• Skip to main content
  • Skip to secondary menu
  • Skip to primary sidebar
  • Skip to footer
  • NCERT Solutions
    • NCERT Books
  • Class 10
  • Class 9
  • Class 8
  • Class 7
  • Class 6
  • Class 11
  • Class 12
  • MCQ Questions
    • CBSE MCQ

Learn Insta

RD Sharma Solutions , RS Aggarwal Solutions and NCERT Solutions

  • Extra Questions
  • CBSE Notes
  • RD Sharma Solutions
    • RD Sharma Class 12 Solutions
    • RD Sharma Class 11 Solutions
    • RD Sharma Class 10 Solutions
    • RD Sharma Class 9 Solutions
    • RD Sharma Class 8 Solutions
  • RS Aggarwal Solutions
    • RS Aggarwal Solutions Class 10
    • RS Aggarwal Solutions Class 9
    • RS Aggarwal Solutions Class 8
    • RS Aggarwal Solutions Class 7
    • RS Aggarwal Solutions Class 6
  • ML Aggarwal Solutions
    • ML Aggarwal Class 10 Solutions
    • ML Aggarwal Class 9 Solutions
    • ML Aggarwal Class 8 Solutions
    • ML Aggarwal Class 7 Solutions
    • ML Aggarwal Class 6 Solutions
  • English Grammar
    • Words with Letters
    • English Summaries
    • Unseen Passages

Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers

May 28, 2022 by Prasanna

Here we are providing Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Answers Solutions, Extra Questions for Class 9 Maths was designed by subject expert teachers.

Extra Questions for Class 9 Maths Heron’s Formula with Answers Solutions

Extra Questions for Class 9 Maths Chapter 12 Heron’s Formula with Solutions Answers

Heron’s Formula Class 9 Extra Questions Very Short Answer Type

Question 1.
Find the area of an equilateral triangle having side 6 cm.
Solutioin:
Area of an equilateral triangle = \(\frac{\sqrt{3}}{4}\) × (side)2 = \(\frac{\sqrt{3}}{4}\) × 6 × 6 = 9√3 cm2

Question 2.
If the perimeter of an equilateral triangle is 90 m, then find its area.
Solutioin:
Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers 1

Question 3.
If every side of a triangle is doubled, then find the percent increase in area of triangle so formed.
Solutioin:
Let the sides of the given triangle be, a units, b units and c units.
Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers 2
Hence, percent increase = 300%

Question 4.
If the length of a median of an equilateral triangle is x cm, then find its area.
Solutioin:
Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers 3
Let each equal sides of given equilateral triangle be 2y2. We know that median is also perpendicular bisector.
∴ y2 + x2 = 4y2
⇒ x2 = 3y2
⇒ x = √3y
or
⇒ y = \(\frac{x}{\sqrt{3}}\)
Now, area of given triangle = \(\frac{1}{2}\)
× 2y × X = y × x = \(\frac{x}{\sqrt{3}}\)× x = \(\frac{x^{2}}{\sqrt{3}}\)

Heron’s Formula Class 9 Extra Questions Short Answer Type 2

Question 1.
Find the area of a triangle whose sides are 11 m, 60 m and 61 m.
Solutioin:
Let a = 11 m, b = 60 m and c = 61 m :
Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers 4

Question 2.
Suman has a piece of land, which is in the shape of a rhombus. She wants her two sons to work on the land and produce different crops. She divides the land in two equal parts by drawing a diagonal. If its perimeter is 400 m and one of the diagonals is of length 120 m, how much area each of them will get for his crops ?
Solutioin:
Here, perimeter of the rhombus is 400 m.
∴ Side of the rhombus = \(\frac{400}{4}\) = 100 m
Let diagonal BD = 120 m and this diagonal divides the rhombus ABCD into two equal parts.
Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers 5
Hence, area of land allotted to two sons for their crops is 4800 m2 each.

Question 3.
The perimeter of a triangular field is 144 m and its sides are in the ratio 3:4:5. Find the length of the perpendicular from the opposite vertex to the side whose length is 60 m.
Solutioin:
Let the sides of the triangle be 3x, 4x and 5x
∴ The perimeter of the triangular field = 144 m
⇒ 3x + 4x + 5x = 144
⇒ 12x = 144
Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers 6

Question 4.
Find the area of the triangle whose perimeter is 180 cm and two of its sides are of lengths 80 cm and 18 cm. Also, calculate the altitude of the triangle corresponding to the shortest side.
Solutioin:
Perimeter of given triangle = 180 cm
Two sides are 18 cm and 80 cm
∴ Third side = 180 – 18 – 80 = 82 cm
Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers 7
Hence, area of triangle is 720 cm2 and altitude of the triangle corresponding to the shortest side is 80 cm.

CBSE Class 10 Maths Formulas for Chapter 4 Quadratic Equations are mentioned in this article.

Heron’s Formula Class 9 Extra Questions Long Answer Type

Question 1.
Calculate the area of the shaded region.
Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers 8
Solutioin:
Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers 9
= 2 × 2 × 3 × 7 = 84 cm2
Area of shaded region = Area of ∆ABC – Area of ∆AOB
= 84 cm2 – 30 cm2 = 54 cm2

Question 2.
The sides of a triangular park are 8 m, 10 m and 6 m respectively. A small circular area of diameter 2 m is to be left out and the remaining area is to be used for growing roses. How much area is used for growing roses ? (use n = 3.14)
Solutioin:
The sides of the triangular park are 8 m, 10 m and 6 m.
Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers 10
Radius of the circle = \(\frac{2}{2}\) = 1 m
Area of the circle = πr2 = 3.14 × 1 × 1 = 3.14 m2
∴ Area to be used for growing roses = Area of the park – area of the circle
= 24 – 3.14 = 20.86 m2

Heron’s Formula Class 9 Extra Questions HOTS

Question 1.
OPQR is a rhombus, whose three vertices P, Q and R lie on the circle with centre 0. If the radius of the circle is 12 cm, find the area of the rhombus.
Solutioin:
Since diagonals bisect each other at 90°.
∴ In right ∆QLR, (LR)2 + (LQ)2 = (QR)2
Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers 11

Question 2.
How much paper of each shade is needed to make a kite given in the figure, in which ABCD is a square with diagonal 60 cm?
Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers 12
Solutioin:
Since diagonals of a square are of equal length and bisect each other at right angles, therefore,
Area of ∆AOD = \(\frac{1}{2}\) × 30 × 30 = 450 cm2
Area of ∆AOD = Area of ∆DOC = Area of ∆BOC
= Area of ∆AOB = 450 cm2
[∵ ∆AOD = ∆AOB ≅ ∆BOC ≅ ∆COD, ∵ they2 have equal area]
Now, area of ACEF (by Heron’s formula)
Here a = 20 cm, b = 20 cm and c = 30 cm

Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers 13

Now, area of orange shaded paper in kite
= Area of ∆AOD + Area of ∆CEF
= 450 cm2 + 198.4 cm2
= 648.4 cm2
Area of blue shaded paper in kite
= Area of ∆AOB + Area of ∆COD
= 450 cm2 + 450 cm2 = 900 cm2
Area of black shaded paper in kite = Area of ∆BOC = 450 cm2.

Heron’s Formula Class 9 Extra Questions Value Based (VBQs)

Question 1.
Sister Nivedita has trapezium shaped plot which she divided into three triangular portion for different purposes. I – for providing free education for orphan children, II – for providing dispensary for the needy villagers and III – for the library for villagers. Find the area of trapezium plot given in the figure. Which qualities of sister Nivedita are being depicted in question ?
Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers 14
Solutioin:
Here, ABCD is the trapezium with AB || DC.
Through C, Draw CF ⊥ AB
For – I: For area of ∆EBC
Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers 15
But, we know, Area of ∆EBC = \(\frac{1}{2}\)(base × height)
⇒ \(\frac{1}{2}\)1 × 30 × CF = 336
15 × CF = 336
⇒ CF = \(\frac{336}{15}\) = 22.4 m
Now, area of trapezium shaped plot = \(\frac{1}{2}\) (20 + 50)(22.4)
= 35 x 22.4 = 784 m2
Caring, kind, social, generous and visionary lady.

Question 2.
In an exhibition, an umbrella is made by stiching 10 triangular pieces of cloth with same message written on two triangular pieces. If each piece of cloth measures 60 cm, 60 cm and 20 cm, find how much cloth is required for each message.
Why we should respect women, educate women, save women and empower women ?
Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers 16
Solutioin:
Here, each triangular piece is an isosceles triangle with sides 60 cm, 60 cm and 20 cm.
Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers 17
Now, there are 2 triangular pieces with same message.
∴ Total area of cloth for each message = 2 × 591.61 = 1183.22 cm2
We should respect women, educate women, save women and empower women to serve humanity and give them equal opportunity so far they deprived.

Filed Under: CBSE Class 9

Primary Sidebar

  • Maths NCERT Solutions
  • Science NCERT Solutions
  • Social Science NCERT Solutions
  • English NCERT Solutions
  • Hindi NCERT Solutions
  • Physics NCERT Solutions
  • Chemistry NCERT Solutions
  • Biology NCERT Solutions
RS Aggarwal Solutions RD Sharma Solutions
RS Aggarwal Class 10 RD Sharma Class 10
RS Aggarwal Class 9 RD Sharma Class 9
RS Aggarwal Class 8 RD Sharma Class 8
RS Aggarwal Class 7 RD Sharma Class 11
RS Aggarwal Class 6 RD Sharma Class 12

Recent Posts

  • Unscramble Words With Letters | Make Words with These Letters
  • Fractions and Decimals Class 7 Notes Maths Chapter 2
  • MCQ Questions for Class 7 Maths Chapter 2 Fractions and Decimals with Answers
  • MCQ Questions for Class 6 Maths Chapter 8 Decimals with Answers
  • MCQ Questions for Class 6 Maths Chapter 1 Knowing Our Numbers with Answers
  • RD Sharma Class 10 Solutions Chapter 12 Heights and Distances Ex 12.1
  • MCQ Questions for Class 10 Science Chapter 10 Light Reflection and Refraction with Answers
  • NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields
  • MCQ Questions for Class 6 Maths Chapter 3 Playing with Numbers with Answers
  • MCQ Questions for Class 10 Maths Chapter 1 Real Numbers with Answers
DMCA.com Protection Status

Footer

CBSE Library
NCERT Library
NCERT Solutions for Class 12
NCERT Solutions for Class 11
NCERT Solutions for Class 10
NCERT Solutions for Class 9
NCERT Solutions for Class 8
NCERT Solutions for Class 7
NCERT Solutions for Class 6
ML Aggarwal Class 10 ICSE Solutions
Concise Mathematics Class 10 ICSE Solutions
CBSE Sample Papers
cbse
ncert
English Summaries
English Grammar
Biology Topics
Microbiology Topics
Chemistry Topics
Like us on Facebook Follow us on Twitter
Watch Youtube Videos Follow us on Google Plus
Follow us on Pinterest Follow us on Tumblr
Percentage Calculator

Copyright © 2022 Learn Insta