Question

The perimeter of a triangle is 45 CM. The longest side exceed The shortest by 8 cm and the sum of the length of the longest and shortest sides is twice the length of the other side. Find the length of the sides. Give me answer by using s
Determinants.

Answer 1

Answer:

$\textbf{The lengths of the triangle are 19 cm, 15 cm and 11 cm}$

Step-by-step explanation:

$\text{Let x,y and be the lengths of the given triangle with }x> y >z$

$\text{As per given data, we have }$

$x+y+z=45$

$x=z+8$

$x+z=2y$

$\text{Rearranging terms, we get}$

$x+y+z=45$

$x+0y-z=8$

$x-2y+z=0$

$\triangle=\left|\begin{array}{ccc}1&1&1\\1 &0&-1\\1&-2&1\end{array}\right|$

$\triangle=1(0-2)-1(1+1)+1(0-2)=-6$

$\triangle_x=\left|\begin{array}{ccc}45&1&1\\8 &0&-1\\0&-2&1\end{array}\right|$

$\triangle_x=45(0-2)-1(8)+1(-16-0)$

$\triangle_x=-90-8-16=-114$

$\triangle_y=\left|\begin{array}{ccc}1&45&1\\1 &8&-1\\1&0&1\end{array}\right|$

$\triangle_y=1(8-0)-45(1+1)+1(0-8)$

$\triangle_y=8-90-8$

$\triangle_y=-90$

$\triangle_z=\left|\begin{array}{ccc}1&1&45\\1 &0&8\\1&-2&0\end{array}\right|$

$\triangle_z=1(0+16)-1(0-8)+45(-2-0)$

$\triangle_z=24-90=-66$

$\text{By cramer's rule}$

$x=\frac{\triangle_x}{\triangle}=\frac{-114}{-6}=19$

$y=\frac{\triangle_y}{\triangle}=\frac{-90}{-6}=15$

$z=\frac{\triangle_x}{\triangle}=\frac{-66}{-6}=11$

$\therefore\textbf{The lengths of the triangle are 19 cm, 15 cm and 11 cm}$