Find the area of a triangle with sides a=5, b=8 and c=11

Mathematics

2 answers:

Lynna [10]3 years ago

5 0

Answer:

440

Step-by-step explanation:

5x8=40

40x11=440

Have a good day :P :P

Debora [2.8K]3 years ago

5 0

ANSWER:

The area of the triangle with sides a=5, b=8 and c=11 is $4 \sqrt{21} \text { square units. }$

SOLUTION:

Given, a = 5, b = 8, c = 11.

We need to find the area of the triangle with sides as given.

Let us find the area of triangle using heron’s formula:

$\begin{array}{l}{\text { Area }=\sqrt[2]{s(s-a)(s-b)(s-c)}} \\\\ {\text { Where } s=\frac{a+b+c}{2}}\end{array}$

Now, let us calculate value of s and put it in heron’s formula.

$s=\frac{5+8+11}{2}=\frac{24}{2}=12$

Substitute s value in heron’s formula along with a, b, c values.

$\begin{array}{l}{\text { Area }=\sqrt{12(12-5)(12-8)(12-11)}} \\ {=\sqrt{12(7)(4)(1)}} \\ {=\sqrt{4^{2} \times 3 \times 7}=4 \sqrt{21} \text { sq units. }}\end{array}$

Thus the area of the triangle with sides a=5, b=8 and c=11 is $4 \sqrt{21} \text { square units. }$

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lora16 [44]

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We have 4 vertical rectangles from left to right:

$A_ {1} = 14 * (18-12) = 14 * 6 = 84 \ cm ^ 2\\A_ {2} = (3 + 3 + 3) * 4 = 9 * 4 = 36 \ cm ^ 2\\A_ {3} = (3 + 3) * 4 = 6 * 4 = 24 \ cm ^ 2\\A_ {4} = 3 * 4 = 12 \ cm ^ 2$