Find the area of the shaded region in figure, where arcs drawn with centres A, B, C and D intersect in pairs at mid-point P, Q, R and 5 of the sides AB, BC, CD and DA, respectively of a square ABCD. (use π = 3.14) areas related to circles ncert class-10 1 Answer 0 votes answered Aug 24, 2018 by BharatLal (28.2k points)

Five different formulas are used to calculate the area of the quadrilateral. In parallelogram.

Solution Verified by Toppr Given, cyclic quadrilaterals ABCD. and ∠A:∠C=5:4. Let the ∠A=5x, ∠B=4x. We know, in a cyclic quadrilateral, the sum of opposite angles are supplementary, i.e. 180o. Then, ∠A+∠C=180o 5x+4x=180o 9x=180o x=20o. Now, ∠A=5x =5×20o =100o. Hence, ∠A=100o. Was this answer helpful? 0 0 Similar questions

A park in the shape of a quadrilateral ABCD has AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m and ∠C = 90°. Find the area of the park. Find the area of the park. Q.

Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar(BOC). Prove that ABCD is a trapezium. Solution It is given that, Area(ΔAOD) =Area(ΔBOC) Area(ΔAOD)+Area(ΔAOB) =Area(ΔBOC)+Area(ΔAOB) [Adding Area (ΔAOB) to both sides] Area(ΔADB) =Area(ΔACB)

A park in the shape of a quadrilateral ABCD has AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m and ∠C = 90°. Find the area of the park.

Ex 9.4, 6 (Optional) Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC). [Hint : From A and C, draw perpendiculars to BD.]

A (x1,y1), B (x2,y2), C (x3,y3), D (x4,y4) are verticies of a quadrilateral either convex or concave (one of the internal angle greater than 180 degrees) taken in order, then we can use the following elegant formula for calculating it's area. This does not provide an answer to the question.

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Explain how a square is a quadrilateral Answer: A square is 4 sided, so it is a quadrilateral. Easy. View solution > Explain how a square is a parallelogram. Medium.

Solution: Using the angle sum property of quadrilaterals, we can find the unknown angles of quadrilateral. So, 85° + 90°+ 65° = 240. We know that the sum of the interior angles of a quadrilateral is 360°. Therefore, the 4th angle = 360 - 240 = 120° Exterior Angles of a Quadrilateral

Byju's Answer Standard IX Mathematics Sum of Pair of Opposite Angles in Quadrilateral ABCD is a cyc. Question ABCD is a cyclic quadrilateral such that ∠ADB=30o and ∠DCA=80o, then ∠DAB = A 70o B 100o C 125o D 150o Solution The correct option is A 70o ABCD is a cyclic quadrilateral. ∠DCA=80o and ∠ADB=30o ∵ ∠ADB= ∠ACB (Angles in the same segment)