143 + Interview Questions in CONSTRUCTIONS IN GENERAL AWARENESS Page 1 InterviewSolution

1.	To divide a line segment AB in the ratio 4:7, a ray AX is drawn first such that `angleBAX` is an acute angle and then points `A_(1),A_(2),A_(3),…..` are located at equal distance on the ray AX and the point B is joined toA. `A_(12)`B. `A_(11)`C. `A_(10)`D. `A_(9)`
Answer» Correct Answer - B (b) Here, minimum 4+7=11 points are located at equal distances on the ray AX, and then B is joined to last point is `A_(11)`.

Discussion

2.	To divide a line segment AB in the ratio 5:7, first a ray AX is drawn, so that `/_BAX` is an acute angle and then at equal distances point are marked on the ray AX such that the minimum number of these points isA. 8B. 10C. 11D. 12
Answer» Correct Answer - D (d) We know that to divide a line segment AB in the ratio m:n, first draw a ray AX which makes an acute angle `angleBAX`, then marked m+n points at equal distance. Here, m=5,n=7 So, minimum number of these points =m+n=5+7=12

Discussion

3.	Identify acute angle from the following. A) 60° B) 180° C) 90° D) 210°
Answer» Correct option is (A) 60° Acute angle is an angle whose measures is less than \(90^\circ.\) In given angles, only \(60^\circ\) is less than \(90^\circ.\) Thus, \(60^\circ\) is an acute angle. Correct option is A) 60°

3.

Identify acute angle from the following. A) 60° B) 180° C) 90° D) 210°

Answer»

Correct option is (A) 60°

Acute angle is an angle whose measures is less than \(90^\circ.\)

In given angles, only \(60^\circ\) is less than \(90^\circ.\)

Thus, \(60^\circ\) is an acute angle.

Correct option is A) 60°

Discussion

4.	Angles like 0°, 30°, 45°, 60°, 90°, 120° and 180° are called ………………. angles. A) Dependant B) Standard C) Equal D) Perpendicular
Answer» Correct option is B) Standard Answer :- Angle between 0° to 89° it is called acute angle. Angle Fix 90° called right angle. Angle between 90° to 179° it is called obtuse angle. Angle Fix 180° called straight angle. so the answer of this question is ( B ) straight angle.

Discussion

5.	Write whether True or False and justify your answer.ΔABC can be constructed in which BC = 6 cm, ∠C = 30° and AC – AB =4 cm.
Answer» True We know that, a triangle can be constructed if sum of its two sides is greater than third side. i.e., in ΔABC, AB + BC > AC => BC > AC – AB => 6 > 4, which is true, so ΔABC with given conditions can be constructed.

5.

Write whether True or False and justify your answer.ΔABC can be constructed in which BC = 6 cm, ∠C = 30° and AC – AB =4 cm.

Answer»

True

We know that, a triangle can be constructed if sum of its two sides is greater than third side.

i.e., in ΔABC, AB + BC > AC

=> BC > AC – AB

=> 6 > 4, which is true, so ΔABC with given conditions can be constructed.

Discussion

6.	State wether following statements are true or false. Justify your answer if possible :(i) Circumcircle and incircle of an quadrilateral triangle can be drawn taking the same center.(ii) All three sides of a triangle touch its in circle.(iii) If the given triangle ¡s obtuse, its circumcenter lies on one of its sides.(iv) The circumcentre of triangle situated inside the triangle when It is an acute triangle.(v) To construct an incircle of a triangle, we have draw the perpendicular bisector of an two sides of the triangle to obtain the incircle.
Answer» (i) True, since the in centre circumcentre and orthocentre of an equilateral triangle are concurrent. (ii) True, since radius of the in circle of a triangle is equal to the perpendicular drawn from in centre to any of the side of triangle. (iii) False, only right angled triangle has its circumcentre only on its hypotenuse. (iv) True. (v) False, because the in centre of a triangle is the intersecting point of bisectors of any two angles of the triangle.

6.

State wether following statements are true or false. Justify your answer if possible :(i) Circumcircle and incircle of an quadrilateral triangle can be drawn taking the same center.(ii) All three sides of a triangle touch its in circle.(iii) If the given triangle ¡s obtuse, its circumcenter lies on one of its sides.(iv) The circumcentre of triangle situated inside the triangle when It is an acute triangle.(v) To construct an incircle of a triangle, we have draw the perpendicular bisector of an two sides of the triangle to obtain the incircle.

Answer»

(i) True, since the in centre circumcentre and orthocentre of an equilateral triangle are concurrent.

(ii) True, since radius of the in circle of a triangle is equal to the perpendicular drawn from in centre to any of the side of triangle.

(iii) False, only right angled triangle has its circumcentre only on its hypotenuse.

(iv) True.

(v) False, because the in centre of a triangle is the intersecting point of bisectors of any two angles of the triangle.

Discussion

7.	Write whether True or False and justify your answer.A ΔABC can be constructed in which ∠B = 60°, ∠C =45°, and AB + BC + CA = 12 cm.
Answer» True We know that, sum of angles of a triangle is 180°. ∠A + ∠B + ∠C = 180° Here, ∠B + ∠C = 60°+ 45° = 105°< 180°, Thus, ΔABC with given conditions can be constructed.

7.

Write whether True or False and justify your answer.A ΔABC can be constructed in which ∠B = 60°, ∠C =45°, and AB + BC + CA = 12 cm.

Answer»

True

We know that, sum of angles of a triangle is 180°.

∠A + ∠B + ∠C = 180°

Here, ∠B + ∠C = 60°+ 45° = 105°< 180°,

Thus, ΔABC with given conditions can be constructed.

Discussion

8.	Draw a line segment `AB` of length `7cm`. Using ruler and compasses, find a point `P` on `AB` such that `(AP)/(AB)=(3)/(5)`.
Answer» Divide the line segement `AB` in the ratio `3 : 2`.

Discussion

9.	Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it where sides are `2/3`of the corresponding sides of the first triangle.
Answer» `(AC)/(CB) = 3/4` here,`(AP)/(PQ) = (AC)/(CB)= 3/4` and, `(AD)/(DB) = (AE)/(EC)` `2/3 : 1/3` Ratio will be `2:1`

Discussion

10.	If the lengths Of four sides and a diagonal is given then the figure is (a) triangle (b) quadrilateral (c) pentagon (d) hexagon
Answer» If the lengths of four sides and a diagonal is given then the figure is quadrilateral.

Discussion

11.	If the lengths of three sides and two diagonals, then the figure is(a) pentagon(b) triangle(c) hexagon(d) quadrilateral
Answer» If the lengths of three sides and two diagonals, then the figure is quadrilateral.

Discussion

12.	Sum of the angles of a quadrilateral is(a) 240°(b) 360°(c) 180°(d) 280°
Answer» Sum of the angles of a quadrilateral is 360°.

Discussion

13.	Construct, if possible, a quadrilateral ABCD in which AB = 6 cm, BC = 7 cm, CD = 3 cm, AD = 5.5 cm and AC = 11 cm. Give reasons for not being able to construct, if you cannot. (Not possible, because in triangle ACD, AD + CD<AC).
Answer» The given details are AB = 6 cm, BC = 7 cm, CD = 3 cm, AD = 5.5 cm and AC = 11 cm. Such a Quadrilateral cannot be constructed because, in a triangle, the sum of the length of its two sides must be greater than that of the third side. In triangle ACD, AD + CD = 5.5 + 3 = 8.5 cm Given, AC = 11 cm So, AD + CD < AC which is not possible. ∴ The construction is not possible

13.

Construct, if possible, a quadrilateral ABCD in which AB = 6 cm, BC = 7 cm, CD = 3 cm, AD = 5.5 cm and AC = 11 cm. Give reasons for not being able to construct, if you cannot. (Not possible, because in triangle ACD, AD + CD<AC).

Answer»

The given details are AB = 6 cm, BC = 7 cm, CD = 3 cm, AD = 5.5 cm and AC = 11 cm.

Such a Quadrilateral cannot be constructed because, in a triangle, the sum of the length of its two sides must be greater than that of the third side.

In triangle ACD,

AD + CD = 5.5 + 3 = 8.5 cm

Given, AC = 11 cm

So, AD + CD < AC which is not possible.

∴ The construction is not possible

Discussion

14.	The length of a rectangle is 8 cm and each of its diagonals measures 10 cm. The breadth of the rectangle is A. 5 cm B. 6 cm C. 7 cm D. 9 cm
Answer» A rectangle can be divided into two triangles. Sides of each triangle will be 8cm and 10 cm. According to Pythagoras theorem, a² = b² + c² 10² = 8² + c² c = √ (10² – 8²) c = \(\sqrt{36}\) c = 6 cm So, breadth of rectangle is 6 cm.

14.

The length of a rectangle is 8 cm and each of its diagonals measures 10 cm. The breadth of the rectangle is A. 5 cm B. 6 cm C. 7 cm D. 9 cm

Answer»

A rectangle can be divided into two triangles.

Sides of each triangle will be 8cm and 10 cm.

According to Pythagoras theorem,

a² = b² + c²

10² = 8² + c²

c = √ (10² – 8²)

c = \(\sqrt{36}\)

c = 6 cm

So, breadth of rectangle is 6 cm.

Discussion

15.	To draw a pair of tangents to a circle which are inclined to each other at an angle of 60°, it is required to draw tangents at endpoints of those two radii of the circle, the angle between them should be A. 135° B. 90° C. 60° D. 120°
Answer» D. 120° The angle between them should be 120° because in that case the figure formed by the intersection point of pair of tangent, the two end points of those-two radii tangents are drawn) and the centre of the circle is a quadrilateral. From figure it is quadrilateral, ∠POQ + ∠PRQ = 180° [∴ sum of opposite angles are 180°] 60°+ θ = 180° θ = 120 Hence, the required angle between them is 120°.

15.

To draw a pair of tangents to a circle which are inclined to each other at an angle of 60°, it is required to draw tangents at endpoints of those two radii of the circle, the angle between them should be A. 135° B. 90° C. 60° D. 120°

Answer»

D. 120°

The angle between them should be 120° because in that case the figure formed by the intersection point of pair of tangent, the two end points of those-two radii tangents are drawn) and the centre of the circle is a quadrilateral.

From figure it is quadrilateral,

∠POQ + ∠PRQ = 180° [∴ sum of opposite angles are 180°]

60°+ θ = 180°

θ = 120

Hence, the required angle between them is 120°.

Discussion

16.	To divide the line segment AB in the ratio 2 : 3, a ray AX is drawn such that ∠BAX is acute, AX is then marked at equal intervals. Find minimum number of these marks.
Answer» Minimum number of marks = 2 + 3 = 5

Discussion

17.	What is the ratio of division of the line segment,AB by the point P from A ?
Answer» The ratio of division of the line segment AB by the point P from A is AP : AB = 3 : 5

Discussion

18.	In Rhombus PQRS ∠P + ∠Q + ∠R + ∠S = ……………… A) 180° B) 300° C) 360° D) 190°
Answer» Correct option is (C) 360° Sum of angles in a quadrilateral is \(360^\circ.\) \(\therefore\) \(\angle P+\angle Q+\angle R+\angle S\) = \(360^\circ\) \((\because\) Rhombus is a special type of quadrilateral) Correct option is C) 360°

18.

In Rhombus PQRS ∠P + ∠Q + ∠R + ∠S = ……………… A) 180° B) 300° C) 360° D) 190°

Answer»

Correct option is (C) 360°

Sum of angles in a quadrilateral is \(360^\circ.\)

\(\therefore\) \(\angle P+\angle Q+\angle R+\angle S\) = \(360^\circ\)

\((\because\) Rhombus is a special type of quadrilateral)

Correct option is C) 360°

Discussion

19.	From the figure, ΔABD ΔCBD byA) S.S.S B) A.S.A C) S A SD) R.H.S
Answer» Correct option is: A) S.S.S

Discussion

20.	From the figure, ΔABD≅ ΔCBD byA) S.S.SB) A.S.A C) S.A.S D) R.H.S
Answer» Correct option is (A) S.S.S. In triangles \(\triangle ABD\;and\;\triangle CBD,\) AB = BC (radii of equal arcs) AD = CD (radii of equal arcs) BD = BD (Common side) \(\therefore\) \(\triangle ABD\cong\triangle CBD\) (By SSS congruence criteria) Correct option is A) S.S.S

20.

From the figure, ΔABD≅ ΔCBD byA) S.S.SB) A.S.A C) S.A.S D) R.H.S

Answer»

Correct option is (A) S.S.S.

In triangles \(\triangle ABD\;and\;\triangle CBD,\)

AB = BC (radii of equal arcs)

AD = CD (radii of equal arcs)

BD = BD (Common side)

\(\therefore\) \(\triangle ABD\cong\triangle CBD\) (By SSS congruence criteria)

Correct option is A) S.S.S

Discussion

21.	In the figure, \(\overrightarrow{BF}\) is the bisector of ∠ABC thenA) ∠ABF = ∠CBF B) ∠ABF = ∠ABC C) ∠ABF + ∠CBF = 90° D) ∠ABF + ∠CBF = 180°
Answer» Correct option is (A) ∠ABF = ∠CBF \(\overrightarrow{BF}\) is bisector of \(\angle ABC,\) \(\therefore\) \(\angle ABF=\angle CBF=\frac{\angle ABC}2\) A) ∠ABF = ∠CBF

21.

In the figure, \(\overrightarrow{BF}\) is the bisector of ∠ABC thenA) ∠ABF = ∠CBF B) ∠ABF = ∠ABC C) ∠ABF + ∠CBF = 90° D) ∠ABF + ∠CBF = 180°

Answer»

Correct option is (A) ∠ABF = ∠CBF

\(\overrightarrow{BF}\) is bisector of \(\angle ABC,\)

\(\therefore\) \(\angle ABF=\angle CBF=\frac{\angle ABC}2\)

A) ∠ABF = ∠CBF

Discussion

22.	In figure, ΔADE is constructed similar to ΔABC, write down the scale factor.
Answer» Scale factor = 3/4.

Discussion

23.	When are the two triangles said to be similar ?
Answer» Two triangles are said to be similar when their corresponding sides are proportional or angles are equal.

Discussion

24.	Look at the picture and answer the following questions :1. Name some things from your daily life that look like triangles.2. What are the types of triangles can see in the picture?3. Do you think all triangles shown are similar with their properties? What are they?
Answer» 1. Samosa, chapathi, window elevations, house tops, bridge trusses, floor tessellations. 2. Right triangles, Equilateral triangles. 3. Yes. All right triangles are similar. All triangles are similar because they have (a) right angle (b) equal side and (c) equal hypotenuse.

24.

Look at the picture and answer the following questions :1. Name some things from your daily life that look like triangles.2. What are the types of triangles can see in the picture?3. Do you think all triangles shown are similar with their properties? What are they?

Answer»

1. Samosa, chapathi, window elevations, house tops, bridge trusses, floor tessellations.

2. Right triangles, Equilateral triangles.

3. Yes. All right triangles are similar. All triangles are similar because they have

(a) right angle

(b) equal side and

(c) equal hypotenuse.

Discussion

25.	{ By similar triangles or by basic proportionality theorem (as in the book)}?
Answer» I think you should follow the steps of similar triangles or by basic proportionality theorem for justification. This can help you understand the thing: https://www.sarthaks.com/31405/draw-triangle-abc-which-4cm-6cm-and-9cm-construct-triangle-similar-%CE%B4abc-with-scale-factor https://www.sarthaks.com/31398/construct-triangle-similar-given-triangle-with-sides-equal-corresponding-sides-triangle

25.

{ By similar triangles or by basic proportionality theorem (as in the book)}?

Answer»

I think you should follow the steps of similar triangles or by basic proportionality theorem for justification.

This can help you understand the thing:

https://www.sarthaks.com/31405/draw-triangle-abc-which-4cm-6cm-and-9cm-construct-triangle-similar-%CE%B4abc-with-scale-factor

https://www.sarthaks.com/31398/construct-triangle-similar-given-triangle-with-sides-equal-corresponding-sides-triangle

Discussion

26.	In figure find ∠X, if XY = YZ and value of ∠Y = 58° :(A) 71°(B) 60°(C) 61°(D) 90°
Answer» The correct option is (C) 61°.

Discussion

27.	The sum of all three internal angles of a triangle is :(A) 80°(B) 180°(C) 100°(D) 90°
Answer» The sum of all three internal angles of a triangle is 180°.

Discussion

28.	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» C) geometrical construction

Discussion

29.	No. of circles drawn from a given point is A) 1B) 2 C) 3 D) Infinite
Answer» Correct option is (D) Infinite We can draw infinitely many circles from a given point. Correct option is D) Infinite

29.

No. of circles drawn from a given point is A) 1B) 2 C) 3 D) Infinite

Answer»

Correct option is (D) Infinite

We can draw infinitely many circles from a given point.

Correct option is D) Infinite

Discussion

30.	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 Circumcentre is the point of concurrence of perpendicular bisectors of all three sides of given triangle. \(\therefore\) The circumcentre O shown in the figure is formed by the concurrence of perpendicular bisectors. A) Perpendicular bisectors

30.

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

Circumcentre is the point of concurrence of perpendicular bisectors of all three sides of given triangle.

\(\therefore\) The circumcentre O shown in the figure is formed by the concurrence of perpendicular bisectors.

A) Perpendicular bisectors

Discussion

31.	Using the ruler, measure the lengths of seg AB and seg PQ. Are they of same length? Trace the seg AB on a sheet of transparent paper. Now place this new segment on PQ verify that if point A is placed on point P, then B falls on Q.l(AB) = ___ l(PQ) = ___
Answer» l(AB) = 4 cm l(PQ) = 4 cm Since the length of two segments is the same, if placed on one another, they will coincide.

31.

Using the ruler, measure the lengths of seg AB and seg PQ. Are they of same length? Trace the seg AB on a sheet of transparent paper. Now place this new segment on PQ verify that if point A is placed on point P, then B falls on Q.l(AB) = _ l(PQ) = _

Answer»

l(AB) = 4 cm

l(PQ) = 4 cm

Since the length of two segments is the same, if placed on one another, they will coincide.

Discussion

32.	Can a triangle be drawn if the three angles are given, but not any side? How many such triangles can be drawn?
Answer» Yes a triangle can be drawn. Since the length of side is not given, any length of side can be selected and then triangle can be constructed. We will get different triangles for different length of sides.

Discussion

33.	If ∆ABC ~ ∆PQR and AB/PQ = 7/5, then ______(A) AABC is bigger. (B) APQR is bigger. (C) both triangles will be equal (D) can not be decided
Answer» Correct answer is (A) AABC is bigger.

33.

If ∆ABC ~ ∆PQR and AB/PQ = 7/5, then ______(A) AABC is bigger. (B) APQR is bigger. (C) both triangles will be equal (D) can not be decided

Answer»

Correct answer is

(A) AABC is bigger.

Discussion

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

Discussion

35.	In the above figure, ΔAXB is A) equilateral B) isosceles C) scalene D) right triangle
Answer» B) isosceles

Discussion

36.	If in the figure, \(\overline{XY}\)is the perpendicular bisector to AB then AX =A) AY B) BY C) BX D) XY
Answer» Correct option is: C) BX

Discussion

37.	∠DAC =..............A) ∠B + ∠A B) ∠A + ∠C C) ∠B + ∠C D) ∠A + ∠B + ∠C
Answer» Correct option is (C) ∠B + ∠C \(\angle DAC\) is exterior angle of \(\triangle ABC\) And angles \(\angle B\;\&\;\angle C\) are interior opposite angles to exterior angle \(\angle DAC\) in \(\triangle ABC.\) \(\therefore\) \(\angle DAC\) = \(\angle B+\angle C\) Correct option is C) ∠B + ∠C

37.

∠DAC =..............A) ∠B + ∠A B) ∠A + ∠C C) ∠B + ∠C D) ∠A + ∠B + ∠C

Answer»

Correct option is (C) ∠B + ∠C

\(\angle DAC\) is exterior angle of \(\triangle ABC\)

And angles \(\angle B\;\&\;\angle C\) are interior opposite angles to exterior angle \(\angle DAC\) in \(\triangle ABC.\)

\(\therefore\) \(\angle DAC\) = \(\angle B+\angle C\)

Correct option is C) ∠B + ∠C

Discussion

38.	Angles in same circle segment are A) Complementary B) Supplementary C) Equal D) Unequal
Answer» Correct option is (C) Equal Angles in same circle segment are equal. Correct option is C) Equal option c) equal

38.

Angles in same circle segment are A) Complementary B) Supplementary C) Equal D) Unequal

Answer»

Correct option is (C) Equal

Angles in same circle segment are equal.

Correct option is C) Equal

option c)

equal

Discussion

39.	Given reasons to show that the construction of △ ABC is not possible AB = 6cm, ∠A = 40° and (BC + AC) = 5.8cm.
Answer» It is given that AB = 6cm, ∠A = 40^o and (BC + AC) = 5.8cm The sum of any two sides of a triangle is greater than the third side It can be written as BC + AC < AB Therefore, construction of △ ABC for the measurements given is not possible.

39.

Given reasons to show that the construction of △ ABC is not possible AB = 6cm, ∠A = 40° and (BC + AC) = 5.8cm.

Answer»

It is given that AB = 6cm, ∠A = 40^o and (BC + AC) = 5.8cm

The sum of any two sides of a triangle is greater than the third side

It can be written as

BC + AC < AB

Therefore, construction of △ ABC for the measurements given is not possible.

Discussion

40.	In the figure, \(\overrightarrow{BF}\) is the bisector of ∠ABC thenA) ∠ABF = ∠CBF B) ∠ABF = ∠ABC C) ∠ABF + ∠CBF = 90° D) ∠ABF + ∠CBF = 180°
Answer» Correct option is: A) ∠ABF = ∠CBF

Discussion

41.	The angels of a quadrilateral are in the ration 1:2:3:4. Find the measure of each angle.
Answer» 36°,72°,108°,144° Let x be the common multiple. As per question, \(\angle A\) = x \(\angle B\) = 2x \(\angle C\) = 3x \(\angle D\) = 4x As we know that, Sum of all four angles of quadrilateral is 360°. \(\angle A\) + \(\angle B\) + \(\angle C\) + \(\angle D=360^\circ\) x + 2x + 3x + 4x = 360° 10x = 360° X = 360/10 = 36° \(\angle A\) = 1 X 36° = 36° \(\angle B\) = 2 X 36° = 72° \(\angle C\) = 3 X 36° = 108° \(\angle D\) = 4 X 36° = 144° So, Angles of quadrilateral are 36°, 72°, 108° and 144°.

41.

The angels of a quadrilateral are in the ration 1:2:3:4. Find the measure of each angle.

Answer»

36°,72°,108°,144°

Let x be the common multiple.

As per question,

\(\angle A\) = x

\(\angle B\) = 2x

\(\angle C\) = 3x

\(\angle D\) = 4x

As we know that, Sum of all four angles of quadrilateral is 360°.

\(\angle A\) + \(\angle B\) + \(\angle C\) + \(\angle D=360^\circ\)

x + 2x + 3x + 4x = 360°

10x = 360°

X = 360/10

= 36°

\(\angle A\) = 1 X 36° = 36°

\(\angle B\) = 2 X 36° = 72°

\(\angle C\) = 3 X 36° = 108°

\(\angle D\) = 4 X 36° = 144°

So, Angles of quadrilateral are 36°, 72°, 108° and 144°.

Discussion

42.	The sides of rectangle are in the ration 4:5 and its perimeter is 180 cm. Find its sides.
Answer» 40 cm, 50 cm Let x be the common multiple. As per question, Length = 4x Width = 5x As per formula, Perimeter = 2× (l + w) 180 = 2× (4x + 5x) 180 = 18x x = 10 So, Length = 40 cm Width = 50 cm

42.

The sides of rectangle are in the ration 4:5 and its perimeter is 180 cm. Find its sides.

Answer»

40 cm, 50 cm

Let x be the common multiple.

As per question,

Length = 4x

Width = 5x

As per formula,

Perimeter = 2× (l + w)

180 = 2× (4x + 5x)

180 = 18x

x = 10

So,

Length = 40 cm

Width = 50 cm

Discussion

43.	Have you looked at lamp posts on the roadside? How do they stand?
Answer» The lamp posts on the road side are standing erect or vertical.

Discussion

44.	Which of the following steps is INCORRECT to construct a tangent to the circle of radius 5 cm at the point P on it without using the centre of the circle.Steps of ConstructionStep I : Draw a circle of radius 5 cm.Step II : Mark a point P on it.Step III: Draw any chord PQ.Step IV : Take a point R in the minor arc QP.Step V : Join PR and RQ.Step VI : Make ∠QPT = ∠PRQ.Step VII : Produce TP to T'. Then, PT is the required tangent at P.(A) Step II(C) Step VI(B) Step IV(D) None of these
Answer» The correct option is: (B) Step IV Explanation: Step IV is incorrect as the point R will be taken in the major arc QP.

44.

Which of the following steps is INCORRECT to construct a tangent to the circle of radius 5 cm at the point P on it without using the centre of the circle.Steps of ConstructionStep I : Draw a circle of radius 5 cm.Step II : Mark a point P on it.Step III: Draw any chord PQ.Step IV : Take a point R in the minor arc QP.Step V : Join PR and RQ.Step VI : Make ∠QPT = ∠PRQ.Step VII : Produce TP to T'. Then, PT is the required tangent at P.(A) Step II(C) Step VI(B) Step IV(D) None of these

Answer»

The correct option is: (B) Step IV

Explanation:

Step IV is incorrect as the point R will be taken in the major arc QP.

Discussion

45.	Which of the following steps is INCORRECT to construct a circle of radius 2 cm with centre O and then drawing two tangents to the circle from P where P is a point outside the circle such that OP = 4.5 cm.Steps of constructionStep I : Draw a circle with O as centre and radius 2 cm.Step ll : Mark a point P outside the circle such that OP = 2.25 cm.Step III : Join OP = 4.5 cm and bisect it at M.Step IV : Draw a circle with M as centre and radius equal to MP to intersect the given circle at the points T and 7.Step V : Joint PT and PT'. Then, PT and PT' are the required tangents.(A) Step V(C) Step II(B) Step IV(D) None of these
Answer» The correct option is: (C) Step II Explanation: Step II is incorrect. Mark the point P outside the circle such that OP = 4.5 cm is correct.

45.

Which of the following steps is INCORRECT to construct a circle of radius 2 cm with centre O and then drawing two tangents to the circle from P where P is a point outside the circle such that OP = 4.5 cm.Steps of constructionStep I : Draw a circle with O as centre and radius 2 cm.Step ll : Mark a point P outside the circle such that OP = 2.25 cm.Step III : Join OP = 4.5 cm and bisect it at M.Step IV : Draw a circle with M as centre and radius equal to MP to intersect the given circle at the points T and 7.Step V : Joint PT and PT'. Then, PT and PT' are the required tangents.(A) Step V(C) Step II(B) Step IV(D) None of these

Answer»

The correct option is: (C) Step II

Explanation:

Step II is incorrect. Mark the point P outside the circle such that OP = 4.5 cm is correct.

Discussion

46.	Arrange the following steps of construction while constructing a triangle of scale AB = 2.3 cm, BC = 5 cm and AC = 2.9 cm such that each of its sides is 2/3 rd of the corresponding side of the ΔABC.Steps of ConstructionStep I : On BE, cut off 3 equal parts making B1, B2 and B3.Step ll : Now, draw C'A' parallel to CA. Then, ΔA'BC' is the required Δ whose sides are 2/3rd of the corresponding sides of the ΔABC.Step III : From point B draw an arc of f.3 cm and from point C draw an arc of 2.9 cm cutting each other at point A.Step IV : Take BC = 5 cm.Step V : Join B3C and from B, draw B2C' parallel to B3C, such that BC' is 2/3 of BC.Step VI: On B make an acute ∠CBE downwards.Step VII : Join AB and AC. Then ABC is the required triangle.(A) IV, III, VII, I , VI, V, II(B) IV, V, I, VI, III, VII, II(C) IV, III, VII, VI, I, V, II(D) IV, VII, III, VI, V, I, II
Answer» The correct option is: (C) IV, III, VII, VI, I, V, II Explanation: Correct sequence of steps is IV, III, VII, VI, I, V, II

46.

Arrange the following steps of construction while constructing a triangle of scale AB = 2.3 cm, BC = 5 cm and AC = 2.9 cm such that each of its sides is 2/3 rd of the corresponding side of the ΔABC.Steps of ConstructionStep I : On BE, cut off 3 equal parts making B1, B2 and B3.Step ll : Now, draw C'A' parallel to CA. Then, ΔA'BC' is the required Δ whose sides are 2/3rd of the corresponding sides of the ΔABC.Step III : From point B draw an arc of f.3 cm and from point C draw an arc of 2.9 cm cutting each other at point A.Step IV : Take BC = 5 cm.Step V : Join B3C and from B, draw B2C' parallel to B3C, such that BC' is 2/3 of BC.Step VI: On B make an acute ∠CBE downwards.Step VII : Join AB and AC. Then ABC is the required triangle.(A) IV, III, VII, I , VI, V, II(B) IV, V, I, VI, III, VII, II(C) IV, III, VII, VI, I, V, II(D) IV, VII, III, VI, V, I, II

Answer»

The correct option is: (C) IV, III, VII, VI, I, V, II

Explanation:

Correct sequence of steps is

IV, III, VII, VI, I, V, II

Discussion

47.	Arrange the steps of construction while constructing pair of tangents to a circle of radius 5 cm from a point 12 cm away from its centre.Steps of ConstructionStep I : Join OA and bisect it. Let P is the mid-point of OA.Step II : Join AB and AC. AB and AC are the required tangents. Length of tangents = 11 cm.Step III : With O as centre, draw a circle of radius 5 cm.Step IV : Taking P as centre and PO as radius, draw a circle intersecting the given circle at the points B and C.Step V : Take a point A at a distance of 12 cm from O.(A) III, V, I, IV, II(B) III, V, IV, I , II(C) II, V, IV, III, I(D) III, IV, II, I , III
Answer» The correct option is: (A) III, V, I, IV, II Explanation: Correct sequence of steps is III, V, I, IV, II

47.

Arrange the steps of construction while constructing pair of tangents to a circle of radius 5 cm from a point 12 cm away from its centre.Steps of ConstructionStep I : Join OA and bisect it. Let P is the mid-point of OA.Step II : Join AB and AC. AB and AC are the required tangents. Length of tangents = 11 cm.Step III : With O as centre, draw a circle of radius 5 cm.Step IV : Taking P as centre and PO as radius, draw a circle intersecting the given circle at the points B and C.Step V : Take a point A at a distance of 12 cm from O.(A) III, V, I, IV, II(B) III, V, IV, I , II(C) II, V, IV, III, I(D) III, IV, II, I , III

Answer»

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

Explanation:

Correct sequence of steps is

III, V, I, IV, II

Discussion

48.	Fill in the blanks:1. If the diagonals of a parallelogram are perpendicular to each other than it is a ……..2. If the diagonals of a quadrilateral bisect each other, then it is a ………3. Each angle of rectangle and square is a ……4. Opposite sides of a parallelogram are ……5. A parallelogram with sides of equal length is called ……….
Answer» 1. rhombus 2. parallelogram 3. 90° 4. equal 5. rhombus.

Discussion

49.	Define square.
Answer» A square is a rectangle with equal sides.

Discussion

50.	Write the properties of parallelogram.
Answer» Opposite sides are equal. Opposite angles are equal. Diagonals bisect one another.

Discussion

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