Question
In a cyclic quadrilateral, the diagonals intersect at an
angle of 60°. If the sides of the quadrilateral are 6 cm, 8 cm, 10 cm, and 12 cm, what is the area of the quadrilateral?ÂSolution
The area of a cyclic quadrilateral can be calculated using formula: Area = √[(s - a)(s - b)(s - c)(s - d)], where s = (a + b + c + d) / 2. Here, s = (6 + 8 + 10 + 12) / 2 = 18 cm. Area = √[(18 - 6)(18 - 8)(18 - 10)(18 - 12)] = √[12 × 10 × 8 × 6] = √5760 = 75.89 cm² ≈ 76 cm². Correct answer: c) 76 cm²
Simplify: (23×32) ÷6
?% of 320 = 40% of 40 × 5
125% of 80 – 6 × 4 = ? × 13 – 54
(60 × 8 ÷ 10) × 5 = ?
Find the HCF of 15x2 + 8x – 12, 3x² + x – 2, 3x² - 2x, 9x² - 12x + 4
140% of 1270 + 60% of 2085 = 1881 + ‘?’% of 287
[∛(3375/19683  )- ∛(125/1728  )  ] ÷ ∛(64/729)  = ? - 3/8Â
1885 ÷ 64.98 + 7.29 + ? = 69.09
(〖(0.4)〗^(1/3) × 〖(1/64)〗^(1/4) × 〖16〗^(1/6) × 〖(0.256)〗^(2/3))/(〖(0.16)〗^(2/3) × 4^(-1/2) ×〖1024〗^(-1/4) ) = ?
(1/2) – (3/5) + 3(1/3) = ? + (5/6)
