Class 9

Multiple Choice Questions of Heron's Formula

1.) An isosceles right triangle has area 8 cm². The length of its hypotenuse is

(A) √32 cm
(B) √16 cm
(C) √48 cm
(D) √24 cm

2. The perimeter of an equilateral triangle is 60 m. The area is

(A) 10√3 m²

(B) 15√3 m²

(D) 100√3 m²

3. The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then the area of the

triangle is

(A) 1322 cm²

(B) 1311 cm²

(D) 1392 cm²

4. The area of an equilateral triangle with side 2√3 cm is

(A) 5.196 cm²

(B) 0.866 cm²

(D) 1.732 cm²

5.) The length of each side of an equilateral triangle having an area of 9√3 cm² is

(A) 8 cm

(B) 36 cm

(D) 6 cm

6.) If the area of an equilateral triangle is 16√3 cm², then the perimeter of the triangle is

(A) 48 cm

(B) 24 cm

(D) 36 cm

7.) The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude

(A) 16√5 cm

(B) 10√5 cm

(D) 28 cm

8.) The area of an isosceles triangle having base 2 cm and the length of one of the equal

sides 4 cm, is

9.) The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per cm² is

(A) Rs 2.00

(B) Rs 2.16

(D) Rs 3.00

10.) The base of a right triangle is 8 cm and hypotenuse is 10 cm. Its area will be

(A) 24 cm²

(B) 40 cm²

(D) 80 cm²

Answer Key

1.) (A) √32 cm

2.) (D) 100√3 m²

3.) (C) 1344 cm²

4.) (A) 5.196 cm²

5.) (D) 6 cm

6.) (B) 24 cm

7.) (C) 24√5 cm

8.) (A) √15 cm²

9.) (B) Rs 2.16

10.) (A) 24 cm²

While selling clothes for making flags, a shopkeeper claims . . .

Case Study Based Question on Heron's Formula

While selling clothes for making flags, a shopkeeper claims to sell each piece of cloth in the shape of an equilateral triangle of each side 10 cm while actually he was selling the same in the shape of an isosceles triangle with sides 10 cm, 10 cm and 8 cm. [Use √3=1.73 and √21 = 4.56]

(i) Find the area of an equilateral triangular flag. [1 Mark]
(ii) What is the semi-perimeter of an isosceles triangular Flag. [1 Mark]

(iii) How much cloth was he saving in selling each isosceles flag instead of equilateral flag? [2 Marks]
OR

(iv) What is the altitude of the isosceles triangle with respect to its smallest side? [2 Marks].

Doubt by Veer

Solution :

(i) Find the area of an equilateral triangular flag.
Ans :
a = 10 cm
Area= √3a²/4
=√3×(10)²/4
= √3×100/4
= √3×25
= 25√3
= 25×1.73
=43.25 cm²

(ii) What is the semi-perimeter of an isosceles triangular Flag.
Ans :
a=10 cm
b=10 cm
c=8 cm

s=(a+b+c)/2
s=(10+10+8)/2
s=28/2
s=14 cm

(iii) How much cloth was he saving in selling each isosceles flag instead of equilateral flag?
Ans :
a=10 cm
b=10 cm
c=8 cm

s=14 cm

Using Heron's Formula
Area of isosceles Flag
=√[s(s-a)(s-b)(s-c)]
=√[14(14-10)(14-10)(14-8)]
=√[14(4)(4)(6)]
=√[7×2×4×4×2×3]
=4×2√21
=8√21 cm²
=8×4.56
=36.48 cm²

Area of cloth the shopkeeper was saving in selling each isosceles flag instead of equilateral flag
= 43.25-36.48
= 6.77 cm²

OR

(iv) What is the altitude of the isosceles triangle with respect to its smallest side?

Ans :

Area = 8√21 cm²
Base = 8 cm

Area = ½×Base×Height
2Area/Base = Height
Height = 2Area/Base
=[2×8√21]/8
=2√21
=2×4.56
=9.12 cm

The area of a given isosceles triangle is 12cm2 . . .

Question : The area of a given isosceles triangle is 12cm². If one of the equal sides is 5cm. Find its base.

Doubt by Jugal

Solution :

Area of isosceles triangle = 12 cm²
Length of each equal sides (a) = 5 cm
Base of the isosceles triangle (b) = ?

12 = (b/4)√[4(5)²-b²]
SBS
(12)²=(b²/16)[100-b²]
144×16=b²(100-b²)
2304=100b²-b⁴
b⁴-100b²+2304=0
(b²)²-100b²+2304=0
Let b²=x
x²-100x+2304=0
x²-(64+36)x+2304=0
x²-64x-36x+2304=0
x(x-64)-36(x-64)=0
(x-36)(x-64)=0

x-36=0
x=36
b²=36
b=±√36

b=±6
but base can't be -ve
so b=6 cm

OR

(x-64)=0
x=64
b²=64
b=±√64
b=±8
but base can't be -ve
so b=8 cm

Hence, the base of the isosceles triangle is either 6 cm or 8 cm.

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.

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