Question

) In ∆, ∠ = 90°. Find the sides of the triangle, if: AB=(x-3) cm, BC= (x+4) cm and AC=

(x+6) cm.​

Answer 1

Answer:

as the triangle is a right angle triangle so

AC^2=AB^2+BC^2

=> (x+6)^2=(x+4)^2+(x-3)^2

=> x^2+12x+36=x^2+8x+16+x^2-6x+9

=> x^2+12x+36=2x^2+2x+25

=> 2x^2+2x+25-x^2-12x-36=0

=> x^2-10x-11=0

=> x^2-11x+x-11=0

=> x(x-11)+1(x-11)=0

=> (x-11)(x+1)=0

x-11=0 ,x=11

x+1=0,x= -1

x is not negative as it is the side of the triangle

So x=11

so the sides of the triangle is (11-3)cm=8cm,

(11+4)cm=15cm,(11+6)cm÷17cm

Answer 2

$\bf{\underline{Correct \; Question\;:}}$

In a ∆ABC, ∠B = 90°. Find the sides of the triangle, if AB = (x - 3)cm, BC = (x + 4)cm and AC = (x + 6)cm.

$\bf{\underline{Given\;:}}$

• AB = (x - 3)cm
• BC = (x + 4)cm
• AC = (x + 6)cm
• ∠B = 90°

$\bf{\underline{To \; Find\;:}}$

• The sides of the triangle.

$\bf{\underline{Solution\;:}}$

Reference of figure :

$\setlength{\unitlength}{1.9cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{B}}\put(10.6,1){\large\sf{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(8.6,0.7){\sf{\large{(X+4)cm}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\put(7.3,1.8){\sf{(x-3)cm}}\put(9.4,2.2){\sf {(x+6)cm}}\end{picture}$

$\bf{\underline{We \; Have\;:}}$

• Perpendicular = (x - 3)cm
• Base = (x + 4)cm
• Hypotenuse = (x + 6)cm

Using Pythagoras Theorem : H² = P² + B²

$\longrightarrow$ (x + 6)² = (x - 3)² + (x + 4)²

$\longrightarrow$ (x² + 6² + 2.x.6) = (x² + 3² - 2.x.3) + (x² + 4² + 2.x.4)

$\longrightarrow$ x² + 36 + 12x = x² + 9 - 6x + x² + 16 + 8x

$\longrightarrow$ x² + 36 + 12x = 2x² + 25 + 2x

$\longrightarrow$ 2x² - x² + 2x - 12x + 25 - 36 = 0

$\longrightarrow$ x² - 10x - 11 = 0

$\longrightarrow$ x² - (11 - 1)x - 11 = 0

$\longrightarrow$ x² - 11x + x - 11 = 0

$\longrightarrow$ x(x - 11) + 1(x - 11) = 0

$\longrightarrow$ (x - 11)(x + 1) = 0

⠀⠀ x = 11 , x = -1

• Sides of a triangle can't be negative. So, we're taking (x = 11)

$\longrightarrow$ AB = (x - 3)cm = (11 - 3)cm = 8 cm

$\longrightarrow$ BC = (x + 4)cm = (11 + 4)cm = 15 cm

$\longrightarrow$ AC = (x + 6)cm = (11 + 6)cm = 17 cm

• Hence, the sides of the triangle are 8 cm, 15 cm and 17 cm.