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All questions of Construction for SSS 1 Exam
If TP and TQ are two tangents to a circle with centre O so that angle POQ = 110° then angle PTQ is equal to
- a)600
- b)800
- c)700
- d)900
Correct answer is option 'C'. Can you explain this answer?
If TP and TQ are two tangents to a circle with centre O so that angle POQ = 110° then angle PTQ is equal to
a)
600
b)
800
c)
700
d)
900
|
Krishna Iyer answered |
We know that the sum of angle made by two tangents joined by an exterior point and the angle made at the centre of the circle is equal to 180 degrees
So, Angle POQ + Angle PTQ = 180
110 + Angle PTQ = 180
Angle PTQ = 180- 110 = 70 degrees
So, Angle POQ + Angle PTQ = 180
110 + Angle PTQ = 180
Angle PTQ = 180- 110 = 70 degrees
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In which of the following is AD not the bisector of angle A?
- a)AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm
- b)AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm
- c)AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm
- d)AB = 8 cm, AC = 24 cm, BD = 6 cm and CD = 24 cm
Correct answer is option 'C'. Can you explain this answer?
In which of the following is AD not the bisector of angle A?
a)
AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm
b)
AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm
c)
AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm
d)
AB = 8 cm, AC = 24 cm, BD = 6 cm and CD = 24 cm
|
Amit Sharma answered |
The angle bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.So in this question,
So , Substituting the values of option C
LHS=5/1.5=1/0.5
RHS=10/3.5=2/0.7
So, RHS ≠ LHS
So , Substituting the values of option C
LHS=5/1.5=1/0.5
RHS=10/3.5=2/0.7
So, RHS ≠ LHS
Length of the tangent to a circle from a point 26 cm away from the centre is 24 cm. What is the radius of the circle?- a)11cm
- b)13 cm
- c)10 cm
- d)12cm
Correct answer is option 'C'. Can you explain this answer?
Length of the tangent to a circle from a point 26 cm away from the centre is 24 cm. What is the radius of the circle?
a)
11cm
b)
13 cm
c)
10 cm
d)
12cm
|
Kashish Juneja answered |
Length of tangent = 24 cmLength from the point to the centre of circle = 26 cm.
Join the point of contact and the centre of the circle.
We know that line drawn from centre to the tangent is always perpendicular.
So by using Pythagoras Theorem,
(26)^2 = (24)^2 + Square of other side.
(Other side)^2 = 676 - 576.
Other side = √100.
Other side = 10 cm
To draw tangents to each of the circle with radii 3 cm and 2 cm from the centre of the other circle, such that the distance between their centres A and B is 6 cm, a perpendicular bisector of AB is drawn intersecting AB at M. The next step is to draw- a)a circle with AB as diameter
- b)a circle with MB as diameter
- c)a circle with AM as diameter
- d)extend AB to P such that BP = MB and draw a circle with MP as diameter
Correct answer is option 'A'. Can you explain this answer?
To draw tangents to each of the circle with radii 3 cm and 2 cm from the centre of the other circle, such that the distance between their centres A and B is 6 cm, a perpendicular bisector of AB is drawn intersecting AB at M. The next step is to draw
a)
a circle with AB as diameter
b)
a circle with MB as diameter
c)
a circle with AM as diameter
d)
extend AB to P such that BP = MB and draw a circle with MP as diameter
|
Isha Dasgupta answered |
According to the question, the next step is to draw is a circle with AB as diameter.
A point O is at a distance of 10 cm from the centre of a circle of radius 6 cm. How many tangents can be drawn from point O to the circle?- a)1
- b)3
- c)Infinite
- d)2
Correct answer is option 'D'. Can you explain this answer?
A point O is at a distance of 10 cm from the centre of a circle of radius 6 cm. How many tangents can be drawn from point O to the circle?
a)
1
b)
3
c)
Infinite
d)
2
|
|
Anita Menon answered |
Two tangents can be drawn to a circle from a point outside the circle.
No tangent line can be drawn through a point within a circle, since any such line must be a secant line.
No tangent line can be drawn through a point within a circle, since any such line must be a secant line.
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 the end points to those two radii of the side, the angle between which is- a)90°
- b)30°
- c)120°
- d)45°
Correct answer is option 'C'. Can you explain this answer?
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 the end points to those two radii of the side, the angle between which is
a)
90°
b)
30°
c)
120°
d)
45°
|
Vikram Kapoor answered |
∠OQP=90o=∠ORP since the angle, between a tangent to a circle and the radius of the same circle passing through the point of contact, is 90o.
∴ By angle sum property of quadrilaterals, we get ∠OQP+∠RPQ+∠ORP+∠ROQ=360o⟹90o+60o+90o+∠ROQ=360o⟹∠ROQ=120o.
The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of first triangle is 9 cm, the corresponding side of the other triangle is
- a)5.6 cm
- b)3 cm
- c)4.5 cm
- d)5.4 cm
Correct answer is option 'D'. Can you explain this answer?
The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of first triangle is 9 cm, the corresponding side of the other triangle is
a)
5.6 cm
b)
3 cm
c)
4.5 cm
d)
5.4 cm
|
|
Aditya Shah answered |
In similar triangles, the perimeter of the triangles will be in the ratio of their corresponding sides.
To divide a line segment AB internally in the ratio 4 : 7 , first a ray AX is drawn so that ∠BAX is an acute angle and then at equal distances points are marked on ray AX such that the minimum number of these points is- a)9
- b)10
- c)11
- d)12
Correct answer is option 'C'. Can you explain this answer?
To divide a line segment AB internally in the ratio 4 : 7 , first a ray AX is drawn so that ∠BAX is an acute angle and then at equal distances points are marked on ray AX such that the minimum number of these points is
a)
9
b)
10
c)
11
d)
12
|
Snehal Choudhury answered |
According to the question, the minimum number of those points which are to be marked should be (Numerator + Denominator) i.e., 4 + 7 = 11
To divide a line segment AB in the ratio 3 : 7 , draw a ray AX such that ∠BAX is an acute angle, then draw a ray BY parallel to AX and the points A1,A2,A3, … and B1,B2,B3,… are located at equal distances on ray AX and BY respectively. Then the points joined are :- a) A4 and B3
- b)A7 and B3
- c)A5 and B5
- d)A3 and B7
Correct answer is option 'D'. Can you explain this answer?
To divide a line segment AB in the ratio 3 : 7 , draw a ray AX such that ∠BAX is an acute angle, then draw a ray BY parallel to AX and the points A1,A2,A3, … and B1,B2,B3,… are located at equal distances on ray AX and BY respectively. Then the points joined are :
a)
A4 and B3
b)
A7 and B3
c)
A5 and B5
d)
A3 and B7
|
Krithika Iyer answered |
If to divide the line segment AB in the ratio m : n, then we draw a ray AX such that ∠BAX is an acute angle, then draw a ray BY parallel to AX and the points A1,A2,A3, …….. Am-1, Am and B1,B2,B3, ….., Bn are located at equal distances on ray AX and BY respectively. Then the points joined are Am and Bn. Therefore, according to the question, the points joined are A3andB7.
Which theorem criterion we are using in giving the just the justification of the division of a line segment by usual method ?- a)SSS criterion
- b)Area theorem
- c)BPT
- d)Pythagoras theorem
Correct answer is option 'C'. Can you explain this answer?
Which theorem criterion we are using in giving the just the justification of the division of a line segment by usual method ?
a)
SSS criterion
b)
Area theorem
c)
BPT
d)
Pythagoras theorem
|
Srishti Singh answered |
Basic Proportionality Theorem (BPT) criterion we are using in giving the just the justification of the division of a line segment by usual method.
Two circles touch each other externally at C and AB is a common tangent to the circles. Then, ∠ACB =- a)30°
- b)45°
- c)60°
- d)90°
Correct answer is option 'D'. Can you explain this answer?
Two circles touch each other externally at C and AB is a common tangent to the circles. Then, ∠ACB =
a)
30°
b)
45°
c)
60°
d)
90°
|
Rohit Chauhan answered |
Here OA ⊥ AB and O'B ⊥ AB, then ∠OAB = ∠O'BA = 90° And ∠AOO' = ∠BO'O = 90°
∴ ABO'O is a rectangle.
Also, OC = OA, then ∠OCA = ∠OAC= x [Let] ∴ 90° (x + x) = 180°
⇒ x = 45° Again, O'C = O'B, then ∠OC'B = ∠O'BC = y [Let] ∴ 90° (y + y) = 180°
⇒ y = 45°
Now, ∠OCA + ∠ACB + ∠BCO' = 180° [Straight angle]
⇒ 45° + ∠ACB + 45° = 180°
⇒ ∠ACB = 90°
∴ ABO'O is a rectangle.
Also, OC = OA, then ∠OCA = ∠OAC= x [Let] ∴ 90° (x + x) = 180°
⇒ x = 45° Again, O'C = O'B, then ∠OC'B = ∠O'BC = y [Let] ∴ 90° (y + y) = 180°
⇒ y = 45°
Now, ∠OCA + ∠ACB + ∠BCO' = 180° [Straight angle]
⇒ 45° + ∠ACB + 45° = 180°
⇒ ∠ACB = 90°
In the given figure, AB and DE are perpendiculars to BC. If AB = 9cm, DE = 3cm and AC = 24cm then AD is equal to
- a)20 cm
- b)16 cm
- c)18 cm
- d)13 cm
Correct answer is option 'B'. Can you explain this answer?
In the given figure, AB and DE are perpendiculars to BC. If AB = 9cm, DE = 3cm and AC = 24cm then AD is equal to
a)
20 cm
b)
16 cm
c)
18 cm
d)
13 cm
|
|
Amit Sharma answered |
Since AB||DE so Angle A= Angle D,Angle C is common
So by AA criterion, ΔABC - ΔDEC
So,the corresponding sides are in proportion
DC=8
AD=24-8=16cm
So by AA criterion, ΔABC - ΔDEC
So,the corresponding sides are in proportion
DC=8
AD=24-8=16cm
ABCD is a quadrilateral with BC = 8 cm. Quadrilateral AB’C’D’ is constructed with a scale factor 3/4 . What will be the length of B’C’?- a)8 cm
- b)12 cm
- c)2 cm
- d)6 cm
Correct answer is option 'D'. Can you explain this answer?
ABCD is a quadrilateral with BC = 8 cm. Quadrilateral AB’C’D’ is constructed with a scale factor 3/4 . What will be the length of B’C’?
a)
8 cm
b)
12 cm
c)
2 cm
d)
6 cm
|
|
Krishna Iyer answered |
The quadrilateral is dilated by a scale factor of 3/4.The shape of image will be same as pre image but size will vary means it will be smaller .If Dilation factor is , means 0< Dilation factor<1, the image is be smaller than preimage.
So new length =k*original size=¾*8 = 6cm
So B’C’ = 6cm
So new length =k*original size=¾*8 = 6cm
So B’C’ = 6cm
A divides the line segment PQ in the ratio : 
- a)1 : 2
- b)1 : 1
- c)1 : 4
- d)1 : 5
Correct answer is option 'C'. Can you explain this answer?
A divides the line segment PQ in the ratio :
a)
1 : 2
b)
1 : 1
c)
1 : 4
d)
1 : 5
|
Anagha Yadav answered |
In the given figure, A divides the line segment PQ in the ratio 1 : 4.
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 points are marked on the ray AX such that the minimum number of these points is :- a)8
- b)10
- c)11
- d)12
Correct answer is option 'D'. Can you explain this answer?
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 points are marked on the ray AX such that the minimum number of these points is :
a)
8
b)
10
c)
11
d)
12
|
|
Snehal Choudhury answered |
According to the question, the minimum number of those points which are to be marked should be (Numerator + Denominator) i.e., 5 + 7 = 12
A point O is at a distance of 10 cm from the centre of a circle of radius 6 cm. How many tangents can be drawn from point O to the circle?- a)2
- b)1
- c)Infinite
- d)0
Correct answer is option 'A'. Can you explain this answer?
A point O is at a distance of 10 cm from the centre of a circle of radius 6 cm. How many tangents can be drawn from point O to the circle?
a)
2
b)
1
c)
Infinite
d)
0
|
|
Ananya Das answered |
Only two tangents can be drawn from an external point.No tangent can be drawn which touches points inside the circle.
To draw a pair of tangents to a circle which are inclined to each other at an angle of 45° it is required to draw tangents at the end point of those two radii of the circle, the angle between which is :- a)105°
- b)135°
- c)145°
- d)70°
Correct answer is option 'B'. Can you explain this answer?
To draw a pair of tangents to a circle which are inclined to each other at an angle of 45° it is required to draw tangents at the end point of those two radii of the circle, the angle between which is :
a)
105°
b)
135°
c)
145°
d)
70°
|
Kajal Mehta Mehta answered |
135 degrees
To construct a triangle similar to given ΔABC with its sides 7/5 of the corresponding side of ΔABC,draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side of A with respect to BC. Then, locate points X1,X2,X3… at equal distances on BX. The points to be joined in the next step are- a)X5 and C
- b)X2 and C
- c)X7 and C
- d)X12 and C
Correct answer is option 'A'. Can you explain this answer?
To construct a triangle similar to given ΔABC with its sides 7/5 of the corresponding side of ΔABC,draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side of A with respect to BC. Then, locate points X1,X2,X3… at equal distances on BX. The points to be joined in the next step are
a)
X5 and C
b)
X2 and C
c)
X7 and C
d)
X12 and C
|
Deepika Shah answered |
According to the question, the points to be joined are X5 and C.
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, points are marked on the ray AX such that the minimum number of these points is :- a)8
- b)10
- c)11
- d)12
Correct answer is option 'D'. Can you explain this answer?
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, points are marked on the ray AX such that the minimum number of these points is :
a)
8
b)
10
c)
11
d)
12
|
Ananya Kumar answered |
According to the question, the minimum number of those points which are to be marked should be (Numerator + Denominator) i.e., 5 + 7 = 12
To divide a line segment PQ in the ratio 2 : 7, first a ray PZ is drawn so that ∠QPX is an acute angle and then at equal distances points are marked on the ray PX such that the minimum number of these points is :- a)2
- b)5
- c)7
- d)9
Correct answer is option 'D'. Can you explain this answer?
To divide a line segment PQ in the ratio 2 : 7, first a ray PZ is drawn so that ∠QPX is an acute angle and then at equal distances points are marked on the ray PX such that the minimum number of these points is :
a)
2
b)
5
c)
7
d)
9
|
Akshay Nair answered |
According to the question, the minimum number of those points which are to be marked should be (Numerator + Denominator) i.e., 2 + 7 = 9
To divide a line segment LM in the ratio 4 : 3, a ray LX is drawn firs such that ∠MLX is an acute angle and then points L1, L2, L3, … are located at equal distances on the ray LX and the points M is joined to :- a)L4
- b)L2
- c)L3
- d)L7
Correct answer is option 'D'. Can you explain this answer?
To divide a line segment LM in the ratio 4 : 3, a ray LX is drawn firs such that ∠MLX is an acute angle and then points L1, L2, L3, … are located at equal distances on the ray LX and the points M is joined to :
a)
L4
b)
L2
c)
L3
d)
L7
|
Pranav Menon answered |
To divide a line segment AB in the ratio m : n, a ray AX is drawn in such that ∠BAX is an acute angle and then points A1, A2, A3, …….., Am, …….. An are located at equal distances on the ray AX and then the point B is joined to AnB.
• Therefore, according to question, the point M is joined to L7M.
• Therefore, according to question, the point M is joined to L7M.
To divide a line segment LM in the ratio a : b, where a and b are positive integers, draw a ray LX so that ∠MLX is an acute angle and then mark points on the ray LX at equal distances such that the minimum number of these points is :- a)greater of a and b
- b)a + b
- c)ab
- d)a + b – 1
Correct answer is option 'B'. Can you explain this answer?
To divide a line segment LM in the ratio a : b, where a and b are positive integers, draw a ray LX so that ∠MLX is an acute angle and then mark points on the ray LX at equal distances such that the minimum number of these points is :
a)
greater of a and b
b)
a + b
c)
ab
d)
a + b – 1
|
Lekshmi Mishra answered |
According to the question, the minimum number of those points which are to be marked should be (Numerator + Denominator) i.e., a + b
To construct a triangle similar to given ΔPQR with its sides 5/8 of the corresponding sides of ΔPQR, first a ray PXis drawn such that ∠QPX is an acute angle and X lies on the opposite side of R with respect to PQ. Then locate points P1, P2, P3…. OnPX at equal distances and next step is to join :- a)P5 to Q
- b)P8 to Q
- c)P3 to Q
- d)P6 to Q
Correct answer is option 'B'. Can you explain this answer?
To construct a triangle similar to given ΔPQR with its sides 5/8 of the corresponding sides of ΔPQR, first a ray PXis drawn such that ∠QPX is an acute angle and X lies on the opposite side of R with respect to PQ. Then locate points P1, P2, P3…. OnPX at equal distances and next step is to join :
a)
P5 to Q
b)
P8 to Q
c)
P3 to Q
d)
P6 to Q
|
Sushant Sen answered |
To construct a triangle similar to given ΔABC with its sides mnmn of the corresponding sides of ΔABC, first a ray AX is drawn such that ∠BAX is an acute angle and X lies on the opposite side of C with respect to AB. Then locate points A1, A2, A3…. Am ….. An on AX at equal distances and next step is to join AnB. Therefore, according to question, the next step is to join P8 to Q.
If the construction of a triangle ABC in which AB = 6 cm, ∠A = 70° and ∠B = 40° is possible then find the measure of ∠C.- a)40°
- b)70°
- c)80°
- d)none
Correct answer is option 'B'. Can you explain this answer?
If the construction of a triangle ABC in which AB = 6 cm, ∠A = 70° and ∠B = 40° is possible then find the measure of ∠C.
a)
40°
b)
70°
c)
80°
d)
none
|
|
Sagnik Joshi answered |
To construct a triangle ABC with AB = 6 cm, BC = 8 cm, and AC = 10 cm, follow these steps:
1. Draw a line segment AB of length 6 cm.
2. Place the compass at point A and draw an arc with a radius of 8 cm to intersect the line segment AB. Label this point as C.
3. Place the compass at point B and draw an arc with a radius of 10 cm to intersect the line segment AB. Label this point as C.
4. Connect points A and C with a line segment to form side AC.
5. Connect points B and C with a line segment to form side BC.
6. Label the final triangle as ABC.
Note: It is important to ensure that the lengths of the constructed sides match the given measurements.
1. Draw a line segment AB of length 6 cm.
2. Place the compass at point A and draw an arc with a radius of 8 cm to intersect the line segment AB. Label this point as C.
3. Place the compass at point B and draw an arc with a radius of 10 cm to intersect the line segment AB. Label this point as C.
4. Connect points A and C with a line segment to form side AC.
5. Connect points B and C with a line segment to form side BC.
6. Label the final triangle as ABC.
Note: It is important to ensure that the lengths of the constructed sides match the given measurements.
To divide a line segment AB in the ratio 4 : 7, a ray AX is drawn first such that ∠BAX is an acute angle and then points A1,A2,A3…… are located at equal distances on the ray AX and the point B is joined to:- a) A10
- b)A9
- c)A11
- d)A12
Correct answer is option 'C'. Can you explain this answer?
To divide a line segment AB in the ratio 4 : 7, a ray AX is drawn first such that ∠BAX is an acute angle and then points A1,A2,A3…… are located at equal distances on the ray AX and the point B is joined to:
a)
A10
b)
A9
c)
A11
d)
A12
|
|
Khushi singh answered |
According to the question, point B is joined to A11.
If two tangents are drawn at the end points of two radii of a circle which are inclined at 120° to each other, then the pair of tangents will be inclined to each other at an angle of- a)60°
- b)90°
- c)100°
- d)120°
Correct answer is option 'A'. Can you explain this answer?
If two tangents are drawn at the end points of two radii of a circle which are inclined at 120° to each other, then the pair of tangents will be inclined to each other at an angle of
a)
60°
b)
90°
c)
100°
d)
120°
|
Navya Patel answered |
According to the question, the pair of tangents will be inclined to each other at an angle of 180°−120° = 60°
To divide a line segment AP in the ration 2 : 9, a ray AX is drawn first such that ∠BAX is an acute angle and then points A1,A2,A3... are located of equal distances on the ray AX and the points P is joined to- a)A3
- b)A8
- c)A11
- d)A12
Correct answer is option 'C'. Can you explain this answer?
To divide a line segment AP in the ration 2 : 9, a ray AX is drawn first such that ∠BAX is an acute angle and then points A1,A2,A3... are located of equal distances on the ray AX and the points P is joined to
a)
A3
b)
A8
c)
A11
d)
A12
|
Harshad Majumdar answered |
According to the question, the points P is joined to A11.
To divide a line segment AB in the ration 2 : 3, first a ray AX is drawn so that ∠BAX is an acute angle and then at equal distances, points are marked on the ray AX, such tha the minimum number of these points is- a)2
- b)3
- c)5
- d)6
Correct answer is option 'C'. Can you explain this answer?
To divide a line segment AB in the ration 2 : 3, first a ray AX is drawn so that ∠BAX is an acute angle and then at equal distances, points are marked on the ray AX, such tha the minimum number of these points is
a)
2
b)
3
c)
5
d)
6
|
Ashish Choudhary answered |
According to the question, the minimum number of those points which are to be marked should be (Numerator + Denominator) i.e., 2 + 3 = 5
To divide a line segment AB in the ratio p : q ( p, q are positive integers), draw a ray AX so that ∠BAX s an acute angle and then mark points on ray AX at equal distances such that the minimum number of these points is :- a)p + q
- b)pq
- c)p + q – 1
- d)greater of p and q
Correct answer is option 'A'. Can you explain this answer?
To divide a line segment AB in the ratio p : q ( p, q are positive integers), draw a ray AX so that ∠BAX s an acute angle and then mark points on ray AX at equal distances such that the minimum number of these points is :
a)
p + q
b)
pq
c)
p + q – 1
d)
greater of p and q
|
Aditi Choudhury answered |
According to the question, the minimum number of those points which are to be marked should be (Numerator + Denominator) i.e., p + q
To draw a pair of tangents to a circle which are inclined to each other at an angle of 35°, it is required to draw tangents at the end-points of those two radii of the circle, the angle between which is- a)90°
- b)135°
- c)130°
- d) 145°
Correct answer is option 'D'. Can you explain this answer?
To draw a pair of tangents to a circle which are inclined to each other at an angle of 35°, it is required to draw tangents at the end-points of those two radii of the circle, the angle between which is
a)
90°
b)
135°
c)
130°
d)
145°
|
|
Neha Tanu answered |
To divide line segment AB in the ration m : n (m, n are positive integers), draw a ray AX so that ∠BAX is an acute angle and then mark points on ray AX at equal distances such that the minimum number of these points is- a)greater of m and n
- b)mn
- c)m + n
- d)m + n - 1
Correct answer is option 'C'. Can you explain this answer?
To divide line segment AB in the ration m : n (m, n are positive integers), draw a ray AX so that ∠BAX is an acute angle and then mark points on ray AX at equal distances such that the minimum number of these points is
a)
greater of m and n
b)
mn
c)
m + n
d)
m + n - 1
|
|
Krithika Iyer answered |
To divide line segment AB in the ration m : n (m, n are positive integers), draw a ray AX so that ∠BAX is an acute angle and then mark points on ray AX at equal distances such that the minimum number of these points is m + n.
To divide a line segment AB internally in the 5 : 2, first a ray AX is drawn so that ∠BAX is an acute angle and then points A1,A2,A3.... are located at equal distance on ray AX and point B is joined to- a)A6
- b)A2
- c)A3
- d)A7
Correct answer is option 'D'. Can you explain this answer?
To divide a line segment AB internally in the 5 : 2, first a ray AX is drawn so that ∠BAX is an acute angle and then points A1,A2,A3.... are located at equal distance on ray AX and point B is joined to
a)
A6
b)
A2
c)
A3
d)
A7
|
Deepika Kumar answered |
To divide a line segment AB in the ratio m : n (when m > n), a ray AX is drawn in such that ∠BAX is an acute angle and then points A1,A2,A3...Am,…..An…. are located at equal distances on the ray AX and then the point B is joined to AnB.
Therefore, according to the question, point B is joined toA7 .
Therefore, according to the question, point B is joined toA7 .
Two distinct tangents can be constructed from a point P to a circle of radius 2r situated at a distance:- a)r from the centre
- b)2r from the centre
- c)more than 2r from the centre
- d)less than 2r from the centre
Correct answer is option 'C'. Can you explain this answer?
Two distinct tangents can be constructed from a point P to a circle of radius 2r situated at a distance:
a)
r from the centre
b)
2r from the centre
c)
more than 2r from the centre
d)
less than 2r from the centre
|
Megha Roy answered |
If we have to draw the tangents from any external point of the circle, then the distance of the external point from the centre should be more than the radius of the circle.
Therefore, two distinct tangents can be constructed from a point P to a circle of radius 2r situated at a distance more than 2r from the centre.
Therefore, two distinct tangents can be constructed from a point P to a circle of radius 2r situated at a distance more than 2r from the centre.
Sides of two similar triangles are in the ratio 3:8. Areas of these triangles are in the ratio
- a)16: 81
- b)9: 64
- c)64: 9
- d)27: 512
Correct answer is option 'B'. Can you explain this answer?
Sides of two similar triangles are in the ratio 3:8. Areas of these triangles are in the ratio
a)
16: 81
b)
9: 64
c)
64: 9
d)
27: 512
|
|
Rising Star answered |
Let
the first triangle side be 3 units while other be 8units..
we know area of triangle is = 1/2 *base * height.
IT IS STATED THAT TRIANGLES ARE SIMILAR SO AS PER BPT THEOREM
PQ/LN = QR/LM= PR/MN
Ratio of ∆ PQR and ∆ LMN = 1/2 *3*3/1/2*8*8..
.here 1/2 will be cut .
THUS,
∆ PQR/ ∆ LMN = 9/64 ...
THUS ,THE RATIO IS 9:64
the first triangle side be 3 units while other be 8units..
we know area of triangle is = 1/2 *base * height.
IT IS STATED THAT TRIANGLES ARE SIMILAR SO AS PER BPT THEOREM
PQ/LN = QR/LM= PR/MN
Ratio of ∆ PQR and ∆ LMN = 1/2 *3*3/1/2*8*8..
.here 1/2 will be cut .
THUS,
∆ PQR/ ∆ LMN = 9/64 ...
THUS ,THE RATIO IS 9:64
To divide a line segment AB in the ration 4 : 7, a ray AX is drawn first such that ∠BAX is an acute angle and then points A1,A2,A3,…. are located at equal distances on the ray AX and the point B is joined to- a)A11
- b)A10
- c)A12
- d)A9
Correct answer is option 'A'. Can you explain this answer?
To divide a line segment AB in the ration 4 : 7, a ray AX is drawn first such that ∠BAX is an acute angle and then points A1,A2,A3,…. are located at equal distances on the ray AX and the point B is joined to
a)
A11
b)
A10
c)
A12
d)
A9
|
Akash Mehta answered |
According to the question, the point B is joined to A11.
The ____________ is the term used to denote a method in which case is driven into the ground and the material inside the casing is washed out and brought to the surface for inspection.- a)Deep boring
- b)Percussion boring
- c)Rotary drilling
- d)Wash boring
Correct answer is option 'D'. Can you explain this answer?
The ____________ is the term used to denote a method in which case is driven into the ground and the material inside the casing is washed out and brought to the surface for inspection.
a)
Deep boring
b)
Percussion boring
c)
Rotary drilling
d)
Wash boring
|
Hiral Chavan answered |
The process of wash boring consists in driving an inner tube of diameter 25 mm to 50 mm, inside an outer tube of diameter 100 mm to 150 mm. Whereas, percussion boring and rotary drilling are the types of Deep boring.
To draw a pair of tangents to a circle which are inclined to each other at angle x°, it is required to draw tangents at the end points of those two radii of the circle, the angle between which is- a)180°−x°
- b) 90°+x°
- c) 90°−x°
- d)180°+x°
Correct answer is option 'A'. Can you explain this answer?
To draw a pair of tangents to a circle which are inclined to each other at angle x°, it is required to draw tangents at the end points of those two radii of the circle, the angle between which is
a)
180°−x°
b)
90°+x°
c)
90°−x°
d)
180°+x°
|
Anisha Mukherjee answered |
To draw a pair of tangents to a circle which are inclined to each other at angle x°, it is required to draw tangents at the end points of those two radii of the circle, the angle between them is180°−x°
In division of a line segment AB, any ray AX making angle with AB is- a)right angle
- b)obtuse angle
- c)any arbitrary angle
- d)acute angle
Correct answer is option 'D'. Can you explain this answer?
In division of a line segment AB, any ray AX making angle with AB is
a)
right angle
b)
obtuse angle
c)
any arbitrary angle
d)
acute angle
|
Abhiram Malik answered |
In division of a line segment AB, any ray AX making angle with AB is an acute angle always.
To draw tangents to a circle of radius ‘p’ from a point on the concentric circle of radius ‘q’, the first step is to find- a)midpoint of q
- b)midpoint of p
- c)midpoint of q – r
- d)midpoint of p + q
Correct answer is option 'A'. Can you explain this answer?
To draw tangents to a circle of radius ‘p’ from a point on the concentric circle of radius ‘q’, the first step is to find
a)
midpoint of q
b)
midpoint of p
c)
midpoint of q – r
d)
midpoint of p + q
|
Prisha Dey answered |
To draw tangents to a circle of radius ‘p’ from a point on the concentric circle of radius ‘q’, the first step is to find midpoint of q.
If triangles PQR and DEF are similar, then which of the similarity criterion is true
- a)SAS criterion
- b)AAA criterion
- c)SSS criterion
- d)AAS criterion
Correct answer is option 'C'. Can you explain this answer?
If triangles PQR and DEF are similar, then which of the similarity criterion is true
a)
SAS criterion
b)
AAA criterion
c)
SSS criterion
d)
AAS criterion
|
|
Nishtha answered |
C is the correct ans because the ratio of all the corresponding sides of the 2 similar ∆s is same . So SSS criteria is being used here .
In the given figure, AC: CB is
- a)3:2
- b)2:3
- c)4:3
- d)3:4
Correct answer is option 'C'. Can you explain this answer?
In the given figure, AC: CB is
a)
3:2
b)
2:3
c)
4:3
d)
3:4
|
|
Navya Shree N answered |
If u notice that the line from A have 4 cuts and the line from B have 3 cuts there fore A:B ---> 4:3
hope can u understand!!!
hope can u understand!!!
In the figure, DE is parallel to BC such that AE = one fourth of AC. If AB = 6cm, then AD is
- a)1.5 cm
- b)3.5 cm
- c)15 cm
- d)2.5 cm
Correct answer is option 'A'. Can you explain this answer?
In the figure, DE is parallel to BC such that AE = one fourth of AC. If AB = 6cm, then AD is
a)
1.5 cm
b)
3.5 cm
c)
15 cm
d)
2.5 cm
|
Teesha Add answered |
Let AC be x. then *(1÷4) + (3÷4)=x AC =1Now In ∆ABC . (by BPL)AD/AB = AE/ACAD=1/4×6AD=1.5 cm
To construct a triangle similar to given ΔABC with its sides 7/4 of the corresponding sides of ΔABC, draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX is :- a)2
- b)3
- c)4
- d)7
Correct answer is option 'D'. Can you explain this answer?
To construct a triangle similar to given ΔABC with its sides 7/4 of the corresponding sides of ΔABC, draw a ray BX such that ∠CBX is an acute angle and X is on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX is :
a)
2
b)
3
c)
4
d)
7
|
|
Deepika Shah answered |
When numerator is greater than denominator, then number of arcs should be drawnlarger of m and n. Therefore, according to question, the minimum number of points to be located at equal distances on ray BX is 7.
To construct a triangle similar to a given triangle ABC with its sides 2/3 of the corresponding sides side of A with respect to BC. Then locate points X1,X2,X3… at equal distance on BX. The points to be joined in the next step are- a)X4 and C
- b)X2 and C
- c)X3 and C
- d)X1 and C
Correct answer is option 'C'. Can you explain this answer?
To construct a triangle similar to a given triangle ABC with its sides 2/3 of the corresponding sides side of A with respect to BC. Then locate points X1,X2,X3… at equal distance on BX. The points to be joined in the next step are
a)
X4 and C
b)
X2 and C
c)
X3 and C
d)
X1 and C
|
Ashwini Chawla answered |
The points to be joined in the next step are X3 and C.
Chapter doubts & questions for Construction - Mathematics for SSS 1 2024 is part of SSS 1 exam preparation. The chapters have been prepared according to the SSS 1 exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for SSS 1 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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Mathematics for SSS 1
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