NCERT Solutions for Class 9 Maths Chapter 7 Heron’s Formula Ex 7.1

NCERT Solutions for Class 9 Maths Chapter 7 Heron’s Formula Ex 7.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 7 Heron’s Formula Ex 7.1

BoardCBSE
TextbookNCERT
ClassClass 9
SubjectMaths
ChapterChapter 7
Chapter NameHeron’s Formula
ExerciseEx 7.1
Number of Questions Solved6
CategoryNCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 7 Heron’s Formula Ex 7.1

Question 1.
A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side a. Find the area of the signal board, using Heron’s formula.If its perimeter is 180 cm, what will be the area of the signal board?
Solution:
We know that, an equilateral triangle has equal sides. So, all sides are equal to a.
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.1 img 1

Question 2.
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see figure). The advertisements yield an earning of ₹5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.1 img 2
Solution:
Leta = 122m,
b = 22m
c = 120m
We have, b2 + c2 = (22)2 + (120)2 = 484 + 14400 = 14884= (122)2 = a2
Hence, the side walls are in right triangled shape.
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.1 img 3

Question 3.
There is a slide in a park. One of its side Company hired one of its walls for 3 months.walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see figure). If the sides of the wall are 15 m, 11 m and 6m, find the area painted in colour.
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.1 img 4
Solution:
The given figure formed a triangle whose sides are
a = 15m
b = 11m,
c = 6m
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.1 img 5

Question 4.
Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Solution:
Solution Let the sides of a triangle, a = 18cm b = 10 cm and c
We have, perimeter = 42 cm
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.1 img 6

Question 5.
Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
Solution:
Suppose that the sides in cm, are 12x, 17x and 25x.
Then, we know that 12x + 17x + 25x = 540 (Perimeter of triangle)
⇒ 54x = 540
⇒ x=10
So, the sides of the triangle are 12 x 10 cm, 17 x 10 cm 25 x 10 cm
i.e., 120 cm, 170 cm, 250 cm
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.1 img 7

Question 6.
An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Solution:
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.1 img 8
Let in isosceles ∆ ABC,
AB = AC = 12 cm (Given)
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.1 img 9

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NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3.

BoardCBSE
TextbookNCERT
ClassClass 9
SubjectMaths
ChapterChapter 5
Chapter NameTriangles
ExerciseEx 5.3
Number of Questions Solved5
CategoryNCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3

Question 1.
∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 img 1
(i) ∆ABD = ∆ACD
(ii) ∆ABP = ∆ACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC.
Solution:
Given ∆ABC and ∆DBC are two isosceles triangles having common
base BC, suchthat AB=AC and DB=OC.
To prove:
(i) ∆ABD = ∆ACD
(ii) ∆ABP = ∆ACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC.
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 img 2
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 img 3

Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A
Solution:
In ∆ ABD and ∆ ACD, we have
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 img 4
AB = AC (Given)
∠ADB = ∠ADC = 90° (∵ Given AD ⊥BC)
AD = AD (Common)
∴ ∆ ABD ≅ ∆ ACD (By RHS congruence axiom)
BD=DC (By CPCT)
⇒ AD bisects BC.
∠ BAD = ∠ CAD (By CPCT)
∴ AD bisects ∠A .

Question 3.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and OR and median PN of ∆PQR (see figure). Show that
(i) ∆ABC ≅ ∆PQR
(ii) ∆ABM ≅ ∆PQN
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 img 5
Solution:
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 img 6

Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Solution:
In ∆BEC and ∆CFB, we have
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 img 7

Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Solution:
In ∆ABP and ∆ACP, We have
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 img 8
AB = AC (Given)
AP = AP (Common)
and ∠APB = ∠APC = 90° (∵ AP ⊥ BC)
∴ ∆ABP ≅ ∆ACP (By RHS congruence axiom)
⇒ ∠B = ∠C (By CPCT)

We hope the NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3.

BoardCBSE
TextbookNCERT
ClassClass 9
SubjectMaths
ChapterChapter 4
Chapter NameLines and Angles
ExerciseEx 4.3
Number of Questions Solved6
CategoryNCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3

Question 1.
In figure, sides QP and RQ of APQR are produced to points S and T, respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 1
Solution:
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 2
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 3

Question 2.
In figure, ∠X – 62°, ∠XYZ = 54°, if YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 4
Solution:
In ∆XYZ,
∵ ∠X+ ∠Y+ ∠Z = 180°
(Sum of all angles of triangle is equal to
∴ 62° + ∠Y + ∠Z = 180° [YZX = 62° (Given)]
⇒ ∠Y + ∠Z = 118° .
⇒ \(\frac { 1 }{ 2 }\)∠Y + \(\frac { 1 }{ 2 }\) ∠Z = \(\frac { 1 }{ 2 }\) x 118°
⇒ ∠OYZ + ∠OZY = 59°
(∵ YO and ZO are the bisectors of ∠XYZ and ∠XZY)
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 5
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 6

Question 3.
In figure, if AB || DE, ∠BAC = 35° and ∠CDE = 539 , find ∠DCE.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 7
Solution:
We have AB || DE
⇒ ∠AED= ∠ BAE (Alternate interior angles)
Now, ∠ BAE = ∠ BAC
⇒ ∠ BAE = 35° [ ∵ ∠ BAC = 35° (Given) ]
∴ ∠ AED = 35°
In ∆DCE,
∵ ∠DCE + ∠CED+ ∠EDC= 180° ( ∵Sum of all angles of triangle is equal to 180°)
⇒ ∠ DCE + 35°+ 53° = 180° ( ∵∠ AED = ∠ CED = 35°)
⇒ ∠ DCE = 180° – (35° + 53°)
⇒ ∠ DCE = 92°

Question 4.
In figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠TSQ = 75°, find ∠ SQT.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 8
Solution:
∵∠PTS = ∠RPT + ∠PRT (Exterior angle = Sum of interior opposite angles)
∠ PTS = 95° + 40° [∵ ∠PPT = 95° (Given)]
⇒ ∠ PTS = 135° [and ∠PRT = 40°]
Also, ∠ TSQ + ∠ SQT = ∠ PTS (Exterior angle = Sum of interior opposite angles)
⇒ 75°+ ∠ SQT = 135°
⇒ ∠ SQT = 60° [∵ ∠ TSQ = 75° (Given)]

Question 5.
In figure, if PQ ⊥ PS, PQ||SR, ∠SQR = 2S° and ∠QRT = 65°, then find the values of x and y.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 9
Solution:
Here, PQ || SR
⇒ ∠ PQR = ∠ OPT (Alternate interior angles)
⇒ x + 28° = 65° ⇒ x = 37°
Now, in right angled ASPQ, we have ∠P = 90°
∴ ∠P + x + y = 180° (∵ Sum of all angles of a triangle is equal to 180°)
⇒ 90°+ 37°+ y= 180°
⇒ 127°+ y=180°
⇒ y = 53°

Question 6.
In figure, the side QR of A PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 10
Solution:
In ∆ PQP,
∵ ∠QPR + ∠PQR = ∠PRS …(i)
(∵ Sum of interior opposite angles = Exterior angle)
Now, in ∆ TOR,
∵ ∠QTR + ∠TQR = ∠TRS …..(ii)
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 img 11

We hope the NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2.

BoardCBSE
TextbookNCERT
ClassClass 9
SubjectMaths
ChapterChapter 4
Chapter NameLines and Angles
ExerciseEx 4.2
Number of Questions Solved6
CategoryNCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2

Question 1.
In figure, find the values of x and y and then show that AB || CD.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 1
Solution:
∵ x + 50° = 180° (Linear pair)
⇒ x = 130°
∴ y = 130° (Vertically opposite angle)
Here, ∠x = ∠COD = 130°
These are corresponding angles for lines AB and CD.
Hence, AB || CD

Question 2.
In figure, if AB || CD, CD || EF and y: z = 3:7, find x.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 2
Solution:
Given
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 3
⇒ Let y = 3k, z = 7k
x = ∠CHG (Corresponding angles)…(i)
∠CHG = z (Alternate angles)…(ii)
From Eqs. (i) and (ii), we get
x = z …….(iii)
Now, x+y = 180°
(Internal angles on the same side of the transversal)
⇒ z+y = 180° [From Eq. (iii)]
7k + 3k = 180°
⇒ 10k = 180°
⇒ k = 18
∴ y = 3 x 18° = 54°
and z = 7x 18°= 126°
∴ x = z
Now, x + y = 180°
(Internal angles on the same side of the transversal)
⇒ z + y = 180° [From Eq. (iii)]
7k + 3k = 180°
⇒ 10k = 180°
⇒ k = 18
∴ y = 3 x 18° = 54°
and z = 7x 18°= 126°
x = z
⇒ x = 126°

Question 3.
In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 4
Solution:
∵ ∠AGE = ∠GED (Alternate interior angles)
But ∠GED = 126°
⇒ ∠AGE = 126° ….(i)
∴ ∠GEF + ∠FED= 126°
⇒ ∠GEF + 90° =126° (∵ EF ⊥ CD)
⇒ ∠GEF = 36°
Also, ∠AGE + ∠FGE = 180° (Linear pair axiom)
⇒ 126° + ∠FGE =180°
⇒ ∠FGE = 54°

Question 4.
In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 5
Solution:
Drawing a tine parallel to ST through R.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 6
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 7

Question 5.
In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 8
Solution:
We have, AB || CD
⇒ ∠APQ = ∠PQR (Alternate interior angles)
⇒ 50° = x
⇒ x = 50°
Now, ∠PQR + ∠QPR = 127°
(Exterior angle is equal to sum of interior opposite angles of a triangle)
⇒ 50°+ ∠QPR = 127°
⇒ y = 77°.

Question 6.
In figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 9
Solution:
NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 img 10

We hope the NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 7 Heron’s Formula Ex 7.2

NCERT Solutions for Class 9 Maths Chapter 7 Heron’s Formula Ex 7.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 7 Heron’s Formula Ex 7.2.

BoardCBSE
TextbookNCERT
ClassClass 9
SubjectMaths
ChapterChapter 7
Chapter NameHeron’s Formula
ExerciseEx 7.2
Number of Questions Solved9
CategoryNCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 7 Heron’s Formula Ex 7.2

Question 1.
A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9m,BC = 12m,CD = 5m and AD = 8 m.
How much area does it occupy?
Solution:
In right ∆ BCD

NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 1
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 2

Question 2.
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Solution:
Area of quadrilateral ABCD = Area of ∆ABC+ Area of ∆ACD
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 3
In ∆ ABC,
We have, AB = 3 cm, BC = 4 cm, CA = 5 cm
Therefore, AB2 + BC2 = 32 + 42 = 9 + 16 = 25 = (5)2 = CA2
Hence, ∆ ABC is a right triangle

Question 3.
Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.

Solution:
For part I
It is a triangle with sides 5 cm, 5cm and 1cm
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 6
For part II
It is a rectangle with sides 6.5 cm and 1 cm
∴ Area of part II = 6.5 x 1
(∵ Area of rectangle = Length x Breadth)
= 6.5 cm2

For part III
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 7
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 8

Question 4.
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram
Solution:
Let ABC be a triangle with sides
AB = 26 cm, BC = 28 cm, CA = 30 cm
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 9
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 10
We know that,
Area of parallelogram = Base x Height …(i)
We have,
Area of parallelogram = Area of ∆ ABC (Given)
= 336 cm2
From Eq. (i), we have
Base x Height = 336
⇒ 28 x Height = 336
⇒ Height = \(\frac { 336 }{ 28 }\)
⇒ Height =12cm

Question 5.
A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Solution:
Let ABCD be a rhombus.
Area of the rhombus ABCD = 2x area of ∆ ABD …(i)
(Since, in a rhombus diagonals divides two equal parts)
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 11
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 12

Question 6.
An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 13
Solution:
In an umbrella, each triangular piece is an isosceles triangle with sides 50 cm, 50 cm, 20 cm.
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 14
Since, there are 10 triangular piece, in those of them 5.5 are of different colours.
Hence, total area of cloth of each colour = 5 x 200√6 cm2 = 1000√6 cm2

Question 7.
A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it?
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 15
Solution:
Since, the kite is in the shape of a square
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 16
Each diagonal of square =32 cm (Given)
We know that, the diagonals of a square bisect each other at right angle.
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 17
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 18
Hence, paper of I colour has been used = 256 cm2
Paper of II colour has been used = 256 cm2
Paper of III colour has been used = 17.92 cm2

Question 8.
A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). Find the cost of polishing the tiles at the rate of 50 paise per cm .
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 19
Solution:
Given, the sides of a triangular tiles are 9 cm, 28 cm and 35 cm.
For each triangular tile, we have
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 20

Question 9.
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
Solution:
Here, ABCD is a trapezium and AB || DC.
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 21
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 22

We hope the NCERT Solutions for Class 9 Maths Chapter 7 Heron’s Formula Ex 7.2 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 7 Heron’s Formula Ex 7.2, drop a comment below and we will get back to you at the earliest.