A triangle $ˆ A B C$ $hat(ABC)$ has vertices of $A (1, 3); B (\frac{1}{2}, \frac{3}{2}); C (2, 1)$ $A(1,3);B(1/2,3/2);C(2,1)$ . Verify that the triangle is isosceles and calculate the area and perimeter?

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Answer 1 · 2017-07-07T13:53:38

$Area= \frac{5}{4}, the Perimeter= \sqrt{5} (1 + \sqrt{2}) .$ $" Area="5/4," the Perimeter="sqrt5(1+sqrt2).$

Let us first verify that the given points are non-collinear.

We use the following necessary and sufficient condition for

collinearity of the points :

$A (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}) are collinear \Leftrightarrow$ $A(x_1,y_1), B(x_2,y_2) and C(x_3,y_3)" are collinear "iff$

$|(x_1,y_1,1),(x_2,y_2,1),(x_3,y_3,1)|=0.$

We have, $D=|(1,3,1),(1/2,3/2,1),(2,1,1)|,$

$=1(3/2-1)-3(1/2-2)+1(1/2-3),$

$=1/2+9/2-5/2=5/2.$

Thus, the points are not collinear, and, hence, form $DeltaABC.$

Knowing that, the Area of $DeltaABC$ is $1/2|D|,$

The Reqd. Area = $5/4.$

Using the Distance Formula, we have,

$AB^2=(1-1/2)^2+(3-3/2)^2=1/4+9/4 rArr AB=sqrt10/2.$

$BC^2=(1/2-2)^2+(3/2-1)^1=9/4+1/4 rArr BC=sqrt10/2.$

$AC^2=(1-2)^2+(3-1)^2=1+4 rArr AC=sqrt5.$

$because," in "DeltaABC, AB=BC, :., Delta "is isisceles,"$ having

$"perimeter="AB+BC+AC=sqrt5+sqrt10=sqrt5(1+sqrt2).$

Enjoy Maths.!

A triangle hat(ABC)ˆABChat(ABC) has vertices of A(1,3);B(1/2,3/2);C(2,1)A(1,3);B(12,32);C(2,1)A(1,3);B(1/2,3/2);C(2,1). Verify that the triangle is isosceles and calculate the area and perimeter?