Answer:
Question 1
Pythagoras' Theorem: a² + b² = c²
(where a and b are the legs, and c is the hypotenuse, or a right triangle)
Given:
- a = BC
- b = AB = 8cm
- c = AC = 10 cm
⇒ BC² + 8² = 10²
⇒ BC² = 36
⇒ BC = 6 cm
Question 2
Given:
⇒ MB = AB - AM = 8 - x
As ΔABC ~ ΔMBN
⇒ AB : AC = MB : MN
⇒ 8 : 10 = (8 - x) : MN
[tex]\sf \implies \dfrac{8}{10}=\dfrac{(8-x)}{MN}[/tex]
[tex]\sf \implies \dfrac45=\dfrac{(8-x)}{MN}[/tex]
[tex]\sf \implies MN=\dfrac54(8-x)[/tex]
Question 3
D = (6, 8)
As ΔBAD ~ ΔMAP
⇒ AB : AD : BD = AM : AP : MP
⇒ 8 : 6 : 10 = x : AP : MP
[tex]\sf \implies \dfrac{8}{6}=\dfrac{x}{AP}[/tex]
[tex]\sf \implies AP=\dfrac34x[/tex]
[tex]\sf \implies \dfrac{8}{10}=\dfrac{x}{MP}[/tex]
[tex]\sf \implies MP=\dfrac54x[/tex]
Question 4
Perimeter of MNCDP = MN + NC + CD + PD + MP
As ΔABC ~ ΔMBN
⇒ AB : BC : AC = MB : BN : MN
⇒ 8 : 6 = (8 - x) : BN
[tex]\sf \implies \dfrac86=\dfrac{(8-x)}{BN}[/tex]
[tex]\sf \implies BN=\dfrac34(8-x)[/tex]
NC = BC - BN
[tex]\sf \implies NC=6-\dfrac34(8-x)[/tex]
PD = 6 - AP
[tex]\sf \implies PD=6-\dfrac34x[/tex]
Perimeter of MNCDP = MN + NC + CD + PD + MP
[tex]\sf \implies perimeter=\dfrac54(8-x)+6-\dfrac34(8-x)+8+6-\dfrac34x+\dfrac54x[/tex]
[tex]\sf \implies perimeter=10-\dfrac54x+6-6+\dfrac34x+8+6-\dfrac34x+\dfrac54x[/tex]
[tex]\sf \implies perimeter=10+8+6=24[/tex]
