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1. The vertices of △ABC are at A(-a, 0), B(0, a) and C(a, 0). Determine whether △ABC is scalene, isosceles, or equilateral.
2. Given: Quadrilateral MATH; M(2, 1), A(4, 4), T(11, 4) and H(9, 1). Show that quadrilateral MATH is a parallelogram.
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Answers
1.
Given that
The vertices of a triangle ABC = A(-a,0) , B(0,a) and C (a,0)
We know that
The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units
Finding AB :-
Let (x1, y1) = A(-a,0) => x1 = -a and y1 = 0
Let (x2, y2) = B(0,a) => x2 = 0 and y2 = a
The distance between two points A and B
=> AB = √[(0-(-a))²+(a-0)²]
=> AB = √[(0+a)²+(a-0)²]
=> AB = √(a²+a²)
=> AB = √(2a²)
=> AB = √2 a units ---------------(1)
Finding BC :-
Let (x1, y1) = B(0,a) => x1 = 0 and y1 = a
Let (x2, y2) = C(a,0) => x2 = a and y2 = 0
The distance between two points B and C
=> BC = √[(a-0)²+(0-a)²]
=> BC = √(a²+a²)
=> BC = √(2a²)
=> BC = √2 a units ---------------(2)
Finding AC :-
Let (x1, y1) = A(-a,0) => x1 = -a and y1 = 0
Let (x2, y2) = C(a,0) => x2 = a and y2 = 0
The distance between two points A and C
=> AC = √[(a-(-a))²+(0-0)²]
=> AC = √[(a+a)²+(0)²]
=> AC = √(2a)²+0)
=> AC = √(4a²)
=> AC = 2a units ---------------(3)
We have,
AB = √2 a units
BC = √2 a units
AC = 2a units
We notice that AB = BC
The lengths of two sides of the given ∆ ABC are equal.
We know that
If two sides of a triangle are equal then the triangle is called an Isosceles triangle.
Therefore, ∆ABC is an Isosceles triangle.
_______________________________
2.
Given that
The vertices of a quadrilateral MATH are : M(2, 1), A(4, 4), T(11, 4) and H(9, 1)
To show that The Quadrilateral MATH is a Parallelogram ,we have to show that
•Two pairs of opposite sides are equal.
• Diagonals are not equal.
We know that
The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units
Finding MA :-
Let (x1, y1) = M(2,1) => x1 = 2 and y1 = 1
Let (x2, y2) = A(4,4) => x2 = 4 and y2 = 4
The distance between two points M and A
=> MA = √[(4-2)²+(4-1)²]
=> MA = √(2²+3²)
=> MA = √(4+9)
=> MA = √13 units ---------------(1)
Finding AT :-
Let (x1, y1) = A(4,4) => x1 = 4 and y1 = 4
Let (x2, y2) = T(11,4) => x2 = 11 and y2 = 4
The distance between two points A and T
=> AT = √[(11-4)²+(4-4)²]
=> AT = √(7²+0²)
=> AT = √7²
=> AT = √49
=> AT = 7 units ---------------(2)
Finding TH :-
Let (x1, y1) = T(11,4) => x1 = 11 and y1 = 4
Let (x2, y2) = H(9,1)=> x2 = 9 and y2 = 1
The distance between two points T and H
=> TH = √[(9-11)²+(1-4)²]
=> TH = √[(-2)²+(-3)²]
=> TH = √(4+9)
=> TH = √13 units ---------------(3)
Finding HM :-
Let (x1, y1) = H(9,1)=> x1 = 9 and y1 = 1
Let (x2, y2) = M(2,1) => x2 = 2 and y2 = 1
The distance between two points H and M
=> HM = √[(2-9)²+(1-1)²
=> HM = √[(-7)²+0²]
=> HM = √(49+0)
=> HM = √49
=> HM = 7 units ---------------(4)
Finding AH :-
Let (x1, y1) = A(4,4) => x1 = 4 and y1 = 4
Let (x2, y2) = H(9,1)=> x2 = 9 and y2 = 1
The distance between two points A and H
=> AH = √[(9-4)²+(1-4)²]
=> AH = √[5²+(-3)²]
=> AH = √(25+9)
=> AH = √34 units ---------------(5)
Finding MT:-
Let (x1, y1) = M(2,1) => x1 = 2 and y1 = 1
Let (x2, y2) = T(11,4) => x2 = 11 and y2 = 4
The distance between two points M and T
=> MT = √[(11-2)²+(4-1)²]
=> MT = √(9²+3²)
=> MT = √(81+9)
=> MT = √90
=> MT = √(9×10)
=> MT = 3√10 units ---------------(6)
We have,
MA = √13 units
AT = 7 units
TH = √13 units
HM = 7 units
AH = √34 units
MT = 3√10 units
We notice that
MA = TH
AT = HM
AH ≠ MT
Therefore, MATH is a Parallelogram.
Alternative Method:-
We know that
The diagonals bisect each other in a Parallelogram.
We have, AH and MT are the diagonals in MATH
We know that
The coordinates of the mid point of the line segment joining the points (x1, y1) and (x2, y2) is ((x1+x2)/2, (y1+y2)/2)
Mid point of AH :-
Let (x1, y1) = A(4,4) => x1 = 4 and y1 = 4
Let (x2, y2) = H(9,1)=> x2 = 9 and y2 = 1
The mid point of AH = ( ( 4+9)/2, (4+1)/2)
=> (13/2,5/2)
The mid point of AH = (13/2,5/2)---(1)
Midpoint of MT :-
Let (x1, y1) = M(2,1) => x1 = 2 and y1 = 1
Let (x2, y2) = T(11,4) => x2 = 11 and y2 = 4
=> Mid point of MT
=> ( (11+2)/2 ,(1+4)/2 )
=> (13/2,5/2)
The mid point of MT = (13/2,5/2)---(2)
From (1)&(2)
Mid point AH = Mid point of MT
Therefore, MATH is a Parallelogram.
Step-by-step explanation:
1.
Given that
The vertices of a triangle ABC = A(-a,0) , B(0,a) and C (a,0)
We know that
The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units
Finding AB :-
Let (x1, y1) = A(-a,0) => x1 = -a and y1 = 0
Let (x2, y2) = B(0,a) => x2 = 0 and y2 = a
The distance between two points A and B
=> AB = √[(0-(-a))²+(a-0)²]
=> AB = √[(0+a)²+(a-0)²]
=> AB = √(a²+a²)
=> AB = √(2a²)
=> AB = √2 a units ---------------(1)
Finding BC :-
Let (x1, y1) = B(0,a) => x1 = 0 and y1 = a
Let (x2, y2) = C(a,0) => x2 = a and y2 = 0
The distance between two points B and C
=> BC = √[(a-0)²+(0-a)²]
=> BC = √(a²+a²)
=> BC = √(2a²)
=> BC = √2 a units ---------------(2)
Finding AC :-
Let (x1, y1) = A(-a,0) => x1 = -a and y1 = 0
Let (x2, y2) = C(a,0) => x2 = a and y2 = 0
The distance between two points A and C
=> AC = √[(a-(-a))²+(0-0)²]
=> AC = √[(a+a)²+(0)²]
=> AC = √(2a)²+0)
=> AC = √(4a²)
=> AC = 2a units ---------------(3)
We have,
AB = √2 a units
BC = √2 a units
AC = 2a units
We notice that AB = BC
The lengths of two sides of the given ∆ ABC are equal.
We know that
If two sides of a triangle are equal then the triangle is called an Isosceles triangle.
Therefore, ∆ABC is an Isosceles triangle.
_______________________________
2.
Given that
The vertices of a quadrilateral MATH are : M(2, 1), A(4, 4), T(11, 4) and H(9, 1)
To show that The Quadrilateral MATH is a Parallelogram ,we have to show that
•Two pairs of opposite sides are equal.
• Diagonals are not equal.
We know that
The distance between two points (x1, y1) and (x2, y2) is √[(x2-x1)²+(y2-y1)²] units
Finding MA :-
Let (x1, y1) = M(2,1) => x1 = 2 and y1 = 1
Let (x2, y2) = A(4,4) => x2 = 4 and y2 = 4
The distance between two points M and A
=> MA = √[(4-2)²+(4-1)²]
=> MA = √(2²+3²)
=> MA = √(4+9)
=> MA = √13 units ---------------(1)
Finding AT :-
Let (x1, y1) = A(4,4) => x1 = 4 and y1 = 4
Let (x2, y2) = T(11,4) => x2 = 11 and y2 = 4
The distance between two points A and T
=> AT = √[(11-4)²+(4-4)²]
=> AT = √(7²+0²)
=> AT = √7²
=> AT = √49
=> AT = 7 units ---------------(2)
Finding TH :-
Let (x1, y1) = T(11,4) => x1 = 11 and y1 = 4
Let (x2, y2) = H(9,1)=> x2 = 9 and y2 = 1
The distance between two points T and H
=> TH = √[(9-11)²+(1-4)²]
=> TH = √[(-2)²+(-3)²]
=> TH = √(4+9)
=> TH = √13 units ---------------(3)
Finding HM :-
Let (x1, y1) = H(9,1)=> x1 = 9 and y1 = 1
Let (x2, y2) = M(2,1) => x2 = 2 and y2 = 1
The distance between two points H and M
=> HM = √[(2-9)²+(1-1)²
=> HM = √[(-7)²+0²]
=> HM = √(49+0)
=> HM = √49
=> HM = 7 units ---------------(4)
Finding AH :-
Let (x1, y1) = A(4,4) => x1 = 4 and y1 = 4
Let (x2, y2) = H(9,1)=> x2 = 9 and y2 = 1
The distance between two points A and H
=> AH = √[(9-4)²+(1-4)²]
=> AH = √[5²+(-3)²]
=> AH = √(25+9)
=> AH = √34 units ---------------(5)
Finding MT:-
Let (x1, y1) = M(2,1) => x1 = 2 and y1 = 1
Let (x2, y2) = T(11,4) => x2 = 11 and y2 = 4
The distance between two points M and T
=> MT = √[(11-2)²+(4-1)²]
=> MT = √(9²+3²)
=> MT = √(81+9)
=> MT = √90
=> MT = √(9×10)
=> MT = 3√10 units ---------------(6)
We have,
MA = √13 units
AT = 7 units
TH = √13 units
HM = 7 units
AH = √34 units
MT = 3√10 units
We notice that
MA = TH
AT = HM
AH ≠ MT
Therefore, MATH is a Parallelogram.