Class 10 Area of a Triangle (Coordinates) — practice questions with answers

Geometry — Coordinates · 5 sample questions from the school syllabus. Try each one, then open the answer and the worked solution.

1. What is the formula for the area of a triangle with vertices (x1,y1), (x2,y2), (x3,y3)?

A $\frac{1}{2}$ |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
B $\frac{1}{2}$ |x1(y2 + y3) + x2(y3 + y1) + x3(y1 + y2)|
C|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
D $\frac{1}{2}$ |x1 + x2 + x3 - y1 - y2 - y3|

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Answer: A)

\frac{1}{2}

|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|

The standard coordinate area formula is Area =

\frac{1}{2}

|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|.

2. The vertices (0,0), (a,0) and (0,b) form a triangle of area:

A $\frac{1}{2}$ |ab|
B|ab|
C $\frac{1}{2}$ (a+b)
D $\frac{ab}{4}$

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Answer: A)

\frac{1}{2}

|ab|

Right triangle with legs a and b. Area =

\frac{1}{2}

|a||b| =

\frac{1}{2}

|ab|.

3. Find the area of the triangle with vertices (0,0), (1,2) and (2,1).

A1
B2
C3
D1.5

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Answer: D) 1.5

Area =

\frac{1}{2}

|0(2-1) + 1(1-0) + 2(0-2)| =

\frac{1}{2}

|0 + 1 - 4| =

\frac{1}{2}

(3) = 1.5 sq units.

4. The vertices of a triangle are A(0,-1), B(2,1) and C(0,3).
Find the area of triangle ABC.

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Answer: D) 4

AC is vertical from (0,-1) to (0,3): length = 4 = base. Horizontal distance of B from the y-axis = 2 = height. Area =

\frac{1}{2}

(4)(2) = 4 sq units.

5. Triangle ABC has A(1,1), B(7,1), C(3,5). A point P lies on AB so that area of triangle APC = 6 sq units.
Given area of ABC = 12 sq units, what fraction of AB is AP?

A $\frac{1}{2}$
B $\frac{1}{3}$
C $\frac{2}{3}$
D $\frac{1}{4}$

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Answer: A)

\frac{1}{2}

Triangles APC and ABC share the same height from C to line AB, so their areas are proportional to bases AP and AB. AP/AB = area(APC)/area(ABC) =

\frac{6}{12}

\frac{1}{2}

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