An isosceles Δ ADB with base angles ∠DAB = ∠ABD = 80°. Another isosceles ΔCDB is constructed taking with base angles ∠CDB =∠CBD = 40°.Find AC2 BD2 if it is given that area of cyclic quadrilateral ABCD so formed has area 12 square units.a)786b)768c)876d)None of theseCorrect answer is option 'B'. Can you explain this answer?

Question

Answer

To find AC^2 and BD^2, we need to use the properties of cyclic quadrilaterals and the given information.

Given:
- ΔADB is an isosceles triangle with base angles ∠DAB = ∠ABD = 80°.
- ΔCDB is an isosceles triangle with base angles ∠CDB = ∠CBD = 40°.
- The area of the cyclic quadrilateral ABCD is 12 square units.

Let's break down the solution into steps:

Step 1: Finding the angles of the cyclic quadrilateral ABCD.
- Since ABCD is a cyclic quadrilateral, the opposite angles add up to 180°.
- ∠DAB + ∠CDB = 80° + 40° = 120°
- ∠ABD + ∠CBD = 80° + 40° = 120°
- The angles ∠DAB and ∠CDB are supplementary angles, as are ∠ABD and ∠CBD. This means ΔACD is an isosceles triangle with base angles ∠CAD = ∠CDA = 60°.

Step 2: Finding the angles of ΔACD.
- Since ΔACD is an isosceles triangle, the angles ∠CAD and ∠CDA are equal and each is 60°.
- The remaining angle ∠ACD can be found using the property that the sum of angles in a triangle is 180°.
- ∠ACD = 180° - 2 * 60° = 60°

Step 3: Finding the angles of ΔBDC.
- Since ΔBDC is an isosceles triangle, the angles ∠CDB and ∠CBD are equal and each is 40°.
- The remaining angle ∠BDC can be found using the property that the sum of angles in a triangle is 180°.
- ∠BDC = 180° - 2 * 40° = 100°

Step 4: Using the Law of Sines to find AC and BD.
- In triangles ΔACD and ΔBDC, we can use the Law of Sines to find the lengths of AC and BD.
- AC/AD = sin(∠CAD)/sin(∠ACD)
- AC/AD = sin(60°)/sin(60°) = 1
- AC = AD

- BD/CD = sin(∠CBD)/sin(∠BDC)
- BD/CD = sin(40°)/sin(100°)
- BD = CD * sin(40°)/sin(100°)

Step 5: Finding AC^2 and BD^2.
- The area of cyclic quadrilateral ABCD can be written as:
- Area = AC * BD * sin(∠CAD)
- 12 = AC * BD * sin(60°)
- AC * BD = 12/sin(60°) = 12/(√3/2) = 8√3

- From Step 4, we know AC = AD and BD = CD * sin(40°)/sin(100°)
- AC^2 = AD^2 = (8√3)^2 = 64 * 3 = 192
- BD^2 =

Answer

Now Area of ADC + ABC = 12

1/2xz sin⁡60º + 1/2xy sin⁡120º = 12

√3(xz+xy)/4 = 12

xz + xy = 48/√3 = 16√3

Using Ptolemy's theorem product of diagonals

(BD)(AC) = xz + yx

Thus AC²BD² = (xz + yx)² = (16√3)2 = 768

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