Questions on Polynomials class 9 SET -1

Extra Questions on Polynomials Class 9

Q1. Find the remainder when \({y^3} + {y^2} - 2y + 5\) is divided by y - 5.

Q2. Determine the remainder when p(x) = \({x^3} + 3{x^2} - 6x + 15\) is divided by x - 2.

Q3.  When \(f(x) = {x^4} - 2{x^3} + 3b{x^2} - ax\) is divided by x+1 and x - 1, we get  remainder as 19 and 5 respectively. Find the remainder if f(x) is divided by x - 3.

Q4.  What must be subtracted from \(4{x^4} - 2{x^3} - 6{x^2} + x - 5\) so that the result is exactly divisible by \(2{x^2} + 3x - 2\) ?

Q5.  If (x + 1) and (x - 1) both are factors of \(a{x^3} + {x^2} - 2x + b\), find a and b.

Q6.  Factorize each of the following expressions:

1. \(48{x^3} - 36{x^2}\)
2. \(5{x^2} - 15xy\)
3. \(15{x^3}{y^2}z - 25x{y^2}{z^2}\)

Q7. Factorize:

1. \(2{x^2}(x + y) - 3(x + y)\)
2. \(5xy(5x + y) - 5y(5x + y)\)
3. \(x({x^2} + {y^2} - {z^2}) + y({x^2} + {y^2} - {z^2}) + z({x^2} + {y^2} - {z^2})\)
4. \(ab({a^2} + {b^2} - {c^2}) + bc({a^2} + {b^2} - {c^2}) + ca({a^2} + {b^2} - {c^2})\)

Q8. Factorize each of the following expressions:

1. \(25{x^2}{y^2} - 20x{y^2}z + 4{y^2}{z^2}\)
2. \(4{x^2} - 4\sqrt 7 x + 7\)
3. \(\dfrac{{{a^2}}}{{{b^2}}} + 2 + \dfrac{{{b^2}}}{{{a^2}}}\)
4. \(4{a^2} + 12ab + 9{b^2} - 8a - 12b\)

Q9. Factorize each of the following:

1. \(25{x^2} - 36{y^2}\)
2. \(2ab - {a^2} - {b^2} + 1\)
3. \(36{a^2} - 12a + 1 - 25{b^2}\)
4. \({a^4} - 81{b^4}\)
5. \({a^{12}}{b^4} - {a^4}{b^{12}}\)
6. \(4{x^2} - 9{y^2} - 2x - 3y\)

Q10. Factorize by completing the square.

1. \({a^4} + {a^2} + 1\)
2. \({y^4} + 5{y^2} + 9\)
3. \({x^4} + 4\)
4. \({x^4} + 4{x^2} + 3\)

Q11. Factorize by completing the square.

1. \({a^3} - 27\)
2. \(1 - 27{x^3}\)
3. \(8{x^3} - {(2x - 3y)^3}\)
4. \({a^8} - {a^2}{b^6}\)
5. \({a^3} - 5\sqrt 5 {b^3}\)

Q12. Factorize the following:

1. \(16{p^3}{q^2} + 54{r^3}\)
2. \(\dfrac{{{a^3}}}{8} + 8{b^3}\)
3. \(2\sqrt 2 {a^3} + 3\sqrt 3 {b^3}\)
4. \(8{a^4}b + \dfrac{1}{{125}}a{b^4}\)
5. \({a^7} - 64a\)

Q13.Q13. Factorize:

1. \({x^3} + 9{x^2} + 27x + 27\)
2. \({x^3} - 9{x^2}y + 27x{y^2} - 27{y^3}\)

Q14. Using identities, find the value of

1. 101²
2. 98²
3. (0.98)²
4. 101 × 99
5. 190 × 190 - 10 × 10

Q15. Expand using suitable identity

1. (x + 5y + 6z)²
2. (2a - 3b + 4c)²
3. ( - a + 6b + 5c)²
4. (- p + 4q - 3r)²

Q16. Expand using suitable identity

1. (2x + 5y)³
2. (5p - 3q)³
3. (- a + 2b)³

Q17. Evaluate using identities

1. 102³
2. 99³

Q18. Simplify : 

1. (2a + b)³ + (2a - b)³
2. (4x + 5y)³ - (4x - 5y)³

Q19. Factorize:

1. 30x³y + 24x²y² - 6xy
2. 5x(a - b) + 6y(a - b)

Q20. Factorize:

1. 9x² - y²
2. (3 - x)² - 36x²
3. (2x - 3y)² - (3x + 4y)²
4. 16x⁴ - y⁴

.....

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Class 9 Chapter 9 (Areas of parallelogram and triangles) Notes

Notes of Chapter 9 Area of Parallelograms and Triangles

Topics in the Chapter

• Area of Plane Figures
• Fundamentals
• Theorems

Areas of Plane Figures

• Two congruent figures have equal areas.
• A diagonal of a parallelogram divides it into two triangles of equal area.
• Parallelograms on the same base (or equal base) and between the same parallels are equal in area.
• Triangles on the same base (or equal bases) and between the same parallels are equal in area.
• Area of a parallelogram = base × height.
• Area of a triangle = 1/2 ×base × height
• A median of a triangle divides it into two triangles of equal area.
• Diagonals of a parallelogram divides it into four triangles of equal area.

Fundamentals

• Two figures are said to be on the same base and between the same parallels, if they have a common base (side) and the vertices (or the vertex) opposite to the common base of each figure lie on a line parallel to the base. 

Examples:

 

 

• Area of triangle is half the product of its base (or any side) and the corresponding altitude (or height).
• Two triangles with same base (or equal bases) and equal areas will have equal corresponding altitudes.
• A median of a triangle divides it into two triangles of equal areas. 

Theorems:

(1) Prove that a diagonal of a parallelogram divides it into two triangles of equal area.

Given:  A parallelogram ABCD in which BD is one of the diagonals.

To prove:   

Proof: Since two congruent geometrical figures have equal area. Therefore, in order to prove that it is sufficient to show that

ΔABD ≅ ΔCDB

In Δs ABD and CDB, we have

AB = CD

AD = CB

And, BD = DB

So, by SSS criterion of congruence, we have

ΔABD ≅ ΔCDB

Hence, ar(ΔABD) = ar(ΔCDB)

(2) Prove that parallelograms on the same base and between the same parallels are equal in area.

Given: Two parallelograms ABCD and ABEF, which have the same base AB and which are between the same parallel lines AB and FC.

To prove: ar(||gm ABCD) = ar(||gm ABCD)

Proof: In Δs ADF and BCE, we have

AD = BC

AF = BE

And, ∠DAF = ∠CBE         [ ⸪ AD ∥ BC and AF ∥ BE]

So, by SAS criterion of congruence, we have

ΔADF ≅ ΔBCE

ar(ΔADF) = ar(ΔBCE)   …..(i)

Now, ar(||gm ABCD) = ar(square ABED) + ar(ΔBCE)

ar(||gm ABCD) = ar(square ABED) + ar(ΔADF)    [Using(i)]

ar(||gm ABCD) = ar(||gm ABEF)

Hence, ar(||gm ABCD) = ar(||gm ABEF)

(3) Prove that the area of a parallelogram is the product of its base and the corresponding altitude.

Given: A parallelogram ABCD in which AB is the base and AL the corresponding altitude.

To prove: ar(parallelogram ABCD) = A B × AL

Construction: Complete the rectangle ALMB by drawing BM⊥CD.

Proof: Since ar(parallelogram ABCD) and rectangle ALMB are on the same base and between the same parallels.

ar(parallelogram ABCD)

= ar(rect.ALMB)

=AB × AL    [By rect. Area axiom area of a rectangle = Base \(\times\) Height]

Hence, ar(parallelogram ABCD) = AB×AL

(4) Prove that parallelograms on equal bases and between the same parallels are equal in area.

Given: Two parallelograms ABCD and PQRS with equal bases AB and PQ and between the same parallels AQ and DR.

To prove: ar(parallelogram ABCD) = ar(parallelogram PQRS)

Construction: Draw AL ⊥ DR and PM ⊥ DR

Proof: Since AB ⊥ DR, AL ⊥ DR and PM ⊥ DR

AL = PM

Now, ar(parallelogram ABCD) = AB × AL

ar(parallelogram ABCD) = PQ × PM  [AB = PQ and AL = PM]

ar(parallelogram ABCD) = ar(parallelogram PQRS)

(5) Prove that triangles on the same bases and between the same parallels are equal in area.

Proof: We have,

BD∥CA

And, BC∥DA

sq.BCAD is a parallelogram.

Similarly, sq.BCQP is a parallelogram.

Now, parallelograms ECQP and BCAD are on the same base BC, and between the same parallels.

ar(parallelogram BCQP) = ar(parallelogram BCAD)    ….(i)

We know that the diagonals of a parallelogram divides it into two triangles of equal area.

ar(ΔPBC) = \(\dfrac{1}{2}\)ar(parallelogram BCQP)   …..(ii)

And, ar(ΔABC) = \(\dfrac{1}{2}\)ar(parallelogram BCAQ)  ....(iii)

Now, ar(parallelogram BCQP) = ar(parallelogram BCAD) [ From (i)]

\(\dfrac{1}{2}\)ar(parallelogram BCQP) = \(\dfrac{1}{2}\)ar(parallelogram BCAD)

ar(ΔABC) = ar(ΔPBC)    [From (ii) and (iii)]

Hence, ar(ΔABC) = ar(ΔPBC)

(6) Prove that the area of a triangle is half the product of any of its sides and the corresponding altitude.

Given: A ΔABC in which AL is the altitude to the side BC.

To prove:

Construction: Through C and A draw CD ∥ BA and AD ∥ BC respectively, intersecting each other at D.

Proof: We have,

BA∥CD and AD ∥ BC

So, BCDA is a parallelogram.

Since AC is a diagonal of parallelogram BCDA.

[BC is the base and AL is the corresponding altitude of parallelogram BCDA]

(7) Prove that if a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to the half of the parallelogram.

Given: A ΔABC and a parallelogram BCDE on the same base BC and between the same parallel BC and AD.

To prove: ar(ΔABC) = \(\dfrac{1}{2}\)ar(||gm BCDE)

Construction: Draw AL ⊥ BC and DM ⊥ BC, meeting BC produced in M.

Proof: Since A, E and D are collinear and BC ∥ AD

AL = DM    …..(i)

Now,

ar(ΔABC) = \(\dfrac{1}{2}\)(BC × AL)

ar(ΔABC) = \(\dfrac{1}{2}\)(BC × DM)   [AL = DM (from (i)]

ar(ΔABC) = \(\dfrac{1}{2}\)ar(parallelogram BCDE)

 

(8) Prove that the area of a trapezium is half the product of its height and the sum of parallel sides.

Given: A trapezium ABCD in which AB ∥ DC; AB = a, DC = b and AL = CM = h, where AL⊥DC and CM ⊥ AB.

To prove: ar(trap. ABCD) = \(\dfrac{1}{2}\)h × (a + b)

Construction: Join AC

Proof: We have,

ar(trap. ABCD) = ar(ΔABC) + ar(ΔACD)

ar(trap. ABCD) = \(\dfrac{1}{2}\)(AB × CM) + \(\dfrac{1}{2}\)(DC × AL)

ar(trap. ABCD) = \(\dfrac{1}{2}\)ah + \(\dfrac{1}{2}\)bh  [AB = a and DC = b]

ar(trap. ABCD) = \(\dfrac{1}{2}\)h×(a+b)

(9) Prove that triangles having equal areas and having one side of one of the triangles, equal to one side of the other, have their corresponding altitudes equal.

Given: Two triangles ABC and PQR such that:

ar(ΔABC) = ar(ΔPQR)

AB = PQ

CN and RT are the altitudes corresponding to AB and PQ respectively of the two triangles.

To prove: CN = RT

Proof: In ΔABC, CN is the altitude corresponding to side AB.

ar(ΔABC) = \(\dfrac{1}{2}\)(AB × CN)    ….(i)

Similarly, we have,

ar(ΔPQR) = \(\dfrac{1}{2}\)(PQ × RT)   …..(ii)

Now, ar(ΔABC) = ar(ΔPQR)

\(\dfrac{1}{2}\)(AB × CN) = \(\dfrac{1}{2}\)(PQ × RT)

(AB × CN) = (PQ × RT)

(PQ × CN) = (PQ × RT)    [ AB = PQ (Given)]

CN = RT

(10) Prove that if each diagonal of a quadrilateral separates it into two triangles of equal area, then the quadrilateral is a parallelogram.

Given: A quadrilateral ABCD such that its diagonals AC and BD are such that

ar(ΔABD) = ar(ΔCDB) and ar(ΔABC) = ar(ΔACD)

To prove: Quadrilateral ABCD is a parallelogram.

Proof: Since diagonal AC of the quadrilateral ABCD separates it into two triangles of equal area. Therefore,

ar(ΔABC) = ar(ΔACD)    …..(i)

But, ar(ΔABC) + ar(ΔACD) = ar(ABCD)

2ar(ΔABC) = ar(ABCD)    [Using (i)]

ar(ΔABC) = \(\dfrac{1}{2}\)ar(ABCD) ….(ii)

Since diagonal BD of the quadrilateral ABCD separates it into triangles of equal area.

ar(ΔABD) = ar(ΔBCD)   ….(iii)

But, ar(ΔABD) + ar(ΔBCD) = ar(ABCD)

2ar(ΔABD) = ar(ABCD)    [Using(iii)]

ar(ΔABD) = \(\dfrac{1}{2}\)ar(ABCD)   …..(iv)

From (ii) and (iv), we get

ar(ΔABC) = ar(ΔABD)

Since Δs ABC and ABD are on the same base AB. Therefore they must have equal corresponding altitudes.

i.e. Altitude from C of ΔABC = Altitude from D of ΔABD

DC∥AB

Similarly, AD∥BC

Hence, quadrilateral ABCD is a parallelogram.

(11) Prove that the area of a rhombus is half the product of the lengths of its diagonals.

Given: A rhombus ABCD whose diagonals AC and BD intersect at O.

To prove: ar(rhombus ABCD) = \(\dfrac{1}{2}\)(AC×BD)

Proof: Since the diagonals of a rhombus intersect at right angles. Therefore,

OB ⊥ AC and OD ⊥ AC

ar(rhombus) = ar(ΔABC) + ar(ΔADC)

ar(rhombus) = \(\dfrac{1}{2}\)(AC × BO) + \(\dfrac{1}{2}\)(AC × DO)

ar(rhombus) = \(\dfrac{1}{2}\)(AC × (BO + DO))

ar(rhombus) =\(\dfrac{1}{2}\)(AC×BD)

(12) Prove that diagonals of a parallelogram divide it into four triangles of equal area.

Given: A parallelogram ABCD. The diagonals AC and BD intersect at O.

To prove: ar(ΔOAB) = ar(ΔOBC) = ar(ΔOCD) = ar(ΔAOD)

Proof: Since the diagonals of a parallelogram bisect each other at the point of intersection.

OA = OC and OB = OD

Also, the median of a triangle divides it into two equal parts.

Now, in ΔABC, BO is the median.

ar(ΔOAB) = ar(ΔOBC)  ….(i)

In ΔBCD, CO is the median

ar(ΔOBC) = ar(ΔOCD)    …..(ii)

In ΔACD, DO is the median

ar(ΔOCD) = ar(ΔAOD)   ….(iii)

From (i), (ii) and (iii), we get

ar(ΔOAB) = ar(ΔOBC) = ar(ΔOCD) = ar(ΔAOD)

(13) Prove that if the diagonals AC and BD of a quadrilateral ABCD, intersect at O and separate the quadrilateral into four triangles of equal area, then the quadrilateral ABCD is parallelogram.

Given: A quadrilateral ABCD such that its diagonals AC and BD intersect at O and separate it into four parts such that

ar(ΔOAB) = ar(ΔOBC) = ar(ΔOCD) = ar(ΔAOD)

To prove: Quadrilateral ABCD is a parallelogram.

Proof: We have,

ar(ΔAOD) = ar(ΔBOC)

ar(ΔAOD) + ar(ΔAOB) = ar(ΔBOC) + ar(ΔAOB)

ar(ΔABD) = ar(ΔABC)

Thus, Δs ABD and ABC have the same base AB and have equal areas. So, their corresponding altitudes must be equal.

Altitude from ΔABD Altitude from C of ΔABC

DC ∥ AB

Similarly, we have, AD ∥ BC.

Hence, quadrilateral ABCD is a parallelogram.

(14) Prove that a median of a triangle divides it into two triangles of equal area.

Given: A ΔABC in which AD is the median.

To prove: ar(ΔABD) = ar(ΔADC)

Construction: Draw AL ⊥ BC.

Proof: Since AD is the median of ΔABC. Therefore, D is the mid point of BC.

BD = DC

BD × AL = DC × AL  [ Multiplying both sides by AL]

\(\dfrac{1}{2}\)(BD × AL) = \(\dfrac{1}{2}\)(DC × AL)

ar(ΔABD) = ar(ΔADC)

ALITER, Since Δs ABD and ADC have equal bases and the same altitude AL.

ar(ΔABD) = ar(ΔADC)

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