**NCERT Solutions for Class 9 Maths Chapter 12 Heron’s
Formula Ex 12.2**

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s
Formula Ex 12.2 are the part
of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for
Chapter 12 Heron’s
Formula Ex 12.2.

**Ex 12.2 Class 9 Maths Question 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?

**Solution:****
**Given, a quadrilateral ABCD in
which ∠C = 90°, AB = 9 m, BC = 12 m, CD =
5 m and AD = 8 m.

Let us join B and D, such that ΔBCD is a right-angled triangle.

In right-angled ∆BCD, BD

^{2}= 50

^{2}+ CD

^{2 }(by Pythagoras theorem)

⇒ BD

^{2}= 12

^{2}+ 5

^{2}

⇒ BD

^{2}= 144 + 25 = 169

⇒ BD = 13 m

Now, for ∆ABD, we have

a = AB = 9 m, b = AD = 8 m, c = BD = 13 m

= 30 m^{2} + 35.5 m^{2}

= 65.5 m^{2} (approx.)

**Ex 12.2 Class 9 Maths 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:
**Given a quadrilateral ABCD with AB = 3 cm, BC = 4
cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

For
∆ABC, a = AB = 3 cm, b = BC = 4 cm and c = AC = 5 cm

Now, area of quadrilateral ABCD = area of ∆ABC + area of ∆ACD

= 6 cm

^{2}+ 9.2 cm

^{2}= 15.2 cm

^{2}(approx.)

**Ex 12.2 Class 9 Maths 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 surface I:**

It is an isosceles triangle whose sides are a = 5 cm, b = 5 cm, c = 1 cm

=
(0.75 × 3.3) cm^{2}

= 2.475 cm^{2} (approx.)

**For
surface II:**

It is a rectangle with length 6.5 cm and breadth
1 cm.

∴ Area of surface II = Length ×
Breadth

= (6.5 × 1) cm^{2} = 6.5 cm^{2}

**For surface III:**

It is a trapezium whose parallel sides are 1 cm
and 2 cm as shown in the figure given below:

**For
surface IV and V:**

Surface V is a right-angled triangle with base 6 cm and height 1.5 cm.

Also, area of surface IV = area of surface V

= ½ × base x height

= (½ × 6 × 1.5) cm^{2} = 4.5 cm^{2}

Thus, the total area of the paper used = (area of surface I) + (area of surface
II) + (area of surface III) + (area of surface IV) + (area of surface V) =
[2.475 + 6.5 + 1.3 + 4.5 + 4.5] cm^{2}

= 19.275 cm^{2}

= 19.3 cm^{2} (approx.)

**Ex 12.2 Class 9 Maths 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:
**For the given triangle, we have a = 28 cm, b = 30
cm, c = 26 cm

Area
of the given parallelogram = Area of the given triangle

∴ Area of the parallelogram = 336 cm^{2}

⇒ base × height = 336

⇒ 28 × h = 336, where ‘h’ be the height of the parallelogram.

⇒ h = 336/28 = 12
cm

Thus, the required height of the parallelogram is 12 cm.

**Ex 12.2 Class 9 Maths 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:
**Here, each side of the rhombus = 30 m.

Let ABCD be the given rhombus and the diagonal, BD = 48 m

Since,
a diagonal divides the rhombus into two congruent triangles.

∴ Area of triangle II = 432 m^{2}

Now, the total area of the rhombus = Area of triangle I + Area of triangle II

= 432 m^{2} + 432 m^{2}= 864 m^{2}

Area of grass for 18 cows to graze = 864 m^{2}

⇒ Area of grass for 1 cow to graze = 864/18 m^{2}

= 48 m^{2}

**Ex 12.2 Class 9 Maths 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?

**Solution:
**Let the sides of each triangular piece be a = 20 cm, b = 50 cm, c = 50 cm

**Ex 12.2 Class 9 Maths 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?

**Solution:**

Each shade of paper is divided into 3 triangles, i.e., I, II, III

**For
triangle I:**

ABCD is a square [Given]

∵ Diagonals of a square are equal and bisect each other.

∴ AC = BD = 32 cm

Height of ΔABD = OA = (½ × 32) cm = 16 cm

Area of triangle I = (½ × 32 × 16) cm^{2 }= 256 cm^{2}

**For triangle II:**

Since, diagonal of a square divides it into two congruent triangles.

So, area of triangle II = area of triangle I

∴ Area of triangle II = 256 cm^{2}

**For triangle III:**

The sides are given as a = 8 cm, b = 6 cm and c = 6 cm

Thus, the area of different shades are:

Area of shade I = 256 cm^{2}

Area of shade II = 256 cm^{2}

and area of shade III = 17.92 cm^{2}

**Ex 12.2 Class 9 Maths 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

^{2}.

**Solution:
**Let the sides of the triangle be a = 9 cm, b = 28
cm, c = 35 cm

Total area of all the 16 triangles = (16 × 88.2) cm

^{2}= 1411.2 cm

^{2}(approx.)

Cost of polishing the tiles = Rs. 0.5 per cm

^{2}∴ Cost of polishing all the tiles = Rs. (0.5 × 1411.2) = Rs. 705.60 (approx.)

**Ex 12.2 Class 9 Maths 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:
**The given field is in the shape of a trapezium ABCD
such that parallel sides are AB = 10 m and DC = 25 m

Non-parallel sides are AD = 13 m and BC = 14 m.

We draw BE || AD, such that BE = 13 m.

The given field is divided into two shapes (i) ∆BCE, (ii) parallelogram ABED

Sides of the triangle are a = 13 m, b = 14 m, c = 15 m

(ii)
For parallelogram ABED:

Let the height of the ∆BCE corresponding to the side EC be h m.

Area of a triangle = ½ × base × height

∴ ½ × 15 × h = 84

⇒ 7.5 × h = 84

⇒ h = 84/7.5 = 11.2 m^{2}

Now, area of a parallelogram = base × height

= (10 × 11.2) = 112
m^{2}

So, area of the field = area of ∆BCE + area of parallelogram ABED

= 84
m^{2} + 112 m^{2} = 196 m^{2}

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