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The statement “diagonals of a parallelogram are perpendicular to each other” is false.

Justification:

Diagonals of a parallelogram bisect each other but not at 90°.

So, they are not perpendicular to each other.

Hence, this statement is false.

(D) 40°

2x = 3x – 40 … [Pythagoras theorem] 

∴ x = 40° 

Correct option is (B) Supplementary

\(\because\) Sum of opposite angles in a cyclic quadrilateral is \(180^\circ)\)

\(\therefore\) Opposite angles of a cyclic quadrilateral are supplementary.

B) Supplementary

Sum of angles of a quadrilateral is 360°

Let angle is x°

Therefore each angle is 3x, 5x, 7x and 9x

3x° + 5x° + 7x° + 9x° = 360°

24x° = 360°

x° = \(\frac{360°}{24}\)

x° = 15°

Therefore angles are: 3x = 3 × 15 = 45°

5x = 5 × 15 = 75°

7x = 7 × 15 = 105°

9x = 3× 15 = 45°

Area of triangles between same base and same parallel lines are equal.

∴ ΔCBE area = ΔDBE area 

ΔBAE area = ΔBEF area 

ΔCBE + ΔBAE = ΔDBE + ΔBEF 

∴ Area of ΔAEC = area of ΔDBF

Sum of angles of a quadrilateral is 360°

Let each equal angle is x°

150° + x° + x° + x° = 360°

3x° = 360° - 150°

x° = \(\frac{210°}{3}\)

x° = 70°

Therefore other equal angles are 70° each.

(a) 27°

∠PSR = ∠PQR = 68°        (opp. ∠s of a ||gm are equal) 

∠PTS = 180° – 139° = 41°      (PTQ is a straight line) 

∴ ∠RST = ∠PTS = 41° (SR || PQ alt. ∠s are equal) 

∴ y = ∠PSR – ∠RST = 68° – 41° = 27°

Ratio of angles = 1 : 2 : 3 : 3 

Angles of quadrilateral x, 2x, 3x and 3x. 

x + 2x + 3x + 3x = 360° 

9x = 360°

x = 360°/9 = 40°

First angle = x = 40° 

Second angle = 2x = 80° 

Third angle = 3x = 120° 

Fourth angle = 3x = 120°

(c) 

In parallelogram ABCD, ∠A+∠D = 180° 

⇒ In ΔASD,∠ADS +∠SAD = \(\frac{180°}{2}\) = 90° 

(∵ DS and AS bisect ∠s D and A respectively) 

∴ ∠ASD = 180° – ( ∠ADS + ∠SAD) 

= 180° – 90° = 90° 

⇒ ∠PSR = 90° (vert. opp. ∠s)

Similarly we can show that ∠SPQ = ∠PQR 

=∠QRS = 90°. 

∴ Quadrilateral PQRS is a rectangle.

(D) diagonals of PQRS are equal.

Explanation:

Since, ABCD is a rhombus

We have,

AB = BC = CD = DA

Now,

Since, D and C are midpoints of PQ and PS

By midpoint theorem,

We have,

DC = ½ QS

Also,

Since, B and C are midpoints of SR and PS

By midpoint theorem

We have,

BC = ½ PR

Now, again, ABCD is a rhombus

∴ BC = CD

⇒ ½ QS = ½ PR

⇒ QS = PR

Hence, diagonals of PQRS are equal

Therefore, option (D) is the correct answer.

(d) 18°

Diagonals of a rectangle are equal and bisect each other. 

∴ In ΔAOB ,∠AOB = 162° – 3\(x\)        (vert. opp. ∠s) 

∠OBA =∠OAB = 2x (∵OA = OB) 

∴ ∠AOB +∠OBA+∠OAB = 180° 

⇒ 162°– 3\(x\) + 2\(x\) + 2\(x\) = 180° ⇒ \(x\) = 18°

(C) trapezium

Explanation:

As angle A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio 3: 7: 6: 4,

We have the angles A, B, C and D = 3x, 7x, 6x and 4x.

Now, sum of the angle of a quadrilateral = 360^o.

3x + 7x + 6x + 4x = 360^o

⇒20x = 360^o

⇒ x = 360 ÷ 20 =18^o

So, the angles A, B, C and D of quadrilateral ABCD are,

∠A = 3×18^o = 54^o,

∠B = 7×18^o = 126^o

∠C = 6×18^o = 108^o

∠D = 4×18^o = 72^o

AD and BC are two lines cut by a transversal CD

Now, sum of angles ∠C and ∠D on the same side of transversal,

∠C +∠D =108^o + 72^o =180

Hence, AD|| BC

So, ABCD is a quadrilateral in which one pair of opposite sides are parallel.

Hence, ABCD is a trapezium.

Therefore, option (C) is the correct answer.

Correct option is (C) \(\frac{1}{4}ar(\triangle ABC)\)

\(\because\) D, E and F are the mid-points of the sides AB, BC and AC respectively.

\(\therefore\) DF = \(\frac12BC,\) FE \(=\frac12AB\) and DE \(=\frac12AC\)

\(\Rightarrow\) \(\frac{DF}{BC}=\frac{1}{2},\frac{FE}{AB}=\frac{1}{2}\;and\;\frac{DE}{AC}=\frac{1}{2}\)

\(\therefore\) \(\frac{FE}{AB}=\frac{DE}{AC}=\frac{DF}{BC}=\frac{1}{2}\)

\(\therefore\) \(\frac{ar(\triangle DEF)}{ar(\triangle ABC)}=(\frac{DF}{BC})^2\)

\(=(\frac12)^2=\frac14\)

\(\therefore\) \(ar(\triangle DEF)=\) \(\frac{1}{4}ar\,(\triangle ABC)\)

Correct option is  C) 1/4 ar(ΔABC)

Correct option is (D) 2 cm

In a triangle, line segment joining mid-points of two sides is parallel to third side and exactly half of third side.

\(\therefore\) DE = \(\frac12BC\)    \((\because\) D & E are mid-points of sides AB & AC)

\(=\frac12\times8\) = 4 cm

\(\because\) AF = \(\frac12AD\) and AG \(=\frac12AE\)

\(\therefore\) F & G are mid-points of sides AD & AE of triangle ADE.

\(\therefore\) FG = \(\frac12DE\)

\(=\frac12\times4\) = 2 cm

Correct option is  D) 2 cm

∠BCD = 90° [Angle of a rectangle] 

∠OCD + ∠OCB = 90° 

30° + ∠QCB = 90° 

∠OCB = 90° – 30° 

∠OCB = 60° 

∠OCB = ∠OBC = 60° 

OC = OB 

[Diagonals of a rectangle bisect each other] 

∴ ∠BOC = 180° – (60° + 60°) 

= 180° -120° 

∠BOC = 60° 

In Δ BOC, 

∠BOC = ∠OBC = ∠OCB = 60° 

& equiangular triangle 

∴ BOC is an equiangular triangle.

The ratio of the sides is 2 : 1 

Let the sides be 2x and x 

Perimeter = 30 

2x + x + 2x + x = 30 

6x = 30 

x = \(\frac{30}{6}\) = 5 cm 

2x = 2 x 5 = 10cm 

∴The sides aer 10 cm, 5 cm, 10 cm, 5 cm.

Name of the above figure:

i. Pentagon (5 sides) 

ii. Hexagon (6 sides) 

iii. Heptagon (7 sides) 

iv. Octagon (8 sides)

Polygon: A closed plane figure bounded by line segments is called a polygon. Poly means 'many', so a polygon is a closed many sided figure.

Polygons are classified as follows with reference to the number of sides they have:

(a) A pair of opposite sides is parallel.

We know that, in a trapezium, a pair of opposite sides are parallel.

Given, 

ABCD is a rectangle 

∠ECD = 146° 

= ∠ECD +∠DCO = 180° 

(Linear pair of angles)

=146°+∠DCO = 180° 

= ∠DCO = 180° - 146° = 34° 

And, 

∠BOC = 180°- ∠OCB -∠OBC 

∵ ∠OCB = ∠OBC 

=∠BOC = 180°- 34°- 34° = 112° 

∠BOC +∠AOB = 180° 

(Linear pair) 

= 112°+ ∠AOB = 180° 

= ∠AOB = 180°- 112° = 68°

(i) parallelogram 

(ii) trapezium 

(iii) parallel to

(iv) rectangle 

(v) supplementary 

(vi) parallelogram 

(vii) parallelogram 

(viii) rhombus

Correct option is (D) All trapeziums are parallelogram

All rectangles, all squares and all rhombus are parallelogram but all trapeziums are not parallelogram.

D) All trapeziums are parallelogram

Given: D and E are the midpoints of side AB and side AC respectively. 

ED = DF 

To prove: AFBE is a parallelogram. 

Proof: 

seg AB and seg EF are the diagonals of AFBE. 

seg AD ≅ seg DB [Given] 

seg DE ≅ seg DF [Given] 

∴ Diagonals of AFBE bisect each other. 

∴ AFBE is a parallelogram. [ By test of parallelogram]

Given,

In figure,

AB = AC

CP || BA

AP is bisector of exterior angle ∠CAD

∠C = ∠B

Now,

∠CAD = ∠B + ∠C

2∠CAP = 2∠C

∠CAP = ∠C

AP || BC

But,

AB || CP (given)

Hence, 

ABCP is a parallelogram

(b) 162°.

AE || BC and AE = BC ⇒ AECB is a parallelogram. 

(For a parallelogram one pair of opposite sides can be shown equal and parallel) 

∴ ∠BCE = ∠BAE = 102° 

(Opposite angles of a parallelogram are equal) 

AB = EC 

(Opposite sides of a parallelogram) 

ED = CD = EC                (∵ AB = ED = CD) 

∴ ∆EDC is equilateral ⇒ ∠ECD = 60° 

∴ ∠BCD = ∠BAE + ∠ECD = 102° + 60° = 162°.

Correct option is (B) parallelogram

If the diagonals of a quadrilateral bisect each other then it is a parallelogram.

B) parallelogram

Correct option is: C) Square

A) Both 1 & 2 are true

Area = \(\frac{1}{2}\) x DE x (AB + CD)

= \(\frac{1}{2}\) x 4 x (8 + 10) = \(\frac{1}{2}\) x 4 x 18

= 4 x 9 = 36 cm²

Area = \(\frac{1}{2}\) x DE x (AB + CD) = \(\frac{1}{2}\) x 8 x (10 + 12) = 4 x 22 = 88 cm²

Rectangle has the maximum area among the parallelograms with same sides.

Correct option is (A) bh

Area of a parallelogram = Base \(\times\) Height

= bh square units

Correct option is  A) bh

Correct option is (B) Area of square is more than area of rectangle.

Let length and breadth of rectangle are \(l\;\&\;b\) respectively and side of square be a.

\(\therefore\) \(2(l+b)\) = 4a    \((\because\) Both have equal parameters)

\(\Rightarrow\) \(l+b\) = 2a

\(\Rightarrow\) \((l+b)^2=4a^2\)   (By squaring both sides)

\(\therefore\) Area of square \(=a^2=\frac{(l+b)^2}4\)     _________(1)

Area of rectangle \(=lb\)

Now \(a^2-lb=\frac{(l+b)^2}4-lb\)

\(=\frac{(l+b)^2-4lb}4\)

\(=\frac{(l-b)^2}4\geq0\)

\(\Rightarrow\) \(a^2\geq lb\)

\(\Rightarrow\) Area of square \(\geq\) Area of rectangle.

B) Area of square is more than area of rectangle.

Correct option is  D) KN