PRELIMINARY STAGE 2018
OLA Girls SHS, Ho: 31 points
Ghanata SHS: 29 points
Mim SHS: 10 points
SPEED RACE
QUESTION
QUESTION
Determine if triangle ABC is right-angled, given $A(-3,-4)$, $B(2,5)$ and $C(-5,3)$.
ANSWER: Yes, it is right-angled at C
SOLUTION
Find the squared lengths of all three sides using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$:
$AB^2 = (2 - (-3))^2 + (5 - (-4))^2 = 5^2 + 9^2 = 25 + 81 = 106$
$BC^2 = (-5 - 2)^2 + (3 - 5)^2 = (-7)^2 + (-2)^2 = 49 + 4 = 53$
$AC^2 = (-5 - (-3))^2 + (3 - (-4))^2 = (-2)^2 + 7^2 = 4 + 49 = 53$
Check Converse of Pythagoras' Theorem ($BC^2 + AC^2 = AB^2$):
$53 + 53 = 106$
Since $106 = 106$, the triangle satisfies Pythagoras' theorem, meaning it is right-angled at C.
PRACTICE QUESTIONS
1. Determine whether triangle ABC is right-angled, given $A(1,1)$, $B(4,1)$ and $C(1,5)$.
ANSWER: Yes, it is right-angled at A
SOLUTION 1
$AB^2=(4-1)^2+(1-1)^2=9$
$BC^2=(1-4)^2+(5-1)^2=25$
$CA^2=(1-1)^2+(1-5)^2=16$
$AB^2+CA^2=9+16$
$AB^2+CA^2=25=BC^2$
The triangle is right-angled at A.
SOLUTION 2
Take the vectors from A:
$\overrightarrow{AB}=(3,0)$
$\overrightarrow{AC}=(0,4)$
$\overrightarrow{AB}\cdot\overrightarrow{AC}=(3)(0)+(0)(4)$
$=0+0$
$=0$
The vectors are perpendicular, so the triangle is right-angled at A.
2. Determine whether triangle ABC is right-angled, given $A(0,0)$, $B(4,3)$ and $C(-3,4)$.
ANSWER: Yes, it is right-angled at A
SOLUTION 1
$AB^2=(4-0)^2+(3-0)^2=25$
$BC^2=(-3-4)^2+(4-3)^2=50$
$CA^2=(0-(-3))^2+(0-4)^2=25$
$AB^2+CA^2=25+25$
$AB^2+CA^2=50=BC^2$
The triangle is right-angled at A.
SOLUTION 2
Take the vectors from A:
$\overrightarrow{AB}=(4,3)$
$\overrightarrow{AC}=(-3,4)$
$\overrightarrow{AB}\cdot\overrightarrow{AC}=(4)(-3)+(3)(4)$
$=-12+12$
$=0$
The vectors are perpendicular, so the triangle is right-angled at A.
3. Determine whether triangle ABC is right-angled, given $A(2,1)$, $B(5,2)$ and $C(3,5)$.
ANSWER: No, it is not right-angled
SOLUTION
$AB^2=(5-2)^2+(2-1)^2=10$
$BC^2=(3-5)^2+(5-2)^2=13$
$CA^2=(2-3)^2+(1-5)^2=17$
$AB^2+BC^2=10+13=23$
$CA^2=17$
Since 23 is not equal to 17, the triangle is not right-angled.
4. Determine whether triangle ABC is right-angled, given $A(0,1)$, $B(2,5)$ and $C(-4,3)$.
ANSWER: Yes, it is right-angled at A
SOLUTION 1
$AB^2=(2-0)^2+(5-1)^2=20$
$BC^2=(-4-2)^2+(3-5)^2=40$
$CA^2=(0-(-4))^2+(1-3)^2=20$
$AB^2+CA^2=20+20$
$AB^2+CA^2=40=BC^2$
The triangle is right-angled at A.
SOLUTION 2
Take the vectors from A:
$\overrightarrow{AB}=(2,4)$
$\overrightarrow{AC}=(-4,2)$
$\overrightarrow{AB}\cdot\overrightarrow{AC}=(2)(-4)+(4)(2)$
$=-8+8$
$=0$
The vectors are perpendicular, so the triangle is right-angled at A.
5. Determine whether triangle ABC is right-angled, given $A(1,2)$, $B(5,2)$ and $C(1,6)$.
ANSWER: Yes, it is right-angled at A
SOLUTION 1
$AB^2=(5-1)^2+(2-2)^2=16$
$BC^2=(1-5)^2+(6-2)^2=32$
$CA^2=(1-1)^2+(2-6)^2=16$
$AB^2+CA^2=16+16$
$AB^2+CA^2=32=BC^2$
The triangle is right-angled at A.
SOLUTION 2
Take the vectors from A:
$\overrightarrow{AB}=(4,0)$
$\overrightarrow{AC}=(0,4)$
$\overrightarrow{AB}\cdot\overrightarrow{AC}=(4)(0)+(0)(4)$
$=0+0$
$=0$
The vectors are perpendicular, so the triangle is right-angled at A.
6. Determine whether triangle ABC is right-angled, given $A(0,0)$, $B(6,2)$ and $C(2,7)$.
ANSWER: No, it is not right-angled
SOLUTION
$AB^2=(6-0)^2+(2-0)^2=40$
$BC^2=(2-6)^2+(7-2)^2=41$
$CA^2=(0-2)^2+(0-7)^2=53$
$AB^2+BC^2=40+41=81$
$CA^2=53$
Since 81 is not equal to 53, the triangle is not right-angled.