Area Under Curves — CBSE Class 12 · 2007 Board Question with Solution

Find the area of the triangle whose vertices are

(2, 7)

(1, 1)

and

(10, 8)

Answer: $\frac{47}{2}$ square units

Solution

Label the vertices

A (1, 1)

B (2, 7)

C (10, 8)

(ordered by increasing

x

).
Equation of

A B

: slope

= \frac{7 - 1}{2 - 1} = 6

, so

y = 6 x - 5

.
Equation of

A C

: slope

= \frac{8 - 1}{10 - 1} = \frac{7}{9}

, so

y = \frac{7 x + 2}{9}

.
Equation of

B C

: slope

= \frac{8 - 7}{10 - 2} = \frac{1}{8}

, so

y = \frac{x + 54}{8}

.
As

x

runs from

1

10

the lower edge is always

A C

; the upper edge is

A B

for

1 \leq x \leq 2

and

B C

for

2 \leq x \leq 10

A = \int_{1}^{2} [(6 x - 5) - \frac{7 x + 2}{9}] d x + \int_{2}^{10} [\frac{x + 54}{8} - \frac{7 x + 2}{9}] d x .

Evaluating the two integrals gives

\frac{47}{18}

and

\frac{188}{9}

respectively, whose sum is

A = \frac{47}{18} + \frac{376}{18} = \frac{423}{18} = \frac{47}{2} .

Hence the area of the triangle is

\frac{47}{2}

square units.

CBSE Class 12 Board · 2007 · Subjective