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13. PRACTICAL GEOMETRY

289733 In which of the following figures the adjacent sides are not necessarily be equal?

1 Parallelogram

2 Rhombus

3 Rectangle

4 Square

13. PRACTICAL GEOMETRY

289734 In Fig. $PQ ⊥ RQ, PQ = 5 cm$ and $QR = 5 cm .$ Then $△ PQR$ is:
7

1 A right triangle but not isosceles.

2 An isosceles right triangle.

3 Isosceles but not a right triangle.

4 Neither isosceles nor right triangle.

Explanation:

An isosceles right triangle.
Since, PQ Perpendicular to RQ,
So, $△ PQR = 90^{\circ}$
$∴ △ PQR$ is right angled triangle.
Also, in $△ PQR,$
$PQ = QR$
$△ PQR$ is an isosceles triangle.
Hence, $△ PQR$ is an isosceles right angled triangle.

13. PRACTICAL GEOMETRY

289735 The number of diagonals in a septagon is:

1 21

2 42

3 7

4 14

Explanation:

14
We know that, if a polygon has n sides, then Number of diagonals $= \frac{n (3 - 2)}{2}$
A septagon is a polygon having seven sides, i.e. n = 7
Number of diagonals in septagon $= \frac{7 (7 - 3)}{2} = 14$
Note: A diagonal is a line segment joining two non-consecutive vertices of a polygon.

13. PRACTICAL GEOMETRY

289736 The steps of construction of an $∠ AOB = 45^{\circ}$ is given in jumbled order below:
Place compass on intersection point.
Place ruler on start point and where arc intersects perpendicular line.
Adjust compass width to reach start point.
Construct a perpendicular line.
Draw 45 degree line.
Draw an arc that intersects perpendicular line.
The third step in process is:

1 2

2 3

3 4

4 5

13. PRACTICAL GEOMETRY

289737 $\vec{MN}$ is the perpendicular bisector of $\overset{\leftrightarrow}{AB}$ Which of the given statements is correct?
$∠ ANM + ∠ MNB = 90^{\circ}$
$\overset{―}{AN} = \overset{―}{NB}$
$\overset{―}{AN} = 2 \overset{―}{NB}$
$∠ MNB = \frac{1}{2} ∠ ANM$

1 (i) and (iii) only

2 (ii) and (iv) only

3 (i) and (ii) only

4 (ii) and (iii) only

Explanation:

(i) and (ii) only
Solution :
$\vec{NM} ⊥ \overset{\leftrightarrow}{AB}$ and $\vec{NM}$ divides $\overset{\leftrightarrow}{AB}$ into two congruent parts.
8
Clearly $∠ ANM + ∠ MNB = 90^{\circ}$ is true.
$\overset{―}{AN} = \overset{―}{NB}$ is true since $\vec{NM} ⊥ \overset{\leftrightarrow}{AB}$ $\overset{―}{AN} = 2 \overset{―}{NB}$ is false, and $∠ MNB = \frac{1}{2} ∠ ANM$ is false.
Thus, only (i) and (ii) are correct.

13. PRACTICAL GEOMETRY

289733 In which of the following figures the adjacent sides are not necessarily be equal?

1 Parallelogram

2 Rhombus

3 Rectangle

4 Square

13. PRACTICAL GEOMETRY

289734 In Fig. $PQ ⊥ RQ, PQ = 5 cm$ and $QR = 5 cm .$ Then $△ PQR$ is:
7

1 A right triangle but not isosceles.

2 An isosceles right triangle.

3 Isosceles but not a right triangle.

4 Neither isosceles nor right triangle.

Explanation:

An isosceles right triangle.
Since, PQ Perpendicular to RQ,
So, $△ PQR = 90^{\circ}$
$∴ △ PQR$ is right angled triangle.
Also, in $△ PQR,$
$PQ = QR$
$△ PQR$ is an isosceles triangle.
Hence, $△ PQR$ is an isosceles right angled triangle.

13. PRACTICAL GEOMETRY

289735 The number of diagonals in a septagon is:

1 21

2 42

3 7

4 14

Explanation:

14
We know that, if a polygon has n sides, then Number of diagonals $= \frac{n (3 - 2)}{2}$
A septagon is a polygon having seven sides, i.e. n = 7
Number of diagonals in septagon $= \frac{7 (7 - 3)}{2} = 14$
Note: A diagonal is a line segment joining two non-consecutive vertices of a polygon.

13. PRACTICAL GEOMETRY

289736 The steps of construction of an $∠ AOB = 45^{\circ}$ is given in jumbled order below:
Place compass on intersection point.
Place ruler on start point and where arc intersects perpendicular line.
Adjust compass width to reach start point.
Construct a perpendicular line.
Draw 45 degree line.
Draw an arc that intersects perpendicular line.
The third step in process is:

1 2

2 3

3 4

4 5

13. PRACTICAL GEOMETRY

289737 $\vec{MN}$ is the perpendicular bisector of $\overset{\leftrightarrow}{AB}$ Which of the given statements is correct?
$∠ ANM + ∠ MNB = 90^{\circ}$
$\overset{―}{AN} = \overset{―}{NB}$
$\overset{―}{AN} = 2 \overset{―}{NB}$
$∠ MNB = \frac{1}{2} ∠ ANM$

1 (i) and (iii) only

2 (ii) and (iv) only

3 (i) and (ii) only

4 (ii) and (iii) only

Explanation:

(i) and (ii) only
Solution :
$\vec{NM} ⊥ \overset{\leftrightarrow}{AB}$ and $\vec{NM}$ divides $\overset{\leftrightarrow}{AB}$ into two congruent parts.
8
Clearly $∠ ANM + ∠ MNB = 90^{\circ}$ is true.
$\overset{―}{AN} = \overset{―}{NB}$ is true since $\vec{NM} ⊥ \overset{\leftrightarrow}{AB}$ $\overset{―}{AN} = 2 \overset{―}{NB}$ is false, and $∠ MNB = \frac{1}{2} ∠ ANM$ is false.
Thus, only (i) and (ii) are correct.

13. PRACTICAL GEOMETRY

289733 In which of the following figures the adjacent sides are not necessarily be equal?

1 Parallelogram

2 Rhombus

3 Rectangle

4 Square

13. PRACTICAL GEOMETRY

289734 In Fig. $PQ ⊥ RQ, PQ = 5 cm$ and $QR = 5 cm .$ Then $△ PQR$ is:
7

1 A right triangle but not isosceles.

2 An isosceles right triangle.

3 Isosceles but not a right triangle.

4 Neither isosceles nor right triangle.

Explanation:

An isosceles right triangle.
Since, PQ Perpendicular to RQ,
So, $△ PQR = 90^{\circ}$
$∴ △ PQR$ is right angled triangle.
Also, in $△ PQR,$
$PQ = QR$
$△ PQR$ is an isosceles triangle.
Hence, $△ PQR$ is an isosceles right angled triangle.

13. PRACTICAL GEOMETRY

289735 The number of diagonals in a septagon is:

1 21

2 42

3 7

4 14

Explanation:

14
We know that, if a polygon has n sides, then Number of diagonals $= \frac{n (3 - 2)}{2}$
A septagon is a polygon having seven sides, i.e. n = 7
Number of diagonals in septagon $= \frac{7 (7 - 3)}{2} = 14$
Note: A diagonal is a line segment joining two non-consecutive vertices of a polygon.

13. PRACTICAL GEOMETRY

289736 The steps of construction of an $∠ AOB = 45^{\circ}$ is given in jumbled order below:
Place compass on intersection point.
Place ruler on start point and where arc intersects perpendicular line.
Adjust compass width to reach start point.
Construct a perpendicular line.
Draw 45 degree line.
Draw an arc that intersects perpendicular line.
The third step in process is:

1 2

2 3

3 4

4 5

13. PRACTICAL GEOMETRY

289737 $\vec{MN}$ is the perpendicular bisector of $\overset{\leftrightarrow}{AB}$ Which of the given statements is correct?
$∠ ANM + ∠ MNB = 90^{\circ}$
$\overset{―}{AN} = \overset{―}{NB}$
$\overset{―}{AN} = 2 \overset{―}{NB}$
$∠ MNB = \frac{1}{2} ∠ ANM$

1 (i) and (iii) only

2 (ii) and (iv) only

3 (i) and (ii) only

4 (ii) and (iii) only

Explanation:

(i) and (ii) only
Solution :
$\vec{NM} ⊥ \overset{\leftrightarrow}{AB}$ and $\vec{NM}$ divides $\overset{\leftrightarrow}{AB}$ into two congruent parts.
8
Clearly $∠ ANM + ∠ MNB = 90^{\circ}$ is true.
$\overset{―}{AN} = \overset{―}{NB}$ is true since $\vec{NM} ⊥ \overset{\leftrightarrow}{AB}$ $\overset{―}{AN} = 2 \overset{―}{NB}$ is false, and $∠ MNB = \frac{1}{2} ∠ ANM$ is false.
Thus, only (i) and (ii) are correct.

13. PRACTICAL GEOMETRY

289733 In which of the following figures the adjacent sides are not necessarily be equal?

1 Parallelogram

2 Rhombus

3 Rectangle

4 Square

13. PRACTICAL GEOMETRY

289734 In Fig. $PQ ⊥ RQ, PQ = 5 cm$ and $QR = 5 cm .$ Then $△ PQR$ is:
7

1 A right triangle but not isosceles.

2 An isosceles right triangle.

3 Isosceles but not a right triangle.

4 Neither isosceles nor right triangle.

Explanation:

An isosceles right triangle.
Since, PQ Perpendicular to RQ,
So, $△ PQR = 90^{\circ}$
$∴ △ PQR$ is right angled triangle.
Also, in $△ PQR,$
$PQ = QR$
$△ PQR$ is an isosceles triangle.
Hence, $△ PQR$ is an isosceles right angled triangle.

13. PRACTICAL GEOMETRY

289735 The number of diagonals in a septagon is:

1 21

2 42

3 7

4 14

Explanation:

14
We know that, if a polygon has n sides, then Number of diagonals $= \frac{n (3 - 2)}{2}$
A septagon is a polygon having seven sides, i.e. n = 7
Number of diagonals in septagon $= \frac{7 (7 - 3)}{2} = 14$
Note: A diagonal is a line segment joining two non-consecutive vertices of a polygon.

13. PRACTICAL GEOMETRY

289736 The steps of construction of an $∠ AOB = 45^{\circ}$ is given in jumbled order below:
Place compass on intersection point.
Place ruler on start point and where arc intersects perpendicular line.
Adjust compass width to reach start point.
Construct a perpendicular line.
Draw 45 degree line.
Draw an arc that intersects perpendicular line.
The third step in process is:

1 2

2 3

3 4

4 5

13. PRACTICAL GEOMETRY

289737 $\vec{MN}$ is the perpendicular bisector of $\overset{\leftrightarrow}{AB}$ Which of the given statements is correct?
$∠ ANM + ∠ MNB = 90^{\circ}$
$\overset{―}{AN} = \overset{―}{NB}$
$\overset{―}{AN} = 2 \overset{―}{NB}$
$∠ MNB = \frac{1}{2} ∠ ANM$

1 (i) and (iii) only

2 (ii) and (iv) only

3 (i) and (ii) only

4 (ii) and (iii) only

Explanation:

(i) and (ii) only
Solution :
$\vec{NM} ⊥ \overset{\leftrightarrow}{AB}$ and $\vec{NM}$ divides $\overset{\leftrightarrow}{AB}$ into two congruent parts.
8
Clearly $∠ ANM + ∠ MNB = 90^{\circ}$ is true.
$\overset{―}{AN} = \overset{―}{NB}$ is true since $\vec{NM} ⊥ \overset{\leftrightarrow}{AB}$ $\overset{―}{AN} = 2 \overset{―}{NB}$ is false, and $∠ MNB = \frac{1}{2} ∠ ANM$ is false.
Thus, only (i) and (ii) are correct.

13. PRACTICAL GEOMETRY

289733 In which of the following figures the adjacent sides are not necessarily be equal?

1 Parallelogram

2 Rhombus

3 Rectangle

4 Square

13. PRACTICAL GEOMETRY

289734 In Fig. $PQ ⊥ RQ, PQ = 5 cm$ and $QR = 5 cm .$ Then $△ PQR$ is:
7

1 A right triangle but not isosceles.

2 An isosceles right triangle.

3 Isosceles but not a right triangle.

4 Neither isosceles nor right triangle.

Explanation:

An isosceles right triangle.
Since, PQ Perpendicular to RQ,
So, $△ PQR = 90^{\circ}$
$∴ △ PQR$ is right angled triangle.
Also, in $△ PQR,$
$PQ = QR$
$△ PQR$ is an isosceles triangle.
Hence, $△ PQR$ is an isosceles right angled triangle.

13. PRACTICAL GEOMETRY

289735 The number of diagonals in a septagon is:

1 21

2 42

3 7

4 14

Explanation:

14
We know that, if a polygon has n sides, then Number of diagonals $= \frac{n (3 - 2)}{2}$
A septagon is a polygon having seven sides, i.e. n = 7
Number of diagonals in septagon $= \frac{7 (7 - 3)}{2} = 14$
Note: A diagonal is a line segment joining two non-consecutive vertices of a polygon.

13. PRACTICAL GEOMETRY

289736 The steps of construction of an $∠ AOB = 45^{\circ}$ is given in jumbled order below:
Place compass on intersection point.
Place ruler on start point and where arc intersects perpendicular line.
Adjust compass width to reach start point.
Construct a perpendicular line.
Draw 45 degree line.
Draw an arc that intersects perpendicular line.
The third step in process is:

1 2

2 3

3 4

4 5

13. PRACTICAL GEOMETRY

289737 $\vec{MN}$ is the perpendicular bisector of $\overset{\leftrightarrow}{AB}$ Which of the given statements is correct?
$∠ ANM + ∠ MNB = 90^{\circ}$
$\overset{―}{AN} = \overset{―}{NB}$
$\overset{―}{AN} = 2 \overset{―}{NB}$
$∠ MNB = \frac{1}{2} ∠ ANM$

1 (i) and (iii) only

2 (ii) and (iv) only

3 (i) and (ii) only

4 (ii) and (iii) only

Explanation:

(i) and (ii) only
Solution :
$\vec{NM} ⊥ \overset{\leftrightarrow}{AB}$ and $\vec{NM}$ divides $\overset{\leftrightarrow}{AB}$ into two congruent parts.
8
Clearly $∠ ANM + ∠ MNB = 90^{\circ}$ is true.
$\overset{―}{AN} = \overset{―}{NB}$ is true since $\vec{NM} ⊥ \overset{\leftrightarrow}{AB}$ $\overset{―}{AN} = 2 \overset{―}{NB}$ is false, and $∠ MNB = \frac{1}{2} ∠ ANM$ is false.
Thus, only (i) and (ii) are correct.

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