Question 36 of 50advanced🔧 ApplyNumerical5 marks

A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.

Correct Answer

Step 1: Drawing the Diagonal Let the cyclic quadrilateral be ABCD with sides AB = 5, BC = 5, CD = 12, and DA = 12. We draw the diagonal BD, which divides the quadrilateral into two triangles: △ABD and △CBD.

Step 2: Proving Triangles Congruent In △ABD and △CBD, we have AB = BC = 5 units, AD = CD = 12 units, and BD is a common side. By the SSS congruency rule, △ABD ≅ △CBD. Therefore, their corresponding angles are equal, meaning ∠A = ∠C.

Step 3: Using the Cyclic Quadrilateral Property According to Theorem 11, the sum of two opposite angles of a cyclic quadrilateral is 180°. Thus, ∠A + ∠C = 180°. Since we established that ∠A = ∠C from congruency, we get 2∠A = 180°, which means ∠A = ∠C = 90°.

Step 4: Calculating Area of Triangles Since ∠A = 90°, △ABD is a right-angled triangle with base AB = 5 and height AD = 12. Its area = ½ × 5×125 \times 12 = 30 sq units. Similarly, ∠C = 90° makes △CBD a right-angled triangle with base BC = 5 and height CD = 12. Its area = ½ × 5×125 \times 12 = 30 sq units.

Step 5: Finding Total Area The total area of the cyclic quadrilateral ABCD is the sum of the areas of △ABD and △CBD. Total Area = 30 + 30 = 60 square units.

Exercise: END-OF-CHAPTER EXERCISES | Q: 10 | (Page 23)
For More Understanding

Explanation

The solution cleverly uses the SSS congruency of the two triangles formed by drawing a diagonal across the equal sides of the quadrilateral. This establishes that the opposite angles are equal. By combining this with Theorem 11 from the provided context (which states that opposite angles of a cyclic quadrilateral sum to 180°), it is deduced that both opposite angles must be 90°. This transforms the area calculation into simply finding the area of two right-angled triangles, bypassing the need for Brahmagupta's formula or trigonometric identities, which are not provided in the context.

Solution Steps

  1. Step 1: Draw diagonal BD in cyclic quadrilateral ABCD (AB=5, BC=5, CD=12, DA=12), forming △ABD and △CBD.

  2. Step 2: AB=BC=5, AD=CD=12, BD=BD, so △ABD ≅ △CBD by SSS. Thus, ∠A = ∠C.

  3. Step 3: By Theorem 11, ∠A + ∠C = 180°. Since ∠A = ∠C, ∠A = ∠C = 90°.

  4. Step 4: Area of right △ABD = ½ × 5×125 \times 12 = 30. Area of right △CBD = ½ × 5×125 \times 12 = 30.

  5. Step 5: Total area = 30 + 30 = 60 square units.

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