#### Class nine Chap 12 Heron Formula NCERT Solution

## Chapter 12: Heron’s Formula NCERT Solutions

#### TEXTBOOK’S EXERCISE – 12.2

**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?**

**Sol.
**ABCD is the park as shown in the figure.

Join BD.

In ΔDBC, we have

DB^{2}
= BC^{2} + CD^{2} [Pythagoras theorem]

⇒ DB^{2} = (12)^{2} + 5^{2}

⇒ DB = 13 m.

⇒ DB = 13 m.

Area of ΔDBC = 1/2 × base × height

= 1/2 × 12 × 5 m2 = 30 m^{2}

In ΔABD, *a *= 9 m, *b *= 8 m, *c *= 13 m

=
35.5 m^{2} (approx.)

∴ Area of the park = area of ΔDBC + area of ΔABD

=
(30 + 35.5) m^{2} = 65.5 m^{2}

**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. **

**Sol.
**In ΔABC, we have

AB^{2}
+ BC^{2} = 9 + 16 = 25 = AC^{2 }

Hence, ABC is a right triangle, right angled at B [By converse of Pythagoras theorem]

∴ Area of ΔABC = 1/2 × base × height = 1/2 × 3 × 4 cm2 = 6 cm2.

In ΔACD, *a *= 5 cm, *b *= 4 cm, *c *= 5 cm.

∴ S = a+b+c/2 = 5+4+5/2 cm = 7 cm

∴ Area of ΔACD

∴ Area of the quadrilateral = area of ΔABC + area of ΔACD = (6 + 9.2) cm2 = 15.2 cm^{2}

####

TEXTBOOK’S EXERCISE – 12.2

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

Sol. For the triangle marked I :

a = 5 cm, b = 5 cm, c = 1 cm

S= a+b+c/2 = 5+5+1/2 cm= 11/2cm = 5.5 cm

Area of the triangle

**For
the rectangle marked II : **

Length = 6.5 cm, Breadth = 1 cm

Area of the rectangle = 6.5 × 1 cm^{2} = 6.5 cm^{2 }

**For
the trapezium marked III : **

Draw AF || DC and AE ⊥ BC.

AD = FC = 1 cm, DC = AF = 1 cm

∴ BF = BC – FC = (2 – 1) cm = 1 cm

Hence, ΔABF is equilateral.

Also, E is the mid-point of BF.

∴ BE = 1/2 cm = 0.5 cm

Also, AB^{2} = AE^{2} + BE^{2} [Pythagoras theorem]

⇒ AE^{2} = 1^{2} – (0.5)^{2} = 0.75

⇒ AE = 0.9 cm (approx.)

Area of the trapezium = 1/2 (sum of the parallel sides) × distance between them.

= 1/2 × (BC + AD) × AE

= 1/2 × (2 + 1) × 0.9 cm^{2} = 1.4 cm^{2}.

**For the triangle marked IV : **

It is a right-triangle

∴ Area of the triangle = 1/2 × base × height

= 1/2 × 6 × 1.5 cm^{2} = 4.5 cm^{2}.

**For
the triangle marked V : **

This triangle is congruent to the triangle marked IV.

Hence, area of the triangle = 4.5 cm^{2}

Total area of the paper used = (2.5 + 6.5 + 1.4 + 4.5 + 4.5) cm^{2} = 19.4 cm^{2}