Class 9

Find the value of x : (4)2x-1-(16)x-1=384 . . .

Question : Find the value of x : (4)^2x-1-(16)^x-1=384

Doubt by Balvinder

Solution :

(4)^2x-1 - (16)^x-1 = 384

(4)^2x-1 - (4²)^x-1 = 384
(4)^2x-1 - (4)^2(x-1) = 384

(4)^2x-1 - (4)^2x-2 = 384

(4)^2x-1 - (4)^2x-1-1 = 384
(4)^2x-1 - (4)^2x-1.4^-1 = 384

(4)^2x-1 [1-1/4] = 384
(4)^2x-1 [(4-1)/4] = 384

(4)^2x-1 [3/4] = 384
(4)^2x-1= (384×4)/3
(4)^2x-1= 128×4
(4)^2x-1= 512
(2²)^2x-1=2⁹
2^2(2x-1)=2⁹
2^4x-2=2⁹

equating the power of 2 both sides
4x-2=9
4x=9+2
4x=11
x=11/4

If xyz=1 then simplify (1+x+y-1)-1 . . .

Question : If xyz=1 then simplify (1+x+y^-1)^-1×(1+y+z^-1)^-1×(1+z+x^-1)^-1

Doubt by Balvinder

Solution :
xyz=1
xy=1/z
1/z=xy— (1)

Also
z=1/xy — (2)

Now,

(1+x+y^-1)^-1
=[1+x+(1/y)]^-1
=[(y+xy+1)/y]^-1
=[y/(y+xy+1)]
=[y/(1+y+xy)] — (3)

(1+y+z^-1)^-1
=[1+y+(1/z)]^-1
=[1+y+xy]^-1 [Using equation (1)]
=[1/(1+y+xy)] — (4)

(1+z+x^-1)^-1
=[1+z+(1/x)]^-1
=[1+(1/xy)+(1/x)]^-1 [Using equation (2)]
=[(xy+1+y)/xy]^-1
=[xy/(xy+1+y)]
=[xy/(1+y+xy)] — (5)

Adding equation (3), (4) and (5)

(1+x+y^-1)^-1×(1+y+z^-1)^-1×(1+z+x^-1)^-1
= [y/(1+y+xy)] + [1/(1+y+xy)] + [xy/(1+y+xy)]
=[(y+1+xy)/(1+y+xy)]
=[(1+y+xy)/(1+y+xy)]
= 1

Hence
(1+x+y^-1)^-1×(1+y+z^-1)^-1×(1+z+x^-1)^-1 = 1

In the figure, we have BX=1/2AB, BY=1/2BC . . .

Question : In the figure, we have BX=½AB, BY=½BC and AB=BC show that BX=BY. Also write the axiom used.

Doubt by Purvi

Solution :

Given :
BX=½AB
BY=½BC
AB=BC

To Prove : BX=BY

Proof :

BX=½AB (Given)
2BX=AB
AB=2BX — (1)

BY=½BC (Given)

2BY=BC
BC=2BY — (2)

AB=BC (Given)
2BX=2BY
BX=BY
[Things which are double of the same things are equal to one another]

Hence Proved.

Plants bring positively and calmness in environment. Ruhi bought a . . .

Question : Plants bring positively and calmness in environment. Ruhi bought a glass pyramid planter with square base for her office. She grew cacti in it and used it as a centre piece in her office.

i) What is the area of each isosceles triangular glass frame ? [1 Mark]

ii) A student pugged in the value a = 15cm, b = 14cm, c = 13cm and s=42 cm in the formula

A=√[s(s-a)(s-b)(s-c)]

What does A stand for in the formula? a, b, c and s have their usual meaning in the formula. [1 Marks]

iii) Find the total length of the brass frame used to connect all sides walls and base. [2 Marks]

Doubt by Purvi

Solution :

(i) Here
Base (b) = 10 cm
Each Equal side (a) = 13 cm

Area of each isosceles triangle
=(b/4)[√(4a²-b²)]
=(10/4)[√(4×(13)²-(10)²)]
=2.5[√(676-100)]
=2.5√576
=2.5×24
=60 cm²

(ii) A stands for Area of the Triangle by Heron's Formula.

(iii) Total length of the brass frame used
= 4×base (b) + 4×Equal Side (a)
= 4b+4a
=4(b+a)

=4(10+13)
=4×23
=92 cm

[CSBQ] If we look closely at computer screen, we see that the screen is made . . .

Question : If we look closely at computer screen, we see that the screen is made up of millions of tiny squares called pixels. Number of pixels together form a digital image on screen every clickable point represents x-co-ordinates as number of pixels along horizontal axis i.e., x-axis and y co-ordinate as number of pixels along vertical axis i.e. y-axis.

Find

i) What is the perpendicular distance of A (165, 250) from x axis. [1 Marks]

ii) In which quadrant the abscissa of a point is negative. [1 Marks]

iii) What is the sum of abscissa of A and B. [2 Marks]

Doubt Question by Purvi

Solution :
(i) 250 units.
(ii) II and III Quadrant

(iii)
A(165, 250)

B(300, 250)

Sum of abscissa of A and B
=165+300
= 465

Along a path, 100 conical pillars are constructed. Each . . .

Question : Along a path, 100 conical pillars are constructed. Each pillar has base radius 7 cm and height 24 cm. Find

(i) Curved surface area of one pillar. [2 Marks]

(ii) Find the total cost of painting 100 pillars at the rate of Rs. 120 per cm².

Doubt by Question by Purvi

Solution :

No. of Pillars = 100

radius (r) = 7 cm
height (h) = 24 cm

Slant Height (l)
l=√(r²+h²)
l=√[(7)²+(24)²]
l=√[49+576]
l=√[625]
l=25 cm

(i) CSA of one pillar
= πrl
= (22/7)×7×25
= 22×25
= 550 cm²

(ii) CSA of one pillar = 550 cm²
CSA of 100 Pillars
= 550 × 100
=55000 cm²

Cost of Painting 1 cm² = Rs. 120
Cost of Painting 55000 cm²
= 55000 ×120
= Rs 6600000

If a+b+c=root(42) and a2+b2+c2=16 . . .

Question : If a+b+c=√42 and a²+b²+c²=16, then find the value of ab+bc+ca

Doubt by Parth

Solution :

a+b+c=√42 — (1)

a²+b²+c²=16 — (2)

We know,
(a+b+c)²=a²+b²+c²+2ab+2bc+2ca
(√42)²=16+2(ab+bc+ca)
42=16+2(ab+bc+ca)
42-16=2(ab+bc+ca)

26=2(ab+bc+ca)
26/2=(ab+bc+ca)
13=ab+bc+ca

Hence, the value of ab+bc+ca=13

In a square ABCD, P is the midpoint of AD. BP and . . .

Question : In a square ABCD, P is the midpoint of AD. BP and CP are joined. Prove that ∠PCB=∠PBC. Also prove that if ∠PCB=45° then PC is perpendicular to PB.

Doubt by Parth

Solution :

Given : ABCD is a square.
AP=DP

∠PCB=45°

To Prove : (i) ∠PCB=∠PBC
(ii) PC is perpendicular to PB.

Proof :

(i) In ΔBAP and ΔCDP
AB=DC (opposite sides of square)
∠BAP=∠CDP [Each 90°]
AP=DP (Given)
ΔBAP≅ΔCDP (By SAS)
PB=PC (By CPCT)

In ΔBPC
PB=PC(Proved above)
∠PCB=∠PBC (Angles opposite to equal sides of a triangle are equal)

(ii) In ΔBPC

∠PCB+∠PBC+∠BPC=180°(ASP)
45°+45°+∠BPC=180°
90°+∠BPC=180°

∠BPC=180°-90°
∠BPC=90°

So, PC is perpendicular to PB.

Hence Proved.

If the bisector of the exterior vertical angle of a triangle . . .

Question : If the bisector of the exterior vertical angle of a triangle is parallel to the base, show that the triangle is isosceles.

Doubt by Veer

Solution :

Given : AD is the angle bisector of ∠ EAC
AD|| BC

To Prove : ABC is an isosceles triangle.

Proof :

∠1=∠2 (AD is angle bisector) — (1)
∠1=∠3 (Corresponding Angles) — (2)
∠2=∠4 (Alternate Interior Angles) — (3)

From equations (1) , (2) and (3)

∠3=∠4

AB=AC (Sides opposite to equal angles of a triangle are equal)

⇒ABC is an isosceles triangle.

Hence Proved.

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

Pages