Maths Class 10 Exercise 7.3
1. Find the area of the triangle whose
vertices are:
(i) (2, 3), (–1, 0), (2, –4)
(ii) (–5, –1), (3, –5), (5, 2)
Solution: (i) The vertices of the triangle are
(2, 3), (–1, 0), (2, –4).
Area
of a triangle = 
= ½
[2(0 + 4) – 1(−7) + 2(3)]
= ½ (8 + 7 + 6) = 21/2 sq. units
(ii) The vertices of the triangle are (–5,
–1), (3, –5), (5, 2).
Area
of a triangle =
=
½ [−5(−5 − 2) + 3{2 − (−1)} + 5{−1 − (−5)}]
= ½
[−5(−7) + 3(3) + 5(4)]
= ½
(35 + 9 + 20)
= ½
(64)
=
32 sq. units
2. In each of the following, find the value of
‘k’, for which the points are
collinear.
(i) (7, –2), (5, 1), (3, k)
(ii) (8, 1), (k, –4), (2, –5)
Solution: (i) The vertices of the triangle are
(7, –2), (5, 1), (3, k).
Since the given
points are collinear, no triangle will be formed and hence the area of the triangle
formed by them is equal to zero.
Area
of a triangle = 
⇒ ½ (7 − 7k + 5k + 10 − 9) = 0
⇒ ½ (7 − 7k + 5k + 1) = 0
⇒ ½ (8 − 2k) = 0
⇒ 8 − 2k = 0
⇒ 2k
= 8
⇒ k
= 4
(ii) The vertices of the triangle are (8, 1),
(k, –4), (2, –5).
Since the given
points are collinear, no triangle will be formed and hence the area of the triangle
formed by them is equal to zero.
Area of a triangle
=
⇒ ½ [8{−4 − (−5)} + k(−5 − 1) + 2{1 − (−4)}] = 0
= ½
(8 − 6k + 10) = 0
⇒ ½ (18 − 6k) = 0
⇒18 − 6k = 0
⇒ 18 = 6k
⇒ k
= 3
3. Find the area of the triangle formed by joining
the mid-points of the sides of the triangle whose vertices are (0, –1), (2, 1)
and (0, 3). Find the ratio of this area to the area of the given triangle.
Solution: Let A(0, –1) = (x1, y1),
B(2, 1) = (x2, y2) and C(0, 3) = (x3, y3) are the vertices of the given triangle.
Area
of △ABC =
⇒ Area of △ABC = ½ [0(1 − 3) + 2{3 − (−1)} +
0(−1 − 1)] = ½ × 8
= 4 sq. units
Let P, Q and R be the mid-points of the sides AB, AC and BC respectively.
Applying
the section formula to find the coordinates of P, Q and R, we get
Area
of △PQR = ½ [1(1 − 2) + 0(2 − 0) + 1(0 − 1)] = ½ (−2)
= 1 sq. units (Neglecting the minus sign)
Now, 
4. Find the area of the quadrilateral whose
vertices, taken in order, are (–4, –2), (–3, –5), (3, –2) and (2, 3).
Solution: Area of quadrilateral ABCD = Area of triangle
ABD + Area of triangle BCD ….. (i)
Using the formula to find the area of triangle ∆ABD:
Area
of ∆ABD =
=
½ [−4(−5 − 3) – 3{3 − (−2)} + 2{−2 − (−5)}]
= ½
(32 – 15 + 6)
= ½
(23) = 11.5 sq. units …… (ii)
Again, using the
formula to find the area of triangle △BCD:
Area
of △BCD =
= ½
[−3(−2 − 3) + 3{3 − (−5)} + 2{−5 − (−2)}]
= ½
(15 + 24 − 6)
= ½
(33) = 16.5 sq. units …… (iii)
Putting
the values from equations (ii) and (iii) in equation (i), we get
Area
of quadrilateral ABCD = 11.5 + 16.5 = 28 sq. units
5. We know that the median of a triangle
divides it into two triangles of equal areas. Verify this result for △ABC whose vertices are A(4,
–6), B(3, –2) and C(5, 2).
Solution: We are given a △ABC whose vertices are A(4, –6), B(3, –2) and C(5, 2),
and AD is one of its medians.
We
need to show that ar(△ABD) = ar(△ACD).
Using
the section formula to find the coordinates of point D, we get
Therefore,
the coordinates of point D are (4, 0).
Using
the formula to find the area of triangle △ABD:
Area
of △ABD =
= ½
[4(−2 − 0) + 3{0 − (−6)} + 4{−6 − (−2)}]
=
½ (−8 + 18 −16)
=
½ (−6)
=
−3 sq. units
Therefore,
area of △ABD = 3 sq. units (Area cannot be negative) …… (i)
Again
using the formula to find the area of triangle △ACD:
Area
of △ACD =
= ½ [4(2 − 0) + 5{0 − (−6)} + 4(−6 −2)]
= ½ (8 + 30 − 32)
=
½ (6)
Therefore,
area of △ACD = 3 sq. units …… (ii)
From
equations (i) and (ii), we get ar(△ABD) = ar(△ACD)
Hence
Proved.