**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).

= 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).

=
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.

*k*) + 5{

*k*− (−2)} + 3(−2 − 1)] = 0

⇒ ½ (7 − 7*k* + 5*k* + 10 − 9) = 0

⇒ ½ (7 − 7*k* + 5*k* + 1) = 0

⇒ ½ (8 − 2*k*) = 0

⇒ 8 − 2*k* = 0

⇒ 2*k*
= 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.

⇒ ½ [8{−4 − (−5)} + *k*(−5 − 1) + 2{1 − (−4)}] = 0

= ½
(8 − 6*k* + 10) = 0

⇒ ½ (18 − 6*k*) = 0

⇒18 − 6k = 0

⇒ 18 = 6*k*

⇒ *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) = (*x*_{1}, *y*_{1}),
B(2, 1) = (*x*_{2}, *y*_{2}) and C(0, 3) = (*x*_{3}, *y*_{3}) are the vertices of the given triangle.

⇒ 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)

**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:

=
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:

= 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).

*x*,

*y*).

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:

= 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:

= 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.