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°

(c) 112°

(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

If 5^2x-1-25^x-1=2500, find the value . . .

Question : If 5^2x-1-25^x-1=2500, find the value of x.

Doubt by Bhumika

Solution : 

5^2x-1-25^x-1=2500
5^2x-1-(5²)^x-1=2500
5^2x-1-5^2x-2=2500
5^2x-1-5^2x-1-1=2500
5^2x-1-5^2x-1÷5¹=2500
5^2x-1(1-1/5)=2500
5^2x-1(4/5)=2500
5^2x-1=[2500×5]/4
5^2x-1=3125
5^2x-1=5⁵
Comparing the exponents both sides
2x-1=5
2x=5+1
2x=6
x=6/2
x=3

Given below is a circle with O. Find the value . . .

Question : Given below is a circle with O. Find the value of ∠BCD in terms of a and b.

(a) 180°+(a-b/2)
(b) 180°-(a+b/2)
(c) 180°-(a-b/2)
(d) None of these. 

Doubt by Vikrant

Solution : 

∠1+b=180° (Linear Pair)

∠1=180°-b — (1)
In ΔOAD

OA=OD (Radii of the same circle)
∠2=∠3 (Angles opposite to equal sides of a triangle are equal)
Also,

∠1+∠2+∠3=180° (ASP)

∠1+∠3+∠3=180° [∵∠2=∠3]
∠1+2∠3=180°
2∠3=180°-∠1

2∠3=180°-(180°-b) [From equation (1)]
2∠3=180°-180°+b
2∠3=b
∠3=b/2 — (2)

In ΔADB

∠3+∠4=a [Exterior Angle Property of the Triangle]

b/2+∠4=a [Using equation (2)]
∠4=a-b/2 — (3)

Now, 

ABCD is a cyclic Quadrilateral

∠BCD+∠4=180° [Sum of opposite angles of a cyclic quadrilateral is supplementary]

∠BCD=180°-∠4

∠BCD=180°-(a-b/2) [Using equation (3)]

Hence, c) 180°-(a-b/2), would be the correct option.