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QUESTION IMAGE

the graph shows triangles jkl and opq. is jkl similar to opq? justify y…

Question

the graph shows triangles jkl and opq. is jkl similar to opq? justify your answer.

Explanation:

Step1: Find the lengths of the sides

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) or count the units for vertical/horizontal sides.
For \(JK\): From \(J(-3,1)\) to \(K(2,-3)\), using the distance formula \(d=\sqrt{(2 + 3)^2+(-3 - 1)^2}=\sqrt{25 + 16}=\sqrt{41}\). For \(KL\): From \(K(2,-3)\) to \(L(-3,-2)\), \(d=\sqrt{(-3 - 2)^2+(-2+ 3)^2}=\sqrt{25 + 1}=\sqrt{26}\). For \(LJ\): From \(L(-3,-2)\) to \(J(-3,1)\), \(d=\sqrt{(-3+3)^2+(1 + 2)^2}=3\).
For \(OP\): From \(O(-3,1)\) to \(P(7,-7)\), \(d=\sqrt{(7 + 3)^2+(-7 - 1)^2}=\sqrt{100+64}=\sqrt{164}=2\sqrt{41}\). For \(PQ\): From \(P(7,-7)\) to \(Q(-3,-6)\), \(d=\sqrt{(-3 - 7)^2+(-6 + 7)^2}=\sqrt{100 + 1}=\sqrt{101}\). For \(QO\): From \(Q(-3,-6)\) to \(O(-3,1)\), \(d=\sqrt{(-3+3)^2+(1 + 6)^2}=7\).

Step2: Check the ratio of sides

\(\frac{JK}{OP}=\frac{\sqrt{41}}{2\sqrt{41}}=\frac{1}{2}\), \(\frac{LJ}{QO}=\frac{3}{7}
eq\frac{1}{2}\).

Answer:

No. The ratios of the corresponding sides are not equal.