A quadratic polynomial, whose zeroes are -3 and 4, is (a) x²- x + 12 (b) x² + x + 12 (c) (? 2 ) 2 − ? 2 − 6 (d) 2x² + 2x – 24

  x² - x + 12=0

(x - 4) (x + 3)= 0

x = 4 , -3

So, answer (a)

Steps of construction:

1. Draw circle with centre O and radius OA = 5 cm, .
2. Mark another point B on the circle such that ∠AOB = 120°, supplementary to the angle between the tangents. Since the angle between the tangents to be constructed is 60°.

          ∴ ∠AOB = 180° – 60° = 120°.  

      3.Construct angles of 90° at A and B and extend the lines so as to intersect at point P.

      4. Thus, AP and BP are the required tangents to the circle.

Area of shaded region=π theta/360°(R²-r²)
=22×40°/7×360°(14²-7²)
=22/7×9(196-49)
=22×147/7×9
=22×7/3
=154/3cm²

The diameter of three concentric circles are in the ratio of 1 : 2 : 3
Let the diameter of the three circles be 1 cm, 2 cm and 3 cm respectively.
So, the radii of three concentric circle will be
1/2 cm , 2/2 cm = 1 cm and 3/2 cm
Ratio of their areas = πr₁² : πr₂² : πr₃²
= π*(1/2)² : π*1² : π*(3/2)²
= 1/4 : 1 : 9/4
So the ratio of their respective ratios are 1/4 : 1 : 9/4

Area of sectors = 

= 39.25 cm2

Since, 14 cm, 48 cm and 50 cm is a Pythagorean triplet, it is a right 

Area = 

Area of shaded portion = (336 - 39.25) cm2

= 296.75 cm2

Let the speed of train be "x km/hr" and speed of taxi be "y km/hr"

here , distance travelled in both cases is same ,

Total distance = 300 + 200 = 260 + 240 = 500 km

----------------------------------------------------------------------------------

Case 1 :-

Time taken by train to travel 300 km = distance/speed
= 300/x

Time taken by taxi to travel 200 km = 200/y

so,
300/x + 200/y = 5.5 hour

Let 1/x = a , 1/y = b

300 a + 200 b =5.5 -----> (1)

-------------------------------------------------------------------------

Case 2 :-

Time taken by train to travel 260 km = 260/x
Time taken by taxi to travel 240 km = 240/x

It is said , that , it took 6 min more ,


6min = 0.1 hour

time taken = 5.5 + 0.1 = 5.6 hours

so ,

260/x + 240/y = 5.6

260a + 240b = 5.6 ----> (2)
-------------------------------------------------------------

Solving equations 1 and 2,

260a + 240b = 5.6 ---------> multiply by 5


300 a + 200 b =5.5 ---------> multiply by 6

========================
1300 a + 1200b = 28
1800a + 1200b = 33
----------------------------------
500a = 5

a = 5/500
= 1/100



260a + 240b = 5.6

260/100 + 240b = 5.6

2.6 + 240b = 5.6

240b = 3

b = 3/240
= 1/80

-----------------------------------------------

we know that

1/a = x , 1/b = y

hence,


x = 100
y = 80


speed of train = 100 km/hr
speed of taxi = 80 km/hr
 

Given , a circle with centre O and radius 8 cm. An external point P from where  a tangent is drawn to meet  the circle at T. 
as OP = 10 cm ; radius OT = 8 cm  As  OT⊥PT = 8 cm 
In right ΔOTP, we  have 
OP2=OT2+PT2 
102=82+PT2
PT2=100−64=36 
So, PT = 6 
Therefore , length of tangent = 6 cm

We know that the line drawn from the centre of the circle to the tangent is perpendicular to the tangent.

∴OP ⊥ PQ

By applying Pythagoras theorem in ΔOPQ,

∴OP² + PQ² = OQ² 

5² + PQ² =12² 

PQ² =144 − 25

PQ =  underroot 119

Hence, the correct answer is  underroot 119 cm test.

∠BQA=∠BCA=60∘
∠CQA=∠CBA=60∘
∴QP is angle bisector of BC
∴BP:PC=4:3     (By angle bisector theorem)
Now △BPQ∼△APC
PQ/BQ ​= PC​/AC
⇒ PQ/4 ​=3/7BC /​BC​

⇒PQ= 12​/7

False, the given pair of linear equations
x-2y-8=0
5x-10y=c

But if c = 40 (real value), then the ratio c₁/c₂ becomes  1/5 and then the system of linear equations has an infinitely many solutions.
Hence, ate = 40, the system of linear equations does not have a unique solution.

Construction Procedure:

1. Draw a line segment BC with the measure of 8 cm.

2. Now draw the perpendicular bisector of the line segment BC and intersect at the point D

3. Take the point D as centre and draw an arc with the radius of 4 cm which intersect the perpendicular bisector at the point A

4. Now join the lines AB and AC and the triangle is the required triangle.

5. Draw a ray BX which makes an acute angle with the line BC on the side opposite to the vertex A.

6. Locate the 3 points B₁, B₂ and B₃ on the ray BX such that BB₁ = B₁B₂ = B₂B₃

7. Join the points B₂C and draw a line from B₃ which is parallel to the line B₂C where it intersects the extended line segment BC at point C’.

8. Now, draw a line from C’ the extended line segment AC at A’ which is parallel to the line AC and it intersects to make a triangle.

9. Therefore, ΔA’BC’ is the required triangle.

Justification:

The construction of the given problem can be justified by proving that

A’B = (3/2)AB

BC’ = (3/2)BC

A’C’= (3/2)AC

From the construction, we get A’C’ || AC

∴ ∠ A’C’B = ∠ACB (Corresponding angles)

In ΔA’BC’ and ΔABC,

∠B = ∠B (common)

∠A’BC’ = ∠ACB

∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Therefore, A’B/AB = BC’/BC= A’C’/AC

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

A’B/AB = BC’/BC= A’C’/AC = 3/2

Hence, justified.

5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.

Construction Procedure:

1. Draw a ΔABC with base side BC = 6 cm, and AB = 5 cm and ∠ABC = 60°.

2. Draw a ray BX which makes an acute angle with BC on the opposite side of vertex A.

3. Locate 4 points (as 4 is greater in 3 and 4), such as B1, B2, B3, B4, on line segment BX.

4. Join the points B4C and also draw a line through B3, parallel to B4C intersecting the line segment BC at C’.

5. Draw a line through C’ parallel to the line AC which intersects the line AB at A’.

6. Therefore, ΔA’BC’ is the required triangle.

Justification:

The construction of the given problem can be justified by proving that

Since the scale factor is 3/4 , we need to prove

A’B = (3/4)AB

BC’ = (3/4)BC

A’C’= (3/4)AC

From the construction, we get A’C’ || AC

In ΔA’BC’ and ΔABC,

∴ ∠ A’C’B = ∠ACB (Corresponding angles)

∠B = ∠B (common)

∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

Therefore, A’B/AB = BC’/BC= A’C’/AC

So, it becomes A’B/AB = BC’/BC= A’C’/AC = 3/4

Hence, justified.

6. Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC.

To find ∠C:

Given:

∠B = 45°, ∠A = 105°

We know that,

Sum of all interior angles in a triangle is 180°.

∠A+∠B +∠C = 180°

105°+45°+∠C = 180°

∠C = 180° − 150°

∠C = 30°

So, from the property of triangle, we get ∠C = 30°

Construction Procedure:

The required triangle can be drawn as follows.

1. Draw a ΔABC with side measures of base BC = 7 cm, ∠B = 45°, and ∠C = 30°.

2. Draw a ray BX makes an acute angle with BC on the opposite side of vertex A.

3. Locate 4 points (as 4 is greater in 4 and 3), such as B1, B2, B3, B4, on the ray BX.

4. Join the points B3C.

5. Draw a line through B4 parallel to B3C which intersects the extended line BC at C’.

6. Through C’, draw a line parallel to the line AC that intersects the extended line segment at C’.

7. Therefore, ΔA’BC’ is the required triangle.

Justification:

The construction of the given problem can be justified by proving that

Since the scale factor is 4/3, we need to prove

A’B = (4/3)AB

BC’ = (4/3)BC

A’C’= (4/3)AC

From the construction, we get A’C’ || AC

In ΔA’BC’ and ΔABC,

∴ ∠A’C’B = ∠ACB (Corresponding angles)

∠B = ∠B (common)

∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

Therefore, A’B/AB = BC’/BC= A’C’/AC

So, it becomes A’B/AB = BC’/BC= A’C’/AC = 4/3

Hence, justified.

7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

Given:

The sides other than hypotenuse are of lengths 4cm and 3cm. It defines that the sides are perpendicular to each other

Construction Procedure:

The required triangle can be drawn as follows.

1. Draw a line segment BC =3 cm.

2. Now measure and draw ∠= 90°

3. Take B as centre and draw an arc with the radius of 4 cm and intersects the ray at the point B.

4. Now, join the lines AC and the triangle ABC is the required triangle.

5. Draw a ray BX makes an acute angle with BC on the opposite side of vertex A.

6. Locate 5 such as B1, B2, B3, B4, on the ray BX such that such that BB₁ = B₁B₂ = B₂B₃= B₃B₄ = B₄B₅

7. Join the points B3C.

8. Draw a line through B5 parallel to B3C which intersects the extended line BC at C’.

9. Through C’, draw a line parallel to the line AC that intersects the extended line AB at A’.

10. Therefore, ΔA’BC’ is the required triangle.

Justification:

The construction of the given problem can be justified by proving that

Since the scale factor is 5/3, we need to prove

A’B = (5/3)AB

BC’ = (5/3)BC

A’C’= (5/3)AC

From the construction, we get A’C’ || AC

In ΔA’BC’ and ΔABC,

∴ ∠ A’C’B = ∠ACB (Corresponding angles)

∠B = ∠B (common)

∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

Therefore, A’B/AB = BC’/BC= A’C’/AC

So, it becomes A’B/AB = BC’/BC= A’C’/AC = 5/3

Hence, justified.

Put x = 3 , y = 4

4x+3y-30

= 4(3) + 3(4) - 30

= 12 + 12 - 30

= 24 - 30

= - 6

So, x=3,y=4 is not a  solution of the linear equation 4x+3y-30=0