Question

for what value of x is △abc ~ △def?

x =

Explanation:

Step1: Use the angle - sum property of triangles

For \(\triangle ABC\), the sum of angles is \(180^{\circ}\). So, \(\angle A+\angle B+\angle C = 180^{\circ}\). For \(\triangle DEF\), \(\angle D+\angle E+\angle F=180^{\circ}\). Since \(\triangle ABC\sim\triangle DEF\), \(\angle A=\angle D\) and \(\angle B=\angle E\) and \(\angle C=\angle F\). We know that \(\angle A = 48^{\circ}\), \(\angle E=84^{\circ}\), \(\angle B=(x^{2}-5x)^{\circ}\), \(\angle D=(x^{2}-8x)^{\circ}\)

Since \(\triangle ABC\sim\triangle DEF\), \(\angle A=\angle D\)

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Factor the quadratic equation: \(x^{2}-8x - 48=(x - 12)(x+4)=0\)

Step2: Solve for \(x\)

Set each factor equal to zero:
If \(x - 12=0\), then \(x = 12\)
If \(x+4=0\), then \(x=- 4\)

We check for the non - negative value of angles.
For \(\angle B=(x^{2}-5x)\)
When \(x = 12\), \(\angle B=(12^{2}-5\times12)=144 - 60=84^{\circ}\)
When \(x=-4\), \(\angle B=((-4)^{2}-5\times(-4))=16 + 20=36^{\circ}\), but if \(x=-4\), \(\angle D=(x^{2}-8x)=16+32 = 48^{\circ}\), and \(\angle E = 84^{\circ}\), but for \(\triangle ABC\) with \(x=-4\), \(\angle B = 36^{\circ}\), \(\angle A=48^{\circ}\), \(\angle C=96^{\circ}\); for \(\triangle DEF\) with \(x=-4\), \(\angle D = 48^{\circ}\), \(\angle E=84^{\circ}\), \(\angle F=48^{\circ}\), which does not satisfy the similarity (angle - angle criterion).

Answer:

\(x = 12\)