abcd is a cyclic trapezium?
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ABCD is a cyclic trapezium and AB∥CD IF AB be the diameter of the circle and ∠CAB=30∘ then the value of ∠ADC is. A .
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ABCD is a rectangle formed by the points A(–1, –1), B(– 1, 4), C(5, 4) and D(5, – 1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.
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ABCD is a parallelogram whose diagonals intersect each other at O with 90°. so it's a rhombus as diagonals of rhombus interest each other at right angles.
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In the adjoining figure ABCD is a rhombus and EDC is an equilateral triangle. If ∠DAB = 48°, find. (1) ∠BEC (2) ∠DEB (3) ∠BFC. [Hint- ∠BCE = 48° + 60° = 180° , BC = EC]
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Answer Expert Verified For trapezoid ABCD, line AB equals to line CD and line AD is parallel to BC. The two diagonals, AC and BD, also has equal length.
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In the given figure, ABCD is a cyclic quadrilateral, in which AC and BD are its diagonals. ... In the given figure, ABCD is a cyclic quadrilateral, in which AC and BD are its diagonals. If ∠DBC = 60° and ∠BAC = 30°, then find ∠BCD.
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ABCD is a quadrilateral P, Q, R, S are the points of trisection of the sides AB, CB, CD and AD respectively. Prove that PQRS is a parallelogram. ABCD is a quadrilateral in which P, Q, R, S are the points of trisection of the sides AB, CB, CD and AD respectively. ... Hence, PQRS is a parallelogram.
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ABC is a straight line with AB = BC = 0.60 nm. BP isperpendicular to ABC and BP = 0.80 nm. Charges of +3.2× 10–19 C are placed at A and C, and acharge of –3.2 × 10–19 C is placed atB.
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Since abcd is a cyclic quadrilateral whose diagonal ac and bd intersect at p if ab = dc, our aim is to prove that triangle pas is congruent to triangle pdc ,pa=pb ,pc=pb ,ad is parallel to bc. ... In third part, we are given that diagonals are bisectors, so ABCD is a parallelogram, therefore AD║BC.
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Given: ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. To Prove: (i) ABCD is a square. (ii) diagonal BD bisects ∠B as well as ∠D. and transversal AC intersects them.
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PQRS is a trapezium with PQ parallel to SR and m angle at P is equal to m angle at Q of 50 degrees. ... Consider the cyclic trapezium whose circumcentre is on one of the sides of the trapezium. If the ratio of two parallel sides is 1:4.
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AB is the diameter and AC is a chord of a circle with centre O such that angle BAC = 30°. The tangent to the circle at C intersects AB produced in D. show that BC = BD.
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As ABCD is a cyclic quadrilateral, so the sum of a pair of two opposite angles will be 180o. The value of angle D is 120o.
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Abcd is a rhombus whose diagonal ac makes an angle alpha with ab. if cos alpha is 2/3 and ob =3cm, find the length of its diagonals ac and bd?/
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Suppose AB is a diameter of a circle, whose centre is at O and C be any point on the circle. If CD ⊥ AB and CD = 12 cm, AD = 16 cm, then BD is equal to. 3). In the figure given below, AO = CD where O is the centre of the circle.
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Thus, "line segments AC and BD are perpendicular bisectors of each other" means that AC cuts BD into two equal parts at 90° and BD cuts AC into two equal parts at 90°. For (2): AB = BC = CD = AD, means that ABCD is either a rhombus or square (so still a rhombus).
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In the figure below, ABCD is a square. Points are connected on each pair of adjacent sides of ABCD to form 4 congruent right triangles, as shown below. Each of these triangles has one leg that is twice as long as the other leg.