsto \bf BC =\: 20\: cm\\\end{gathered}

↦BC=20cm

And, the height is :

\begin{gathered}\mapsto \bf AD =\: Height\\\end{gathered}

↦AD=Height

Now,

\implies \bf BD =\: \dfrac{1}{2} BC⟹BD=

2

1

BC

\implies \sf BD =\: \dfrac{1}{2} \times BC⟹BD=

2

1

×BC

We have :

BC = 20 cm

So, by putting this value we get,

\implies \sf BD =\: \dfrac{1}{2} \times 20⟹BD=

2

1

×20

\implies \sf BD =\: \dfrac{1 \times 20}{2}⟹BD=

2

1×20

\implies \sf BD =\: \dfrac{\cancel{20}}{\cancel{2}}⟹BD=

2

20

\implies \sf BD =\: \dfrac{10}{1}⟹BD=

1

10

\implies \sf\bold{BD =\: 10\: cm}⟹BD=10cm

Now, we have to find the value of height (AD) :

Given :

AB = 26 cm

BD = 10 cm

According to the question by using the Pythagoras Theorem we get,

\implies \bf (AB)^2 =\: (AD)^2 + (BD)^2⟹(AB)

2

=(AD)

2

+(BD)

2

\implies \sf (26)^2 =\: (AD)^2 + (10)^2⟹(26)

2

=(AD)

2

+(10)

2

\begin{gathered}\implies \sf (26 \times 26) =\: (AD)^2 + (10 \times 10)\\\end{gathered}

⟹(26×26)=(AD)

2

+(10×10)

\begin{gathered}\implies \sf (676) =\: (AD)^2 + (100)\\\end{gathered}

⟹(676)=(AD)

2

+(100)

\implies \sf 676 =\: (AD)^2 + 100⟹676=(AD)

2

+100

\begin{gathered}\implies \sf 676 - 100 =\: (AD)^2\\\end{gathered}

⟹676−100=(AD)

2

\implies \sf 576 =\: (AD)^2⟹576=(AD)

2

\begin{gathered}\implies \sf \sqrt{576} =\: AD\\\end{gathered}

⟹

576

=AD

\begin{gathered}\implies \sf \sqrt{\underline{24 \times 24}} =\: AD\\\end{gathered}

⟹

24×24

=AD

\implies \sf 24 =\: AD⟹24=AD

\implies \sf\bold{AD =\: 24\: cm}⟹AD=24cm

Hence, the height is 24 cm .

Now, we have to find the value of area of an isosceles triangle :

Given :

Height (AD) = 24 cm

Base (BC) = 20 cm

According to the question by using the formula we get,

\begin{gathered}\small \dashrightarrow \sf\boxed{\bold{Area_{(Triangle)} =\: \dfrac{1}{2} \times Height \times Base}}\\\end{gathered}

⇢

Area

(Triangle)

=

2

1

×Height×Base

\begin{gathered}\dashrightarrow \sf Area_{(Triangle)} =\: \dfrac{1}{2} \times 24 \times 20\\\end{gathered}

⇢Area

(Triangle)

=

2

1

×24×20

\begin{gathered}\dashrightarrow \sf Area_{(Triangle)} =\: \dfrac{1 \times 24 \times 20}{2}\\\end{gathered}

⇢Area

(Triangle)

=

2

1×24×20

\begin{gathered}\dashrightarrow \sf Area_{(Triangle)} =\: \dfrac{24 \times 20}{2}\\\end{gathered}

⇢Area

(Triangle)

=

2

24×20

\begin{gathered}\dashrightarrow \sf Area_{(Triangle)} =\: \dfrac{\cancel{480}}{\cancel{2}}\\\end{gathered}

⇢Area

(Triangle)

=

2

480

\begin{gathered}\dashrightarrow \sf Area_{(Triangle)} =\: \dfrac{240}{1}\\\end{gathered}

⇢Area

(Triangle)

=

1

240

\begin{gathered}\dashrightarrow \sf\bold{\underline{Area_{(Triangle)} =\: 240\: cm^2}}\\\end{gathered}

⇢

Area

(Triangle)

=240cm

2

\therefore∴ The area of an isosceles triangle is 240 cm² .

[Note :- Please refer that attachment for the figure. ]

Step-by-step explanation:

thanks for points