© 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

The missing figure is attached down

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

If two triangles are similar, then their corresponding 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

∴

∵  

→ 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

answer: 42

use a calculator to find that 845/73 = 11.575

the whole part of the answer is 11. multiply this by 73 to get 11*73 = 803.

subtract that result from 845 to get 845 - 803 = 42

ddd

step-by-step explanation: