Question

find the area of an isosceles triangle whose base is 6 cm and perimeter is 16 cm

Answer 1

in an isosceles triangle,
any two sides are equal

let each of the equal side be x
perimeter=16cm
base=6cm

therefore,
x+x+6=16
2x=16-6
2x=10
$x = \frac{10}{2}$
x=5

area of triangle =½×base×height

since we need the height,
we draw a perpendicular from the base which will divide this triangle into two right angled triangles
with each base = to 6/2=3cm
hypotenuse is the side of isosceles triangle which is 5cm
let the height be x


Using Pythagoras theorem,
we have,

${3}^{2} + {h}^{2} = {5}^{2}$
$9 + {h}^{2} = 25$
${h}^{2} = 25 - 9$
${h}^{2} = 16$
$h = \sqrt{16}$
h=4
height is 4cm

area of triangle= ½×base×height
$= \frac{1}{2} \times 6 \times 4$
$= 12 {cm}^{2}$
is the area

Answer 2

It is given that

Base = 6 cm

Perimeter = 16 cm

Consider ABC as an isosceles triangle in which

AB = AC = x

So BC = 6 cm

We know that

Perimeter of △ABC = AB + BC + AC

Substituting the values

16 = x + 6 + x

By further calculation

16 = 2x + 6

16 – 6 = 2x

10 = 2x

So we get

x = 10/2 = 5

Here AB = AC = 5 cm

BC = ½ × 6 = 3 cm

In △ABD

AB2 = AD2 + BD2

Substituting the values

52 = AD2 + 32

25 = AD2 + 9

By further calculation

AD2 = 25 – 9 = 16

So we get

AD = 4 cm

Here

Area of △ABC = ½ × base × height

Substituting the values

= ½ × 6 × 4

= 3 × 4

= 12 cm2

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