In a triangle, the sum of any two sides exceed the third side by 6 cm. Find its area (in sq. cm)

Question

In a triangle, the sum of any two sides exceed the third side by 6 cm. Find its area (in sq. cm) (A) 12 √3 (B) 9 √3 (C) 15 √3 (D) 18 √3

Tardigrade Admin · Accepted Answer

Let the sides of the triangle be a cam , b cm, and c cm. a+b-c=a+c-b=b+c-a=6 ⇒ a+b-c+a+c-b+b+c-a=18 a+b+c=18 Area of the triangle = <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/093ca224af5eae5aca1e1d479e84c221-.jpeg" /> =(1/4) √(18)(6)(6)(6)=9 √3 cm 2 .

Q. In a triangle, the sum of any two sides exceed the third side by 6 cm. Find its area (in sq. cm)

$123$

$93$

$153$

$183$

Let the sides of the triangle be $a c am, b c m$ , and $cc m$ .
$a + b - c = a + c - b = b + c - a = 6$
$\Rightarrow a + b - c + a + c - b + b + c - a = 18$
$a + b + c = 18$
Area of the triangle $=$

$= \frac{1}{4} (18) (6) (6) (6) = 93 c m^{2} .$

Q. In a triangle, the sum of any two sides exceed the third side by 6 cm. Find its area (in sq. cm)

123​

93​

153​

183​

Let the sides of the triangle be acam,bcm, and ccm. a+b−c=a+c−b=b+c−a=6 ⇒a+b−c+a+c−b+b+c−a=18 a+b+c=18 Area of the triangle = =41​(18)(6)(6)(6)​=93​cm2.

$123$

$93$

$153$

$183$

Let the sides of the triangle be $a c am, b c m$ , and $cc m$ .
$a + b - c = a + c - b = b + c - a = 6$
$\Rightarrow a + b - c + a + c - b + b + c - a = 18$
$a + b + c = 18$
Area of the triangle $=$

$= \frac{1}{4} (18) (6) (6) (6) = 93 c m^{2} .$