143 + Interview Questions in CONSTRUCTIONS IN GENERAL AWARENESS Page 3 InterviewSolution

101.	Write True or False and give reasons for your answer.To construct a triangle similar to a given ∆ ABC with its sides 7/3 of the corresponding sides of ∆ABC, draw a ray BX making an acute angle with BC and X lies on the opposite side of A with respect of BC. The points B1, B2,...........B1 are located at equal distances on BX, B3 is joined to C and then a where C’ lines on BC produced. Finally line segment A’C’ is drawn parallel to AC.
Answer» False Steps of construction 1. Draw a line segment BC. 2. B and C as centers draw two arcs of suitable radius intersecting each other at A. 3. Join BA and CA. ∆ABC is required triangle. 4. From B draw any ray BX downwards making an acute angle CBX. 5. Marked seven points B₁,B₂,B₃,............B₇ on BX (BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇). 6. Join B₃C and from B₇ draw a line B₇C’\|\| B₃C intersecting the extended line segment BC at C’. 7. Draw C’ A’\|\|CA intersecting the extended line segment BA at A’. Then, ∆A’BC’ is the required triangle whose sides are 7/3 of the corresponding sides of ∆ABC. Given that, segment B₆C’ \|\| B₃C. But since our construction is never possible that segment B₆C’ \|\| B₃C because the similar triangle A’BC’ has its sides 7/3 of the corresponding sides of triangle ABC. So, B₇C’ is parallel to B₃C.

101.

Write True or False and give reasons for your answer.To construct a triangle similar to a given ∆ ABC with its sides 7/3 of the corresponding sides of ∆ABC, draw a ray BX making an acute angle with BC and X lies on the opposite side of A with respect of BC. The points B1, B2,...........B1 are located at equal distances on BX, B3 is joined to C and then a where C’ lines on BC produced. Finally line segment A’C’ is drawn parallel to AC.

Answer»

False

Steps of construction

1. Draw a line segment BC.

2. B and C as centers draw two arcs of suitable radius intersecting each other at A.

3. Join BA and CA. ∆ABC is required triangle.

4. From B draw any ray BX downwards making an acute angle CBX.

5. Marked seven points B₁,B₂,B₃,............B₇ on BX (BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇).

6. Join B₃C and from B₇ draw a line B₇C’|| B₃C intersecting the extended line segment BC at C’.

7. Draw C’ A’||CA intersecting the extended line segment BA at A’.

Then, ∆A’BC’ is the required triangle whose sides are 7/3 of the corresponding sides of ∆ABC.

Given that,

segment B₆C’ || B₃C. But since our construction is never possible that segment B₆C’ || B₃C because the similar triangle A’BC’ has its sides 7/3 of the corresponding sides of triangle ABC.

So, B₇C’ is parallel to B₃C.

Discussion

102.	Each interior angle of a polygon is 135. How many sides does it have? A. 10 B. 8 C. 6 D. 5
Answer» Interior Angle = 135 So, Exterior Angle = 180 – 135 = 45° Sum of exterior angles of polygon is 360° No. of Sides = \(\frac{360}{45}\) = 8

102.

Each interior angle of a polygon is 135. How many sides does it have? A. 10 B. 8 C. 6 D. 5

Answer»

Interior Angle = 135

So, Exterior Angle = 180 – 135

= 45°

Sum of exterior angles of polygon is 360°

No. of Sides = \(\frac{360}{45}\)

= 8

Discussion

103.	Fill in the blanks. For a convex polygon of n sides, we have: i. Sum of all exterior angles = ...... . ii. Sum of all interior angles = ...... . iii. Number of diagonals = ...... .
Answer» i. 4 right angles = 360° Convex Polygon is also a polygon and sum of all exterior angles of any polygon is 360° ii. (2n - 4) right angles Convex Polygon is also a polygon and sum of all interior angles of any polygon is (n - 2)× 180° Here, n represents the no of sides of polygon. iii. \(\frac{1}{2}n(n\,-\,3)\) No. of diagonals = \(\frac{n(n\,-\,3)}{2}\) [n is No. of Sides]

103.

Fill in the blanks. For a convex polygon of n sides, we have: i. Sum of all exterior angles = ...... . ii. Sum of all interior angles = ...... . iii. Number of diagonals = ...... .

Answer»

i. 4 right angles = 360°

Convex Polygon is also a polygon and sum of all exterior angles of any polygon is 360°

ii. (2n - 4) right angles

Convex Polygon is also a polygon and sum of all interior angles of any polygon is

(n - 2)× 180°

Here, n represents the no of sides of polygon.

iii. \(\frac{1}{2}n(n\,-\,3)\)

No. of diagonals = \(\frac{n(n\,-\,3)}{2}\) [n is No. of Sides]

Discussion

104.	Fill in the blanks. For a regular polygon of n sides, we have: i. Sum of all exterior angles = ...... . ii. Sum of all interior angles = ...... .
Answer» i. 360° Sum of all exterior angles of any polygon is 360° ii \(\{{180^\circ-(\frac{360}{n}^\circ)}\}\) Exterior Angle = \(\frac{360}{n}\) [n represents no of sides of polygon] So, Interior Angle = \((180-\frac{360}{n}){^\circ}\)

104.

Fill in the blanks. For a regular polygon of n sides, we have: i. Sum of all exterior angles = ...... . ii. Sum of all interior angles = ...... .

Answer»

i. 360°

Sum of all exterior angles of any polygon is 360°

ii \(\{{180^\circ-(\frac{360}{n}^\circ)}\}\)

Exterior Angle = \(\frac{360}{n}\) [n represents no of sides of polygon]

So, Interior Angle = \((180-\frac{360}{n}){^\circ}\)

Discussion

105.	Fill in the blanks.i. Each interior angles of a regular octagon is (......)°.ii. The sum of all interior angle of a regular hexagon is (......)°.iii. Each exterior angle of a regular polygon is 60°. This polygon is a ...... . iv. Each interior angle of a regular polygon is 108°. This polygon is a ...... . v. A pentagon has ...... diagonals.
Answer» i. 135° Exterior Angle = \(\frac{360}{8}\) [n represents no of sides of polygon] = 45° Interior Angle + Exterior Angle = 180o Interior Angle = 180 – 45 = 135° ii. 720° Sum of Interior Angle = (n - 2)× 180° = (6 - 2) × 180 ° = 720° iii. Hexagon Exterior Angle = \(\frac{360}{n}\) 60 = \(\frac{360}{n}\) N = \(\frac{360}{60}\) = 6 No. of Sides is 6. So, it is a hexagon. iv. Pentagon Interior Angle = 108° Exterior Angle = 180° - 108° = 72° No. of Sides = \(\frac{360}{72}\) = 5 So, it is a pentagon. v. 5 No. of diagonals = \(\frac{n(n\,-\,3)}{2}\) [n is No. of Sides] = \(\frac{5\times(5\,-\,3)}{2}\) = 5

105.

Fill in the blanks.i. Each interior angles of a regular octagon is (......)°.ii. The sum of all interior angle of a regular hexagon is (......)°.iii. Each exterior angle of a regular polygon is 60°. This polygon is a ...... . iv. Each interior angle of a regular polygon is 108°. This polygon is a ...... . v. A pentagon has ...... diagonals.

Answer»

i. 135°

Exterior Angle = \(\frac{360}{8}\) [n represents no of sides of polygon]

= 45°

Interior Angle + Exterior Angle = 180o

Interior Angle = 180 – 45 = 135°

ii. 720°

Sum of Interior Angle = (n - 2)× 180°

= (6 - 2) × 180 °

= 720°

iii. Hexagon

Exterior Angle = \(\frac{360}{n}\)

60 = \(\frac{360}{n}\)

N = \(\frac{360}{60}\)

= 6

No. of Sides is 6.

So, it is a hexagon.

iv. Pentagon

Interior Angle = 108°

Exterior Angle = 180° - 108° = 72°

No. of Sides = \(\frac{360}{72}\)

= 5

So, it is a pentagon.

v. 5

No. of diagonals = \(\frac{n(n\,-\,3)}{2}\) [n is No. of Sides]

= \(\frac{5\times(5\,-\,3)}{2}\)

= 5

Discussion

106.	Write ‘T’ for true and ‘F’ for false of each of the following: i. The diagonals of a parallelogram are equal. ii. The diagonals of a rectangle are perpendicular to each other. iii. The diagonals of a rhombus bisect each other at right angles. iv. Every rhombus is a kite.
Answer» i. F The diagonals of square and rectangle only are equal. Rest all the parallelograms like Rhombus etc. do not have diagonals equal in size. ii. F Diagonals of Rectangle do not intersect in right angle hence it is not perpendicular to each other. Only in case of Square, diagonal intersects at right angle. iii. T In rhombus, diagonals bisect the angles and are the perpendicular bisector of each other. iv. F In rhombus, every side has equal length but it in kite only pair of adjacent sides are equal in length.

106.

Write ‘T’ for true and ‘F’ for false of each of the following: i. The diagonals of a parallelogram are equal. ii. The diagonals of a rectangle are perpendicular to each other. iii. The diagonals of a rhombus bisect each other at right angles. iv. Every rhombus is a kite.

Answer»

i. F

The diagonals of square and rectangle only are equal. Rest all the parallelograms like Rhombus etc. do not have diagonals equal in size.

ii. F

Diagonals of Rectangle do not intersect in right angle hence it is not perpendicular to each other. Only in case of Square, diagonal intersects at right angle.

iii. T

In rhombus, diagonals bisect the angles and are the perpendicular bisector of each other.

iv. F

In rhombus, every side has equal length but it in kite only pair of adjacent sides are equal in length.

Discussion

107.	With the help of a ruler and a compass it is not possible to construct an angle ofA. `37.5^(@)`B. `40^(@)`C. `22.5^(@)`D. `67.5^(@)`
Answer» Correct Answer - B With the help of a ruler and a compass, we can construct the angels, `90^(@), 60^(@), 45^(@), 22.5^(@), 30^(@)` etc., and its bisector of an angle . So, it is not possible to construct an angle of `40^(@)`.

Discussion

108.	Try to draw triangles with the following data. Can you draw these triangles. If not, look for the reason why you could not draw so.∆ABC in which m∠A = 85°, m∠B = 115°, l(AB) = 5cm.
Answer» m∠A + m∠B = 85° + 115° = 200°>180° But the sum of the measures of the angles of a triangle is 180° Hence, ∆ABC cannot be drawn.

108.

Try to draw triangles with the following data. Can you draw these triangles. If not, look for the reason why you could not draw so.∆ABC in which m∠A = 85°, m∠B = 115°, l(AB) = 5cm.

Answer»

m∠A + m∠B = 85° + 115°

= 200°>180°

But the sum of the measures of the angles of a triangle is 180°

Hence, ∆ABC cannot be drawn.

Discussion

109.	Draw ∆ABC such that l(AB) = 4 cm, and l(BC) = 3 cm.1. Can this triangle be drawn? 2. A number of triangles can be drawn to fulfill these conditions. Try it out. 3. Which further condition must be placed if we are to draw a unique triangle using the above information?
Answer» 1. ∆ABC triangle cannot be drawn as length of third side is not given. 2. For ∆ABC to draw l(AC) > l(AB) + l(BC) i.e., l(AC) > 4 + 3 i.e., l(AC) > 7 cm ∴ number of triangles can be drawn if l(AC) > 7 cm 3. l(AC) > l(AB) + l(BC) is the required condition to draw a unique triangle.

109.

Draw ∆ABC such that l(AB) = 4 cm, and l(BC) = 3 cm.1. Can this triangle be drawn? 2. A number of triangles can be drawn to fulfill these conditions. Try it out. 3. Which further condition must be placed if we are to draw a unique triangle using the above information?

Answer»

1. ∆ABC triangle cannot be drawn as length of third side is not given.

2. For ∆ABC to draw l(AC) > l(AB) + l(BC)

i.e., l(AC) > 4 + 3

i.e., l(AC) > 7 cm

∴ number of triangles can be drawn if l(AC) > 7 cm

3. l(AC) > l(AB) + l(BC) is the required condition to draw a unique triangle.

Discussion

110.	Write whether True or False and justify your answer.An angle of 42.5° can be constructed.
Answer» False We know that, 42.5° = ½ x 85° and an angle of 85° cannot be constructed with the help of ruler and compass.

110.

Write whether True or False and justify your answer.An angle of 42.5° can be constructed.

Answer»

False

We know that, 42.5° = ½ x 85° and an angle of 85° cannot be constructed with the help of ruler and compass.

Discussion

111.	An angle of `42.5^(@)` can be constructed.
Answer» we know that, `42.5^(@)=(1)/(2)xx85^(@)` and an angle of `85^(@)` cannot be constructed with the help of ruler and compass.

Discussion

112.	`DeltaABC` can be constructed in which `AB = 5cm, angleA = 45^(@) and BC + AC = 5cm`.
Answer» we know that, a triangle can be constructed, if sum of its two sides is greater than third side. Here, ` BC + AC = AB = 5cm` So, `DeltaABC` cannot be constructed.

Discussion

113.	Define trapezium.
Answer» Trapezium is a quadrilateral with a pair of parallel sides.

Discussion

114.	Define rhombus.
Answer» A rhombus is a parallelogram with sides of equal length.

Discussion

115.	Arrange the following steps of construction while dividing a line segment of length 8 cm internally in the ratio 3 : 4.Steps of ConstructionStep I : Draw a ray BY parallel to AX by making ∠ABY equal to ∠BAX.Step II : Join A3B4. Suppose it intersects AB at a point P. Then, P is the point dividing AB internally in the ratio 3 : 4.Step III : Draw the line segment AB of length 8 cm.Step IV : Mark of three point A1, A2, A3, on AX and 4 points B1, B2, B3, B4 on BY such that AA1 = A1A2 = A2A3 = BB1 = B1B2 = B2B3 = B3B4.Step V : Draw any ray AX making an acute angle ∠BAX with AB.(A) III, V, I , II, IV(B) III, IV, I , V, II(C) III, I, V, IV, II(D) III, V, I , IV, II
Answer» The correct option is: (D) III, V, I, IV, II Explanation: Correct sequence of steps is III, V, I, IV, II

115.

Arrange the following steps of construction while dividing a line segment of length 8 cm internally in the ratio 3 : 4.Steps of ConstructionStep I : Draw a ray BY parallel to AX by making ∠ABY equal to ∠BAX.Step II : Join A3B4. Suppose it intersects AB at a point P. Then, P is the point dividing AB internally in the ratio 3 : 4.Step III : Draw the line segment AB of length 8 cm.Step IV : Mark of three point A1, A2, A3, on AX and 4 points B1, B2, B3, B4 on BY such that AA1 = A1A2 = A2A3 = BB1 = B1B2 = B2B3 = B3B4.Step V : Draw any ray AX making an acute angle ∠BAX with AB.(A) III, V, I , II, IV(B) III, IV, I , V, II(C) III, I, V, IV, II(D) III, V, I , IV, II

Answer»

The correct option is: (D) III, V, I, IV, II

Explanation:

Correct sequence of steps is III, V, I, IV, II

Discussion

116.	Can you draw a parallelogram BATS where BA = 5 cm, AT = 6 cm and AS = 6.5 cm? Why?
Answer» Yes, the parallelogram BATS with the given data can be drawn as it can be divided into two triangles BAS and SAT.

Discussion

117.	Which of the following steps of construction is INCORRECT while drawing a tangent to a circle of radius 5 cm and making an angle of 30° with a line passing through the centre.Steps of ConstructionStep I : Draw a circle with centre O and radius 2.5 cm.Step II : Draw a radius OA of this circle and produce it to B.Step III : Construct an angle ∠AOP equal to the complement of 30° i.e. equal to 150°.Step IV : Draw perpendicular to OP at P which intersects OA produced at Q.Clearly, PQ is the desired tangent such that ∠OQP = 30°(A) Both I and III(B) Only III(C) Both III and IV(D) Only I
Answer» The correct option is: (A) Both I and III Explanation: Step I and III are incorrect. Draw a circle with centre O and radius 5 cm and construct ∠AOP equal to complement of 30° i.e., equal to 60° are correct steps.

117.

Which of the following steps of construction is INCORRECT while drawing a tangent to a circle of radius 5 cm and making an angle of 30° with a line passing through the centre.Steps of ConstructionStep I : Draw a circle with centre O and radius 2.5 cm.Step II : Draw a radius OA of this circle and produce it to B.Step III : Construct an angle ∠AOP equal to the complement of 30° i.e. equal to 150°.Step IV : Draw perpendicular to OP at P which intersects OA produced at Q.Clearly, PQ is the desired tangent such that ∠OQP = 30°(A) Both I and III(B) Only III(C) Both III and IV(D) Only I

Answer»

The correct option is: (A) Both I and III

Explanation:

Step I and III are incorrect. Draw a circle with centre O and radius 5 cm and construct ∠AOP equal to complement of 30° i.e., equal to 60° are correct steps.

Discussion

118.	Given below are the steps of construction of a pair of tangents to a circle of radius 6 cm which are inclined to each other at an angle of 60°. Find which of the following step is wrong?Steps of ConstructionI. With centre O and radius = 6 cm, draw a circle.II. Taking a point A on the circle and draw ∠AOB = 120°.III. Draw a perpendicular on OA at A. Draw another perpendicular on OB at B.IV. Let the two perpendiculars meet at C.Thus CA and CB are the two required tangents to the given circle which are inclined to each other at 120°.(A) Only Step I(C) Only Step III(B) Only Step II(D) Only Step IV
Answer» Only Step IV of construction of a pair of tangents to a circle of radius 6 cm which is inclined to each other at an angle of 60°.

Discussion

119.	We saw that 5 measurements of a quadrilateral can determine a quadrilateral uniquely. Do you think any five measurements of the quadrilateral can do this?
Answer» It is necessary to have at least the knowledge of five parts to be able to construct quadrilateral uniquely When four sides and one diagonal are given. When two diagonals and three sides are given. When two adjacent sides and three angles are given. When three sides and two included angles are given. When four sides and one angle are given.

119.

We saw that 5 measurements of a quadrilateral can determine a quadrilateral uniquely. Do you think any five measurements of the quadrilateral can do this?

Answer»

It is necessary to have at least the knowledge of five parts to be able to construct quadrilateral uniquely

When four sides and one diagonal are given.
When two diagonals and three sides are given.
When two adjacent sides and three angles are given.
When three sides and two included angles are given.
When four sides and one angle are given.

Discussion

120.	Arshad has five measurements of a quadrilateral ABCD. These are AB = 5 cm, ∠A = 50°, AC = 4 cm, BD = 5 cm and AD = 6 cm. Can he construct a unique quadrilateral? Give reasons for your answer.
Answer» The quadrilateral ABCD with the given data cannot be constructed because of the following reasons : If he takes diagonal BD first, then he is able to construct one ∆ABD but the second ABCD to complete the quadrilateral ABCD cannot be drawn as its only one side BD is known. If he takes the diagonal AC first, then the construction of ∆ACP and ∆ABC is not possible as the data are insufficient.

120.

Arshad has five measurements of a quadrilateral ABCD. These are AB = 5 cm, ∠A = 50°, AC = 4 cm, BD = 5 cm and AD = 6 cm. Can he construct a unique quadrilateral? Give reasons for your answer.

Answer»

The quadrilateral ABCD with the given data cannot be constructed because of the following reasons :

If he takes diagonal BD first, then he is able to construct one ∆ABD but the second ABCD to complete the quadrilateral ABCD cannot be drawn as its only one side BD is known.
If he takes the diagonal AC first, then the construction of ∆ACP and ∆ABC is not possible as the data are insufficient.

Discussion

121.	To draw a unique quadrilateral, the total measurements are required(a) two(b) three(c) four(d) five
Answer» To draw a unique quadrilateral, the total measurements are required five.

Discussion

122.

Write the number of faces, edges and vertices of each shape in the table.NameCylinderConePentagonalpyramidHexgonalpyramidHexgonalprismPentagonalprismShapeFacesVerticesEdges

Answer»

Name	Cylinder	Cone	Pentagonal pyramid	Hexagonal pyramid	Hexagonal prism	Pentago prism
Faces	3 (2 flat + 1 curved)	2 (1 flat + 1 curved)	6 (5 triangles + 1 pentagon)	7 (6 triangles + 1 hexagon)	8 (6 rectangles + 2 hexagons)	7 (5 rectangle + 2 pentagon
Vertices	0	1	6	7	12	10
Edges	2	1	10	12	18	15

Discussion

123.	Write any two conditions for constructing a unique quadrilateral.
Answer» A quadrilateral can be constructed uniquely if the length of its four sides and a diagonal is given, if its two adjacent sides and three angles are known.

123.

Write any two conditions for constructing a unique quadrilateral.

Answer»

A quadrilateral can be constructed uniquely

if the length of its four sides and a diagonal is given,
if its two adjacent sides and three angles are known.

Discussion

124.	Write any two necessary conditions for constructing a unique square.
Answer» A quadrilateral can be constructed if the length of its four sides and a diagonal is given. the two diagonals and three sides are known.

124.

Write any two necessary conditions for constructing a unique square.

Answer»

A quadrilateral can be constructed if

the length of its four sides and a diagonal is given.
the two diagonals and three sides are known.

Discussion

125.	Arrange the following steps of construction while constructing a pair of tangents to a circle of radius 3 cm from a point 10 cm away from the centre of the circle.Steps of ConstructionStep I : Bisect the line segment OP and let the point of bisection be M.Step II : Taking M as centre and OM as radius, draw a circle. Let it intersect the given circle at the point Q and R.Step III : Draw a circle of radius 3 cm.Step IV : Join PQ and PR.Step V : Take an external point P which is 10 cm away from its centre. Join OP.(A) III, V, I , II, IV(B) III, I, V, I, II(C) III, V, I, IV, II(D) III, V, II, I, IV
Answer» The correct option is: (A) III, V, I , II, IV Explanation: Correct sequence of steps is III, V, I, II, IV

125.

Arrange the following steps of construction while constructing a pair of tangents to a circle of radius 3 cm from a point 10 cm away from the centre of the circle.Steps of ConstructionStep I : Bisect the line segment OP and let the point of bisection be M.Step II : Taking M as centre and OM as radius, draw a circle. Let it intersect the given circle at the point Q and R.Step III : Draw a circle of radius 3 cm.Step IV : Join PQ and PR.Step V : Take an external point P which is 10 cm away from its centre. Join OP.(A) III, V, I , II, IV(B) III, I, V, I, II(C) III, V, I, IV, II(D) III, V, II, I, IV

Answer»

The correct option is: (A) III, V, I , II, IV

Explanation:

Correct sequence of steps is III, V, I, II, IV

Discussion

126.	Arrange the following steps of construction while constructing a pair of tangents to circle, which are inclined to each other at an angle of 60° to a circle of radius 3 cm.Steps of ConstructionStep I : Draw any diameter AOB of this circle.Step II : Draw AM ⊥ AB and CN ⊥ OC.Let AM and CN intersect each other at P. Then PA and PC are the desired tangents to the given circle, inclined at an angle of 60°.Step III : Draw a circle with O as centre and radius 3 cm.Step IV : Construct ∠BOC = 60° such that radius OC meets the circle at C.(A) III, I , IV, II(B) III, II, IV, I(C) II, I, IV, III(D) IV, II, III, I
Answer» The correct option is: (A) III, I, IV, II Explanation: Correct sequence of steps is III, I, IV, II

126.

Arrange the following steps of construction while constructing a pair of tangents to circle, which are inclined to each other at an angle of 60° to a circle of radius 3 cm.Steps of ConstructionStep I : Draw any diameter AOB of this circle.Step II : Draw AM ⊥ AB and CN ⊥ OC.Let AM and CN intersect each other at P. Then PA and PC are the desired tangents to the given circle, inclined at an angle of 60°.Step III : Draw a circle with O as centre and radius 3 cm.Step IV : Construct ∠BOC = 60° such that radius OC meets the circle at C.(A) III, I , IV, II(B) III, II, IV, I(C) II, I, IV, III(D) IV, II, III, I

Answer»

The correct option is: (A) III, I, IV, II

Explanation:

Correct sequence of steps is III, I, IV, II

Discussion

127.	When constructing a building, what is the method used to make sure that a wall is exactly upright? What does the mason in the picture have in his hand? What do you think is his purpose for using it?
Answer» When constructing a building, a weight (usually with a pointed tip at the bottom) suspended from a string called as plummet or plump bob is aligned from the top of the wall to make sure that the wall is built exactly upright. The mason in the picture is holding a plumb bob. The string of the plumb bob is suspended from the top of the wall, such that plumb bob hangs freely. By observing whether the vertical wall is parallel to the string we can check if the constructed wall is vertical.

127.

When constructing a building, what is the method used to make sure that a wall is exactly upright? What does the mason in the picture have in his hand? What do you think is his purpose for using it?

Answer»

When constructing a building, a weight (usually with a pointed tip at the bottom) suspended from a string called as plummet or plump bob is aligned from the top of the wall to make sure that the wall is built exactly upright.

The mason in the picture is holding a plumb bob. The string of the plumb bob is suspended from the top of the wall, such that plumb bob hangs freely. By observing whether the vertical wall is parallel to the string we can check if the constructed wall is vertical.

Discussion

128.	Some angles are given below. Using the symbol of congruence write the names of the pairs of congruent angles in these figures.
Answer» i. ∠AOC ≅ ∠PQR ii. ∠DOC ≅ ∠LMN iii. ∠AOB ≅ ∠BOC ≅ ∠RST

128.

Some angles are given below. Using the symbol of congruence write the names of the pairs of congruent angles in these figures.

Answer»

i. ∠AOC ≅ ∠PQR

ii. ∠DOC ≅ ∠LMN

iii. ∠AOB ≅ ∠BOC ≅ ∠RST

Discussion

129.	Observe the given angles and write the names of those having equal measures.
Answer» i. ∠ABC and ∠SPM ii. ∠NIT and ∠SRI iii. ∠PTQ and ∠RTS

129.

Observe the given angles and write the names of those having equal measures.

Answer»

i. ∠ABC and ∠SPM

ii. ∠NIT and ∠SRI

iii. ∠PTQ and ∠RTS

Discussion

130.	The Perpendicular Bisector.1. A wooden ‘yoke’ is used for pulling a bullock cart. How is the position of the yoke determined? 2. To do that, a rope is used to measure equal distances from the spine/midline of the bullock cart. Which geometrical property is used here? 3. Find out from the craftsmen or from other experienced persons, why this is done.
Answer» 1. For the bullock cart to be pulled in the correct direction by the yoke, its Centre O should be equidistant from the either sides of the cart. 2. The property of perpendicular bisector is used to make the point equidistant from both the ends 3. A rope is used just like a compass to get equal distances from the spine/midline of bullock cart.

130.

The Perpendicular Bisector.1. A wooden ‘yoke’ is used for pulling a bullock cart. How is the position of the yoke determined? 2. To do that, a rope is used to measure equal distances from the spine/midline of the bullock cart. Which geometrical property is used here? 3. Find out from the craftsmen or from other experienced persons, why this is done.

Answer»

1. For the bullock cart to be pulled in the correct direction by the yoke, its Centre O should be equidistant from the either sides of the cart.

2. The property of perpendicular bisector is used to make the point equidistant from both the ends

3. A rope is used just like a compass to get equal distances from the spine/midline of bullock cart.

Discussion

131.	Two adjacent angles of a parallelogram are the ration 2:3. Find the measure of each of its angles.
Answer» \(\angle A=72^\circ\), \(\angle B=108^\circ\), \(\angle C=72^\circ\), \(\angle D=108^\circ\) Let x be the common multiple. As per question, \(\angle A\) = 2x \(\angle B\) = 3x \(\angle C\) = 2x \(\angle D\) = 3x \(\angle A\) + \(\angle B\) = 180° (Adjacent angles of parallelogram is supplementary) 2x + 3x = 180° 5x = 180° X = 180 / 5 = 36° \(\angle A\) = 2 × 36° = 72° \(\angle B\) = 3 × 36° = 108° \(\angle C\) = 2 × 36° = 72° \(\angle D\) = 3 × 36° = 108° So, Angles of quadrilateral are 72°, 108°, 72° and 108°.

131.

Two adjacent angles of a parallelogram are the ration 2:3. Find the measure of each of its angles.

Answer»

\(\angle A=72^\circ\), \(\angle B=108^\circ\), \(\angle C=72^\circ\), \(\angle D=108^\circ\)

Let x be the common multiple.

As per question,

\(\angle A\) = 2x

\(\angle B\) = 3x

\(\angle C\) = 2x

\(\angle D\) = 3x

\(\angle A\) + \(\angle B\) = 180° (Adjacent angles of parallelogram is supplementary)

2x + 3x = 180°

5x = 180°

X = 180 / 5 = 36°

\(\angle A\) = 2 × 36° = 72°

\(\angle B\) = 3 × 36° = 108°

\(\angle C\) = 2 × 36° = 72°

\(\angle D\) = 3 × 36° = 108°

So, Angles of quadrilateral are 72°, 108°, 72° and 108°.

Discussion

132.	Observe the image shown in the adjacent figure and answer the following questions.1. What time does this clock show? 2. What is the measure of the angle between its two hands? 3. At which other times is the angle between the hands congruent with this angle?
Answer» 1. 3 o’ clock. 2. 90°. 3. 9 o’ clock.

132.

Observe the image shown in the adjacent figure and answer the following questions.1. What time does this clock show? 2. What is the measure of the angle between its two hands? 3. At which other times is the angle between the hands congruent with this angle?

Answer»

1. 3 o’ clock.

2. 90°.

3. 9 o’ clock.

Discussion

133.	In the figure, ΔAXO ≅ A) ΔAYO B) ΔBYO C) ΔAXB D) ΔBXO
Answer» Correct option is: D) ΔBXO

Discussion

134.	In the figure, ΔAXB is A) equilateral B) isosceles C) scalene D) right triangle
Answer» Correct option is: B) isosceles

Discussion

135.	Angle in the semi-circle is A) right angle B) 180° C) acute angle D) obtuse angle
Answer» Correct option is (A) right angle Angle in the semi-circle is a right angle. A) right angle

135.

Angle in the semi-circle is A) right angle B) 180° C) acute angle D) obtuse angle

Answer»

Correct option is (A) right angle

Angle in the semi-circle is a right angle.

A) right angle

Discussion

136.	The maximum number of tangents that can be drawn to a circle from a point outside it is ______ (A) 2 (B) 1 (C) one and only one(D) 0
Answer» Correct answer is (A) 2

136.

The maximum number of tangents that can be drawn to a circle from a point outside it is ______ (A) 2 (B) 1 (C) one and only one(D) 0

Answer»

Correct answer is

(A) 2

Discussion

137.	The number of tangents that can be drawn to a circle at a point on the circle is ______(A) 3 (B) 2 (C) 1 (D) 0
Answer» Correct answer is (C) 1

137.

The number of tangents that can be drawn to a circle at a point on the circle is ______(A) 3 (B) 2 (C) 1 (D) 0

Answer»

Correct answer is

(C) 1

Discussion

138.	The process of drawing a geometrical figure by using a compass and straight edge is calledA) geometrical proof B) hypothesis C) geometrical construction D) none
Answer» Correct option is: C) geometrical construction

Discussion

139.	The circumcentre ‘O’ shown in the figure is formed by the concurrence ofA) Perpendicular bisectors B) Angular bisectors C) Medians D) Altitudes
Answer» Correct option is: A) Perpendicular bisectors

Discussion

140.	If in ΔABC \(\overrightarrow{BX}\) and \(\overrightarrow{CY}\) are bisectors to the base angles then ∠BXC = A) 90° + 1/2∠A B) 90° – 1/2∠A C) 180°- 1/2 ∠A D) None
Answer» Correct option is: A) 90° + 1/2∠A

Discussion

141.	Maithili, Shaila and Ajay live in three different places in the city. A toy shop is equidistant from the three houses. Which geometrical construction should be used to represent this? Explain your answer.
Answer» Since, Maithili, Shaila and Ajay live in three different places, lines joining their houses will form a triangle. The position of the toy shop which is equidistant from three houses can be found out by drawing the perpendicular bisector of the sides of the triangle joining the three houses. The shop will be at the point of concurrence of the perpendicular bisectors.

141.

Maithili, Shaila and Ajay live in three different places in the city. A toy shop is equidistant from the three houses. Which geometrical construction should be used to represent this? Explain your answer.

Answer»

Since, Maithili, Shaila and Ajay live in three different places, lines joining their houses will form a triangle. The position of the toy shop which is equidistant from three houses can be found out by drawing the perpendicular bisector of the sides of the triangle joining the three houses.

The shop will be at the point of concurrence of the perpendicular bisectors.

Discussion

142.	Let ABC be a right triangle in which AB = 3 cm, BC = 4 cm and ∠B= 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Given below are the steps of constructions of a pair of tangents from A to this circle. Which of the following steps is INCORRECT?Steps of ConstructionStep I : Draw ΔABC and perpendicular BD from B on AC.Step II : Draw a circle with BC as a diameter. This circle will pass through D.Step III : Let O be the mid-point of BC. Join AO.Step IV : Draw a circle with AO as diameter. This circle cuts the circle drawn in step II at B and P. AO, AP and AB are desired tangents drawn from A to the circle passing through B, C and D.(A) Only Step I(C) Only Step III(B) Only Step II(D) Only Step IV
Answer» The correct option is: (D) Only Step IV Explanation: Only AB and AP are tangents to the given circle. AO is secant to the circle. So, step IV is incorrect.

142.

Let ABC be a right triangle in which AB = 3 cm, BC = 4 cm and ∠B= 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Given below are the steps of constructions of a pair of tangents from A to this circle. Which of the following steps is INCORRECT?Steps of ConstructionStep I : Draw ΔABC and perpendicular BD from B on AC.Step II : Draw a circle with BC as a diameter. This circle will pass through D.Step III : Let O be the mid-point of BC. Join AO.Step IV : Draw a circle with AO as diameter. This circle cuts the circle drawn in step II at B and P. AO, AP and AB are desired tangents drawn from A to the circle passing through B, C and D.(A) Only Step I(C) Only Step III(B) Only Step II(D) Only Step IV

Answer»

The correct option is: (D) Only Step IV

Explanation:

Only AB and AP are tangents to the given circle.

AO is secant to the circle.

So, step IV is incorrect.

Discussion

143.	To construct a triangle similar to a given `DeltaABC` with its sides `(3)/(7)` of the corresponding sides of `DeltaABC`, first draw a ray BX such that `angleCBX` is an acute angle and X lies on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX isA. 5B. 8C. 13D. 3
Answer» Correct Answer - B (b) To construct a triangle similar to a given triangle, with its sides `(m)/(n)` of the corresponding sides of given triangle the minimum number of points to be located at equal distances is equal to the greater of m and n in `(m)/(n)` Here, `(m)/(n)=(8)/(5)` So, the minimum number of point to be located at equal distance on ray BX is 8.

Discussion

Explore topic-wise InterviewSolutions in .

Each interior angle of a polygon is 135. How many sides does it have? A. 10 B. 8 C. 6 D. 5

Fill in the blanks. For a convex polygon of n sides, we have: i. Sum of all exterior angles = ...... . ii. Sum of all interior angles = ...... . iii. Number of diagonals = ...... .

Fill in the blanks. For a regular polygon of n sides, we have: i. Sum of all exterior angles = ...... . ii. Sum of all interior angles = ...... .

Write ‘T’ for true and ‘F’ for false of each of the following: i. The diagonals of a parallelogram are equal. ii. The diagonals of a rectangle are perpendicular to each other. iii. The diagonals of a rhombus bisect each other at right angles. iv. Every rhombus is a kite.

With the help of a ruler and a compass it is not possible to construct an angle ofA. `37.5^(@)`B. `40^(@)`C. `22.5^(@)`D. `67.5^(@)`

Try to draw triangles with the following data. Can you draw these triangles. If not, look for the reason why you could not draw so.∆ABC in which m∠A = 85°, m∠B = 115°, l(AB) = 5cm.

Draw ∆ABC such that l(AB) = 4 cm, and l(BC) = 3 cm.1. Can this triangle be drawn? 2. A number of triangles can be drawn to fulfill these conditions. Try it out. 3. Which further condition must be placed if we are to draw a unique triangle using the above information?

Write whether True or False and justify your answer.An angle of 42.5° can be constructed.

An angle of `42.5^(@)` can be constructed.

`DeltaABC` can be constructed in which `AB = 5cm, angleA = 45^(@) and BC + AC = 5cm`.

Define trapezium.

Define rhombus.

Can you draw a parallelogram BATS where BA = 5 cm, AT = 6 cm and AS = 6.5 cm? Why?

We saw that 5 measurements of a quadrilateral can determine a quadrilateral uniquely. Do you think any five measurements of the quadrilateral can do this?

Arshad has five measurements of a quadrilateral ABCD. These are AB = 5 cm, ∠A = 50°, AC = 4 cm, BD = 5 cm and AD = 6 cm. Can he construct a unique quadrilateral? Give reasons for your answer.

To draw a unique quadrilateral, the total measurements are required(a) two(b) three(c) four(d) five

Write the number of faces, edges and vertices of each shape in the table.NameCylinderConePentagonalpyramidHexgonalpyramidHexgonalprismPentagonalprismShapeFacesVerticesEdges

Write any two conditions for constructing a unique quadrilateral.

Write any two necessary conditions for constructing a unique square.

When constructing a building, what is the method used to make sure that a wall is exactly upright? What does the mason in the picture have in his hand? What do you think is his purpose for using it?

Some angles are given below. Using the symbol of congruence write the names of the pairs of congruent angles in these figures.

Observe the given angles and write the names of those having equal measures.

Two adjacent angles of a parallelogram are the ration 2:3. Find the measure of each of its angles.

Observe the image shown in the adjacent figure and answer the following questions.1. What time does this clock show? 2. What is the measure of the angle between its two hands? 3. At which other times is the angle between the hands congruent with this angle?

In the figure, ΔAXO ≅ A) ΔAYO B) ΔBYO C) ΔAXB D) ΔBXO

In the figure, ΔAXB is A) equilateral B) isosceles C) scalene D) right triangle

Angle in the semi-circle is A) right angle B) 180° C) acute angle D) obtuse angle

The maximum number of tangents that can be drawn to a circle from a point outside it is ______ (A) 2 (B) 1 (C) one and only one(D) 0

The number of tangents that can be drawn to a circle at a point on the circle is ______(A) 3 (B) 2 (C) 1 (D) 0

The process of drawing a geometrical figure by using a compass and straight edge is calledA) geometrical proof B) hypothesis C) geometrical construction D) none

The circumcentre ‘O’ shown in the figure is formed by the concurrence ofA) Perpendicular bisectors B) Angular bisectors C) Medians D) Altitudes

If in ΔABC \(\overrightarrow{BX}\) and \(\overrightarrow{CY}\) are bisectors to the base angles then ∠BXC = A) 90° + 1/2∠A B) 90° – 1/2∠A C) 180°- 1/2 ∠A D) None

Maithili, Shaila and Ajay live in three different places in the city. A toy shop is equidistant from the three houses. Which geometrical construction should be used to represent this? Explain your answer.