Maths

Exercise 13:2

QUESTION 1

The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder. Assume π =

ANSWER

 

Height (h) of cylinder = 14 cm

Let the diameter of the cylinder be d.

Curved surface area of cylinder = 88 cm2

⇒ 2πrh = 88 cm2 (r is the radius of the base of the cylinder)

⇒ πdh = 88 cm2 (d = 2r)

d = 2 cm

Therefore, the diameter of the base of the cylinder is 2 cm.

Surface Areas and Volumes

SDFSGFG

CBSE NCERT SOLUTION MATHS Surface Areas and Volumes EXERCISE 13:1

QUESTION 1

 

A plastic box 1.5 m long, 1.25 m wide and 65 cm deep, is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine:

(i) The area of the sheet required for making the box.

(ii) The cost of sheet for it, if a sheet measuring 1 m2 costs Rs 20.

ANSWER

 

It is given that, length (l) of box = 1.5 m

Breadth (b) of box = 1.25 m

Depth (h) of box = 0.65 m

(i) Box is to be open at top.

Area of sheet required

= 2lh + 2bh + lb

= [2 × 1.5 × 0.65 + 2 × 1.25 × 0.65 + 1.5 × 1.25] m2

= (1.95 + 1.625 + 1.875) m2 = 5.45 m2

(ii) Cost of sheet per m2 area = Rs 20

Cost of sheet of 5.45 m2 area = Rs (5.45 × 20)

= Rs 109

Exercise 7:5

hkhkljklj

CBSE NCERT SOLUTION CLASS 9TH MATHS

 

QUESTION 1

 

ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.

ANSWER

 

Circumcentre of a triangle is always equidistant from all the vertices of that triangle. Circumcentre is the point where perpendicular bisectors of all the sides of the triangle meet together.

In ΔABC, we can find the circumcentre by drawing the perpendicular bisectors of sides AB, BC, and CA of this triangle. O is the point where these bisectors are meeting together. Therefore, O is the point which is equidistant from all the vertices of ΔABC.

 

 

QUESTION 2

In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

ANSWER

 

Exercise 7:4

rytuiguh

CLASS9TH, MATHS, NCERT CBSE SOLUTION EXERCISE 7:4

QUESTION 1

Show that in a right angled triangle, the hypotenuse is the longest side.

ANSWER

Let us consider a right-angled triangle ABC, right-angled at B. In ΔABC, ∠A + ∠B + ∠C = 180° (Angle sum property of a triangle) ∠A + 90º + ∠C = 180° ∠A + ∠C = 90° Hence, the other two angles have to be acute (i.e., less than 90º). ∴ ∠B is the largest angle in ΔABC. ⇒ ∠B > ∠A and ∠B > ∠C ⇒ AC > BC and AC > AB [In any triangle, the side opposite to the larger (greater) angle is longer.] Therefore, AC is the largest side in ΔABC. However, AC is the hypotenuse of ΔABC. Therefore, hypotenuse is the longest side in a right-angled triangle.

 

QUESTION 2

 

In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

ANSWER

 

 

In the given figure,

∠ABC + ∠PBC = 180° (Linear pair)

⇒ ∠ABC = 180° − ∠PBC ... (1)

Also,

∠ACB + ∠QCB = 180°

∠ACB = 180° − ∠QCB … (2)

As ∠PBC < ∠QCB,

⇒ 180º − ∠PBC > 180º − ∠QCB

Exercise 7:3

hkjlkjlk

QUESTION  1

 

ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show that

(i) ΔABD ≅ ΔACD

(ii) ΔABP ≅ ΔACP

(iii) AP bisects ∠A as well as ∠D.

(iv) AP is the perpendicular bisector of BC.

ANSWER

 

(i) In ΔABD and ΔACD,

AB = AC (Given)

BD = CD (Given)

AD = AD (Common)

∴ ΔABD ≅ ΔACD (By SSS congruence rule)

⇒ ∠BAD = ∠CAD (By CPCT)

⇒ ∠BAP = ∠CAP …. (1)

(ii) In ΔABP and ΔACP,

AB = AC (Given)

∠BAP = ∠CAP [From equation (1)]

AP = AP (Common)

∴ ΔABP ≅ ΔACP (By SAS congruence rule)

⇒ BP = CP (By CPCT) … (2)

(iii) From equation (1),

∠BAP = ∠CAP

Hence, AP bisects ∠A.

In ΔBDP and ΔCDP,

BD = CD (Given)

DP = DP (Common)

BP = CP [From equation (2)]

∴ ΔBDP ≅ ΔCDP (By S.S.S. Congruence rule)

⇒ ∠BDP = ∠CDP (By CPCT) … (3)

Hence, AP bisects ∠D.

(iv) ΔBDP ≅ ΔCDP

∴ ∠BPD = ∠CPD (By CPCT) …. (4)

Exercise 7:2

gjkjhhkj

CBSE NCERT Solution Exercise 7:2

QUESTION 1

 

In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:

(i) OB = OC (ii) AO bisects ∠A

ANSWER

 

(i) It is given that in triangle ABC, AB = AC

⇒ ∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are equal)

∠ACB = ∠ABC

⇒ ∠OCB = ∠OBC

⇒ OB = OC (Sides opposite to equal angles of a triangle are also equal)

(ii) In ΔOAB and ΔOAC,

AO =AO (Common)

AB = AC (Given)

OB = OC (Proved above)

Therefore, ΔOAB ≅ ΔOAC (By SSS congruence rule)

⇒ ∠BAO = ∠CAO (CPCT)

⇒ AO bisects ∠A.

 

QUESTION 2

 

In ΔABC, AD is the perpendicular bisector of BC (see the given figure). Show that ΔABC is an isosceles triangle in which AB = AC.

ANSWER

 

In ΔADC and ΔADB,

Exercise 7:1

 

hkjlklkl

CBSE NCERT SOLUTION FOR CLASS 9TH MATHS EXERCISE 7:1

QUESTION 1

 

In quadrilateral ACBD, AC = AD and AB bisects ∠A (See the given figure). Show that ΔABC ≅ ΔABD. What can you say about BC and BD?

ANSWER

 

In ΔABD and ΔBAC,

AD = BC (Given)

∠DAB = ∠CBA (Given)

AB = BA (Common)

∴ ΔABD ≅ ΔBAC (By SAS congruence rule)

∴ BD = AC (By CPCT)

And, ∠ABD = ∠BAC (By CPCT)

 

QUESTION 2

ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (See the given figure). Prove that

(i) ΔABD ≅ ΔBAC

(ii) BD = AC

(iii) ∠ABD = ∠BAC.

Get free NCERT solutions of Class cbse 9th in your email: