AREA OF TRIANGLE WITH THREE VERTICES

Here, we are going to see, how to find area of a triangle when coordinates of the three vertices are given. 

Let us consider the triangle given below. 

Practice Problems

Problem 1 :

Find the area of the triangle whose vertices are 

(1,–1), (–4, 6) and (–3, –5)

Solution :

Area of triangle ABC =

= (1/2)[(6 + 20 + 3) - (4 - 18 - 5)]

= (1/2)[29 - (-19)]

= (1/2)[29 + 19]

= (1/2)48

  = 24 square units

Problem 2 :

Find the area of the triangle whose vertices are 

(-10, -4) (-8, -1) and (-3, -5)

Solution :

= (1/2)[(32 + 50 + 3) - (10 + 12 + 40)]

= (1/2)[85 - 62]

= 23/2

= 11.5 square units

Problem 3 :

Determine whether the sets of points are collinear.

(-1/2, 3), (-5, 6) and (-8, 8)

Solution :

= (1/2)[(-3 - 40 - 24) - (-15 - 48 - 4)]

= (1/2)[(-67) - (-67)]

= (1/2)(-67 + 67)

= 0

Because the area of the triangle is zero, the given points are collinear.

Problem 4 :

Determine whether the sets of points are collinear.

(a, b + c), (b, c + a) and (c, a + b)

Solution :

= (1/2)[a(c+a) + b(a+b) + c(b+c) - b(b+c)-c(c+a)-a(a+b)]

= (1/2)[ac+a²+ba+b²+cb+c²-b²-bc-c²-ac-a²-ab]

= 0

Problem 5 :

Find the value of p in each case. 

Solution (i) :

Area of triangle = 20 square units

(1/2)[(0 + 2p + 0) - (0 + 48 + 0)] = 20

2p - 48 = 40

2p = 40 + 48

2p = 88

p = 44

Solution (ii) :

Area of triangle = 32 square units

(1/2)[(6p - 10 + 5p) - (5p + 30 - 2p)] = 32

(11p - 10) - (3p + 30) = 64

11p - 3p - 10 - 30 = 64

8p = 64 + 40

8p = 104

  p = 104/8

p = 13

Problem 6 :

If the area of the triangle formed by the vertices A(-1 , 2), B (k ,-2) and C(7, 4) (taken in order) is 22 sq. units, find the value of k.

Solution :

Area of triangle ABC = 22 square units

= (1/2)[(2 + 4k + 14) - (2k - 14 - 4)]

= (1/2)[(4k + 16) - (2k - 18)]

= (1/2)[4k + 16 - 2k + 18]

= (1/2)[2k + 34]

= k + 17

Equating this area to 22 square units, we get

k + 17 = 22

k = 22 - 17

k = 5

So, the value of k is 5.

Problem 7 :

If the points P(-1, -4), Q(b, c) and R(5, -1) are collinear and if 2b + c = 4, then find the values of b and c.

Solution :

Since the given point P, Q and R are collinear,

(-c - b - 20) - (-4b + 5c + 1)] = 0

[-c - b - 20 + 4b - 5c - 1] = 0

[-6c + 3b - 21] = 0

-6c + 3b = 21

Dividing by 3, we get

-2c + b = 7----(1)

c + 2b = 4 ----(2)

(1) + (2) x 2 ==> -2c + b + 2c + 4b = 7 + 8

5b = 15

b = 15/5

b = 3

Applying the value of b in (1), we get

-2c + 3 = 7

-2c = 7 - 3

-2c = 4

c = 4/(-2)

c = -2

So, the values of b and c are -2 and 3 respectively.

Problem 8 :

The floor of a hall is covered with identical tiles which are in the shapes of triangles. One such triangle has the vertices at (-3, 2), (-1, -1) and (1, 2). If the floor of the hall is completely covered by 110 tiles, find the area of the floor.

Solution :

Let the vertices be A(-3, 2), B(-1, -1) and C(1, 2)

Area of one triangular tile = 

= (1/2)[(3 - 2 + 2) - (-2 - 1 - 6)]

= (1/2)[(3 - (-9)]

= (1/2)[3 + 9]

= (1/2)(12)

= 6 square units.

Area of 110 tiles = 6(110)

= 660 square units.

Problem 9 :

The given diagram shows a plan for constructing a new parking lot at a campus. It is estimated that such construction would cost $1300 per square feet. What will be the total cost for making the parking lot?

Solution :

Area of quadrilateral =

= (1/2) [(10 + 45 + 28 + 2) - (10 + 20 + 9 + 14)]

= (1/2) [85 - 53]

= (1/2) (32)

= 16 square units 

Cost per square feet = $1300

Total cost = 16(1300)

= $20800

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