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#### Page No 109:

#### Question 1:

With the help of a ruler and a compass it is not possible to construct an angle of :

(A) 37.5°

(B) 40°

(C) 22.5°

(D) 67.5°

#### Answer:

We may make the angels, 90°, 60°, 45°, 22.5°, 30°, and its bisector of an angle, using a ruler and a compass.

As a result, constructing a 40° angle is impossible.

#### Page No 109:

#### Question 2:

The construction of a triangle ABC, given that BC = 6 cm, ∠B = 45° is not possible when difference of AB and AC is equal to:

(A) 6.9 cm

(B) 5.2 cm

(C) 5.0 cm

(D) 4.0 cm

#### Answer:

It is given that, BC = 6 cm and$\angle \mathrm{B}=45\xb0$

If the total sum of two sides is less than or equal to the triangle's third side, the triangle cannot be constructed.

Then, AB + BC < AC

$\Rightarrow $BC < AC – AB

$\Rightarrow $6 < AC – AB

Thus, if AC – AB= 6.9 cm, then with given conditions construction of ΔABC is not possible.

#### Page No 109:

#### Question 3:

The construction of a triangle ABC, given that BC = 3 cm, ∠C = 60° is possible when difference of AB and AC is equal to :

(A) 3.2 cm

(B) 3.1 cm

(C) 3 cm

(D) 2.8 cm

#### Answer:

Given that, BC = 3 cm and $\angle \mathrm{C}=60\xb0$.

If the total sum of two sides is less than or equal to the triangle's third side, the triangle cannot be constructed.

Then, AB + BC > AC

$\Rightarrow $BC > AC – AB

$\Rightarrow $3 > AC – AB

Thus, if AC – AB = 2.8 cm, then with given conditions construction of ΔABC is not possible.

#### Page No 109:

#### Question 1:

Write **True** or **False** in the given question. Give reasons for your answer:

An angle of 52.5° can be constructed.

#### Answer:

The statement is True.

To get a 52.5° angle,

Step 1) Make a 90° angle.

Step 2) Then a 120° angle

Step 3) Finally plot an angle bisector of 120° and 90° to achieve an angle of 105° (90° + 15°).

Step 4) Bisect angle 105° to get an angle of 52.5°.

#### Page No 109:

#### Question 2:

Write **True** or **False** in the given question. Give reasons for your answer:

An angle of 42.5° can be constructed.

#### Answer:

The statement is False.

$42.5\xb0=\frac{1}{2}\times 85\xb0$ and the angle of 85° cannot be constructed with the help of a ruler and compass.

#### Page No 109:

#### Question 3:

Write **True** or **False** in the given question. Give reasons for your answer:

A triangle ABC can be constructed in which AB = 5 cm, ∠A = 45° and BC + AC = 5 cm.

#### Answer:

The statement is False

If the sum of the two sides is more than the third side, a triangle can be formed.

Here, BC + AC = AB = 5 cm

Hence, $\u25b3$ABC cannot be constructed.

#### Page No 109:

#### Question 4:

Write **True** or **False** in the given question. Give reasons for your answer:

A triangle ABC can be constructed in which BC = 6 cm, ∠C = 30° and AC – AB = 4 cm.

#### Answer:

The statement is True.

If the sum of the two sides is more than the third side, a triangle can be formed.

Then, in $\u25b3$ABC,

AB + BC > AC

$\Rightarrow $BC > AC – AB

$\Rightarrow $6 > 4, which is true,

As a result, $\u25b3$ABC with the provided conditions can be constructed.

#### Page No 109:

#### Question 5:

Write **True** or **False** in the given question. Give reasons for your answer:

A triangle ABC can be constructed in which ∠B = 105°, ∠C = 90° and AB + BC + AC = 10 cm.

#### Answer:

The statement is False

Here, $\angle \mathrm{B}=105\xb0,\angle \mathrm{C}=90\xb0$ and AB + BC + CA = 10cm

The sum of a triangle's angles is 180°.

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\xb0$

As, $\angle \mathrm{B}+\angle \mathrm{C}=105\xb0+90\xb0$

$\Rightarrow $195° > 180° which is not true.

As a result, $\u25b3$ABC with the provided conditions cannot be constructed.

#### Page No 109:

#### Question 6:

Write **True** or **False** in the given question. Give reasons for your answer:

A triangle ABC can be constructed in which ∠B = 60°, ∠C = 45° and AB + BC + AC = 12 cm.

#### Answer:

The statement is True

The sum of a triangle's angles is 180°.

$\Rightarrow \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\xb0$

As, $\angle \mathrm{B}+\angle \mathrm{C}=60\xb0+45\xb0$

= 105°< 180°, Which is true

As a result, ΔABC with the provided conditions can be constructed.

#### Page No 110:

#### Question 1:

Draw an angle of 110° with the help of a protractor and bisect it. Measure each angle.

#### Answer:

To draw 110° with the help of a protractor. We use the following steps for the required construction.

Step 1: Taking X as centre and any radius daw an arc to Intersect the rays XA and XB, say at E and D.

Step 2: Taking D and E as centres and with the radius more than 1⁄2 DE, draw arcs to intersect each other, say at F.

Step 3: Draw the ray XF.

As a result, ray XF is the angle BXA's needed bisector. When calculating each angle, we get

$\angle \mathrm{BXC}=\angle \mathrm{AXC}=55\xb0.\phantom{\rule{0ex}{0ex}}[\therefore \angle \mathrm{BXC}=\angle \mathrm{AXC}=\frac{1}{2}\angle \mathrm{BXA}=\frac{1}{2}\times 110\xb0=55\xb0]$

#### Page No 110:

#### Question 2:

Draw a line segment AB of 4 cm in length. Draw a line perpendicular to AB through A and B, respectively. Are these lines parallel?

#### Answer:

To draw a line perpendicular to AB through A and B, respectively. Use the following steps of construction.

Step 1: Draw a line segment with the length AB = 4 cm.

Step 2: Draw an arc with a radius greater than $\frac{1}{2}$AB (i.e. 2 cm) and indicate it intersects AB at E.

Step 3: Draw an arc that intersects the previous arc at F, using E as the centre and the same radius as before.

Step 4: Draw an arc using F as the centre and the same radius as before, intersecting the previous arc (obtained in step 2) at G.

Step 5: Draw arcs that intersect each other at H, using G and F as centres.

Step 6: Join AH. Thus, AX is perpendicular to AB at A. Similarly, draw BY ⊥ AB at B.

When two lines are parallel, the angle formed by them is either 0° or 180°.

Here, $\angle \mathrm{XAB}=90\xb0[\therefore \mathrm{XA}\perp \mathrm{AB}]$

and $\angle \mathrm{YBA}=90\xb0[\therefore \mathrm{YB}\perp \mathrm{AB}]$

$\begin{array}{rcl}\therefore \angle \mathrm{XAB}+\angle \mathrm{YBA}& =& 90\xb0+90\xb0\\ & =& 180\xb0\end{array}$

Thus, lines XA and YS are parallel.

#### Page No 110:

#### Question 3:

Draw an angle of 80° with the help of a protractor. Then construct angles of

(i) 40°

(ii)160° and

(iii) 120°.

#### Answer:

To begin, use a protractor to draw an angle of 80°, such as $\angle \mathrm{QOA}=180\xb0.$

Now, using the procedures below, construct angles of

(1)40°

(2) 160°

(3) 120°

1. Draw an arc with O as the centre and any radius that intersects OA at E and OO at F.

2. Draw arcs that intersect at P, using E and F as centres with a radius greater than $\frac{1}{2}$ EF.

3. Join OP Thus, $\angle \mathrm{POA}=40\xb0[\therefore 40\xb0=\frac{1}{2}\times 80\xb0]$

4. Draw an arc with F as the centre and a radius equal to EF that intersects the previous arc created in step 2 at S.

5. Join OS. Thus, $\angle \mathrm{SOA}=160\xb0[\therefore 160\xb0=2\times 80\xb0]$

6. Taking S and F as centre and radius more than $\frac{1}{2}$SF draw arcs which intersect each other at R.

7. Join OR. Thus, $\angle \mathrm{ROA}=\angle \mathrm{ROQ}=40\xb0+80\xb0=120\xb0.$

#### Page No 110:

#### Question 4:

Construct a triangle whose sides are 3.6 cm, 3.0 cm and 4.8 cm. Bisect the smallest angle and measure each part.

#### Answer:

Use the methods below to make a triangle ABC with AB = 3.6 cm, AC = 3.0 cm, and BC = 4. 8 cm.

1. Draw a line segment BC of length 4.8 cm.

2. The distance between point A and point B is 3.6 cm. Draw an arc with a radius of 3.6 cm with B as the centre.

3. The distance between point C and point A is 3 cm. Using C as the centre, draw an arc with a radius of 3 cm that intersects the preceding arc at A.

4. Join AB and AC.

Thus, ΔABC is the required triangle.

Now, here angle B is the smallest, as AC is the smallest side.

To bisect angle B, we use the following steps.

1. We draw an AB and BC intersecting at D and E, respectively, with B as the centre.

2. We create arcs intersecting at P using D and E as centres.

3. Joining BP, we obtain the angle bisector of $\angle \mathrm{B}.$ Where, $\angle \mathrm{ABC}=39\xb0$

Thus, $\angle \mathrm{ABD}=\angle \mathrm{DBC}=\frac{1}{2}\times 39\xb0=19.5\xb0$.

#### Page No 110:

#### Question 5:

Construct a triangle ABC in which BC = 5 cm, ∠B = 60° and AC + AB = 7.5 cm.

#### Answer:

Given: BC = 5 cm, ∠B = 60° and AC + AB = 7.5 cm

Use the steps below to create the triangle ABC.

1. Draw the line segment BC = 5 cm.

2. At the point B make an $\angle \mathrm{XBC}=60\xb0.$

3. Cut a line segment BD equal to AB + AC = 7.5 cm from the ray BX.

4. Join DC.

5. Make an $\angle \mathrm{DCY}=\angle \mathrm{BDC}.$

6. Let CY intersect BX at A.

Thus, $\u25b3$ABC is the required triangle.

#### Page No 110:

#### Question 6:

Construct a square of side 3 cm.

#### Answer:

Use the steps below to construct a square ABCD with side 3 cm.

1. Draw a line segment AB of length 3 cm.

2. Now, draw an angle XAB = 90° at point A of line segment AB.

3. Join BD with a line segment AD = 3 cm cut from the ray AX.

4. The distance between D and C is now 3 cm. Draw an arc with a radius of 3 cm with D as the centre.

5. The distance between point B and point C is 3 cm. Draw an arc with a radius of 3 cm that intersects the preceding arc (obtained in step 4) at C, using B as the centre.

6. Join DC and BC.

Thus, ABCD is the required square of side 3 cm.

#### Page No 110:

#### Question 7:

Construct a rectangle whose adjacent sides are of lengths 5 cm and 3.5 cm.

#### Answer:

Use the steps below to construct a rectangle ABCD with adjacent sides of 5 cm and 3.5 cm in length.

1. Draw a line segment BC of length 5 cm.

2. Now, draw an ∠XBC = 90° at point B of line segment BC.

3. Mark a line segment AB = 3.5 cm cut from the ray BX.

4. Point D is now 5 cm away from point A. Draw an arc with a radius of 5 cm with A as the centre.

5. The distance between point C and point D is 3.5 cm. Draw an arc of radius 3.5 cm with C as the centre, intersecting the preceding arc (obtained in step 4) at D.

5. Join AD and CD.

ABCD is the needed rectangle, which has adjacent sides of 5 cm and 3.5 cm in length.

#### Page No 110:

#### Question 8:

Construct a rhombus whose side is of length 3.4 cm and one of its angles is 45°.

#### Answer:

To construct a rhombus with a side length of 3.4 cm and a 45° angle on one of its sides, use the following steps

1. Draw a line segment AB of length 3.4 cm.

2. Now, at point A of line segment AB, draw an angle XAB = 45°.

3. Cut a line segment AD = 3.4 cm from the ray AX and join BD.

4. The distance between D and C is now 3.4 cm. Draw an arc with a radius of 3.4 cm with D at the centre.

5. Point C is 3.4 cm away from point B. Draw an arc of radius 3.4 cm with B as the centre, intersecting the previous arc (obtained in step 4) at C.

6. Join CD and BC.

Thus, ABCD is the required rhombus with a side length of 3.4 cm and a 45° angle on one of its sides.

#### Page No 111:

#### Question 1:

Construct the following and give justification :

A triangle if its perimeter is 10.4 cm and two angles are 45° and 120°.

#### Answer:

Let ABC be the triangle. Then, given perimeter = 10.4 cm

i.e., AB + BC + CA = 10.4 cm and two angles are 45° and 120°.

Let $\angle \mathrm{B}=45\xb0\mathrm{and}\angle \mathrm{C}=120\xb0.$

Now, to construct the $\u25b3$ABC use the following steps.

1. Draw a line segment say XY and equal to perimeter i.e., AB + BC + CA = 10.4 cm

2. Make angle $\angle \mathrm{LXY}=\angle \mathrm{B}=45\xb0$ and $\angle \mathrm{MYX}=\angle \mathrm{C}=120\xb0.$

3. Make a point A where the bisectors of $\angle $LXY and $\angle $MYX intersect.

4. Draw perpendicular bisectors PQ and RS of AX and AY, respectively.

5. PQ should intersect XY at B, while RS should intersect XY at C.

6. Join AB and AC.

Thus, ΔABC is the required triangle.

**Justification:**

Since, B lies on the perpendicular bisector PQ of AX.

Thus, AB + BC + CA = XB + BC + CY = XY

Again,$\angle \mathrm{BAX}=\angle \mathrm{AXB}[\therefore \mathrm{in}\mathrm{\Delta AXB},\mathrm{AB}=\mathrm{XB}]...\left(1\right)]$

Also,

$\angle \mathrm{ABC}=\angle \mathrm{BAX}+\angle \mathrm{AXB}$[ $\angle $ABC is an exterior angle of $\u25b3$AXB]

= $\angle \mathrm{AXB}+\angle \mathrm{AXB}[\mathrm{from}(1\left)\right]$

= $2\angle \mathrm{AXB}=\angle \mathrm{LXY}$[ AX is a bisector of $\angle \mathrm{LXB}$]

Also, $\angle \mathrm{CAY}=\angle \mathrm{AYC}$ [∴ in A AYC, AC = CY]

$\angle \mathrm{ACB}=\angle \mathrm{CAY}+\angle \mathrm{AYC}$ [$\angle \mathrm{ACB}$ is an exterior angle of $\u25b3\mathrm{AYC}$]

= $\angle \mathrm{CAY}+\angle \mathrm{CAY}$

= $2\angle \mathrm{CAY}=\angle \mathrm{MYX}$ [∴ AY is a bisector of $\angle \mathrm{MYX}$]

Thus, our construction is justified.

#### Page No 111:

#### Question 2:

Construct the following and give justification :

A triangle PQR given that QR = 3 cm, ∠PQR = 45° and QP – PR = 2 cm.

#### Answer:

In $\u25b3$PQR, QR = 3 cm, $\angle \mathrm{PQR}=45\xb0$ and QP – PR = 2 cm

To construct $\u25b3$PQR, use the following steps.

1. Draw the base QR of length 3 cm.

2. Make an angle XQR = 45° at point Q of base QR.

3. Cut the line segment QS = QP $-$ PR = 2 cm from the ray QX.

4. Join SR. Draw the perpendicular bisector of SR, which is AB.

5. Let bisector AB intersect QX at P.

6. Join PR.

Thus, $\u25b3$PQR is the required triangle.

**Justification:**

Base QR and $\angle \mathrm{PQR}$ are drawn as given.

Since, the point P lies on the perpendicular bisector of SR.

PS = PR

Now, QS = PQ – PS

= PQ – PR

Hence, our construction is justified.

#### Page No 111:

#### Question 3:

Construct the following and give justification :

A right triangle when one side is 3.5 cm and sum of other sides and the hypotenuse is 5.5 cm.

#### Answer:

Let the right triangle be ABC.

If BC = 3.5 cm, B = 90°, and the total of the other side and hypotenuse AB + AC = 5.5 cm.

To construct ΔABC use the following steps:

1. Draw the base BC = 3.5 cm

2. Make an angle XBC = 90° at the point B of base BC.

3. Cut the line segment BD equal to AB + AC i.e., 5.5 cm from the ray XB.

4. Join DC and make a $\angle \mathrm{DCY}$ equal to $\angle \mathrm{BDC}$.

5. Let Y intersect BX at A.

Thus, $\u25b3$ABC is the required triangle.

**Justification:**

Base BC and ∠B are drawn as given.

In $\u25b3$ACD,

$\angle \mathrm{ACD}=\angle \mathrm{ADC}$ [by construction]

AD = AC [sides opposite to equal angles are equal] .....(1)

Now, AB = BD – AD = BD – AC [from (1)]

$\Rightarrow $BD = AB + AC

Thus, our construction is justified.

#### Page No 111:

#### Question 4:

Construct the following and give justification :

An equilateral triangle if its altitude is 3.2 cm.

#### Answer:

In an equilateral triangle, we know that all sides are equal and all angles are equal, i.e., each angle is of 60°.

Given, altitude of an equilateral triangle, say ∆ABC, is 3.2 cm.

Follow the steps below to construct the ∆ABC.

1. Draw a line PQ.

2. Take a point D on PQ and draw a ray DE ⊥ PQ.

3. Cut the line segment AD of length 3.2 cm from DE.

4. Make angles equal to 30° at A on both sides of AD say ∠CAD and ∠BAD, where B and C lie on PQ.

5. Cut the line segment DC from PQ such that DC = BD.

6. Join AC.

Thus, ∆ABC is the required triangle.

**Justification:**

Here, $\angle \mathrm{A}=\angle \mathrm{BAD}+\angle \mathrm{CAD}$

= 30° + 30° = 60°

Also, AD ⊥ SC

$\therefore \angle \mathrm{ADS}=90\xb0.$

In $\u25b3$ABD,

$\angle \mathrm{BAD}+\angle \mathrm{DBA}=180\xb0$ [angle sum property]

$\Rightarrow 30\xb0+90\xb0+\angle \mathrm{DBA}=180\xb0[\angle \mathrm{BAD}=30\xb0]$

$\Rightarrow \angle \mathrm{DBA}=60\xb0$

Similarly, $\angle \mathrm{DCA}=60\xb0$

Thus, $\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=60\xb0$.

Hence, ΔABC is an equilateral triangle.

#### Page No 111:

#### Question 5:

Construct the following and give justification :

A rhombus whose diagonals are 4 cm and 6 cm in lengths.

#### Answer:

We know that the diagonals of a rhombus are perpendicular bisectors of one another and that all sides of a rhombus are equal.

So, to make a rhombus with diagonals of 4 cm and 6 cm, follow the methods below.

1. Draw the diagonal say AC = 4 cm.

2. Taking A and C as centres and radius more than $\frac{1}{2}$AC draw arcs on both sides of the line segment AC to intersect each other.

3. Cut both arcs intersect each other at P and Q, then join PQ.

4. PQ should intersect AC at position O. As a result, PQ is AC's perpendicular bisector.

5. Cut off 3 cm lengths from OP and OQ, then we get points B and D.

6. Now, join AB, BC, CD, and DA .

Thus, ABCD is the required rhombus.

**Justification:**

Since, D and B lie on perpendicular bisector of AC.

DA = DC and BA = BC .....(1)

[since, every point on the perpendicular bisector of line segment is equidistant from end points of line segment]

Now, $\angle \mathrm{DOC}=90\xb0$

Also, OD = OB = 3 cm

Thus, AC is perpendicular bisector or BD.

$\Rightarrow $CD = CB .....(2)

$\Rightarrow $AB = BC = CD = DA

from (1) and (2)

Thus, ABCD is a rhombus.

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