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In figure, the area of triangle ABC (in sq. units) is:

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15

7.5

2.5

10

answer is C.

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Detailed Solution

Given Coordinates of Point

A (1, 3), B (- 1, 0)

and

C (4, 0)

,
Drop a perpendicular from

A

x

-axis, which meets x-axis at

D = (1, 0)

Area of a triangle =

\frac{1}{2} \times base \times height

Now in

ΔADC, AD = 3 cm, DC = 3 cm

Area of

ΔADC = \frac{1}{2} \times DC \times AD

\Rightarrow Area of ΔADC = \frac{1}{2} \times 3 cm \times 3 cm

\Rightarrow Area of ΔADC = \frac{9}{2} c m^{2}

Now in

ΔADB, AD = 3 cm, DB = 2 cm

Area of

ΔADB = \frac{1}{2} \times DB \times AD

\Rightarrow Area of ΔADB = \frac{1}{2} \times 2 cm \times 3 cm

\Rightarrow Area of ΔADB = 3 c m^{2}

Area of

ΔABC =

Area of

ΔADC +

Area of

ΔABD

\Rightarrow Area of ΔABC = \frac{9}{2} c m^{2} + 3 c m^{2}

\Rightarrow Area of ΔABC = \frac{15}{2}

\Rightarrow Area of ΔABC = 7.5 c m^{2}

Thus, the required area of the triangle is 7.5

c m^{2}

.
Therefore, option (3) is correct.

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In figure, the area of triangle ABC (in sq. units) is:

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