Mathematics Part i Solutions Solutions for Class 8 Math Chapter 1 Congruent Triangles are provided here with simple step-by-step explanations. These solutions for Congruent Triangles are extremely popular among class 8 students for Math Congruent Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part i Solutions Book of class 8 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part i Solutions Solutions. All Mathematics Part i Solutions Solutions for class 8 Math are prepared by experts and are 100% accurate.

Question 1:

In the figure below, AP and BQ are equal and parallel.

Prove that AM = MB. Can you suggest a construction to locate the midpoint of a line, using this idea?

Given, AP = BQ, AP || BQ

In ΔAMP and ΔBMQ,

APM = MQB (Alternate interior angles)

AP = BQ (Given)

MAP = MBQ (Alternate interior angles)

As one side and two angles of a triangle are equal to the one side and two angles of the other triangle, ΔAMP ΔBMQ.

Corresponding parts of congruent triangles are also congruent.

AM = MB

To locate the midpoint of a line, we can follow the following steps of construction:

1) Draw a line segment AB of any desired length.

2) Draw an acute angle at point A on the upper side of AB.

3) Mark a point P on ray AT.

4) From point B, draw a ray BY parallel to ray AP on the lower side of AB.

5) Mark a point Q on AY such that AP = BQ.

6) Join PQ and mark its point of intersection with AB as M.

M is the midpoint of line segment AB.

Question 2:

Prove that the point of intersection of the two diagonals of a parallelogram is the midpoint of both the diagonals.

Given: A parallelogram ABCD with diagonals AC and BD intersecting at point O.

To Prove: AO = OC and BO = OD

Proof:

Since, ABCD is a parallelogram, AB || DC and BC || AD.

DAO = OCB (Alternate interior angles)

ADO = OBC (Alternate interior angles formed by parallel lines equal in magnitude)

AD = BC (Opposite sides of a parallelogram are equal in length)

As one side and two angles of a triangle are equal to the one side and two angles of the other triangle, ΔOAD ΔOCB

DO = OB and AO = OC (Corresponding parts of congruent triangles are equal)

O is the midpoint of DB and AC.

Thus, the point of intersection of the diagonals of a parallelogram is the midpoint of both the diagonals.

Question 1:

From isosceles triangles are shown below. One angle of each is given. Find the other two angles of each.

We know that in an isosceles triangle, two sides are of equal lengths.

Also, angles opposite to equal sides of an isosceles triangle are equal.

(1)

ΔABC is an isosceles triangle.

AB = AC

⇒ ∠ACB = ABC

Using angle sum property in ΔABC, we have:

ACB + CBA + BAC = 180°

⇒ ∠BAC + 2ACB = 180°

⇒ ∠BAC + 2 × 33° = 180°

⇒ ∠BAC = 180° 66°

= 114°

Thus, the other two angles of the triangle are of measures 33° and 114°.

(2)

ΔPQR is an isosceles triangle.

PQ = PR

⇒ ∠PRQ = PQR

Using angle sum property in ΔPQR, we have:

QRP + RPQ + PQR = 180°

2PQR + RPQ = 180°

2 × 70° + RPQ = 180°

⇒ ∠RPQ = 180° 140°

= 40°

Thus, the other two angles of the triangle are of measures 70° and 40°.

(3)

ΔSTU is an isosceles triangle.

ST = UT

⇒ ∠TUS = TSU

Using angle sum property in ΔSTU, we have:

STU + TUS + TSU = 180°

17° + 2TSU = 180°

2TSU = 180° 17°

⇒∠TSU = = 81.5°

Thus, the other two angles of the triangle are of measures 81.5° each.

(4)

ΔXYZ is an isosceles triangle.

XY = ZY

⇒ ∠YZX = YXZ

Using angle sum property in ΔXYZ, we have:

XYZ + YZX + ZXY = 180°

115° + 2YZX = 180°

2YZX = 180° 115°

⇒∠YZX = = 32.5°

Thus, the other two angles of the triangle are of measures 32.5° each.

(5)

ΔABC is an isosceles triangle.

AB = AC

⇒ ∠ACB = ABC

Using angle sum property in ΔABC, we have:

ACB + CBA + BAC = 180°

⇒ ∠BAC + 2ABC = 180°

⇒ ∠BAC + 2 × 50° = 180°

⇒ ∠BAC = 180° 100°

= 80°

Thus, the other two angles of the triangle are of measures 80° and 50°.

(6)

ΔPQR is an isosceles triangle.

PQ = PR

⇒ ∠PRQ = PQR

Using angle sum property in ΔPQR, we have:

QPR + PRQ + RQP = 180°

⇒ ∠QPR + 2RQP = 180°

⇒ ∠QPR + 2 × 67° = 180°

⇒ ∠QPR = 180° 134°

= 46°

Thus, the other two angles of the triangle are of measures 67° and 46°.

Question 1:

One angles of an isosceles triangle is 120°. What are the other two angles?

Given: An isosceles triangle ABC with BAC = 120°.

We know that in an isosceles triangle, two sides are of equal length.

In ΔABC, AB = AC

Also, angles opposite to equal sides of an isosceles triangle are equal.

⇒∠ACB = ABC

Using angle sum property in ΔABC, we have:

ABC + BCA + CAB = 180°

2ABC + 120° = 180°

2ABC = 180° 120°

⇒ ∠ABC = = 30°

Thus, the other two angles of the triangle are of measures 30° each.

Question 2:

What are the angles of an isosceles right angled triangles?

Given: ΔPQR is an isosceles right-angled triangle.

We know that in an isosceles triangle, two sides are of equal length.

In ΔPQR, PQ = QR

Also, angles opposite to equal sides of an isosceles triangle are equal.

⇒∠QRP = QPR

Using angle sum property in ΔPQR, we have:

PQR + QRP + RPQ = 180°

90° + 2QRP = 180°

2QRP = 180° 90°

⇒ ∠ QRP = = 45°

Thus, the angles of an isosceles right-angled triangle are 45°, 90° and 45°.

Question 3:

What are the angles of an equilateral triangle?

Given: ΔSTU is an equilateral triangle.

We know that equilateral triangle is a special type of isosceles triangle with all the three sides equal.

ST = TU = US

We know that angles opposite to equal sides are equal in measure.

∴ ∠SUT = UTS = TSU

Using angle sum property in ΔSTU, we have:

SUT + UTS + TSU = 180°

3SUT = 180°

⇒ ∠SUT =

⇒ ∠SUT = 60°

Thus, each angle of an equilateral triangle measures 60°.

Question 1:

Prove that in a parallelogram with all four sides equal, the diagonals are perpendicular bisector of each other.

Given: A parallelogram ABCD with AB = BC = CD = DA

Also, diagonals AC and BD intersect each other at point O.

To prove: AO and BD are perpendicular bisectors of each other.

Proof:

In ΔAOD and ΔCOB:

DAO = BCO (Alternate interior angles as AD || BC)

As one side and two angles of a triangle are equal to the one side and two angles of the other triangle, ΔAOD ΔCOB

Corresponding parts of congruent triangles are also congruent.

AO = CO and DO = BO

Now, in ΔAOD and ΔAOB:

AO = AO (Common side)

DO = BO (Proved above)

As all the sides of one triangle are equal to all the sides of the other triangle, ΔAOD ΔAOB

Corresponding parts of congruent triangles are also congruent.

⇒ ∠AOD = AOB

We know that sum of angles forming a linear pair is 180°.

∴ ∠AOD + AOB = 180°

2AOD = 180°

⇒ ∠AOD = 90°

Similarly, AOB = BOC = COD = 90°

Thus, diagonals of a parallelogram with all sides equal are perpendicular bisectors of each other.

Question 2:

Can you draw a line of length 2.25 centimetres using a ruler? How about using ruler and compass?

Yes, we can draw a line segment of length 2.25 cm using a ruler.

A line segment of length 2.25 cm can be drawn using a ruler and compass by using the concept of perpendicular bisector of a line segment.

The steps of construction are as follows:

1) Draw a line segment AB of length 4.5 cm.

2) Take radius more than half the length of AB and mark arcs on the upper and lower sides of AB by taking A and B as the centres. Name the points of intersection as C and D.

3) Join CD and name the point on which it cuts the line AB as E.

Line segment AE, thus obtained, has length 2.25 cm.

Question 3:

How can we draw a circle with a given line a diameter, without actually measuring the line?

We are given a line as the diameter of the circle and we need to draw a circle without actually measuring the line.

By constructing the perpendicular bisector of the diameter, we can find the centre of the circle and using the centre we can draw a circle.

The steps of construction are as follows:

1) Draw a line segment AB of any length. This line AB is the diameter of the circle.

2) Take radius more than half the length of AB and mark arcs on the upper and lower sides of AB by taking A and B as the centres. Name the points of intersection as C and D.

3) Join CD and name the point on which it cuts AB as E. E is the centre of the required circle.

4) With E as the centre and radius as EB = EA as the centre, draw a circle.

Thus, the required circle is drawn above.

Question 4:

How do we draw an angle of?

An angle of measure can be made by following the given steps of construction:

1) Draw a line segment AB of any length.

2) Take radius more than half the length of AB and mark arcs on the upper side of AB by taking A and B as the centres. Name the point of intersection as C.

3) Join CD, where D is the point on AB such that CDB = 90°.

4) Mark points L and K on CD and DB respectively such that LD = DK. Join LK.

5) Taking D as the centre and LD as the radius, draw an arc on LK passing through both L and K.

6) Taking L and K as the centres and radius more than half of LK, draw arcs cutting each other at point E.

7) Join DE. DE is the bisector of CDB, therefore, EDB = 45°.

8) Similarly, by taking M and K as the centre, draw the bisector of EDB.

FDB so formed is an angle of measure

Question 5:

In the figure, ABCD is a parallelogram and AP = CQ. Prove that PD = BQ. Prove also that the quadrilateral PBQD is a parallelogram.

Given: A parallelogram ABCD

Also, P and Q are points AB and CD respectively such that AP = CQ.

In ΔAPD and ΔCQB,

AD = BC (Opposite sides of a parallelogram are equal)

A = C (Opposite angles of a parallelogram are equal)

AP = CQ (Given)

As the two sides and the included angle are equal to the two sides and the included angle of the other triangle, ΔAPD ΔCQB

Corresponding parts of congruent triangles are congruent.

PD = BQ

AB = DC

AP + PB = CQ + QD

PB = QD (As AP = CQ)

Also, PD = BQ

We know that if opposite sides of a quadrilateral are equal, then the quadrilateral is a parallelogram.

Thus, PBQD is a parallelogram.

Question 6:

Prove that if one pair of opposite sides of a quadrilateral are equal and parallel, then it is a parallelogram.

Consider parallelogram quadrilateral PQRS with PQ = SR and PQ || SR.

In ΔPQS and ΔRSQ:

PQ = SR (Given)

PQS = RSQ (As PQ || SR)

SQ = SQ (Common side)

As the two sides and the included angle are equal to the two sides and the included angle of the other triangle, ΔPQS ΔRSQ.

Corresponding parts of congruent triangles are congruent.

PS = QR

PSQ = RQS

Using converse of alternate angle axiom, we have:

PS || RQ

Thus, both the pairs of quadrilateral PQRS are equal and parallel.

Hence, PQRS is a parallelogram.

Question 1:

Are the triangles below congruent? Give the reason.

In ΔABC, A = 70°, B = 60° and AB = 3.5 cm

By angle sum property, we have:

A + B + C = 180°

70° + 60° + C = 180°

130° + C = 180°

⇒ ∠C = 180° 130° = 50°

In ΔPQR, P = 70°, R = 50° and PQ = 3.5 cm

Using angle sum property, we have:

P + Q + R = 180°

70° + Q + 50° = 180°

120° + Q = 180°

⇒ ∠Q = 180° 120° = 60°

⇒ ∠C = Q …(1)

Now, in ΔABC and ΔPQR:

A = P (Given)

AB = PQ (Given)

C = Q (From equation (1))

As the two angles and the included side are equal to the two angles and the included side of the other triangle, ΔABC ΔPQR.

Question 2:

How many different (non-congruent) isosceles triangles can be drawn with one angle 80° and one side 8 centimetres.

Four different non-congruent isosceles triangles can be drawn with one angle 80° and one side 8 cm.

These triangles are as follows:

I . When AB = AC = 8 cm and B = C = 80°

In ΔABC, using angle sum property, we have:

A + B + C = 180°

⇒ ∠A + 80° + 80° = 180°

⇒ ∠A + 160° = 180°

⇒ ∠A = 180° 160°

⇒ ∠A = 20°

II. When BC = 8 cm and A = 80°

As ΔABC is an isosceles triangle, B = C

In ΔABC, using angle sum property, we have:

A + B + C = 180°

80° + 2B = 180°

2B = 180° 80°

2B = 100°

⇒ ∠B = 50°

⇒ ∠C = 50°

III. When AB = AC = 8 cm and A = 80°

As ΔABC is an isosceles triangle with AB = AC, C = B

In ΔABC, using angle sum property, we have:

A + B + C = 180°

80° + 2B = 180°

2B = 180° 80°

2B = 100°

⇒ ∠B = 50°

⇒ ∠C = 50°

IV. When B = C = 80° and BC = 8 cm

As ΔABC is an isosceles triangle with C = B, AB = AC

In ΔABC, using angle sum property, we have:

A + B + C = 180°

⇒ ∠A + 2B = 180°

⇒ ∠A + 2 × 80° = 180°

⇒ ∠A + 160° = 180°

⇒ ∠A = 180° 160°

⇒ ∠A = 20°

All the triangles shown above are non-congruent isosceles triangles.

Question 3:

In the figure below, PQ = PR. Prove that the point P is one the bisector of ABC.

Given, PQ = PR and BRP =BQP = 90°

Construction: Join BP

In ΔBPQ and ΔBPR:

PQB = PRB = 90° (Given)

PB = PB (Common side)

PQ = PR (Given)

As the hypotenuse and one side of a right-angled triangle are equal to the hypotenuse and one other side of another right-angled triangle, ΔBPQ ΔBPR.

Congruent parts of congruent triangles are congruent.

∴ ∠PBQ = PBR

Thus, BP is the bisector of QBR.

Hence, P lies on the bisector of ABC.

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