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.

#### Page No 16:

#### 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?

#### Answer:

Given, *AP* = *BQ*, *AP* || *BQ*

In Δ*AMP* and Δ*BMQ*,

∠*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, Δ*AMP* ≅ Δ*BMQ.*

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*.

#### Page No 16:

#### Question 2:

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

#### Answer:

**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*.

In Δ*OAD* and Δ*OCB*:

∠*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, Δ*OAD* ≅ Δ*OCB*

⇒ *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.

#### Page No 21:

#### Question 1:

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

#### Answer:

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)**

Δ*ABC* is an isosceles triangle.

∴ *AB* = *AC*

⇒ ∠*ACB* = ∠*ABC*

Using angle sum property in Δ*ABC*, we have:

∠*ACB* + ∠*CBA* + ∠*BAC* = 180°

⇒ ∠*BAC* + 2∠*ACB *= 180°

⇒ ∠*BAC* + 2 × 33° = 180°

⇒ ∠*BAC* = 180° − 66°

= 114°

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

**(2)**

Δ*PQR *is an isosceles triangle.

∴ *PQ* = *PR*

⇒ ∠*PRQ* = ∠*PQR*

Using angle sum property in Δ*PQR*, we have:

∠*QRP* + ∠*RPQ* + ∠*PQR* = 180°

⇒ 2∠*PQR* + ∠*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)**

Δ*STU* is an isosceles triangle.

∴ *ST* = *UT*

⇒ ∠*TUS *= ∠*TSU*

Using angle sum property in Δ*STU*, we have:

∠*STU* + ∠*TUS* + ∠*TSU *= 180°

⇒ 17°* *+ 2∠*TSU* = 180°

⇒ 2∠*TSU* = 180° − 17°

⇒∠*TSU* = = 81.5°

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

**(4)**

Δ*XYZ* is an isosceles triangle.

∴ *XY* = *ZY*

⇒ ∠*YZX *= ∠*YXZ*

Using angle sum property in Δ*XYZ*, we have:

∠*XYZ* + ∠*YZX* + ∠*ZXY* = 180°

⇒ 115° + 2∠*YZX* = 180°

⇒ 2∠*YZX *= 180° − 115°

⇒∠*YZX* = = 32.5°

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

**(5)**

Δ*ABC* is an isosceles triangle.

∴ *AB* = *AC*

⇒ ∠*ACB* = ∠*ABC*

Using angle sum property in Δ*ABC*, we have:

∠*ACB* + ∠*CBA* + ∠*BAC* = 180°

⇒ ∠*BAC* + 2∠*ABC *= 180°

⇒ ∠*BAC* + 2 × 50° = 180°

⇒ ∠*BAC* = 180° − 100°

= 80°

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

**(6)**

Δ*PQR *is an isosceles triangle.

∴ *PQ* = *PR*

⇒ ∠*PRQ* = ∠*PQR*

Using angle sum property in Δ*PQR*, we have:

∠*QPR* + ∠*PRQ *+ ∠*RQP *= 180°

⇒ ∠*QPR *+ 2∠*RQP *= 180°

⇒ ∠*QPR *+ 2 × 67° = 180°

⇒ ∠*QPR *= 180° − 134°

= 46°

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

#### Page No 22:

#### Question 1:

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

#### Answer:

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

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

In Δ*ABC*, *AB* = *AC*

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

⇒∠*ACB* = ∠*ABC*

Using angle sum property in Δ*ABC*, we have:

∠*ABC* + ∠*BCA* + ∠*CAB* = 180°

⇒ 2∠*ABC* + 120° = 180°

⇒ 2∠*ABC* = 180° − 120°

⇒ ∠*ABC* = = 30°

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

#### Page No 22:

#### Question 2:

What are the angles of an isosceles right angled triangles?

#### Answer:

Given: Δ*PQR* is an isosceles right-angled triangle.

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

In Δ*PQR*, *PQ* = *QR*

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

⇒∠*QRP* = ∠*QPR*

Using angle sum property in Δ*PQR*, we have:

∠*PQR* + ∠*QRP* + ∠*RPQ* = 180°

⇒ 90° + 2∠*QRP* = 180°

⇒ 2∠*QRP* = 180° − 90°

⇒ ∠ *QRP* = = 45°

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

#### Page No 22:

#### Question 3:

What are the angles of an equilateral triangle?

#### Answer:

Given: Δ*STU* 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 Δ*STU*, we have:

∠*SUT *+ ∠*UTS* + ∠*TSU* = 180°

⇒ 3∠*SUT *= 180°

⇒ ∠*SUT *=

⇒ ∠*SUT * = 60°

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

#### Page No 25:

#### Question 1:

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

#### Answer:

**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 Δ*AOD* and Δ*COB*:

∠*ADO* = ∠*OBC* (Alternate interior angles as *AD* || *BC*)

*AD* = *BC*

∠*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, Δ*AOD* ≅ Δ*COB*

Corresponding parts of congruent triangles are also congruent.

⇒ *AO *= *CO* and *DO* =* BO*

Now, in Δ*AOD* and Δ*AOB*:

*AO* = *AO * (Common side)

*DO *= *BO * (Proved above)

*AD* = *AB* (Given)

As all the sides of one triangle are equal to all the sides of the other triangle, Δ*AOD* ≅ Δ*AOB*

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°

⇒ 2∠*AOD* = 180°

⇒ ∠*AOD* = 90°

Similarly, ∠*AOB* = ∠*BOC* = ∠*COD* = 90°

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

#### Page No 25:

#### Question 2:

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

#### Answer:

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.

#### Page No 25:

#### Question 3:

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

#### Answer:

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.

#### Page No 25:

#### Question 4:

How do we draw an angle of?

#### Answer:

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

#### Page No 25:

#### 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.

#### Answer:

Given: A parallelogram *ABCD*

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

In Δ*APD* and Δ*CQB*,

*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, Δ*APD* ≅ Δ*CQB*

Corresponding parts of congruent triangles are congruent.

∴*PD *= *BQ*

Now, in quadrilateral* PBQD*:

*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.

#### Page No 25:

#### Question 6:

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

#### Answer:

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

In Δ*PQS *and Δ*RSQ*:

*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, Δ*PQS *≅ Δ*RSQ.*

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.

#### Page No 26:

#### Question 1:

Are the triangles below congruent? Give the reason.

#### Answer:

In Δ*ABC*, ∠*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 Δ*PQR*, ∠*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 Δ*ABC* and Δ*PQR*:

∠*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, Δ*ABC* ≅ Δ*PQR.*

#### Page No 26:

#### Question 2:

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

#### Answer:

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 Δ*ABC*, 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 Δ*ABC* is an isosceles triangle, ∠*B* = ∠*C*.

In Δ*ABC*, using angle sum property, we have:

∠*A* + ∠*B* + ∠*C* = 180°

⇒ 80° + 2∠*B* = 180°

⇒ 2∠*B* = 180° − 80°

⇒ 2∠*B* = 100°

⇒ ∠*B* = 50°

⇒ ∠*C* = 50°

**III.** When *AB* = *AC* = 8 cm and ∠*A* = 80°

As Δ*ABC* is an isosceles triangle with* AB* = *AC*, ∠*C* = ∠*B*.

In Δ*ABC*, using angle sum property, we have:

∠*A* + ∠*B* + ∠*C* = 180°

⇒ 80° + 2∠*B* = 180°

⇒ 2∠*B* = 180° − 80°

⇒ 2∠*B* = 100°

⇒ ∠*B* = 50°

⇒ ∠*C* = 50°

**IV. **When ∠*B* = ∠*C* = 80° and *BC* = 8 cm

As Δ*ABC* is an isosceles triangle with* *∠*C* = ∠*B*, *AB* = *AC*

In Δ*ABC*, using angle sum property, we have:

∠*A* + ∠*B* + ∠*C* = 180°

⇒ ∠*A* + 2∠*B* = 180°

⇒ ∠*A* + 2 × 80° = 180°

⇒ ∠*A* + 160° = 180°

⇒ ∠*A* = 180° − 160°

⇒ ∠*A* = 20°

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

#### Page No 26:

#### Question 3:

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

#### Answer:

Given, *PQ* = *PR* and ∠*BRP *=∠*BQP* = 90°

**Construction:** Join BP

In Δ*BPQ* and Δ*BPR*:

∠*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, Δ*BPQ* ≅ Δ*BPR*.

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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