ABC is a right angled triangle, B being the right angle. Midpoints of ...
Formulas:
Area of a triangle
Area of a trapezium =
Triangle ABC and A’B’C are similar by SAS
By property of similar triangles,
∠B = ∠A’B’C and 2A’B’ = AB
∴ A’B’ is parallel to AB.
Thus quadrilateral AA’ B’B is a trapezium.
In quadrilateral AA’ B’B,
Height = BB’
⇒ Height = BC/2
Sum of parallel sides = (A’B’ + AB) = 3AB/2
Area of quadrilateral AA’ B’B
⇒ Area of quadrilateral AA’ B’B
Area of triangle ABC
Ratio area of quadrilateral AA’ B’B to area of triangle ABC
⇒ Ratio of area of quadrilateral AA’ B’B to area of triangle ABC = 3 : 4
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ABC is a right angled triangle, B being the right angle. Midpoints of ...
1 and M2. Let P be the midpoint of AB. We need to prove that
(i) AM1 = BM2 = PM1 = PM2
(ii) Quadrilateral AM1PB is a square.
Proof:
(i) As P is the midpoint of AB, we have AP = PB. Also, as M1 and M2 are midpoints of BC and AC respectively, we have BM1 = CM1 and AM2 = CM2.
Since ABC is a right angled triangle, we have BM2 = BC/2 and AM1 = AC/2. Therefore, to prove AM1 = BM2, we need to prove that AC = BC. But this is true because ABC is a right angled triangle with B as the right angle.
To prove PM1 = PM2, we use the fact that P is the midpoint of AB. Therefore, PM1 is the median of triangle PBC and PM2 is the median of triangle PAC. But as ABC is a right angled triangle, we have BP = PC and AP = PC. Therefore, PM1 = PM2.
(ii) To prove that AM1PB is a square, we need to prove that all its sides are equal and the diagonals are equal and perpendicular to each other.
As we have already proved that AM1 = BM2 and PM1 = PM2, it follows that AM1PB is a rhombus with equal sides AM1, BM1, PM1, and PM2.
To prove that the diagonals are equal and perpendicular, we use the fact that P is the midpoint of AB and M1 and M2 are midpoints of BC and AC respectively. Therefore, PM1 is perpendicular to BC and PM2 is perpendicular to AC. Also, PB is perpendicular to BC and PA is perpendicular to AC. Therefore, PB || AC and PA || BC. Therefore, PM1 and PM2 are the perpendicular bisectors of AB and PB and PA are the perpendicular bisectors of BC.
Therefore, the diagonals of AM1PB are PB and PM1 and they are equal and perpendicular to each other. Hence, AM1PB is a square.