Question

23. The base of an isosceles
triangle is 16cm, and its area is
48cm.sq. The perimeter of the
triangle is?​

Answer 1

☢️$\:\:\huge\bigstar\rm\blue{GIVEN}\:\:\bigstar$

• The Base of an Isosceles triangle is 16cm.

• The area of triangle is 48cm.sq.

• AB=AC

$\:\:\huge\bigstar\rm\blue{TO\:FIND}\:\:\bigstar$

• The Perimeter if a triangle.

$\:\:\huge\bigstar\tt\red{CONSTRUCTION}\:\:\bigstar$

• Draw a triangle ABC From $\angle{A}$ Draw a perpendicular to Base BC.

$\:\:\huge\bigstar\rm\red{FORMULAE\:USED}\:\:\bigstar$

• ${\boxed{\rm{\blue{Area\:of\:Triangle=\frac{1}{2}\times{Base}\times{Height}}}}}$

• ${\boxed{\rm{\blue{Pythogoras\:Theorem}}}}$

Now,

• we will First find out height AD using the Formula 1,then we will Find out the perimeter of triangle

$\implies\tt\blue{Area\:of\:triangle=\dfrac{1}{2}\times{Base}\times{Height}}$

$\implies\tt\blue{48cm.sq.=\dfrac{1}{\cancel{2}}\times{\cancel{16}}^8\times{h}}$

$\implies\tt\blue{48=8h}$

$\implies\tt\blue{h=\dfrac{\cancel{48}}{\cancel{6}}}$

Thus, The height of triangle is 6cm.(AD)

Now,

In ∆ACD

$\implies\tt\red{CD=\frac{BD}{2}=8cm,AD=6cm,AC?}$

Using pythogoras theorem.

$\implies\rm\blue{(AD)^2+(CD)^2=(AC)^2}$

$\implies\rm\blue{(6)^2+(8)^2=(AC)^2}$

$\implies\rm\blue{(100)=(AC)^2}$

$\implies\rm\blue{AC=10cm}$

Hence, AC=AB=10cm.

Perimeter of Triangle= AB+AC+AD=10+10+16=36cm.

Hence,

The perimeter of triangle is 36cm.

Answer 2

★ See this attachment ...