Question

the base of an icoceles triangle is 48 cm and its perimeter is 98 cm . Find it's area.​

Answer 1

$\huge\mathtt{\underline{\blue{ANSWER:-}}}$

$\sf\implies 168cm{}^{2}$

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

$\bold{Given}\begin{cases}\sf{Base=48cm}\\ \sf{perimeter=98cm}\end{cases}$

• we know that the isosceles triangle have two equal sides ..

Let the equal sides be x

$\sf{\underline{\underline{\purple{According\:to\: question:-}}}}$

we know that the :-

premeter of triangle=Sum of all sides

$\tt\implies 98=48+x+x\\ \\ \tt\longrightarrow 98-48=2x\\ \\ \tt\longrightarrow 50=2x\\ \\ \tt\longrightarrow \frac{\cancel{50}}{\cancel{2}}=x\\ \\ \tt\implies x=25$

• The two sides of isosceles triangle are 25cm and 25cm its 3rd sides is 48cm

$\tt Area\:of\: triangle=\sqrt{s(s-a)(s-b)(s-c)}$

$\bold{where}\begin{cases}\sf{a=25cm}\\ \sf{b=25cm}\\ \sf{c=48cm}\end{cases}$

$\sf s=\frac{a+b+c}{2}\\ \\ \sf s=\frac{25+25+48}{2}\\ \\ \sf s=\frac{\cancel{98}}{\cancel{2}}\\ \\ \sf\implies s = 49$

Now Area :-

$\tt Area=\sqrt{49(49-25)(49-25)(49-1)}\\ \\ \sf\longrightarrow Area =\sqrt{49×24×24×1}\\ \\ \sf\longrightarrow Area=\sqrt{7×7×2×2×2×3×2×2×2×3}\\ \\ \sf\longrightarrow Area\:of\: triangle= 7×2×2×2×3\\ \\ \sf\longrightarrow Area\:of\: triangle=168cm{}^{2}$

$\huge\boxed{\underline{\purple{Area=168cm{}^{2}}}}$

#BAL

#Answerwithquality

Answer 2

$\huge\bold\color{blue}{\underbrace{\overbrace{\mathfrak{\color{red}{♡Answer♧}}}}}</p><p>$

Given :-

• Perimeter = 98 cm
• Base = 48 cm

To find :-

• Area of triangle

Let :-

• A,B,C be the sides
• B is base
• A = C [ ∵ Isosceles triangles has two equal sides.]

Then other two sides are :-

$\mathtt{ \implies \: A+B+C = 98}$

$\mathtt{ \implies \: A+48+A = 98.........(∵A=C)}$

$\mathtt{\implies 2A = 98- 48 }$

$\bold{\mathtt{\implies A = \frac{50}{2} }}$

$\mathtt{\implies A = 25}$

${\mathtt { \bold{ \therefore A = 25 cm \: \: \: and \: \: \: C = 25 cm}}}$

Finding area of triangle,

$\color{brown}{\mathtt{ {\boxed{ \bold {Semiperimeter = \: \frac{a + b + c}{2} }}}}}$

$\mathtt{Semiperimeter = \frac{25 + 25 + 48}{2} }$

$\mathtt{Semiperimeter = 49}$

Area of triangle:-

$\color{brown}{\boxed{\mathtt{ \bold{\bold{ \sqrt{s(s - a)(s - b)(s - c)}}}}}}$

$\implies \sqrt{49(49 - 25)(49 - 25)(49 - 48}$

$\implies \sqrt{49 \times 24 \times 24 \times 1 }$

$\implies \sqrt{7 \times 7 \times 3 \times 2 \times 2 \times 2 \times 3 \times 2 \times 2 \times 2}$

$\implies 7 \times 3 \times 2 \times 2 \times 2$

$\color{green}{\boxed{ \mathtt{ \bold{ \implies \: A\: = 168 \: {cm}^{2} }}}}$

∴Area of triangle is 168 cm².