The area of a triangle with vertices A (3, 0), B (7, 0) and C (8, 4) is
a. 14
b. 28
c. 8
d. 6

Solution:

The vertices of the triangle are

A (3, 0), B (7, 0) and C (8, 4)

The formula to find the area of a triangle is

Area = 1/2 [x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

Substituting the values

Area = 1/2 [3(0 - 4) + 7(4 - 0) + 8(0 - 0)]

By further calculation

Area = 1/2 [3(-4) + 7(4) + 8(0)]

Area = 1/2 [-12 + 28]

So we get

Area = 1/2 [16]

Area = 8 sq. units

Therefore, the area of a triangle is 8 sq. units.

✦ Try This: The area of a triangle with vertices A (2, 0), B (5, 0) and C (6, 3) is

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 7

NCERT Exemplar Class 10 Maths Exercise 7.1 Problem 7

The area of a triangle with vertices A (3, 0), B (7, 0) and C (8, 4) is a. 14, b. 28, c. 8, d. 6

Summary:

The area of a triangle with vertices A (3, 0), B (7, 0) and C (8, 4) is 8 sq. units

☛ Related Questions:

• The points (–4, 0), (4, 0), (0, 3) are the vertices of a a. right triangle, b. isosceles triangle, c . . . .
• The point which divides the line segment joining the points (7, –6) and (3, 4) in ratio 1 : 2 intern . . . .
• The point which lies on the perpendicular bisector of the line segment joining the points A (–2, –5) . . . .