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Find the area (i n square units) of the triangle whose vertices are (a, b + c), (a, b – c) and (–a, c).
Area of triangle
Points (6, 8), (3, 7), (–2, –2) and (1, –1) are joined to form a quadrilateral. What will be the structure of quadrilateral?
Let the points be A(6, 8), B(3, 7), C(–2, –2) and D(1, –1)
Now, AB =
Also, AC =
BD =
Since, AB = DC and BC = DA and AC ≠ BD
∴ It is a parallelogram.
Four vertices of a parallelogram taken in order are (–3, –1), (a, b), (3, 3) and (4, 3). What will be the ratio of a and b?
Let points be A (–3, –1), B(a, b), C(3, 3) and D(4, 3). So, coordinates of the midpoint of AC = coordinates of the midpoint of BD [∵ In parallelogram, diagonals bisect each other]
⇒ a = – 4 and b = –1
Now,
Find the area of the quadrilateral, the coordinates of whose angular points taken in order are (1, 1), (3, 4), (5, –2) and (4, –7).
Let ABCD be a quadrilateral with vertices
A(1, 1), B(3, 4), C(5, –2) and D(4, –7)
Area of quadrilateral ABCD
= Area of ΔABC + Area of ΔADC...(i)
Now, area of ΔABC
Similarly, area of ΔADC =
= (1/2) × 23 = 11.5 sq.units
Hence, area of quadrilateral ABCD
= 9 + 11.5 = 20.5 sq. units
If the points (a, 0), (0, b) and (1, 1) are collinear then which of the following is true?
Since, the given points A(a, 0), B(0, b) and C(1, 1) are collinear
∴ Area of ΔABC = 0
⇒ 1/2a(b − 1) + 0(1− 0) + 1(0 − b) = 0
⇒ ab – a – b = 0 ⇒ a + b = ab
Dividing both sides by ab, we get,
The points (1, 1), (–1, 5), (7, 9) and (9, 5) taken in such order that it will form a
Let the points be A(1, 1), B(–1, 5), C(7, 9) and D(9, 5)
Now, by using distance formula, we get
AB = CD, BC = DA and AC = BD
∴ ABCD forms a rectangle.
The coordinates of the midpoints of the sides of a triangle are (4, 2), (3, 3) and (2, 2). What will be the coordinates of the centroid of the triangle?
Let PQR be a triangle and A (4, 2), B (3, 3) and C (2, 2) be the midpoints of sides PQ, PR and QR respectively.
Now, G is the centroid of triangle PQR. Also, G (x, y) is the centroid of triangle formed by joining A, B and C.
⇒ G =
Three points A(1, –2), B(3, 4) and C(4, 7) form
Distance from A to B =
Distance from B to C =
Distance from C to A =
Since, AB + BC = AC
∴ Points lie on a straight line.
Find the area of triangle whose vertices are (t, t – 2), (t + 2, t + 2) and (t + 3, t).
Let A (t, t – 2), B(t + 2, t + 2) and C(t + 3, t) be the vertices of the given triangle. Then,
Area of ΔABC =1/2{t(t + 2 − t ) + (t + 2)(t − t + 2) + (t + 3)(t – 2 – t – 2)}
⇒ Area of ΔABC =1/2{2t + 2t + 4 − 4t − 12} = −4 = 4sq. units
Area of quadrilateral formed by the vertices (–1, 6), (–3, –9), (5, –8) and (3, 9) is _______ (sq. units).
Let A(–1, 6), B(–3, –9), C(5, –8) and D(3, 9) are the vertices of quadrilateral ABCD.
Then, Area of quadrilateral ABCD
= Area of ΔABC + Area of ΔACD ...(i)
Area of (ABCD) = 59 + 37 = 96 sq. units.
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