Question

find the area of shaded region in the figure how many triangular flower beds of 6 metre square can be made from this area use root 105 equal to 10.25​

Answer 1

The area of the shaded region in the figure is 1074 m² and the no. of triangular flower beds of a 6-metre square that can be made from this area is  179.

Step-by-step explanation:

Hi there,

The measurements in the question is not clear enough if we take the sides to be in cm and convert them in meter and the flower bed measurement in m² then the answer will be in decimal. Therefore, I have considered all the measurements in terms of meter and solving the sum accordingly. Hope this is helpful. Thanks:)

Step 1:

Let's consider ΔABC with AB = 122 m, BC = 22 m and AC = 120 m

Using the Heron's Formula, we get

Semi-perimeter, s = [a+b+c]/2 ....where a,b & c are the three given sides of the triangle.

s = $\frac{122 + 22 + 120}{2}$ = $\frac{264}{2}$ = 132 m

∴ Area of ΔABC is,

= $\sqrt{s(s-a)(s-b)(s-c)}$

$= \sqrt{132 * (132 - 122)(132-22)*(132-120)} \\= \sqrt{132 * 10 * 110 * 12} \\= \sqrt{1742400}$

= 1320 m²

Step 2:

Let's consider ΔBOC with BO = 24 m, BC = 22 m and OC = 26 m

Using the Heron's Formula, we get

s =  $\frac{24 + 22 + 26}{2} = \frac{72}{2}$ = 36 m

∴ Area of ΔOBC is,

= $\sqrt{s(s-a)(s-b)(s-c)}$

$= \sqrt{36 * (36 - 24)(36-22)*(36 - 26)} \\= \sqrt{36 * 12 * 14 * 10} \\= \sqrt{6*6*6*2*2*7*2*5}\\= 6*2 * 2* \sqrt{105$

= 24 * 10.25

= 246 m²

Step 3:

Therefore,

The area of the shaded region is given by,

= [Area of ΔABC] - [Area of ΔBOC]

= 1320 - 246

= 1074 m²

and,

The no. of triangular flower beds of 6 m² each that can be made by an area of 1074 m² is,

= $\frac{1074}{6}$

= 179

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Answer 2

Answer:

Let's consider ΔABC with AB = 122 m, BC = 22 m and AC = 120 m

Using the Heron's Formula, we get

Semi-perimeter, s = [a+b+c]/2 ....where a,b & c are the three given sides of the triangle.

s = \frac{122 + 22 + 120}{2}

2

122+22+120

= \frac{264}{2}

2

264

= 132 m

∴ Area of ΔABC is,

= \sqrt{s(s-a)(s-b)(s-c)}

s(s−a)(s−b)(s−c)

$$\begin{lgathered}= \sqrt{132 * (132 - 122)(132-22)*(132-120)} \\= \sqrt{132 * 10 * 110 * 12} \\= \sqrt{1742400}\end{lgathered}$$

= 1320 m²

Step 2:

Let's consider ΔBOC with BO = 24 m, BC = 22 m and OC = 26 m

Using the Heron's Formula, we get

s = $$\frac{24 + 22 + 26}{2} = \frac{72}{2}$$ = 36 m

∴ Area of ΔOBC is,

= $$\sqrt{s(s-a)(s-b)(s-c)}$$

$$\begin{lgathered}= \sqrt{36 * (36 - 24)(36-22)*(36 - 26)} \\= \sqrt{36 * 12 * 14 * 10} \\= \sqrt{6*6*6*2*2*7*2*5}\\= 6*2 * 2* \sqrt{105\end{lgathered}$$

= 24 * 10.25

= 246 m²

Step 3:

Therefore,

The area of the shaded region is given by,

= [Area of ΔABC] - [Area of ΔBOC]

= 1320 - 246

= 1074 m²

and,

The no. of triangular flower beds of 6 m² each that can be made by an area of 1074 m² is,

= $$\frac{1074}{6}$$

= 179

Step-by-step explanation:

Yes this is the correct answer