Question

The sides of the trianglar ground are 22m, 120m and 122m. Find the area and cost of leveling the ground at the rate of rupees 20 per metre square.

Answer 1

Draw a triangle ABC in which AB = 22m

AC = 120m and BC= 122 m

We are not having the height so we will draw a cordinate AD is drawn as a perpendicular bisector of BC.

Now considering ADC as a right angled triangle in which AC = 120 cm

Now we have to find length of the base DC

DC = BC/2

= 122/2

= 61 cm

(H)² = (P)² + (B)²

(120)²= (P)² + (61)²

14400-3721= (P)²

10679= (P)²

√10679 = (P)²

103.3 = P

Area of triangle= ½ x base x height

= ½ x 61 x 103.3

= 3150.65 m²

cost of leveling 1 m²= ₹ 20

cost of leveling 3150.65 m²= ₹20 x 3150.65

=₹ 63013

Answer 2

$\underline\mathfrak{Given:-}$

• $\: \: \: \: \: the \: \: side \: \: of the \: \: triangular \: \: ground \: \: are \: \: {22} \: m, \: \: {120}\: m,$

$\underline\mathfrak{To \: \: Find:-}$

• $\: \: \: \: \: The \: \: area \: \: and \: \: cast \: \: of \: \: levelling \: \: the \: \: ground \: \: at \: \: the \: \: rate \: \: of \: \: rupees \: \: {20} \: per \: {m}^{2} \: .?$

$\huge\underline\mathfrak{Solutions:-}$

• $\: \: \: \: \: \: \: a \: \: = \: \: {22} \: m$

• $\: \: \: \: \: \: \: b \: \: = \: \: {120} \: m$

• $\: \: \: \: \: \: \: c \: \: = \: \: {122} \: m$

$\underline{By \: \: using \: \: Heroes's \: \: formula:-}$

$\: \: \: \: \: \fbox{\sqrt{s \: ( s \: - \: a) \: (s \: - \: b) \: (s \: - \: c)}}$

$\: \: \: \: \: \leadsto s \: \: = \: \: \frac{a \: + \: b \: + \: c}{2}$

$\: \: \: \: \: \leadsto s \: \: = \: \: \frac{22 \: + \: 120 \: + \: 122}{2}$

$\: \: \: \: \: \leadsto s \: \: = \: \: \frac{264}{2}$

$\: \: \: \: \: \leadsto s \: \: = \: \: {132} \: m$

$\: \: \: \: \: \underline{Now, \: \: Area \: \: of \: \: Triangle.}$

$\: \: \: \: \: \leadsto {\sqrt{s \: ( s \: - \: a) \: (s \: - \: b) \: (s \: - \: c)}}$

$\: \: \: \: \: \leadsto {\sqrt{132 \: ( 132 \: - \: 22) \: (132 \: - \: 120) \: (132 \: - \: 122)}}$

$\: \: \: \: \: \leadsto {\sqrt{132 \: ( 110) \: (12) \: (10)}}$

$\: \: \: \: \: \leadsto {\sqrt{132 \: \times \: 13200}}$

$\: \: \: \: \: \leadsto {\sqrt{1742400}}$

$\: \: \: \: \: \leadsto {1320}$

$\: \: \: \: \: \underline{Area \: \: of \: \: of \: \: triangle \: \: = \: \: {1320} \: {m}^{2}}$

• $\: \: \: \: \: {Now, \: \: cast \: \: of \: \: levelling \: \: {1} \: {m}^{2} = \: \: Rs \: {10}}$

$\: \: \: \: \: \leadsto {17.32051} \: {m}^{2} \: \: = \: \: {10} \: \times \: {17.32051}$

$\: \: \: \: \: \leadsto Rs \: {173.2051}$

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