Question

© a In Fig. 6.28, show that APAX is similar to

AYAQ.
b Hence show that
Pal X JAQI = YAI X laxl.
c IfXal = 14 cm, Pal = 6 cm and
IYA) = 3 cm, calculate (AQI.
X
Р
А
Q
Y​

Answer 1

The missing figure is attached down

Answer:

ΔPAX ≈ ΔYAQ ⇒ proved down

PA × AQ = YA × AX ⇒ proved down

AQ = 7 cm

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

• If ΔABC ≈ Δ XYZ, then

• 
  `(AB)/(XY)=(BC)/(YZ)=(AC)/(XZ)` ⇒⇒⇒ 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

∴
`(PA)/(YA)=(AX)/(AQ)=(PX)/(YQ)`

∵
`(PA)/(YA)=(AX)/(AQ)`

→ 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