The missing figure is attached down
Answer:
ΔPAX ≈ ΔYAQ ⇒ proved down
PA × AQ = YA × AX ⇒ proved down
AQ = 7 cm
Step-by-step explanation:
In two triangles: if their corresponding angles have the same measures, then they are similar
- In triangles ABC and XYZ, if
- m∠ABC = m∠XYZ, m∠BCA = m∠YZX, m∠CAB = m∠ZXY
- Then, ΔABC ≈ ΔXYZ
If two triangles are similar, then their corresponding sides are proportion
- [tex]\frac{AB}{XY}=\frac{BC}{YZ}=\frac{AC}{XZ}[/tex] ⇒⇒⇒ Their sides are proportion
In our question:
From the attached figure
∵ YQ // PX
∴ m∠AYQ = m∠APX ⇒ corresponding angles
∴ m∠AQY = m∠AXP ⇒ corresponding angles
∵ ∠A is a common angle
∴ ΔPAX ≈ ΔYAQ ⇒ proved
∵ ΔPAX ≈ ΔYAQ
∴ Their corresponding sides are proportion
→ side PA is corresponding to side YA
→ side AX is corresponding to side AQ
→ side PX is corresponding to side YG
∴ [tex]\frac{PA}{YA}=\frac{AX}{AQ}=\frac{PX}{YQ}[/tex]
∵ [tex]\frac{PA}{YA}=\frac{AX}{AQ}[/tex]
→ By using cross multiplication
∴ PA × AQ = YA × AX ⇒ proved
∵ AX = 14 cm
∵ PA = 6 cm
∵ YA = 3 cm
→ Substitute these values in the relation above
∴ 6 × AQ = 3 × 14
∴ 6 AQ = 42
→ Divide both sides by 6
∴ AQ = 7 cm
