Question

In the adjacent figure ,AC=6cm,AB=5cm and angle BAC=30° .find the area of the triangle

Answer 1

Given:

• ∠BAC = 30°
• AC = 6cm
• AB = 5,cm

To Find:

Area of the triangle

Solution:

in ∆ABC,

draw BD perpendicular to AC

in ∆ABD,

$\tt{\sin30° = \frac{BD}{AB} }$

$\implies\tt{ \: \frac{1}{2} = \frac{BD}{5}}$

$\implies\tt{ \: BD = \frac{5}{2} cm}$

Now,

$\tt{area \: of \: ∆ABC \: = ( \frac{1}{2} ) \times base \times height}$

$\implies\tt{ \: ( \frac{1}{2} ) \times \: AC \times BD}$

$\implies\tt {\: ( \frac{1}{2} ) \times 6cm \times ( \frac{5}{2} )cm}$

$\implies\tt{ \: \frac{15}{2} {cm}^{2}}$

$\implies\tt {\: 7.5 {cm}^{2}}$

Hence,

area of the triangle is $\tt\underline\red{7.5cm²}$

Answer 2

Answer:

since the height is not given we can solve using another formula

Area of Triangle = base, side, sin 0/2

Here let us take the base to be 6 cm and the side measure to be 5 cm

Substituting in the formula we get,

Area of Triangle 6* 5* sin 30/2

Area =30*(1/2) /2

Area = 30* 114

Area = 30* 0.25=7.5

Hence the area of the Triangle given is 7.5 cm

Step-by-step explanation:

i hope it's help you