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Triangle PQR ≅ Triangle XYZ
PQ = 3a + 4 and XY = 5a – 12. Find a and PQ.

Sagot :

Given:

[tex]\Delta PQR\cong \Delta XYZ[/tex]

[tex]PQ=3a+4[/tex]

[tex]XY=5a-12[/tex]

To find:

The value of a and PQ.

Solution:

We have,

[tex]\Delta PQR\cong \Delta XYZ[/tex]

[tex]PQ\cong XY[/tex]                            (CPCTC)

So,

[tex]PQ=XY[/tex]

[tex]3a+4=5a-12[/tex]

Isolating variable terms, we get

[tex]3a-5a=-4-12[/tex]

[tex]-2a=-16[/tex]

Divide both sides by -2.

[tex]a=\dfrac{-16}{-2}[/tex]

[tex]a=8[/tex]

Now,

[tex]PQ=3a+4[/tex]

[tex]PQ=3(8)+4[/tex]

[tex]PQ=24+4[/tex]

[tex]PQ=28[/tex]

Therefore, the value of a is 8 and the value of PQ is 28.

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