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The area (in sq. units) of the triangle whose vertices are (0,0)(0, 0), (a,0)(a, 0), (0,b)(0, b), is equal to

Solution

✅ Correct Option: 3

The triangle has vertices at (0,0)(0, 0), (a,0)(a, 0), and (0,b)(0, b).

The point (0,0)(0, 0) is at the origin.

The point (a,0)(a, 0) lies on the x-axis at distance ∣a∣|a| from the origin.

The point (0,b)(0, b) lies on the y-axis at distance ∣b∣|b| from the origin.

This forms a right-angled triangle with the right angle at the origin.


The base of the triangle is ∣a∣|a| units.

The height of the triangle is ∣b∣|b| units.

Absolute values are used since aa and bb can be negative, but distances are always positive.


Area =12×base×height= \dfrac{1}{2} \times \text{base} \times \text{height}

Area =12×∣a∣×∣b∣= \dfrac{1}{2} \times |a| \times |b|

Area =12∣ab∣= \dfrac{1}{2}|ab|


Therefore, the area of the triangle is 12∣ab∣\dfrac{1}{2}|ab| square units.

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