Question

Triangle QRS with vertices Q(6, -2), R(4, -7), and S(2, -5), is drawn inside a rectangle, as shown below. What is the area, in square units, of the triangle QRS?
A- 7
B-10
C-13
D-18

Answer 1

The rectangle is 4 x 5. Area of 20. 
The three triangles that are cut off from the rectangle to make QRS have areas of 5, 2, and 6. 13 total cut off. 
Subtract 13 from 20. 
20 - 13 = 7. 
The answer is A, 7. Hope this helps!

Answer 2

Answer:

7 sq.units.

Step-by-step explanation:

Q=(6, -2)

R=(4, -7)

S=(2, -5)

Now find the sides of triangle QR,RS,QS

To find QR use distance formula :

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$(x_1,y_1)=(6,-2)$

$(x_2,y_2)=(4,-7)$

Substitute the values in the formula :

$QR=\sqrt{(4-6)^2+(-7+2)^2}$

$QR=\sqrt{(-2)^2+(-5)^2}$

$QR=\sqrt{4+25}$

$QR=\sqrt{29}$

$QR=5.38516480713$

To Find RS

$(x_1,y_1)=(4,-7)$

$(x_2,y_2)=(2,-5)$

Substitute the values in the formula :

$RS=\sqrt{(2-4)^2+(-5+7)^2}$

$RS=\sqrt{(-2)^2+(2)^2}$

$RS=\sqrt{4+4}$

$RS=\sqrt{8}$

$RS=2.82842712475$

To Find QS

$(x_1,y_1)=(6,-2)$

$(x_2,y_2)=(2,-5)$

Substitute the values in the formula :

$QS=\sqrt{(2-6)^2+(-5+2)^2}$

$QS=\sqrt{(-4)^2+(-3)^2}$

$QS=\sqrt{16+9}$

$QS=\sqrt{25}$

$QS=5$

So, sides of triangle :

a =5.38516480713

b=2.82842712475

c=5

Now to find area:

$Area = \sqrt{s(s-a)(s-b)(s-c)}$  

Where $s = \frac{a+b+c}{2}$  

a,b,c are the side lengths of triangle  

Now substitute the values :  

$s = \frac{5.38516480713+2.82842712475+5}{2}$  

$s =6.60679596594$  

$\sqrt{6.606(6.606-5.38516480713)(6.606-2.82842712475)(6.606-5)}$

$Area = 7$  

Hence the area of the given triangle is 7 sq. units.