Herons Formula Test - 17

Let each side of an equilateral be x.

Then, perimeter of an equilateral triangle = 60 m

x + x + x = 60

3x = 60

x = \(\frac{60}{3}\) = 20 m

Area of an equilateral triangle = \({\sqrt3 \over 4} (side)^2\) = \({\sqrt3 \over 4} \times 20 \times 20\) = \(100 \sqrt3 m^2\)

Thus, the area of triangle is \(100 \sqrt3 m^2\).

Since, the three sides of triangle are a = 56 cm, b = 60 cm and c = 52 cm.

Then the semi-perimeter of triangle,

\(s = {{ a + b + c} \over 2}\) = \({56 + 60 + 52} \over 2\) = \(\frac{168}{2}\) = 84 cm

Area of a triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{84(84-56)(84-60)(84-52)}\)

= \(\sqrt{84 \times 28 \times 24 \times 32}\)

= \(\sqrt {4 \times 7 \times 3 \times 4 \times 7 \times 4 \times 2 \times 3 \times 4 \times 4 \times 2}\)

= \(\sqrt{(4)^5 \times (7)^2 \times (3)^2 \times (2)^2}\)

= \((4)^2 \times 2 \times 7 \times 3 \times 2\)

= 1344 \(cm^2\)

Hence, the area of triangle is 1344 \(cm^2\).

Given, side of an equilateral triangle is \(2\sqrt3\, cm\) .

Area of an equilateral triangle = \({\sqrt 3 \over 4} (side)^2\)

= \({\sqrt 3 \over 4} (2 \sqrt3)^2\)

= \({\sqrt 3 \over 4} \times 4 \times 3\)

= \(3 \sqrt3\)

= 3 x 1.732

= 5.196 \(cm^2\)

Hence, the area of an equilateral triangle is 5.196 \(cm^2\).

Area of equilateral Δ i.e., \(9 \sqrt3 = {\sqrt3 \over 4} (side)^2\)

\((side)^2 = {{9 \sqrt3 \times 4} \over \sqrt3}\) = 36

side = \(\sqrt{36}\) = 6 cm

Given, area of an equilateral triangle = \(16 \sqrt3\, cm^2\)

Area of an equilateral triangle = \({\sqrt3 \over 4} (side)^2\)

\({\sqrt3 \over 4} (side)^2\) = \(16 \sqrt3\)

\((side)^2 = 64\)

Side = 8 cm [taking positive square root because side is always positive]

Perimeter of an equilateral triangle = 3 x Side = 3 x 8 = 24 cm

Hence, the perimeter of an equilateral triangle is 24 cm.

Sides of the triangle are 35 cm, 54 cm and 61 cm

s = \({{35 + 54 + 61} \over 2} = 75 cm\)

Area of Δ = \(\sqrt {75 (75 - 35) (75 - 54) (75 - 61)}\)

= \(\sqrt{75 \times 40 \times 21 \times 14}\)

= \(\sqrt{5 \times 5 \times 3 \times 2 \times 2 \times 2 \times 5 \times 3 \times 7 \times 7 \times 2}\)

= 5 x 3 x 2 x 2 x \(7 \sqrt5\)

= \(420 \sqrt5\, cm^2\)

Now, longest altitude will be the perpendicular on the smallest side of the triangle from the opposite vertex.

Length of longest altitude = \(2( Area of \triangle) \over 35\)

= \(2 \times 420 \sqrt5 \over 35\)

= \(24 \sqrt5\, cm\)

Here, s = \({4 + 4 + 2} \over 2\) = 5 cm

Area of Δ = \(\sqrt{5 (5-2)(5-4)(5-4)}\)

= \(\sqrt{5 \times 3 \times 1 \times 1}\)

= \(\sqrt{15}\, cm^2\)

Here, 2s = 6 + 8 + 10 = 24

s = \(\frac{24}{2}\) = 12 cm

Area of Δs = \(\sqrt{12 (12 -6)(12 - 8)(12-10)}\)

= \(\sqrt {2 \times 6 \times 6 \times 4 \times 2}\)

= 2 x 6 x 2

= 24 \(cm^2\)

Cost of painting at the rate of 9 paise per \(cm^2\) = Rs.0.09

Cost of painting at the rate of 9 paise for 24 \(cm^2\) = Rs (24 x 0.09) = Rs. 2.16

Perimeter = 32 cm

Let one of the equal sides be 3x and other be 2x

3x + 3x + 2x = 32

8x = 32

x = 4

Sides of isosceles triangles are 12 cm, 12cm, 8 cm

S = \(\frac{32}{2}\) = 16 cm

Area = \(\sqrt{16 \times 4 \times 4 \times 8}\)

= \(32 \sqrt2\, cm^2\)

Let the length of the sides be 3x, 4x, 5x meters.

144 = 12x

x = 12

Sides of the triangular field are 36 m, 48 m, 60 m

Δ = \(\sqrt {72 (72 - 36) (72 - 48) (72 -60)}\)

= 864 sq. m

Given perimeter of a rhombus is 52 cm, each side of the rhombus = \(\frac{52}{4} = 13\, cm\)

Area of rhombus = \(2 \sqrt{25(25-24)(25-13)(25-13)}\)

= 120 sq. cm.

Area = \(\frac{1}{2} \times product\, of\, diagonals\)

120 = \(\frac{1}{2}\) x 24 x d

d = 10 cm

The other diagonal to 10 cm.

Perimeter of an equilateral triangle 60 m.

3a = 60 m

a = 20m

Area of an equilateral triangle = \({\sqrt3 \over 4} a^2\) = \(100 \sqrt3\, m^2\)

Here we have perimeter of the triangle = 32 cm

Let a = 8 cm and b = 11 cm

Third side, c = 32 − (8 + 11) = 13cm

S = \(\frac{32}{2}\) = 16 cm

Therefore, area of the triangle

= \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{16 \times 8 \times 5 \times 3}\, cm^2\)

= \(80 \sqrt{30}\, cm^2\)

= \(k \sqrt{30}\, cm^2\)

= \(k\sqrt{30} = 8\sqrt{30}\)

k = 8

We have, sides of triangle 11 cm, 15 cm and 16 cm.

S = \({11 + 15 + 16}\over 2\) = 21

Area of triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{21(21-11)(21-15)(21-16)}\)

= \(30 \sqrt7\, cm^2\)

Let altitude to the largest side be h cm.

\(\frac{1}{2} \times 16 \times h = 30 \sqrt7\)

\(8h = 30\sqrt7\)

h = \(15 \sqrt7\over 4\) cm

Let the other two equal sides of an isosceles triangle be a cm.

Then, S = \({{a + a + 24}\over 2} = (a + 12) cm\)

Area of triangle = \(192\, cm^2\)

\(\sqrt{s(s-a)(s-b)(s-c)}\) = 192

\(\sqrt{(a +12)(a+12-a)(a+12-a)(a+12-24)} = 192\)

\(144(a^2 - 144) = (192)^2\)

\(a^2 = 400\)

a = 20 cm

Perimeter = 20 cm + 20 cm + 24 cm = 64 cm

Given,

Area of equilateral triangle = \({\sqrt3 \over 4} a^2\)

Here side(a) = 2 cm

= \({\sqrt3 \over 4} 2^2\)

= \({\sqrt3 \over 4} \times 2 \times 2\)

= \(\sqrt3\)

= \(\sqrt3\, cm^2\)

Therefore area of an equilateral triangle whose side is 2 cm is \(\sqrt3\, cm^2\).

Given; perimeter = 180 cm

Hence, s = \(\frac{180}{2}\) = 90 cm

Side = \(\frac{180}{3}\) = 60 cm

Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{90 (90-60)^3}\)

= \(\sqrt{90 \times 30^3}\)

= \(\sqrt{30^4 \times 3}\)

= \(30^2 \sqrt3\)

= \(900 \sqrt3\)