Question

Prove that the area of the quadrilateral ABCD is 3(4+3root3)m^.If AB=5m,BC=5m,CD=6m,AD=6m and diagonal AC=6m.

Answer 1

by using equation of heron's
√(s(s-a)(s-b)(s-c)

Answer 2

It has been proved that area of quadrilateral ABCD is $\bf\red{ 3(4 + 3 \sqrt{3} ) {m}^{2}}$

Given:

• In quadrilateral ABCD; If AB=5m,BC=5m,CD=6m,AD=6m and diagonal AC=6m.

To find:

• Prove that the area of the quadrilateral ABCD is 3(4+3root3)m².

Solution:

Concept/formula to be used:

• Apply Heron's formula.
• Area of ∆=√s(s-a)(s-b)(s-c)
• Area of equilateral triangle= $\frac{ \sqrt{3} }{4} {a}^{2}$ where a is side of triangle.

Step 1:

Apply Heron's formula in ∆ABC.

let a= 5 cm

b=5 cm and

c=6 cm

So,

$s = \frac{5 + 5 + 6}{2}$

or

s= 8 m

$Area \triangle(ABC)= \sqrt{8(8 - 5)(8 - 5)(8 - 6)}$

or

$Area \triangle(ABC) = \sqrt{8 \times 3 \times 3 \times 2}$

or

$\bf Area \triangle(ABC) = 12 \: {m}^{2}$

Step 2:

Find Area of ∆ACD.

Here all sides are equal; ACD is an equilateral triangle.

Area of equilateral triangle= $\frac{ \sqrt{3} }{4} {a}^{2}$

where a is side of triangle.

Here a= 6

$Area \triangle(ACD) = \frac{ \sqrt{3} }{4} \times ( {6)}^{2}$

or

$\bf Area \triangle(ACD) = 9\sqrt{3} \: {m}^{2}$

Step 3:

Find Area of ABCD.

Area of ABCD= Area of ∆(ABC)+Area of ∆(ADC)

$\text{Area of ABCD} = 12 + 9 \sqrt{3} {m}^{2}$

or

$\text{\bf Area of ABCD} \bf = 3(4 + 3 \sqrt{3} ) {m}^{2}$

Hence proved.

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