Class 8 Maths Chapter 12 Trapeziums Questions and Answers Kerala Syllabus

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SCERT Class 8 Maths Chapter 12 Solutions Trapeziums

Class 8 Kerala Syllabus Maths Solutions Chapter 12 Trapeziums Questions and Answers

Trapeziums Class 8 Questions and Answers Kerala Syllabus

Sides and Angles (Page 194)

Question 1.
Draw the trapeziums below:

Answer:
(i) Construction Steps:
Draw AB = 6 cm
Mark E on AB, (EB = 2 cm)
Find D by drawing arcs of radius 3 cm with centre A and E.
Find C by drawing an arc of radius 3 cm with centre B and another arc of radius 2 cm with centre D.

(ii) Construction steps
Draw AB = 5 cm
Mark E on AB, (EB = 3 cm)
Locate point D by drawing a line of 60° from A and a perpendicular from E.
The point where they meet is D.
Locate point C by drawing a perpendicular from B and an arc of radius 3 cm from D.
The point where they meet is C.
Draw the line segments AD, DC, and CB to complete the trapezium ABCD.

(iii) Construction Steps:
Draw AB = 6 cm
Mark E on AB, (EB = 4 cm)
Locate point D by drawing a line of 3cm arc from A and a perpendicular line from E.
The point where they meet is D.
Locate point C by drawing a perpendicular from B and an arc of radius 4 cm from D.
The point where they meet is C.
Draw the line segments AD, DC, and CB to complete the trapezium ABCD.

Question 2.
The picture below shows four equal trapeziums joined together:

Draw this picture.
Answer:
Draw a rectangle with a length of 8 cm and a breadth of 4 cm.
Divide this rectangle into two squares with a side of 4 cm.
Consider one square and half a second.
Mark the midpoint of the sides.
Join as in the figure.

Isosceles trapeziums (Page 197)

Question 1.
The picture shows an isosceles trapezium with the midpoint of one base joined to the endpoints of the other base:

Prove that these lines are of the same length.
Answer:
An isosceles trapezium ABCD where AB || CD and the non-parallel sides are equal (AD = BC).
Let M be the midpoint of the base AB.
Consider the two triangles, ∆AMD and ∆BMC.

AD = BC (This is given because it is an isosceles trapezium.)
∠A = ∠B (The base angles of an isosceles trapezium are equal)
AM = BM
Since M is the midpoint of AB.
By the SAS (Side-Angle-Side) congruence rule,
∆AMD ≅ ∆BMC
Therefore, corresponding sides are equal:
DM = CM

Question 2.
Prove that the line joining the midpoints of the bases of an isosceles trapezium is perpendicular to both bases.
Answer:
An isosceles trapezium ABCD with AB || CD.
Let M be the midpoint of AB and N be the midpoint of CD.
Join M and N.

Join DM and CM to form a ∆MDC.
From the proof in Question 1, we know that DM = CM
This makes ∆MDC an isosceles triangle with CD as the base.
N is the midpoint of the base CD.
In an isosceles triangle, the line joining the vertex (M) to the midpoint of the base (N) is perpendicular to the base.
Therefore, MN ⊥ CD.
Since the bases of a trapezium are parallel (AB || CD), a line perpendicular to one base is automatically perpendicular to the other.
Therefore, MN ⊥ AB.

Question 3.
Draw the pictures below:
(i) Three equal isosceles trapeziums:

(ii) Three ether equal isosceles trapeziums:

(iii) Six equal isosceles trapeziums:

Answer:
(i) Three equal isosceles trapeziums:
Usually, this pattern forms a larger equilateral triangle (with a triangular hole in the center) or a linear strip.
Construct one trapezium, then construct the second one sharing a non-parallel side, and the third one sharing the next non-parallel side.

(ii) Three other equal isosceles trapeziums:
Draw an equilateral triangle, ABC. Find its centroid O.
From the centroid, draw a line to side AC parallel to side AB and mark the point as R.
Then, from the center, measure angles of 120° from OR, to mark point P on side AB and point Q on side BC.

(iii) Six equal isosceles trapeziums:
Draw two circles of radius 1 cm and 3 cm with the same centre.
Divide the circumference of both circles into 6 equal parts.
And join them to obtain six equal isosceles trapeziums.

Area (Page 200)

Question 1.
Calculate the area of the trapeziums shown below:

Answer:
(i) Parallel sides (a and b) = 5 cm and 3 cm
Distance between them (h) = 2 cm
Area of the trapezium = \(\frac {1}{2}\)(a + b)h
= \(\frac {1}{2}\) × (3 + 5) × 2
= 8 cm²

(ii) Parallel sides (a and b) = 5 cm and 3 cm
Distance between them (h) = 2 cm
Area of the trapezium = \(\frac {1}{2}\)(a + b)h
= \(\frac {1}{2}\) × (3 + 5) × 2
= 8 cm²

Question 2.
The picture shows four equal trapeziums joined to form a large trapezium:

Calculate the area of the large trapezium.
Answer:

Area of the trapezium = \(\frac {1}{2}\)(a + b)h
= \(\frac {1}{2}\) × (8 + 4) × 4
= 24 cm²

Question 3.
The picture shows a line drawn through the point of intersection of the diagonals of a parallelogram:

Prove that this line splits the parallelogram into two trapeziums of the same area.
Answer:
Let the parallelogram be ABCD.
Let the diagonals intersect at O.
Draw a line passing through O, cutting the side AB at P and the side CD at Q.

We need to prove that the area of trapezium APQD is equal to the area of trapezium PBCQ.
Consider ΔOAP and ΔOCQ:
∠AOP = ∠COQ (Vertically opposite angles).
OA = OC (Diagonals of a parallelogram bisect each other).
∠PAO = ∠QCO (Alternate interior angles, since AB || CD).
By the ASA Congruence Rule,
ΔOAP ≅ ΔOCQ
Therefore, Area(ΔOAP) = Area(ΔOCQ)
Area of Trapezium APQD:
Area(APQD) = Area(ΔADO) + Area(ΔOAP) + Area(ΔODQ)
Area(APQD) = Area(ΔAOD) + Area(ΔPOQ)
Actually, let’s use the half-area property.
The diagonal AC divides the parallelogram into two equal triangles,
so Area(ΔADC) = Area(ΔABC) = \(\frac {1}{2}\) Area(ABCD)
Compare Areas:
Area (APQD) = Area(ADQO) + Area(ΔOAP)
Substitute Area(ΔOCQ) for Area(ΔOAP)
Area(APQD) = Area(ADQO) + Area(ΔOCQ) = Area(ΔADC)
Since Area(ΔADC) is exactly half the area of the parallelogram, the trapezium APQD occupies exactly half the area.
Consequently, the remaining part (Trapezium PBCQ) must also occupy the other half.
The line splits the parallelogram into two trapeziums of equal area.

Class 8 Maths Chapter 12 Kerala Syllabus Trapeziums Questions and Answers

Class 8 Maths Trapeziums Questions and Answers

Question 1.
A trapezium has parallel sides of length 8 cm and 12 cm. The distance between them is 5 cm. What is its area?
(A) 100 sq. cm
(B) 50 sq. cm
(C) 25 sq. cm
(D) 40 sq. cm
Answer:
(B) 50 sq. cm
Area = \(\frac {1}{2}\) × (8 + 12) × 5 = 50

Question 2.
Which of the following statements is true for an Isosceles Trapezium?
(A) All sides are equal.
(B) Diagonals are perpendicular.
(C) Non-parallel sides are equal.
(D) Opposite angles are equal.
Answer:
(C) Non-parallel sides are equal.
This is the definition of an isosceles trapezium.

Question 3.
In a trapezium, the sum of parallel sides is 10 cm and the area is 30 sq. cm. What is the height?
(A) 3 cm
(B) 6 cm
(C) 5 cm
(D) 10 cm
Answer:
(B) 6 cm
30 = \(\frac {1}{2}\) × 10h
⇒ 30 = 5h
⇒ h = 6

Question 4.
If a non-parallel side of a trapezium is perpendicular to the parallel sides, the figure is called a:
(A) Isosceles Trapezium
(B) Right Trapezium
(C) Rhombus
(D) Parallelogram
Answer:
(B) Right Trapezium

Question 5.
The area of a trapezium is calculated as half the product of the distance between parallel sides and the
(A) Product of parallel sides
(B) Difference of parallel sides
(C) Sum of parallel sides
(D) Sum of non-parallel sides
Answer:
(C) Sum of parallel sides

Question 6.
Read the given statements.
Statement I: The base angles of an isosceles trapezium are equal.
Statement II: The diagonals of an isosceles trapezium are of equal length.
Choose the correct answer:
(A) Statement I is true, and Statement II is false.
(B) Statement I is false, and Statement II is true.
(C) Both statements are true.
(D) Both statements are false.
Answer:
(C) Both statements are true.

Question 7.
Read the given statements.
Statement I: Any trapezium can be split into a parallelogram and a triangle.
Statement II: A trapezium is a quadrilateral with two pairs of parallel sides.
Choose the correct answer:
(A) Statement I is true, and Statement II is false.
(B) Statement I is false, and Statement II is true.
(C) Both statements are true.
(D) Both statements are false.
Answer:
(A) Statement I is true and Statement II is false.
Reason:
Statement I is a standard construction method.
Statement II is false because a trapezium has only one pair of parallel sides.

Question 8.
Read the given statements.
Statement I: If the non-parallel sides of a trapezium are equal, it is a cyclic quadrilateral.
Statement II: The opposite angles of an isosceles trapezium sum to 180°.
Choose the correct answer:
(A) Both statements are true and II explains I.
(B) Both statements are true, but II does not explain I.
(C) Statement I is true, Statement II is false.
(D) Both statements are false.
Answer:
(A) Both statements are true, and II explains I.
Isosceles trapezoids are always cyclic because their opposite angles are supplementary.

Question 9.
Read the given statements.
Statement I: To calculate the area of a trapezium, we need the lengths of all four sides.
Statement II: The area formula requires the lengths of parallel sides and the height.
Choose the correct answer:
(A) Statement I is true.
(B) Statement II is true.
(C) Both are true.
(D) Both are false.
Answer:
(B) Statement II is true.
We don’t need all four sides, just the parallel ones and the height.

Question 10.
A line drawn through the intersection of the diagonals of a parallelogram divides it into two shapes. These shapes are:
(A) Triangles of equal area
(B) Trapeziums of equal area
(C) Rectangles of equal area
(D) Rhombuses of equal area
Answer:
(B) Trapeziums of equal area

Question 11.
If you cut an isosceles triangle parallel to its base, the bottom part formed is a:
(A) Parallelogram
(B) Isosceles Trapezium
(C) Rhombus
(D) Square
Answer:
(B) Isosceles Trapezium

Question 12.
Which property distinguishes a parallelogram from a general trapezium?
(A) It has four sides.
(B) It has parallel sides.
(C) Both pairs of opposite sides are parallel.
(D) Sum of angles is 360°.
Answer:
(C) Both pairs of opposite sides are parallel.

Question 13.
In an isosceles trapezium, if one base angle is 60°, what is the other base angle on the same parallel side?
(A) 120°
(B) 90°
(C) 60°
(D) 30°
Answer:
(C) 60°
Base angles of an isosceles trapezium are equal.

Question 14.
In an isosceles trapezium, if one base angle is 70°, what is the angle adjacent to the other parallel side (the angle on the same leg)?
(A) 70°
(B) 110°
(C) 90°
(D) 20°
Answer:
(B) 110°
Angles between parallel lines on the same side of a transversal sum to 180°.

Question 15.
To construct a trapezium when four sides are given, we usually first construct a:
(A) Square
(B) Circle
(C) Triangle
(D) Rectangle
Answer:
(C) Triangle
We split the trapezium into a parallelogram and a triangle to find the third vertex.

Question 16.
A line joining the midpoints of the non-parallel sides of a trapezium is:
(A) Perpendicular to the bases
(B) Equal to the sum of the bases
(C) Parallel to the bases
(D) Equal to the difference of the bases
Answer:
(C) Parallel to the bases

Question 17.
Can a trapezium have three right angles?
(A) Yes
(B) No
(C) Only if it’s a square
(D) Only if it’s a rectangle
Answer:
(D) Only if it’s a rectangle
If three are 90°, the fourth must be 90° (360° – 270°), making it a rectangle, which is a special parallelogram, not a general trapezium.
So technically “No” for a strict trapezium definition, but usually considered a rectangle.

Question 18.
If the parallel sides are 4 cm and 8 cm, and the height is 4 cm, the area is:
(A) 24
(B) 48
(C) 16
(D) 32
Answer:
(A) 24
Area = \(\frac {1}{2}\) × (4 + 8) × 4
= \(\frac {1}{2}\) × 12 × 4
= 24

Question 19.
Draw the figure below.
Four equal isosceles trapeziums.

Answer:
Draw two parallel sides AB = 8 cm, CD = 4 cm.
AE = FB = 2 cm
EF = 4 cm
Also, GD = HC = 2 cm.
∠B + ∠BCD = 180°.
The angles at C are equal.
They are equal to B.
So ∠B = 60°, ∠A = 60°.
Now draw the pattern.

Question 20.
Area of a trapezium is 128 sq. cm, and the distance between its parallel sides is 8 cm. Length of one parallel side is 28 cm. Find the length of the other parallel side.
Answer:
Area of trapezium = \(\frac {1}{2}\)h(a + b)
This is given as 128 sq.cm
⇒ 128 = \(\frac {1}{2}\) × 8 × (28 + b)
⇒ 128 = 4(28 + b)
⇒ 28 + b = 32
⇒ b = 32 – 28
⇒ b = 4
Length of the other Parallel side = 4 cm

Question 21.
Compute the area of the trapezium shown below.

Answer:
Area = \(\frac {1}{2}\) × DE × (AB + CD)
= \(\frac {1}{2}\) × 4 × (8 + 10)
= \(\frac {1}{2}\) × 4 × 18
= 4 × 9
= 36 cm²

Question 22.
Compute the area of the quadrilateral ABCD.

Answer:
Area of the quadrilateral ABCD = \(\frac {1}{2}\) × AC × (PD + BQ)
= \(\frac {1}{2}\) × 5.5 × (2.5 + 1.5)
= \(\frac {1}{2}\) × 5.5 × 4
= 11 cm²

Question 23.
In the figure AB || CD. AD = BC = 13 cm. The distance between the parallel sides is 12 cm. If CD = 20 cm. Find the area of ABCD.

Answer:
To compute AB

APD is a right-angled triangle
13² = AP² + 12²
169 = AP² + 144
AP² = 169 – 144 = 25
AP = 5 cm
Similarly QB = 5 cm
PQ = CD = 20 cm
AB = 20 + 5 + 5 = 30 cm
Area = \(\frac {1}{2}\) × 12 × (30 + 20)
= 6 × 50
= 300 cm²

Question 24.
In the figure ABCD, AB is parallel to CD, and the distance between them is 8 cm.

AB = 12 cm, CD = 10 cm. Compute the area of the quadrilateral (trapezium) ABCD?
Answer:
Area = \(\frac {1}{2}\) × DE × (AB + CD)
= \(\frac {1}{2}\) × 8 × (10 + 12)
= 4 × 22
= 88 cm²

Question 25.
Compute the area of the quadrilateral ABCD in the figure.

Answer:
Area of the quadrilateral ABCD = \(\frac {1}{2}\) × AC × (BX + DY)
= \(\frac {1}{2}\) × 8 × (4.5 + 3.5)
= 4 × 8
= 32 cm²

Question 26.
Calculate the area of the isosceles trapezium drawn below:

Answer:
Area of isoceless trapezium = \(\frac {1}{2}\) × h × (a + b)
= \(\frac {1}{2}\) × 4 × (7 + 3)
= 2 × 10
= 20 cm²

Question 27.
The parallel sides of an isosceles trapezium are 8 centimetres and 4 centimetres long; and non-parallel sides are 5 centimetre long. What is its area?
Answer:
Divide the isosceles trapezium into a rectangle and two triangles.

In ΔAPD,
PD² = AD² – AP² (by Pythagoras’ theorem)
= 5² – 3²
= 16
PD = 4 cm
Area of an isosceles trapezium = \(\frac {1}{2}\) × distance between parallel sides × Sum of parallel sides.
= \(\frac {1}{2}\) × 4 × (14 + 8)
= 2 × 22
= 44 cm²

Question 28.
The lengths of the parallel sides of a trapezoid are 30 centimeters and 10 centimeters, and the distance between them is 20 centimeters. What is its area?
Answer:
Area of the trapezium = \(\frac {1}{2}\)(a + b)d
= \(\frac {1}{2}\) × (30 + 10) × 20
= 400 sq.cm

Question 29.
Compute the area of the trapezium shown below:

Answer:
Consider the ∆PQS, which is a right-angled triangle

PS² = QS² – PQ²
= 13² – 12²
= 169 – 144
= 25
PS = 5 cm
Area of a trapezium = \(\frac {1}{2}\) × (a + b)d
= \(\frac {1}{2}\) × (12 + 4) × 5
= 40 sq.cm

Class 8 Maths Chapter 12 Notes Kerala Syllabus Trapeziums

Trapezium
A trapezium is a quadrilateral with only one pair of opposite sides parallel.
The parallel sides are called bases.
The non-parallel sides are called the other two sides.

Isosceles Trapeziums
A trapezium where the non-parallel sides are equal in length is called an isosceles trapezium.
Base Angles: The base angles (angles on the same parallel side) are equal.
Diagonals: The diagonals of an isosceles trapezium are equal in length.

Consider the isosceles trapezium ABCD.
Only one pair of sides is parallel.
AB || CD
Non-parallel sides (legs) are equal in measure.
AD = BC
The diagonals are equal in measure.
AC = BD
The base angles are equal in measure.
∠D = ∠C
The opposite angles are supplementary.
∠D + ∠B = 180° and ∠C + ∠A = 180°

Drawing a Trapezium
1. Lengths of four sides are given.

Draw AB = 6 cm.
Mark E on AB such that EB = 4.5 cm
Mark D by drawing arcs of radius 4 cm with centre A and 3 cm with centre E
Find C by drawing an arc of radius 3 cm with B and another arc of 4.5 cm with D.
Complete the figure.

2. Three sides and an angle between any two of these sides are given.

Draw a right-angled triangle PST as shown in the figure.
Also, draw a rectangle of TQRS, TQ = 5 cm, and one side is equal to the attitude of the PST.
Complete the trapezium.

3. Two adjacent sides and two angles at the end of any one of these sides are given.

AB and CD are parallel then D = 180° – A = 120°
Draw AB = 5 cm.
Draw an angle of 60° at A and mark D on it, 3 cm away from A.
Mark C by drawing D = 120° & B = 50°

4. Two parallel sides and two angles at the end of any of these sides.

Divide the trapezium into a triangle and a parallelogram.
In PTS, P = 40°, PTS = 60°. You draw it.

Isosceles trapezium
An isosceles trapezium can be drawn if the following measures are given.
1. Two parallel sides and another side
Draw AB = 7 cm
Mark E on AB, (EB = 4 cm)
Find D by drawing arcs of radius 3 cm with centre A and E.
Find C by drawing an arc of radius 3 cm with centre B and another arc of radius 4 cm with centre D.

2. Two adjacent sides and the angle between them are given.
Draw PQ = 6 cm long.
On both ends draw angles ot 55°.
Mark S and R on these lines after taking a distance of 5 cm from P & Q respectively.
Draw SR.

Area of a Trapezium
The area of a trapezium is half the product of the sum of the lengths of the parallel sides and the distance between them.
A = \(\frac {1}{2}\)(a + b)h