Question

\(A(-4,9,k)\), \(B(-1,k,k)\), \(C(0,7,10)\) form an isosceles right-angled triangle. If \(AB=BC\) and \(AC\) is an integer then the perimeter of \(\Delta ABC\) is

• A. \(4(1+\sqrt{2})\)
• B. \(\sqrt{14}(2+\sqrt{2})\)
• C. \(\sqrt{10}(2+\sqrt{2})\)
• D. \(6(1+\sqrt{2})\)

Hint:

For an isosceles right triangle with equal legs \(a\), the hypotenuse is always \(a\sqrt2\).

Answer 1

The Correct Option is D

Concept:
Since the triangle is an isosceles right-angled triangle and \(AB=BC\), the equal sides \(AB\) and \(BC\) form the perpendicular legs, while \(AC\) is the hypotenuse. Therefore, \[ AB=BC,\qquad AC^2=AB^2+BC^2=2AB^2 \] We shall first determine the value of \(k\).

Step 1: Find \(AB^2\). \[ AB^2=(-1+4)^2+(k-9)^2+(k-k)^2 \] \[ =3^2+(k-9)^2 \] \[ =9+k^2-18k+81 \] \[ AB^2=k^2-18k+90 \]

Step 2: Find \(BC^2\). \[ BC^2=(0+1)^2+(7-k)^2+(10-k)^2 \] \[ =1+(49-14k+k^2)+(100-20k+k^2) \] \[ =2k^2-34k+150 \]

Step 3: Use the condition \(AB=BC\). \[ k^2-18k+90=2k^2-34k+150 \] \[ k^2-16k+60=0 \] \[ (k-6)(k-10)=0 \] Hence, \[ k=6 \quad \text{or} \quad k=10 \]

Step 4: Use the condition that \(AC\) is an integer. For \(k=6\), \[ AB^2=36-108+90=18 \] Thus, \[ AB=BC=3\sqrt2 \] Since the triangle is right-angled, \[ AC^2=18+18=36 \] \[ AC=6 \] which is an integer. For \(k=10\), \[ AB^2=100-180+90=10 \] \[ AC^2=10+10=20 \] \[ AC=2\sqrt5 \] which is not an integer. Therefore, \[ k=6 \]

Step 5: Find the perimeter. \[ P=AB+BC+AC \] \[ =3\sqrt2+3\sqrt2+6 \] \[ =6\sqrt2+6 \] \[ P=6(1+\sqrt2) \] \[ \boxed{6(1+\sqrt2)} \]