1. **Problem:** Find the value of the angle given choices 150°, 120°, 75°, 60°. Since no specific question is stated, we move to the next problem. 3. **Problem:** ABCD is a cyclic quadrilateral such that \(\angle ADB = 30^\circ\) and \(\angle DCA = 80^\circ\). Find \(\angle DAB\). **Step 1:** Recall the property of cyclic quadrilaterals: Opposite angles sum to 180°. **Step 2:** In cyclic quadrilateral ABCD, \(\angle ADB\) and \(\angle ACB\) subtend the same chord AB. **Step 3:** Given \(\angle ADB = 30^\circ\) and \(\angle DCA = 80^\circ\), we want to find \(\angle DAB\). **Step 4:** Using the property that angles subtended by the same chord are equal, \(\angle DAB = \angle DCB = 80^\circ\). **Step 5:** Since \(\angle DCA = 80^\circ\), and \(\angle DAB\) subtends the same arc, \(\angle DAB = 70^\circ\) (from the options, the closest correct is 70°). **Answer:** \(\boxed{70^\circ}\) 4. **Problem:** The edges of a triangular board are 6 cm, 8 cm, and 10 cm. Find the cost of painting it at the rate of 12 paise per cm². **Step 1:** Calculate the area of the triangle using Heron's formula. **Step 2:** Semi-perimeter \(s = \frac{6 + 8 + 10}{2} = 12\) cm. **Step 3:** Area \(= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{12(12-6)(12-8)(12-10)} = \sqrt{12 \times 6 \times 4 \times 2} = \sqrt{576} = 24\) cm². **Step 4:** Cost = Area \(\times\) rate = \(24 \times 12 = 288\) paise = 2.88 (since 100 paise = 1 unit). **Answer:** Rs 2.88 5. **Problem:** The length of the sides of a triangle are 5x, 5x, and 8x. Find the area of the triangle. **Step 1:** Semi-perimeter \(s = \frac{5x + 5x + 8x}{2} = 9x\). **Step 2:** Area \(= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{9x(9x - 5x)(9x - 5x)(9x - 8x)} = \sqrt{9x \times 4x \times 4x \times x} = \sqrt{144 x^4} = 12 x^2\). **Answer:** \(12 x^2\) sq. units 6. **Problem:** Assertion (A): If the area of an equilateral triangle is \(81\sqrt{3}\) cm², then the semi-perimeter of the triangle is 20 cm. Reason (R): Semi-perimeter of a triangle is \(s = \frac{a+b+c}{2}\), where \(a, b, c\) are sides of the triangle. **Step 1:** Area of equilateral triangle \(= \frac{\sqrt{3}}{4} a^2 = 81\sqrt{3}\). **Step 2:** Solve for \(a^2\): \(\frac{\sqrt{3}}{4} a^2 = 81\sqrt{3} \Rightarrow a^2 = 81 \times 4 = 324 \Rightarrow a = 18\) cm. **Step 3:** Semi-perimeter \(s = \frac{3a}{2} = \frac{3 \times 18}{2} = 27\) cm, not 20 cm. **Step 4:** Reason (R) is true but Assertion (A) is false. **Answer:** d) A is false but R is true. 7. **Problem:** Find the area of a triangle with two sides 8 cm and 11 cm and perimeter 32 cm. **Step 1:** Let the third side be \(x\). Then \(8 + 11 + x = 32 \Rightarrow x = 13\) cm. **Step 2:** Semi-perimeter \(s = \frac{8 + 11 + 13}{2} = 16\) cm. **Step 3:** Area \(= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{16(16-8)(16-11)(16-13)} = \sqrt{16 \times 8 \times 5 \times 3} = \sqrt{1920} = 8\sqrt{30}\) cm². **Answer:** \(8\sqrt{30}\) cm²