Let the floor of a hall is covered with identical tiles which are in shape of triangle. One of the triangle has the vertices at $(- 3, 2), (- 1, - 1)$ and $(1, 2)$ . If the floor of the hall is completely covered by $110$ tiles, then the area of floor (in sq.units) is

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Solution

Given vertices of the triangle are
$(- 3, 2), (- 1, - 1)$ and $(1, 2)$

Now, the area of triangular tiles
$= \frac{1}{2} ∣ ∣ ∣ \begin{matrix} x_{1} & x_{2} & x_{3} & x_{1} y_{1} & y_{2} & y_{3} & y_{1} \end{matrix} ∣ ∣ ∣ = \frac{1}{2} ∣ ∣ ∣ \begin{matrix} - 3 & - 1 & 1 & - 3 2 & - 1 & 2 & 2 \end{matrix} ∣ ∣ ∣ = \frac{1}{2} | (3 + 2) + (- 2 + 1) + (2 + 6) | = \frac{1}{2} | 12 | = 6 sq. units$

$∴$ Total required area
$= 6 \times 110 = 660 sq.units$

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Let the floor of a hall is covered with identical tiles which are in shape of triangle. One of the triangle has the vertices at (−3,2),(−1,−1) and (1,2). If the floor of the hall is completely covered by 110 tiles, then the area of floor (in sq.units) is

Given vertices of the triangle are (−3,2),(−1,−1) and (1,2) Now, the area of triangular tiles =12∣∣∣x1x2x3x1y1y2y3y1∣∣∣=12∣∣∣−3−11−32−122∣∣∣=12|(3+2)+(−2+1)+(2+6)|=12|12|=6 sq. units ∴ Total required area =6×110=660 sq.units

Let the floor of a hall is covered with identical tiles which are in shape of triangle. One of the triangle has the vertices at $(- 3, 2), (- 1, - 1)$ and $(1, 2)$ . If the floor of the hall is completely covered by $110$ tiles, then the area of floor (in sq.units) is