Question

In a triangle, the average of any two sides is 6cm more than half of the third side. Find area of the triangle

Answer 1

let a,b,c are 3 Sides
so , a+b-c=6
a+c-b=6
b+c-a= 6
add all three questions
a+b+c= 18.
a+b = b+6 (given
so , 2c= 12
c= 6
and So , a&b = 6
hence ,it is an equilateral triangle
so, area = √3/4×6²
= 9√3 ans.

Answer 2

Answer:

Concept:

A triangle is a 3-sided polygon that is occasionally (though not frequently) referred to as the trigon.

Step-by-step explanation:

• The triangle's internal angle, which is 180 degrees, is constructed. It implies that the internal angles of a triangle add up to 180 degrees. It is the polygon with the fewest sides.
• There are three sides and three angles in every triangle, some of which may be the same.
• Triangle's area =$\frac{1}{2} \times b \times h$ square units.

where the triangle's base and height, respectively, are denoted by $b$ and $h$.

Given:

The average of any two sides is 6cm more than half of the third side.

To find:

Area of the triangle

Solution:

• Let the three sides be $s1,s2$ and $s3$
• By given data,

                             $\frac{s1+s2}{2} =6+(\frac{1}{2} \times s3)$

                              $\frac{s1+s2}{2} =6+\frac{s3}{2}$

                              $\frac{s1+s2}{2}-\frac{s3}{2}= 6$

                             $s1+s2-s3=6$      ⇒    A

• In a similar way,

            $s1+s3-s2=6$       ⇒    B        and

            $s2+s3-s1=6$        ⇒  C

• On simplifying  A and B,

                       $2s1=12$  

                        $s1=\frac{12}{2}$

                          $s1=6$

• On simplifying B and C,                  

                         $2s3=12$

                           $s3=\frac{12}{2}$

                             $s3=6$

• On simplifying A and C,

                        $2s2=12$

                         $s2=\frac{12}{2}$

                          $s2=6$        

• All sides are equal i.e. $s1=s2=s3=6 cm$.  Hence, its an equilateral triangle.
• Area of equilateral triangle = $\frac{\sqrt{3} }{4} \times side^{2}$   $sq.cm$
• Substituting value of side,

                             Area =     $\frac{\sqrt{3} }{4} \times 6^{2}$

                                       =  $\frac{\sqrt{3} }{4} \times 36$

                                         =$9\sqrt{3}$

Hence, area of triangle = $9\sqrt{3}$  $sq.cm$      

                             

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