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1. The length of three sides of a triangle are 18 cm, 24 cm and 30 cm. Using this information write
as many problems as you can related to Heron's formula.
2. The answer is 500 m². Frame as many questions as you can on the topic Heron's formula.
3. List several mathematical problems related to the topic Heron's formula that can arise in your
home.
4. A cubical box has each edge 10 cm and another cuboidal box has 12 cm length, 10 cm width and
8 cm height. Using this information write as many problems as you can related to the topic
Surface areas and Volumes.
5. The answer is * 100. Frame as many questions as you can on the topic Surface areas and
Volumes.
6. List several mathematical problems related to the topic Surface areas and Volumes that can
arise in your home.
7. The observations are 45,60,56,82 and 73. Using this information, write as many problems as you
can on the topic Statistics.
8. The answer is 200. Frame as many questions as you can on the topic Statistics.
9. List several mathematical problems related to the topic Statistics that can arise in your home.
10. A bag contains 10 red balls and 5 blue balls. Using this information write as many problems as
you can related to the topic Probability.
11. The answer is 1/12. Frame as many questions as you can on the topic Probability.
12. List several mathematical problems related to the topic Probability that can arise in your home.
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Answer:
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Answer:
According to the Heron's formula, Area (A) of the triangle having sides a,b,c units is
A=
s(s−a)(s−b)(s−c)
where
s=
2
a+b+c
For the given triangle,
a=18 cm
b=24 cm
c=30 cm
s=
2
18+24+30
=36
A=
36(36−18)(36−24)(36−30)
A=
36×18×12×6
A=
216×216
=216 cm
2
Smallest side =18 cm
Area of the triangle =
2
1
×base×altitude=216
2
1
×18× altitude=216
Altitude =
9
216
=24 cm
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Step-by-step explanation:
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?
Ans. For an equilateral triangle with side ‘a’, area
∴ Each side of the triangle = a cm
∴ a + a + a = 180 cm
⇒ 3a = 180 cm
Now, s = Semi–perimeter
∴ Area of a triangle
∴ Area of the given triangle
Thus, the area of the given triangle
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 Fig.). The advertisements yield an earning of Rs. 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?
Ans. The sides of the triangular wall are
a = 122 m, b = 120 m, c = 22 m
3. There is a slide in a park. One of its. side 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 6 m, find the area painted in colour.
Ans. ∴ Sides of the wall are 15 m, 11 m and 6 m.
∴ a = 15 m, b = 11 m, c = 6 m
4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter? is 42 cm.
Ans. Let the sides of the triangle be
a = 18 cm, b = 10 cm and c = ?
∴ Perimeter (2s) = 42 cm
5. Side of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm its area.
Ans. Perimeter of the triangle = 540 cm
⇒ Semi–perimeter of the triangle,
∴ The sides are in the ratio of 12 : 17 : 25.
∴ a = 12x cm, b = 17x cm, c = 25x cm
∴ 12x + 17x + 25x = 540
⇒ 54x = 540
∴ a = 12 × 10 = 120 cm
b = 17 × 10 = 170 cm
c = 25 × 10 = 250 cm
⇒ (s - a) = (270 - 120) cm = 150 cm
(s - b) = (270 - 170) cm = 100 cm
(s - c) = (270 - 250) cm = 20 cm
∴ Area of the triangle
6. An isosceles triangle has perimeter 30 cm and each of the equal side is 12 cm. Find the area of the triangle.
Ans. Equal sides of the triangle are 12 cm each.
Let the third side = x cm.
Since, perimeter = 30 cm
∴ 12 cm + 12 cm + x cm = 30 cm
⇒ x = 30 - 12 - 12 cm
⇒ x = 6 cm
Now, semi–perimeter
∴ Area of the triangle
Exercise 12.2 (Page 206) .
1. A park, in the shape of a quadrilateral ABCD, has �C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Ans. Let us join B and D, such that BCD is a right–angled triangle.
We have area of BCD
Now, to find the area of ABD, we need the length of BD.
∴ In right, BCD,
BD2 = BC2 = CD2 [Pythagoras theorem]
⇒ BD2 = 122 + 52
⇒ BD2 = 144 + 25 = 169 = 132
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.
Ans. For ΔABC,
3. Radha made a picture of an aeroplane with coloured paper as shown in Fig, Find the total area of the paper used.
Ans. Area of surface I
It is an isosceles triangle whose sides are
a = 5 cm, b = 5 cm, c = 1 cm
= 0.75 × 3.3 cm2 (approx.) = 2.475 cm2 (approx.)
Area of surface II
It is a rectangle with length = 6.5 cm and breadth 1 cm.
∴ Area of rectangle II = Length × Breadth = 6.5 × 1 = 6.5 cm2
Area of surface III
It is a trapezium whose parallel sides are 1 cm and 2 cm as shown in the adjoining figure. Its height is given by
Note: The perpendicular distance between the parallel sides is called the height of the trapezium.
Area of surface IV
It is a right triangle with base as 6 cm and height as 1.5 cm.
Area of surface V
∴ Right triangle V Right triangle IV
∴ Area of right triangle V = Area of right triangle IV = 4.5 cm2
Thus, the total area of the paper used
= (area I) + (area II) + (area III) + (area IV) + (area V)
= [2.475 cm2 (approx.)] + [6.5 cm2] + [1.3 cm2 (approx.)] + [4.5 cm2] + [4.5 cm2]
= 19.275 cm2 = 19.3 cm2 (approx.)
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.