Question

10. Area of a right triangular park is 420 m² whose altitude exceeds the base by 11 m. Using the above information, answer the questions given below: (i) Length of base of triangle is​

Answer 1

Answer :

• 76.36 m

Given :

• Area of Right triangular park = 420 m²
• Length of Altitude = 11 m

To find :

• the length of the base.

Solution :

⇒ Let's find the length of the base.

$\sf \longrightarrow Area \: of \: Triangle = \dfrac{1}{2} \times b \times h$

$\sf \longrightarrow 420= \dfrac{1}{2} \times b \times 11$

$\sf \longrightarrow Base = \dfrac{Area \times 2}{Height}$

$\sf \longrightarrow Base = \dfrac{420 \times 2}{11}$

$\sf \longrightarrow Base = \dfrac{840}{11}$

$\sf \longrightarrow Base = 76.36$

$\sf \therefore Base = 76.36 \: m$

• Therefore, the base of the right triangular park is 76.36 m.

Answer 2

Given : area of a  right triangular park is 420m² whose altitude exceeds base by 11 m length  of base of triangle

To Find : Length of base of triangle

Solution:

Base = x   m

altitude = x + 11  m

Area = (1/2) * base * altitude

= (1/2) * x (x + 11)

(1/2) * x (x + 11)  = 420

=> x² + 11x = 840

=>  x² + 11x - 840  = 0

=> x² + 35x - 24x - 840 =0

=> (x + 35)(x - 24) = 0

=> x = 24

Base = 24 m

Altitude = 35 m

Length of base of triangle = 24  m

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