Two parallelogram stand on equal bases and between the same parallels. The ratio of their areas is
In quadrilateral ABCD, if ∠A = 60^{∘} and ∠B : ∠C : ∠D = 2:3:7, then ∠D is :
D and E are the midpoints of the sides AB and AC res. Of △ABC. DE is produced to F. To prove that CF is equal and parallel to DA, we need an additional information which is:
P, Q, R are the mid points of AB, BC, AC res, If AB = 10cm, BC = 8cm, AC = 12cm, Find the perimeter of △PQR.
In Triangle ABC which is right angled at B. Given that AB = 9cm, AC = 15cm and D, E are the midpoints of the sides AB and AC res. Find the length of BC?
Three Statements are given below:
(I) In a, Parallelogram the angle bisectors of 2 adjacent angles enclose a right angle.
(II) The angle bisector of a Parallelogram form a Rectangle.
(III) The Triangle formed by joining the midpoints of the sides of an isosceles triangle is not necessarily an isosceles triangle. Which is True?
If APB and CQD are 2 parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form, square only if
If bisectors of ∠A and ∠B of a quadrilateral ABCD intersect each other at P, of ∠B and ∠C at Q, of ∠C and ∠D at R and of ∠D and ∠A at S, then PQRS is a
To show: ∠PSR + ∠PQR = 180°
∠SPQ + ∠SRQ = 180°
In △DSA,
∠DAS + ∠ADS + ∠DSA = 180° (angle sum property)
+ ∠ SA = 180° (since RD and AP are bisectors of ∠D and ∠A)
∠DSA = 180°
∠PSR = 180°−
(∵ ∠DSA = ∠PSR are vertically opposite angles)
Similarly,
∠PQR = 180°−
Adding (i) and (ii), we get, ∠PSR + ∠PQR = 180°
=360° − 1/2 × (∠A + ∠B + ∠C + ∠D)
=360°− 1/2 × 360° = 180° ∴ ∠PSR + ∠PQR = 180°
In quadrilateral PQRS,
∠SPQ + ∠SRQ + ∠PSR + ∠PQR = 360°
=> ∠SPQ + ∠SRQ + 180° = 360°
=> ∠SPQ + ∠SRQ = 180°
Hence, showed that opposite angles of PQRS are supplementary.
The Diagonals AC and BD of a Parallelogram ABCD intersect each other at the point O such that ∠DAC = 30^{∘} and ∠AOB = 70^{∘}. Then, ∠DBC?
In Parallelogram ABCD, bisectors of angles A and B intersect each other at O. The measure of ∠AOB is
Three statements are given below:
(I) In a Rectangle ABCD, the diagonals AC bisects ∠A as well as ∠C.
(II) In a Square ABCD, the diagonals AC bisects ∠A as well as ∠C.
(III) In rhombus ABCD, the diagonals AC bisects ∠Aas well as ∠C.
Which is True?
D and E are the midpoints of the sides AB and AC of △ABC and O is any point on the side BC, O is joined to A. If P and Q are the midpoints of OB and OC res, Then DEQP is
Given Rectangle ABCD and P, Q, R and S are the midpoints of the sides AB, BC, CD and DA res. If length of a diagonal of Rectangle is 8cm, then the quadrilateral PQRS is a
In the given figure, ABCD is a Rhombus. Then,
D and E are the midpoints of the sides AB and AC. Of △ABC. If BC = 5.6cm, find DE.
In a triangle P, Q and R are the midpoints of the sides BC, CA and AB res. If AC = 21cm, BC = 29cm and AB = 30cm, find the perimeter of the quadrilateral ARPQ?
The bisectors of the angles of a Parallelogram enclose a
Opposite angles of a Quadrilateral ABCD are equal. If AB = 4cm, find the length of CD.
In a Trapezium ABCD, if AB ║ CD, then (AC^{2}+ BD^{2}) = ?
In quadrilateral ABCD, ∠B=90^{∘}, ∠C−∠D = 60^{∘} and ∠A−∠C−∠D = 10^{∘}. Find ∠A, ∠C and ∠D.
If a Quadrilateral ABCD,∠A = 90^{∘} and AB = BC = CD = DA, Then ABCD is a
Rhombus is a quadrilateral
In △ABC, EF is the line segment joining the midpoints of the sides AB and AC. BC = 7.2cm, Find EF.
In the figure, ABCD is a rhombus, whose diagonals meet at 0. Find the values of x and y.
Since diagonals of a rhombus bisect each other at right angle .
∴ In △AOB , we have
∠OAB + ∠x + 90° = 180°
∠x = 180°  90°  35° [∵ ∠OAB = 35°]
= 55°
Also, ∠DAO = ∠BAO = 35°
∴ ∠y + ∠DAO + ∠BAO + ∠x = 180°
⇒ ∠y + 35° + 35° + 55° = 180°
⇒ ∠y = 180°  125° = 55°
Hence the values of x and y are x = 55°, y = 55°.
The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If ∠DAC = 32^{∘} and ∠AOB = 70^{∘} then, ∠DBC is equal to
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