Q1. The area of an equilateral triangle whose side is ‘a’ cm is (√3/4)a^{2 }cm^{2}. Find its height.
Area of the triangle = (√3/4)a^{2} cm^{2}
∵ Area of a triangle = (1/2)x base x height∴ (1/2) x a x height =(√3/4)x a^{2}
⇒ height =(√3/4) a^{2} x (2/a) = (√3/2) a cm
Q2. Find the height of an equilateral triangle whose side is 2 cm.
Since height of an equilateral triangle is given by
height = (√3/2) x side
⇒ height =(√3/2) x 2 cm = 3 cm
Q3. Find the length of a diagonal of a square whose side is 2 cm.
The diagonal of a square = (√2) a cm
∴ Length of the diagonal = (√2) x 2 cm = 2√2cm.
Q4. The diagonal of a square is 9√2 cm. What is the side?
Let side of the square = x cm.
∵ Its diagonal is given by √2 x side.
∴ √2 x x= 9 x √2⇒
Thus, the required length of sides of the square is 9 cm.
Q5. The length of a rectangular plot of land is twice its breadth. If the perimeter of the plot be 180 metres, then find its area.
Let the breadth of the plot be ‘x’ metres.
∴ Its length = 2x metres
Since perimeter of a rectangular plot = 2[Length + Breadth]
∴ Perimeter of the given plot = 2[x + 2x]
= 2[3x]
= 6x metres
⇒ 6x = 180
⇒ x= 180/6= 30 metres
⇒ 2x = 2 x 30 = 60 metres.
∴ Length of the plot = 60 metres and Breadth of the plot = 30 meters.
∴ Area of the plot = Length x Breadth = 60 x 30 m^{2} = 1800 m^{2 }
Q6. The length of the sides containing the right angle in a right triangle differ by 7 cm. The area of the triangle is 60 cm^{2}. Find the length of the hypotenuse.
Let the sides containing the right angle be ‘x’ cm and (x – 7) cm.
i.e. Base = x cm and height = (x – 7) cm∴ Area = (1/2) x base x height=(1/2) x x x (x – 7) cm^{2}
Now (1/2) x (x – 7) = 60
⇒ x(x – 7) = 120
⇒ x^{2} – 7x – 120 = 0
⇒ x^{2} – 15x + 8x – 120 = 0
⇒ x(x – 15) + 8(x – 15) = 0
⇒ (x + 8)(x – 15) = 0
⇒ x = – 8 or x = 15
Rejecting x = –8, we have x – 15 = 0
⇒ x = 15 cm x – 7 = 15 – 7 = 8 cm
Now, Hypotenuse = √289 = 17cm
Thus, the required length of the hypotenuse is 17 cm.
Q7. The lengths of sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 120 cm, find its area.
The sides are in the ratio of 3 : 4 : 5.
Let the sides be 3x, 4x and 5x.
∴ Perimeter = 3x + 4x + 5x = 12x
Now 12x = 120 [Perimeter = 120 cm]
⇒ x =(120/12) = 10
∴ Lengths are: a = 3x = 3 x 10 = 30 cm
b = 4x = 4 x 10 = 40 cm
c = 5x = 5 x 10 = 50 cm
Now, semiperimeter (s) = (120/12)
cm = 60 cm
∵ (s – a) = 60 – 30 = 30 cm
(s – b) = 60 – 40 = 20 cm
(s – c) = 60 – 50 = 10 cmUsing Heron’s formula, we have
Area of the triangle =
= 2 x 30 x 10 cm^{2} = 600 cm^{2} Thus, the required area of the triangle = 600 cm^{2}.
Q8. Find the area of a quadrilateral ABCD in which AB = 8 cm, BC = 6 cm, CD = 8 cm, DA = 10 cm and AC = 10 cm.
In ΔABC, ∠B = 90°
∴ area of right (rt ΔABC) = (1/2) x 8 x 6 cm^{2} = 24 cm^{2}
In ΔACD,
a = AC = 10 cm b = AD = 10 cm c = CD = 8 cm∴Area of ΔACD
= 2 x 4√21 = 8√21 cm^{2}
= 8 x 4,58 cm^{2} = 36.64 cm^{2}
Now, area of quadrilateral ABCD = ar (ΔABC) + ar (ΔACD)
= 24 cm^{2} + 8√21 cm^{2 }
= 24 cm^{2} + 36.64 cm^{2}
= 60.64 cm^{2}
Q9. How much paper of each shade is needed to make a kite given in the figure, in which ABCD is a square of diagonal 44 cm.
∵ The diagonals of a square bisect each other at right angles
∴ OB = OD = OA = OC = (44/2) = 22 cm
Now, ar rt (Δ –I) = (1/2)× OB × OA
= (1/2) × 22 × 22 cm^{2} = 242 cm^{2}
Similarly ar rt (Δ –II) = arrt(Δ–III) = ar rt (Δ –IV) = 242 cm^{2}∵ Sides of ΔCEF are 20 cm, 20 cm and 14 cm
⇒ Area of ΔCEF
Now, area of yellow paper = ar (Δ – I) + ar (Δ – II)
= 242 cm^{2 }+ 242 cm^{2} = 484 cm^{2}
Area of red paper = ar (Δ – IV) = 242 cm^{2}
Area of green paper = ar (Δ – III) + ar ΔCEF
= 242 cm^{2} + 131.14 cm^{2}
= 373.14 cm^{2}
1. What is Heron's formula and how is it used to find the area of a triangle? 
2. How is Heron's formula derived? 
3. Can Heron's formula be used for all types of triangles? 
4. Are there any limitations or conditions for using Heron's formula? 
5. Can Heron's formula be used to find the area of a triangle given only the lengths of two sides and an angle? 
62 videos426 docs102 tests

62 videos426 docs102 tests
