Given below are the **Class 9 Maths** Assignments for Quadrilaterals

(a) Concepts questions

(b) Calculation problems

(c) Long answer questions

(a) Concepts questions

(b) Calculation problems

(c) Long answer questions

Show that the quadrilateral formed by joining the mid- points of adjacent sides of rectangle is a rhombus.

The figure is shown as below

To Prove: quadrilateral PQRS is a rhombus

Proof:

In Δ ABC

P and Q are mid points of sides AB and BC

By Mid point theorem

PQ= 1/2 AC and PQ || AC -(X)

In Δ ACD

S and R are mid points of sides AD and DC

By Mid point theorem

SR= 1/2 AC and SR || AC --(Y)

From (X) and (Y), we have

PQ=SR

PQ ||SR

Hence PQSR is a parallelogram

Now in Δ BCD

Q and R are mid points of sides BC and DC

By Mid point theorem

QR= 1/2 BD

Nowe AC = BC

Hence

PQ=SR=QR

Now a parallelogram whose adjacent sides are equal is a rhombus.

Hence proved

P, Q, R and S are respectively the mid- point of sides AB, BC, CD and DA of a quadrilateral ABCD such that AC=BD. Prove that PQRS is a rhombus.

Given AC=BD

Proof:

In Δ ABC

P and Q are mid points of sides AB and BC

By Mid point theorem

PQ= 1/2 AC and PQ || AC -(X)

In Δ ACD

S and R are mid points of sides AD and DC

By Mid point theorem

SR= 1/2 AC and SR || AC --(Y)

From (X) and (Y), we have

PQ=SR= 1/2 AC ----(1)

Similarly in Δ BCD

Q and R are mid point of BC and CD

By Mid point theorem

QR= 1/2 BD

Similarly in Δ ADB

S and P are mid point of AD and AB

By Mid point theorem

SP= 1/2 BD

Therefore

SP=QR=1/2 BD ---(2)

AC=BD

So from (1) and (2)

PQ=SR=SP=QR

Hence PQRS is a rhombus

Prove that the quadrilateral formed (if possible) by the internal angular bisectors of any quadrilateral is cyclic.

PQ and RS are to equal and parallel line segments. Any point M not lying on PQ or RS is joined to Q and S and lines through P parallel to QM and through R parallel to SM meet at N. Prove that the line segments MN and PQ are equal and parallel to each other.

l, m and n are three parallel lines intersected by transversal's p and q such that l, m and n cut off equal intercepts AB and BC on p. Show that l, m and n cut off equal intercepts DE and EF on q also.

Given

AB=BC

To Prove:

DE=EF

Proof:

Let us join A to F intersecting m at G

The trapezium ACFD is divided into two triangles namely Δ ACF and &Delta AFD

In Δ ACF, it is given that B is the mid-point of AC (AB = BC)

and BG || CF (since m || n).

So, G is the mid-point of AF (by Mid Point Theorem)

Now, in Δ AFD, we can apply the same argument as G is the mid-point of AF,

GE || AD and so by by Mid Point Theorem, E is the mid-point of DF,

i.e., DE = EF.

In other words, l, m and n cut off equal intercepts on q also.

In a quadrilateral ABCD, CO and DO are bisectors of $\angle C$ and $\angle D$ respectively. Prove that $\angle COD = \frac {1}{2} (\angle A + \angle B)$

In a parallelogram ABCD, $\angle D = 135^0$, determine the angles measures of $\angle A \; and \; \angle B$

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 °

ABCD is a parallelogram in which $\angle A = 70^0$. Compute $\angle B \; ,\; \angle C \;and \; \angle D$

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°

ABCD is a parallelogram in which $\angle DAB = 75^0$ and $\angle DBC = 60^0$. Compute $\angle CDB \; and \; \angle ADB$.

The situation is depicted in below figure

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°

ABCD is a parallelogram and X, Y are the mid- points of sides AB and DC respectively. Show that in parallelogram ABCD, AXCY is a parallelogram.

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.

The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.

ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively. Such that AE = BF = CG = DH. Prove that EFGH is a square.

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

ABCD is a rhombus, EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.

ABCD is a parallelogram, AD is produced to E so that DE = DC and EC produced meets AB produced in F. Prove that BF = BC.

ABCD is a parallelogram P is a point on AD such that AP = 1/3 AD and Q is a point on BC such that CQ = 1/3 BC. Prove that AQCP is a parallelogram.

P is the mid- point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R. Prove that

- AR = 2BC
- BR = 2BQ.

ABCD is a kite having AB = AD and BC = CD. Prove that the figure formed by joining the mid- points of the side, in order, is a rectangle.

ABC is a triangle. D is a point on AB such that AD = ¼ AB and E is the point on A such that AE = ¼ AC. Prove that DE = ¼ BC.

If the diagonals of a parallelogram are equal, then show that it is a rectangle.

In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles.

Show that the line segment joining the mid- point of any two sides of a triangle is parallel to third side and is equal to half of it.

A diagonal of a rectangle is inclined to one side of a rectangle at $25^0$. Find the acute angles between the rectangle diagonals.

**Notes****Assignments & NCERT Solutions**

Class 9 Maths Class 9 Science