Question

A floor design is made on a floor of a room by joining 4 triangular tiles of dimensions 12cm, 20cm and 24cm. Find the cost of the tile at the rate of Rs. 14 per square meter.

Answer 1

Hint: All the sides of a triangle are given. Use Heron’s formula to find the area of the triangle. The formula is given by \[A=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}\]. The cost of each tile will be the product of the area found and the cost of each tile.

Complete step-by-step answer:
It is said that a floor design is made by joining 4 triangular tiles. We have been given the dimensions of the triangles as 12cm, 20cm and 24cm each.
Let us consider ABC as the triangle. Let AB = 12cm, BC = 20 cm and AC = 24cm.

We have been given the length of all three sides of the triangle, so let us use Heron’s formula.
The area of the triangle, by Heron’s formula is given as,
\[A=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}\], where S is the semi – perimeter of the triangle equal to,
\[S=\dfrac{a+b+c}{2}\]
\[\therefore \] Semi – perimeter of \[\Delta ABC=S=\dfrac{12+20+24}{2}=\dfrac{56}{2}=28\].
Area of \[\Delta ABC=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}\]
Area of \[\Delta ABC=\sqrt{28\left( 28-12 \right)\left( 28-20 \right)\left( 28-24 \right)}\]
Area of \[\Delta ABC=\sqrt{28\times 16\times 8\times 4}\]
Area of \[\Delta ABC=\left( 4\times 2 \right)\sqrt{28\times 8}=8\sqrt{56}=119.73\].
Hence we got the area of the triangle as \[119.73c{{m}^{2}}\].
We have been given the cost of each tile as Rs.14 per square cm.
\[\therefore \] Cost of each tile = \[119.73\times 14=1676.26\].
Hence we got the cost of each tile as Rs.1676.26.

Note: As three sides are given you can use the Heron’s formula directly. Hence it is important that you know the formula by heart. You have to subtract the sides from semi – perimeter or you will get negative value.