Question

ABC is an isosceles triangle with AB=AC =7.5 cm and BC=9 cm the length of altitude from A to BC is 6 cm then find the area of triangle ABC also fond the length of altitude from C to AB​

Answer 1

Answer:

Finding area of ∆ ABC

using BC as base and

AD as height.

Here,

Base=BC

=9cm

Height= AD

=6cm

$\bold{Area \: of \: ∆ABC= \frac{1}{2} \times base \times height}$

$\bold{ \: \: \: \: \: = \frac{1}{2} \times BC \times AD}$

$\bold{ \implies \cancel \frac{1}{2} \times 9 \times \cancel6 \: \: {}^{3} }$

$\bold{ \implies{9 \times 3}}$

$\bold{ \implies{ 27 {cm}^{2} }}$

Now we have to find EC which is height corresponding to base AB .

So,finding Area using EC as height and AB as base.

Base =AB=7.5cm

Height =EC=?

$\bold{Area \: of \: ∆ABC= \frac{1}{2} \times base \times height}$

$\bold{27 = \frac{1}{2} \times 7.5 \times \: EC}$

$\bold{EC = \frac{27 \times 2}{7.5} }$

$\bold{EC = \frac{27 \times2 \times 10}{75} }$

$\bold{EC = \frac{9 \times 2 \times 10}{25} }$

$\bold{EC = \frac{9 \times 2 \times 2}{5} }$

$\bold{EC = \frac{36}{5} }$

$\bold{EC = 7.2}$

$\bold{EC = 7.2cm}$

$\bold{ \therefore \: area \: of \: \triangle \: ABC \: is \: 27 {cm}^{2} }$

$\bold{and \: height \: CE \: is \: 7.2cm}$

Answer 2

Finding area of ∆ ABC

using BC as base and

AD as height.

Here,

Base=BC

=9cm

Height= AD

=6cm

Now we have to find EC which is height corresponding to base AB .

So,finding Area using EC as height and AB as base.

Base =AB=7.5cm

Height =EC=7.5 cm