In Triangle APB and CQDAnswer
AD=BC ( opposite sides of the Parallelogram)
$\angle ADP=\angle QBC$ ( Alternate angles between parallel lines)
$\angle APB= \angle CQD =90$
Hence by AAS Congruency
$ \Delta APB \cong \Delta CQD$
Now from CPCT of Congruent triangles
AP = CQ
Angles are 3x,5x,7x,9xAnswer
So
$3x+5x+7x+9x=360$
$24x=360$
x=15
So angles are
45, 75,105, 145
Now we know thatAnswer
$ \angle A + \angle B + \angle C + \angle D=360$
$ \angle C + \angle D=360 - (\angle A + \angle B)$
$ \frac {1}{ 2} (\angle C + \angle D)=180 - \frac {1}{2}(\angle A + \angle B)$ -(1)
In Triangle AOB
$\angle AOB + \frac {1}{2} \angle A + \frac {1}{2} \angle B=180$
$\angle AOB= 180 - \frac {1}{2}(\angle A + \angle B)$ -(2)
From (1) and (2)
$\angle AOB = \frac {1}{ 2} (\angle C + \angle D)$
(i) In triangle ADE and BCE ,Answer
$\angle ADC = \angle BCD = 90$
$\angle EDC = \angle ECD$ as triangle ECD is an equilateral triangle.
Therefore
$\angle EDA = \ angle ECB$
AD = BC ( same sides of a square )
DE = CE ( equal sides of an equilateral triangle)
By SAS both are congruent
By CPCT
AE = BE
(ii) In triangle ADE
$ \angle DAE= \angle DEA$ as both the sides are equal
Also
$\angle EDA= 90 + 60=150$
Now
$ \angle DAE + \angle DEA + \angle EDA=180$
Hence $\angle DAE = 15^0$
Opposite angles in a parallelogram are equalAnswer
$(3y - 10)=(2y + 35)$
$y=45$
So opposite angles are 125° and 125°
Other will be 65° and 65°
Answer
$OB^2 + OC^2=BC^2$
$OB^2=BC^2-OC^2= 100-64$
$OB=6$
Therefore BD=12 cm
Sum of the angles of the quadrilateral=360Answer
Here It is not, so this is not possible
Answer
Opposite angles in a parallelogram are equal.
Therefore
∠ B=∠ D=135°
Adjacent angles in a parallelogram are supplementary
∠ A+ ∠ B=180°
∠A=180-135
∠A=45 °
Answer
Opposite angles in a parallelogram are equal.
Therefore
∠ A=∠ C=70°
Adjacent angles in a parallelogram are supplementary
∠ A+ ∠ B=180°
∠B=180-70
∠B=110 °
Again
∠ B=∠ D=110°
The situation is depicted in below figureAnswer
Opposite angles in a parallelogram are equal.
Therefore
∠ BCD = ∠ DAB=75°
In Δ BDC
∠ BCD + ∠ DBC + ∠ CDB =180°
∠ CDB= 180 -75-60=45°
Now alternat angles are equal
∠ ADB= ∠ DBC= 60°
Answer
AB=CD ( Opposite side of parallelogram)
Since X and Y are mid-points of AB and DC respectively.
AX = 1/2 AB and CY = 1/2 DC
Therefore
AX = CY and AX || CY.
One pair of opposite sides is parallel and equal
Hence AECF is a parallelogram.
Answer
Given: AE = BF = CG = DH
Since all the sides are equal in Square
Therefore
BE=CF=DG=AH
Now in Δ AHE and BEF
AE=BF
AH=BE
∠ A = ∠ B
By SAS congruence
Δ AEH ≅ Δ BFE
Therefore by CPCT
HE=EF
∠ 1 = ∠ 2
∠ 3 = ∠ 4
Now but ∠ 1 + ∠ 3=90
As ∠ 3 = ∠ 4
∠ 1 + ∠ 4=90°
i.e ∠ E= 90°
Similarly we can proof for other sides and angles
Hence EFGH is a square
This Quadrilaterals Worksheet for Class 9 Maths with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.