Class 9

A farmer has a triangular plot with sides 26 m, 28 m and 30 m . . .

Case Study based Question
A farmer has a triangular plot with sides 26 m, 28 m and 30 m. He wants to divide it into two equal area parts by constructing a small fence from a vertex perpendicular to the opposite side.

Based on the above information, answer the following questions :
(a) Find the semi-perimeter of the plot.
(b) Find the area of each part of the triangular plot after division.
(c) If fencing costs Rs 12 per mere, then find the cost of fencing the plot.

Doubt by Parth

Solution :

Here
a=26 m
b=28 m
c=30 m

(i) Semiperimeter (s)
s=(a+b+c)/2
s=(26+28+30)/2
s=84/2
s=42 m

(ii) s=42 m
Area of the Plot by using Heron's Formula
Area=√[s(s-a)(s-b)(s-c)
=√[42(42-26)(42-28)(42-30)
=√[42(16)(14)(12)
=√[14×3×4×4×14×3×2×2]
=√[14×14×4×4×3×3×2×2]
=14×4×3×2
=56×6
=336 m²

As per the question the fence is dividing the plot in two parts of equal area so area of each part of the triangular plot
= 336/2
=168 m²

(iii) The fencing is only along the length AD, so first we have to find the length of AD

We know,
Area of ΔABC = ½×Base×Height
336=½×BC×AD
336=½×30×AD
336=15×AD
336/15=AD
AD=336/15
AD=22.4 m

Cost of fencing 1 m = Rs 12
Cost of fencing 22.4 m = Rs 12×22.4
=Rs 268.80

Multiple Choice Questions of Triangle

1.) If ∆ ABC ≅ ∆ PQR and ∆ ABC is not congruent to ∆ RPQ, then which of the following is not true:

(A) BC = PQ
(B) AC = PR
(C) QR = BC
(D) AB = PQ

2.) Which of the following is not a criterion for congruence of triangles?

(A) SAS
(B) ASA
(C) SSA
(D) SSS

3.) If AB = QR, BC = PR and CA = PQ, then

(A) ∆ ABC ≅ ∆ PQR
(B) ∆ CBA ≅ ∆ PRQ
(C) ∆ BAC ≅ ∆ RPQ
(D) ∆ PQR ≅ ∆ BCA

4.) 3. In ∆ ABC, AB = AC and ∠B = 50°. Then ∠C is equal to

(A) 40°
(B) 50°
(C) 80°
(D) 130°

5.) In ∆ ABC, BC = AB and ∠B = 80°. Then ∠A is equal to

(A) 80°
(B) 40°
(C) 50°
(D) 100°

6.) In ∆ PQR, ∠R = ∠P and QR = 4 cm and PR = 5 cm. Then the length of PQ is

(A) 4 cm
(B) 5 cm
(C) 2 cm
(D) 2.5 cm

7.) D is a point on the side BC of a ∆ ABC such that AD bisects ∠BAC. Then

(A) BD = CD
(B) BA > BD
(C) BD > BA
(D) CD > CA

8.) It is given that ∆ ABC ≅ ∆ FDE and AB = 5 cm, ∠B = 40° and ∠A = 80°. Then which of the following is true?

(A) DF = 5 cm, ∠F = 60°
(B) DF = 5 cm, ∠E = 60°
(C) DE = 5 cm, ∠E = 60°
(D) DE = 5 cm, ∠D = 40°

9.) Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side

of the triangle cannot be

(A) 3.6 cm
(B) 4.1 cm
(C) 3.8 cm
(D) 3.4 cm

10.) In ∆ PQR, if ∠R > ∠Q, then

(A) QR > PR
(B) PQ > PR
(C) PQ < PR
(D) QR < PR

11.) In triangles ABC and PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are

(A) isosceles but not congruent
(B) isosceles and congruent
(C) congruent but not isosceles
(D) neither congruent nor isosceles

12.) . In triangles ABC and DEF, AB = FD and ∠A = ∠D. The two triangles will be congruent by SAS axiom if

(A) BC = EF
(B) AC = DE
(C) AC = EF
(D) BC = DE

Answer Keys

1.) (A) BC = PQ

2.) (C) SSA

3.) (B) ∆ CBA ≅ ∆ PRQ

4.) (B) 50°

5.) (C) 50°
6.) (A) 4 cm

7.) (A) BD = CD

8.) (B) DF = 5 cm, ∠E = 60°

9.) (D) 3.4 cm

10.) (B) PQ > PR

11.) (A) isosceles but not congruent

12.) (B) AC = DE

MCQs of Chapter - 2 Polynomials

1.) If x²+kx+6=(x+2)(x+3) for all x, then the value of k is

(A) 1

(B) –1

(D) 3

Ans : (C) 5

2.) Which one of the following is a polynomial?

Ans : (C)

3.) √2 is a polynomial of degree

(A) 2
(B) 0
(C) 1
(D) 1/2

Ans : (B) 0

4.) Degree of the polynomial 4x⁴+0x³+0x⁵+5x+7 is
(A) 4
(B) 5
(C) 3
(D) 7

Ans : (A) 4

5.) Degree of the zero polynomial is

(A) 0
(B) 1
(C) Any natural number

(D) Not defined

Ans : (D) Not defined

6.) If p(x)=x²-2√2x+1, then p(2√2) is equal to

(A) 0

(B) 1

(D) 8√2+1

Ans : (B) 1

7.) The value of the polynomial 5x–4x²+3, when x=–1 is

(A)– 6

(B) 6

(D) –2

Ans : (A) – 6

8.) If p(x)=x+3, then p(x)+ p(–x) is equal to

(A) 3

(B) 2x

(D) 6

Ans : (D) 6

9.) Zero of the zero polynomial is

(A) 0

(B) 1

(D) Not defined

Ans : (C) Any real number

10.) Zero of the polynomial p(x)=2x+5 is

(A) -2/5

(B) -5/2

(D) 5/2

Ans : (B) -5/2

11.) One of the zeroes of the polynomial 2x²+7x–4 is

(A) 2

(B) 1/2

(D) -2

Ans : (B) 1/2

12.) If x^51+51 is divided by x+1, the remainder is

(A) 0

(B) 1

(D) 50

Ans : (D) 50

13.) If x+1 is a factor of the polynomial 2x²+kx, then the value of k is

(A) –3

(B) 4

(D) –2

Ans : (C) 2

14.) x+1 is a factor of the polynomial

(A) x³+x²–x+1

(B) x³+x²+x+1

(D) x⁴+3x³+3x²+x+1

Ans : (B) x³+x²+x+1

15.) One of the factors of (25x²–1)+(1+5x)² is

(A) 5+x

(B) 5–x

(D) 10x

Ans : (D) 10x

16.) The value of 249²–248² is

(A) 12

(B) 477

(D) 497

Ans : (D) 497

17.) The factorisation of 4x²+8x+3 is

(A) (x+1)(x+3)

(B) (2x+1)(2x+3)

(D) (2x–1)(2x–3)

Ans : (B) (2x+1)(2x+3)

18.) Which of the following is a factor of (x+y)³-(x³+y³)?

(A) x²+y²+2xy

(B) x²+y²-xy

(D) 3xy

Ans : (D) 3xy

19.) The coefficient of x in the expansion of (x+3)³ is

(A) 1

(B) 9

(D) 27

Ans : (D) 27

20. )

(A) 1

(B) -1

(D) 1/2

Ans : (C) 0

21.)

(A) 0

(B) 1/√2

(D) 1/2

Ans : (C) 1/4

22.) If a+b+c=0, then a³+b³+c³ is equal to

(A) 0

(B) abc

(D) 2abc

Ans : (C) 3abc

If x=root(a+2b) + root(a-2b) / root(a+2b) - root(a-2b), prove that bx²-ax+b=0

Question : If x={√[a+2b] + √[a-2b]} / {√[a+2b] - √[a-2b]}, prove that bx²-ax+b=0

Doubt by Veer

Solution :

Rationalising the denominator

2bx-a=√(a²-4b²)

SBS

(2bx-a)²=a²-4b²

4b²x²+a²-4abx=a²-4b²

4b²x²-4abx=-4b²

4b²x+4b²-4abx=0

4b(bx²+b-ax)=0

4b(bx²-ax+b)=0

4b=0 OR bx²-ax+b=0

b=0/4 OR bx²-ax+b=0

b=0 OR bx²-ax+b=0

but b≠0

Hence bx²-ax+b=0.

The area of the isosceles triangle is 5√11/4 cm² if the . . .

Question : Write true or false for the given statement.

"The area of the isosceles triangle is 5√11/4 cm² if the perimeter is 11 cm and the base is 5 cm."

Doubt by Purvi

Solution :

Let the equal sides of the isosceles triangle be
a and base be b. So the three sides of the isosceles triangle would be a, a and b.

b=5 cm (Given)
Perimeter (P) = 11 cm
a+a+b=11

2a+5=11
2a=11-5
2a=6
a=6/2
a=3 cm

So the three sides of the isosceles triangle would be a, a and b where a=3 cm and b=5 cm

Now,
Area of isoseceles triangle is given by

Area of isoseceles triangle is 5√11/4 cm². Hence, the above statement is absolutely correct.

If x3+ax3+10-bx is divisible by (x-2) . . .

Question : If x³+ax²+10-bx is divisible by (x-2) and (x-1) then find the values of a and b.

Doubt by Purvi

Solution :

p(x)=x³+ax²+10-bx

p(x)=x³+ax²-bx+10

Case I

g(x)=(x-2)
p(x) is divisible by g(x), it means g(x) must be a factor of p(x).

g(x)=0
x-2=0
x=2

p(2)=0
(2)³+a(2)²-b(2)+10=0
8+4a-2b+10=0
4a-2b+18=0
4a-2b=-18
2(2a-b)=-18
2a-b=-18/2
2a-b=-9 — (1)

Case II

g(x)=(x-1)
p(x) is divisible by g(x), it means g(x) must be a factor of p(x).

g(x)=0
x-1=0
x=1

p(1)=0
(1)³+a(1)²-b(1)+10=0
1+a-b+10=0
a-b+11=0
a-b=-11 — (2)

Subtracting equation (2) from equation (1)

2a-b-(a-b)=-9-(-11)
2a-b-a+b=-9+11
a+0=2
a=2
putting this value of a in equation (1)

2(2)-b=-9
4-b=-9
4+9=b
b=13

Hence, a=2 and b=13

In Fig, if AB || CD || EF, PQ || RS, ∠RQD = 25 . . .

Question : In Fig, if AB || CD || EF, PQ || RS, ∠RQD = 25° and ∠CQP = 60°, then ∠QRS is equal to

(A) 85°
(B) 135°
(C) 145°
(D) 110°

Doubt by veer

Solution :

AB || CD || EF (Given)

∠RQD = 25° and ∠CQP = 60° (Given)

∠QRS=?

∠ARQ=∠RQD =25° [Alternate Interior Angles]
∠ARQ=25° — (1)

∠CQP=∠SRB=60° [Alternate Exterior Angles]
∠SRB=60°

Now,
∠SRB + ∠SRA = 180° [Linear Pair]

60° + ∠SRA = 180°

∠SRA=180°-60°
∠SRA=120° — (2)

Adding equation (1) and (2)

∠ARQ+∠SRA=25°+120°
∠QRS = 145°
Hence, (C) 145°, would be the correct option.

The base of an isosceles triangle is of length 48 cm and its. . .

Question : The base of an isosceles triangle is of length 48 cm and its area is 240 cm². Use Heron's formula and hence find the perimeter of the triangle.

Doubt by Veer

Solution :

Here
a=48 cm

b=x cm

c=x cm

Semiperimeter (s)

s=[a+b+c]/2
s=[48+x+x]/2
s=[48+2x]/2
s=2[24+x]/2
s=[24+x] cm

Area = 240 cm² [Given]

√[s(s-a)(s-b)(s-c)] = 240
√[24+x(24+x-48)(24+x-x)(24+x-x)] = 240
√[24+x(x-24)(24)(24)] = 240
Squaring both sides
[(x+24)(x-24)24×24]=240×240
x²-(24)²=(240×240)/(24×24)
x²-576=100
x²=100+576
x²=676
x=√676
x=26

b=c=26 cm

So,
Perimeter of triangle
P=a+b+c
P=48+26+26
P=48+52
P=100 cm²

If x=1/7-4root3, show that x2-14x+1=0 and hence evaluate . . .

Question : If x=1/(7-4√3), show that x²-14x+1=0 and hence evaluate x³-9x²-69x+7.

Doubt by Purvi

Solution :

x=1/(7-4√3)
Rationising the denominator

x=1/(7-4√3)×(7+4√3)/(7+4√3)
x=(7+4√3)/[(7)²-(4√3)²]
x=(7+4√3)/(49-48)
x=(7+4√3)/1
x=7+4√3

x²-14x+1=0

LHS

x²-14x+1
=[7+4√3]²-14[7+4√3]+1
=(7)²+(4√3)²+2(7)(4√3)-14(7)-14(4)√3+1
=49+48+56√3-98-56√3+1
=50+48-98
=98-98
=0
= RHS

LHS = RHS

Hence Proved.

Now

x²-14x+1=0
x²=14x-1— (1)

Also
x³=x².x
x³=(14x-1)x [∵ Using Equation (1)]

x³=14x²-x — (2)

Now

x³-9x²-69x+7
=14x²-x-9(14x-1)-69x+7
[Using Equation (1) and (2)]

=14x²-x-126x+9-69x+7
=14x²-196x+16
=14x²-196x+14+2
=[14x²-196x+14]+2
=14[x²-14x+1]+2
=14[0]+2 [∵x²-14x+1=0]
=0+2
=2

Hence, x³-9x²-69x+7=2

If one angle of a parallelogram is 24° less than twice the . . .

Question : If one angle of a parallelogram is 24° less than twice the smallest angle, then the measure of the largest angle of a parallelogram is :

(a) 176°

(b) 68°

(d) 102°

Doubt by Noor

Solution :

Let the smallest angle of the parallelogram be x and the largest angle be y

We also know that in parallelogram opposite angles are equal to each other and adjacent angles are supplementary (180°).

ATQ
y=2x-24° — (1)

x+y=180° (Sum of adjacent angles of a parallelogram)

x+2x-24°=180° [From Equation (1)]
3x-24°=180°
3x=180°+24°
3x=204°
x=204°/3
x=68°

Substituting in equation (1)

y=2(68°)-24°
y=136°-24
y=112°

The largest angle of the parallelogram is 112°.

Hence, (c) 112°, would be the correct option.

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