prove the points A(1,-3),B(-3,0) and C(4,1) are the vertices of an isosceles right angled triangle. Also find the area of the triangle
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area of triangle= 25/2=12.5
another method:
Vertices of the triangle are A(-3, 0), B(1, -3), C(4, 1) Distance between two points = (x2 - x1)2+(y2 - y1)2 AB = √[(1 + 3)2+ (-3 -0)2] = 5 BC = √[(4 - 1)2 + (1 + 3)2] = 5AC = √[(4 + 3)2 + (1 -0)2] = 50 = 52 AB = BC Therefore, ΔABC is an isosceles triangle. 52 + 52 = (52 )2 ⇒ 50 = 50 ⇒ (AB)2 + (BC)2 = (AC)2 So, the triangle satisfies the pythagoras theorem and hence it is a right angled triangle. Area of the isosceles right angled triangle = 1 2 ×base × height = 1 2 × 5 × 5 = 12.5 sq units.
another method:
Vertices of the triangle are A(-3, 0), B(1, -3), C(4, 1) Distance between two points = (x2 - x1)2+(y2 - y1)2 AB = √[(1 + 3)2+ (-3 -0)2] = 5 BC = √[(4 - 1)2 + (1 + 3)2] = 5AC = √[(4 + 3)2 + (1 -0)2] = 50 = 52 AB = BC Therefore, ΔABC is an isosceles triangle. 52 + 52 = (52 )2 ⇒ 50 = 50 ⇒ (AB)2 + (BC)2 = (AC)2 So, the triangle satisfies the pythagoras theorem and hence it is a right angled triangle. Area of the isosceles right angled triangle = 1 2 ×base × height = 1 2 × 5 × 5 = 12.5 sq units.
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the answer of thisquestion is 2sq units
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2sq units is the answer of this question
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