Quadrilaterals- 2 - Free MCQ Practice Test with solutions, CTET Mathematics

Consider P1 as the area of first parallelogram and P2 as the area of second parallelogram

We know that

Area of parallelogram = b × h

P1 = b1 × h1

P2 = b2 × h2

From the question, b1 = b2

As the distance between two parallel lines is equal

h1 = h2

P1 = b × h …. (1)

P2 = b × h …. (2)

From equation (1) and (2)

P1 = P2

So we get

P1 : P2 = 1 : 1

Therefore, the ratio of their areas is 1 : 1.

Given in quadrilateral ABCD:

• ∠A = 60°
• Ratio ∠B : ∠C : ∠D = 2:3:7

We let:

• ∠B = 2x
• ∠C = 3x
• ∠D = 7x

Since the sum of the angles in a quadrilateral is 360°, we have:

∠A + ∠B + ∠C + ∠D = 360°
60° + 2x + 3x + 7x = 360°
12x + 60° = 360°
12x = 300°
x = 25°

Substituting back to find ∠D:

∠D = 7x = 7 × 25° = 175°

Thus, ∠D is 175°.

D and E are the mid-points of the sides AB and AC respectively of ∆ABC. DE is produced to F. To prove that CF is equal and parallel to DA, we need an additional information which is DE = EF.

Let us assume that, DE = EF

AE = CE  ......[∵ E is mid-point of AC]

DE = EF  .....[Assumed]

∠AED = ∠FEC  ......[Vertically opposite angles]

∴ By SAS, ∆AED ≅ ∆FEC

∴ By CPCT, AD = CF and ∠ADE = ∠CFE

∵ Alternate interior angles are equal

∴ AD ∥ CF

That proves our assumption was correct.

Given: P, Q, and R are the midpoints of sides AB, BC, and AC of triangle ABC, respectively.

We know that:

• AB = 10 cm
• BC = 8 cm
• AC = 12 cm

Step-by-Step Solution

1. Calculate the Perimeter of ΔABC:

   The perimeter of ΔABC is given by the sum of its sides:

   AB + BC + AC = 10 cm + 8 cm + 12 cm = 30 cm

2. Determine the Perimeter of ΔPQR:

   Since P, Q, and R are midpoints, ΔPQR is similar to ΔABC with a ratio of 1:2.

   Therefore, the perimeter of ΔPQR is half of the perimeter of ΔABC:

   Perimeter of ΔPQR = (1/2) × 30 = 15 cm

The perimeter of ΔPQR is 15 cm.

Correct option: c) 15 cm

In right  ΔABC, ∠B = 90°

By using Pythagoras theorem

In ΔABC

D and E are midpoints of  AB and AC

Given 2 statements.
I. In a parallelogram, the angle bisectors of two adjacent angles enclose a right angle.
II. The angle bisectors of a parallelogram form a rectangle.
III. The triangle formed by joining the mid-points of the sides of an isosceles triangle is not necessarily an isosceles triangle.
As, in a parallelogram the angle bisectors intersect and form a right-angle.
Hence, statement I is true.
The angle bisector forms a rectangle as it also intersects at the right-angle.
The statement II is also correct.
Hence, option 2 is correct.

Line APB is parallel to CQD

when we join PQ it will be transversal

then angleBPQ = angleCQP    (alternate angles)

angleAPQ = anglePQD   

when we will draw bisectors 

then the figure formed will have opposite angles equal

which means that it is a parallelogram

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.

∠ OAD = ​∠ OCB = 30^∘
(Alternate interior angles)
∠ AOB + ∠ BOC = 180^∘
(Linear pair of angles)
∴ ∠ BOC =180^∘ − 70^∘ = 110^∘
(∠ AOB = 70^∘)
In ∆ BOC, we have:
∠ OBC = 180^∘ − (110^∘ + 30^∘) = 40^∘
∴ ​∠ DBC = 40^∘

In a parallelogram ABCD, the bisectors of angles A and B intersect at point O. The following explanation details the calculation of the measure of ∠AOB.

Parallelogram Properties

In parallelogram ABCD, angles A and B are supplementary, which means:

∠A + ∠B = 180°

Angle Bisectors

The bisectors of angles A and B intersect at O, forming angles:

∠OAB = ½∠A and ∠OBA = ½∠B

Forming ∠AOB

Angle ∠AOB can be calculated as:

∠AOB = 360° - (∠OAB + ∠OBA + ∠AOB'), where ∠AOB' is the angle opposite ∠AOB within triangle AOB.

Sum in Triangle AOB

In triangle AOB, the sum of angles is:

∠OAB + ∠OBA + ∠AOB' = 180°

Using Supplementary Angles

Given that ∠A and ∠B are supplementary:

∠AOB' = 180° - (½∠A + ½∠B) = 180° - ½ × 180° = 90°

Calculating ∠AOB

Since ∠AOB and ∠AOB' form a straight line at point O:

∠AOB = 360° - 180° - 90° = 90°

Therefore, the measure of ∠AOB is 90°.

Clearly, statements II and III are true. Statement I is false, as diagonal of a rectangle does not bisect ∠A and ∠C (∴ adjacent sides are not equal).

In ΔABC, D and E are the mid-points of sides AB and AC, respectively.

By mid-point theorem,

DE || BC  .....(i)

⇒ DE = PO + OQ

⇒ DE = PQ

Now, In ΔAOC, Q and E are the midpoints of OC and AC respectively.

From equations (iii) and (iv),

EQ || PD and EQ = PD

From equations (i) and (ii),

 DE || BC (or DE || PQ) and DE = PQ

Hence, DEQP is a parallelogram.

We will use the mid-point theorem here. It states that the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it. 

Let us join AC and BD. In △ABC,

P and Q are the mid-points of AB and BC respectively.

∴ PQ || AC and PQ = 1/2 AC (Mid-point theorem) ... (1)

Similarly, in △ADC,

SR || AC and SR = 1/2 AC (Mid-point theorem) ... (2)

Clearly, PQ || SR and PQ = SR [From equation (1) and (2)]

Since in quadrilateral PQRS, one pair of opposite sides are equal and parallel to each other, it is a parallelogram.

∴ PS || QR and PS = QR (Opposite sides of the parallelogram) ... (3)

In △BCD, Q and R are the mid-points of side BC and CD respectively.

∴ QR || BD and QR = 1/2 BD (Mid-point theorem) ... (4)

However, the diagonals of a rectangle are equal.

∴ AC = BD ... (5)

Thus, QR = 1/2 AC

Also, in △BAD

PS || BD and PS = 1/2 BD

Thus, QR = PS .... (6)

By using Equations (1), (2), (3), (4), and (5), we obtain

PQ = QR = SR = PS

Therefore, PQRS is a rhombus.

Given: A rhombus ABCD
To Prove: 4AB² = AC² + BD²
Proof: The diagonals of a rhombus bisect each other at right angles.

In the problem, points D and E are the midpoints of sides AB and AC respectively in triangle △ABC, and you need to find the length of segment DE given that BC = 5.6 cm.

Since D and E are midpoints, segment DE is a midsegment of the triangle. A midsegment of a triangle connects the midpoints of two sides of the triangle and is parallel to the third side. Importantly, the length of a midsegment is half the length of the side it is parallel to.

In this case, DE is parallel to side BC, and hence its length is half the length of BC. Therefore:

DE = 1/2 * BC = 1/2 * 5.6 cm = 2.8 cm

This matches the given answer D: 2.8 cm.

P, Q, R are the mid-points of sides BC, CA and AB respectively.
AC = 21 cm, BC = 29 cm and AB = 30°
.: P, Q, R and the midpoints of sides BC, CA and AB respectively.
.: PQ || AB and PQ = ½AB

PQ = ½ × 30° = 15cm
Similarly, QR || BC and
QR = ½ × BC = ½ × 29 = 14.5cm
and RP || AC and RP = ½AC
= ½ × 21 = 10.5cm
Now perimeter of quad. ARPQ,
= AR+RP+PQ+AQ
= ½AB + ½AC + ½AB + ½AC
= AB + AC = 30 + 21 = 51 cm

We have ABCD, a parallelogram given below:

Therefore, we have AB || BC

Now,  AD || BC and transversal AB intersects them at A and B respectively. Therefore,

Sum of consecutive interior angle is supplementary. That is;
∠A + ∠B = 180°

We have AR and BR as bisectors of ∠A and ∠B  respectively.

 ∠RAB +∠RBA = 90° …… (i)

Now, in ΔABR, by angle sum property of a triangle, we get:

∠RAB + ∠RBA +∠ARB = 180° 

From equation (i), we get:

90° + ∠ARB = 180°

∠ARB = 90° 

Similarly, we can prove that ∠DPC = 90° .

Now, AB || DC and transversal ADintersects them at A and D respectively. Therefore,

Sum of consecutive interior angle is supplementary. That is;

∠A + ∠D = 180°

We have AR and DP as bisectors of ∠A and ∠D  respectively.

 ∠DAR + ∠ADP = 90° …… (ii)

Now, in ΔADR, by angle sum property of a triangle, we get:

∠DAR + ∠ADP + ∠AQD = 180° 

From equation (i), we get:

 90° +∠AQD = 180°

∠AQD = 90° 

We know that ∠AQD and ∠PQR are vertically opposite angles, thus,

∠PQR = 90°

Similarly, we can prove that ∠PSR = 90° .

Therefore, PQRS is a rectangle.

Hence, the correct choice is (d).

Consider ABCD as a quadrilateral

It is given that the opposite angles are equal hence the given quadrilateral is a parallelogram.

Here the opposite side of AB is CD

If AB = 4 cm then CD = 4 cm

Therefore, CD = 4 cm.

Draw perpendicular from D and C on AB which meets AB at E and F, respectively.
∴ DEFC is a parallelogram and EF = CD .
In Δ ABC, ∠B is acute.
∴ AC² = BC² + AB² - 2AB.AE
In Δ ABD, ∠A is acute.
∴ BD² = AD² + AB² - 2AB.AF

∴ AC² + BD² = (BC² + AD²) + (AB² + AB²) - 2AB(AE + BF)
= (BC² + AD²) + 2AB(AB - AE - BF)
[∵ AB = AE + EF + FB and AB - AE = BE]
= (BC² + AD²) + 2AB(BE - BF)
= (BC² + AD²) + 2AB.EF
∴ AC² + BD² = (BC² + AD²) + 2AB.CD

∠B = 90°
∠C - ∠D = 60°
∠A - ∠C - ∠D = 10°.......(1)
∠A + ∠C + ∠D = 360° - 90° = 270° (from Angle sum property of quadrilateral)

On adding,
∠A + ∠C + ∠D = 270°
∠A - ∠C - ∠D = 10°

2∠A = 280°
∠A = 140°

=> from (1)
∠A = 140°, ∠A - (∠C + ∠D) = 10°
140° - (∠C + ∠D) = 10°
∠C + ∠D = 130° ...........(2)

If ∠C - ∠D = 60° & ∠C + ∠D = 130°
2∠C = 190°, ∠C = 95°

∠D = 130° - 95° - from (2)
∠D = 35°

So, the correct answer is option (a).

Step 1: Given that angle A = 90 degrees, we know that ABCD has one right angle.

Step 2: Since AB = BC = CD = DA, all sides of quadrilateral ABCD are equal in length. Let's denote the length of each side as 's'.

Step 3: Now, we have a quadrilateral with one right angle and all sides equal. This means that ABCD is an equilateral quadrilateral.

Step 4: In an equilateral quadrilateral with one right angle, the other angles must also be right angles. Therefore, angle B = angle C = angle D = 90 degrees.

Step 5: Since all angles are 90 degrees and all sides are equal, quadrilateral ABCD is a square.

Quadrilateral ABCD is a square.

This is because in a rhombus, the diagonals always bisect each other at right angles (90 degrees). While the diagonals do bisect each other (option C) and also bisect the opposite angles (option D), the defining characteristic in the context of a rhombus's diagonal properties is that they are perpendicular.

So, if the question specifically asks about the nature of the diagonals' intersection, option A is the most accurate. Option D, "in which diagonals bisect opposite angles," is also true but is not specifically about the angle at which they bisect each other.

In triangle △ABC, if EF is the line segment joining the midpoints of sides AB and AC, then EF is a midsegment of the triangle. According to the triangle midsegment theorem, a midsegment in a triangle is parallel to the third side and its length is half the length of that side.

Since BC = 7.2 cm, the length of EF, which is a midsegment, would be half of BC. Therefore:

EF = 1/2 * BC = 1/2 * 7.2 cm = 3.6 cm

So, the correct answer is A: 3.6cm.

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°.    

Given,

ABCD is a parallelogram

AC and BD are the diagonals

AD || BC

∠DAC = ∠ACB [Alternate angle]

∠ACB = 32º

We know that

∠AOB + ∠BOC = 180º [Straight line]

Substituting the values

70º + ∠BOC = 180º

∠BOC = 110º

In Δ BOC,

∠OBC + ∠BOC + ∠OCB = 180º

Substituting the values

∠OBC + 110º + 32º = 180º

∠OBC = 38º

∠DBC = 38º

Therefore, ∠DBC is equal to 38º.