Question

Find the area of the
red triangle in terms
of X, if the black
quadrilateral is a
rectangle.
2x
2x
12 cm
9cm​

Answer 1

Answer:

Area of triangle is $35.8cm^{2}$

Step-by-step explanation:

Though the diagram is not given, assumed the diagram for the question you need is same as that i've given here.

Let CEF be the triangle in the rectangle ABCD.

The measures of rectangle is $12cm\times 9cm$

The opposite side of 12cm is divided into two equal parts, each equal to $2x$

Therefore, $2x+2x=12\implies 4x=12\implies x=3cm$

Let a, b and c be the sides of triangle, each side making a hypotenuse side in the rectangle. Applying Pythagorean theorem to each triangle:

1) the triangle AEF,

               $a^{2}= x^{2} +(2x)^{2} =3^{2} +(2\times 3)^{2}=45$

         $\implies a=\sqrt[]{45} =6.7$

2) the triangle BCF,

               $b^{2}= 9^{2} +(2x)^{2} =9^{2} +(2\times 3)^{2}=81+36=117$

         $\implies b=\sqrt[]{117} =10.8$

3) the triangle CEF,

               $c^{2}= 12^{2} +(9-x)^{2} =12^{2} +(9- 3)^{2}=144+36=180$

         $\implies c=\sqrt[]{180} =13.41$

Perimeter of the triangle, $s=\frac{a+b+c}{2}$

                                             $=\frac{6.7+10.8+13.41}{2}=15.45cm$

From Heron's formula of area of triangle

                  $A=\sqrt[]{s(s-a)(s-b)(s-c)}$

                     $=\sqrt[]{15.45(15.45-6.7)(15.45-10.8)(15.45-13.41)}$

                     $=\sqrt[]{15.45(8.75)(4.65)(2.04)}$

                     $=\sqrt[]{1282.39}=35.8cm^{2}$

Hence, area of triangle is $35.8cm^{2}$