Class 9

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.

If two equal chords of a circle . . .

Question : If two equal chords of a circle intersect within a circle. Prove that the segment of one chord are equal to the corresponding segment of another.

Doubt by Samridhi

Solution :

Given : A circle with centre O. AB and CD are two chords such that AB=CD.

To Prove :
(i) AP=DP
(ii) BP=CP

Construction : Draw OM⊥AB and ON⊥CD

Proof :
(i) In ∆OMP and ∆ ONP

∠OMP = ∠ONP (Each 90°)
OP = OP (Common)
OM = ON (Equal chords are equidistant from the centre of a circle)
∆OMP≅∆ONP (By RHS)
MP=NP — (1) (By CPCT)

AM=BM (Perpendicular drawn from the centre to the chord bisect the chord)
Similarly
DN=CN

Also
AB=CD — (2) (Given)
AM+BM=DN+CN
AM+AM=DN+DN
2AM=2DN
AM=DN — (3)

Adding equation (1) and (2)
AM+MP=DN+NP
AP=DP — (4)

(ii) Subtracting equation (4) from (2)

AB-AP=CD-DP

BP=CP

AD, BE, CF are equal altitudes of a a triangle ABC . . .

Question : AD, BE, CF are equal altitudes of a ΔABC. Prove that ΔABC is an equilateral triangle.

Doubt by Suraj

Solution :

Given : AD, BE and CF are altitudes of ΔABC.
AD=BE=CF
To Prove : ABC is an equilateral triangle.

Proof :
Area of ΔABC
= ½×Base×Height
= ½×BC×AD — (1)
Again,
Area of ΔABC
= ½×Base×Height
= ½×AB×CF — (2)

From equation (1) and (2)
½×BC×AD=½×AB×CF
BC×AD=AB×CF
BC×AD=AB×AD [∵AD=CF]
BC=AB — (3)

Similarly
BC=AC — (4)

From equation (3) and (4)

AB=BC=CA

All sides of triangle ABC are equal.

So, ABC is an equilateral triangle.

Hence Proved.

Note : The above question can also be proved by using the concept of congruency. Can you do that by yourself?

Similar Question : In given figure the altitudes AD, BE and CF, the altitudes of triangle ABC are equal. Prove that ABC is an equilateral triangle.

Similar Question : The altitudes of ΔABC, AD, BE and CF are equal. Prove that ΔABC is an equilateral triangle.

The value of root 32 + root 48 / root 8 + root 12 is equal to . . .

Question : The value of (√32 + √48) / (√8 + √12) is equal to
(a) √2
(b) 2
(c) 4
(d) 8

Doubt by Suraj

Solution :
(√32 + √48) / (√8 + √12)
=(√48 + √32) / (√12 + √8)
=(4√3 + 4√2) / (2√3 + 2√2)

Rationalising the denominator

=(4√3 + 4√2) × (2√3 - 2√2) / (2√3 + 2√2)×(2√3 - 2√2)
=[4√3(2√3 - 2√2) + 4√2(2√3 - 2√2)] / [(2√3)² - (2√2)²]
[∵(a+b)(a-b)=(a²-b²)]
=[8×3-8√6+8√6-8×2] / [12-8]
=[24-0-16] / 4
=[8]/4
=2

Hence, (b) 2, would be the correct option.

If x3+mx2+nx+6 has (x-2) as a factor & leaves a remainder. . .

Question : If x³+mx²+nx+6 has (x-2) as a factor & leaves a remainder 3 when divided by (x-3), find the values of m and n.

Doubt by Bhumika

Solution :

Let p(x) = x³+mx²+nx+6

(x-2) is a factor of p(x)

x-2=0
x=2
p(2)=0
(2)³+m(2)²+n(2)+6=0
8+4m+2n+6=0
4m+2n=-14
2(2m+n)=-14
2m+n=-14/2
2m+n=-7 — (1)

Also

When p(x) is divided by (x-3) then remainder is 3.

x-3=0
x=3
p(3)=3
(3)³+m(3)²+n(3)+6=3
27+9m+3n+6=3
9m+3n+33=3
9m+3n=3-33
9m+3n=-30
3(3m+n)=-30
3m+n=-30/3
3m+n=-10 — (2)

Subtracting equation (2) from (1)

2m+n-(3m+n)=-7-(-10 )
2m+n-3m-n=-7+10
-m+0=3

-m=3
m=-3

Putting m=-3 in equation (1)

2(-3)+n=-7
-6+n=-7
n=-7+6
n=-1

∴ m=-3 & n=-1

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