scalene triangle

\(m=\frac { 19 }{ 14 } \)

\(a=2,\ OR\ \left( \triangle ABC \right) =6\ sq.units\)

We have P(\(\sqrt{2}\) , \(\sqrt{2}\)) , Q(-\(\sqrt{2}\),-\(\sqrt{2}\)) and R(-\(\sqrt{6}\),\(\sqrt{6}\))
\(\therefore\)  PQ = \(\sqrt { \left( \sqrt { 2 } +\sqrt { 2 } \right) ^{ 2 }+\left( \sqrt { 2 } +\sqrt { 2 } \right) ^{ 2 } } \)
\(=\sqrt { \left( 2\sqrt { 2 } \right) ^{ 2 }+\left( 2\sqrt { 2 } \right) ^{ 2 } } \)
\(=\sqrt { 4\times 2+4\times 2 } =\sqrt { 8+8 } \)
\(=\sqrt { 16 } \)=4 units
PR = \(\sqrt { \left( \sqrt { 2 } +\sqrt { 6 } \right) ^{ 2 }+\left( \sqrt { 2 } +\sqrt { 6 } \right) ^{ 2 } } \)
\(=\sqrt { 2+6+2\sqrt { 2 } +2+6-2\sqrt { 2 } } \)
\(=\sqrt { 2+6+2+6 } \)
\(=\sqrt { 16 } \) = 4 units
RQ = \(\sqrt { [(-\sqrt { 2 } )+\sqrt { 6 } ^{ 2 }+\left( -\sqrt { 2 } -\sqrt { 6 } \right) ^{ 2 } } \)
\(=\sqrt { 2+6-2\sqrt { 2 } +2+6+2\sqrt { 2 } } \)
\(=\sqrt { 2+6+2+6 } \)
\(=\sqrt { 16 } \) =4 units
Since PQ=PR=RQ = 4 units
∴ PQR is an equilateral triangle.

Given, vertices of  \(\Delta \)ABC are  A(-2, 0), B(0, 2) and C(2,0) and D(- 4, 0), E(4, 0) and F(0, 4).
Now, AB = \(\sqrt { { (0+2) }^{ 2 }+{ (2-0) }^{ 2 } } =\sqrt { 4+4 } =2\sqrt { 2 } \) units   [\(\because \) distance=\(\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }-{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } } \)]
BC = \(\sqrt { { (2-0) }^{ 2 }+{ (0-2) }^{ 2 } } \) \( =\sqrt { 4+4 } =2\sqrt { 2 } \) units
CA = \(\sqrt { { (-2-2) }^{ 2 }+{ (0-0) }^{ 2 } } \) = \(\sqrt { { (-4) }^{ 2 }+0 } \)= 4 units
FD = \(\sqrt { { (0+4) }^{ 2 }+{ (4-0) }^{ 2 } } \) = \(\sqrt { { (4) }^{ 2 }+{ (-4) }^{ 2 } } \) = \(4\sqrt { 2 } \) units
FE = \(\sqrt { { (4-0) }^{ 2 }+{ (0-4) }^{ 2 } } \) = \(\sqrt { { (4) }^{ 2 }+{ (-4) }^{ 2 } } \) = \(4\sqrt { 2 } \) units
and ED = \(\sqrt { { (-4-4) }^{ 2 }+{ (0-0) }^{ 2 } } \) = \(\sqrt { { (-8) }^{ 2 }} \) = \(\sqrt { {64} }\) = 8 units
Here, we see that sides of \(\Delta \) DEF are twice the sides of \(\Delta \)ABC.

Hence, both the triangle are similar.    
Hence proved.

Here, (x₁, y₁) = (a, b + c), (x₂, y₂) = (b,c + a) and (x₃, y₃) = (c, a + b) 
Now, area of a triangle
=\(\frac { 1 }{ 2 } \)[x₁y₂ + x₂y₃ + x₃y₁ - x₁y₃ - x₂y₁ - x₃y₂]
=\(\frac { 1 }{ 2 } \)[a(c + a) + b(a + b) + c(b + c)-a(a + b) - b(b + c) - c(c + a)]
=\(\frac { 1 }{ 2 } \)[ac + a² + ba + b² + cb + c² - a² - ab - b² - bc - c² - ca]
=\(\frac { 1 }{ 2 } \)(0) = 0
Hence, the area of a triangle with vertices (a,b + c), (b, c + a) and (c, a + b) is zero.

Let A = (x₁,y₁) = (-4, 0), B = (x₂,y₂) = (4, 0) and C = (x₃, y₃) = (0, 3)
Now, AB = \(\sqrt { { [4-(-4)] }^{ 2 }+{ (0-0) }^{ 2 } } \) [ using distance formula ]
\(=\sqrt { { (4+4) }^{ 2 } } =\sqrt { { 8 }^{ 2 } } =8\quad units\)
\(BC=\sqrt { (0-4{ ) }^{ 2 }+{ (3-0) }^{ 2 } } =\sqrt { { (-4) }^{ 2 }+{ (3) }^{ 2 } } \)
\(=\sqrt { 16+9 } =\sqrt { 25 } =5\quad units\)
and AC \(=\sqrt { [0-(-4){ ] }^{ 2 }+(3-0{ ) }^{ 2 } } \)
\(=\sqrt { 16+9 } =\sqrt { 25 } =5\quad units\)
\(\because BC = AC\)
So, \(\triangle \) ABC is an isosceles triangle.

Here, coordinates of \(A\equiv ({ x }_{ 1 },{ y }_{ 2 })\) = (-6, 10),
Coordinates of B \(\equiv \) (x₂, y₂) = (-4, 6) and 
Coordinates of C \(\equiv \) (x₃, y₃) = (3, -8).
We know that,
Area of triangle = \(\frac { 1 }{ 2 } \left| { x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }) \right| \)
\(\therefore\) Area of \(\Delta ABC\) = \(\frac { 1 }{ 2 } \left| -6\{ 6-(-8)\} +(-4)(-8-10)+3(10-6) \right| \)
\(=\frac { 1 }{ 2 } \left| -6(14)+(-4)(-18)+2(4) \right| \)
\(=\frac { 1 }{ 2 } \left| -84+72+12 \right| =0\)
Since, area of \(\Delta ABC\)  is zero. So, points A, B and C are collinear.
Now, \(AB=\sqrt { { (-4+6) }^{ 2 }+{ (6-10) }^{ 2 } } \)
[ using distance formula ]
\(=\sqrt { { 2 }^{ 2 }+{ (-4) }^{ 2 } } =\sqrt { 4+16 } =\sqrt { 20 } \)
\(=2\sqrt { 5 } units\)
\(AC=\sqrt { ({ 3+6) }^{ 2 }+(-8-10)^{ 2 } } \)
\(=\sqrt { { 9 }^{ 2 }+({ -18 })^{ 2 } } =\sqrt { 81+324 } \)
\(=\sqrt { 405 } =\sqrt { 81\times 5 } =9\sqrt { 5 } units\)
\(\therefore \quad AB=2\sqrt { 5 } \times \frac { 9 }{ 9 } =\frac { 2 }{ 9 } AC\)      
Hence proved.

Hint : First find the type of triangle using distance formula and hence obtain the angles.
Ans. 60°, equilateral triangle.

Hint : First find the type of triangle using distance formula and hence obtain the angles.
Ans. 60°, equilateral triangle.

It is clear from the figure that,
Perimeter of \(\Delta\)AOB
=Distance (AO) + Distance (OB) + Distance (AB)

\(=\sqrt { { (0-0) }^{ 2 }+{ (4-0) }^{ 2 } } +\sqrt { { (0-3) }^{ 2 }+{ (0-0) }^{ 2 } } +\sqrt { { (0-3) }^{ 2 }+{ (4-0) }^{ 2 } } \)
[ using distance formula]
\(=\sqrt { { ({ x }_{ 1 }-{ x }_{ 2 }) }^{ 2 }+{ ({ y }_{ 1 }-{ y }_{ 2 }) }^{ 2 } } \)
\(=\sqrt { { (4) }^{ 2 } } +\sqrt { { (3) }^{ 2 } } +\sqrt { 9+16 } \)
= 4+3+\(\sqrt { 25 } \) = 4 + 3 + 5 = 12 units
Hence, the perimeter of given triangle is 12 units.