Question

In the given figure if area of triangle Ras is equals to area of triangle RBS and area of triangle pqr b is equals to area of triangle PS then show that both the quadrilateral pqrs and where is ba
re

Answer 1

Answer:

where is the area dear ?

Answer 2

Right Triangles Proves PQSR as Trapezium

Step-by-step explanation:

R & S are extended to points J & K.

Given that area(∆RAS) = area(∆RBS) …(i)

Common base is RS

Let height of ∆RAS be and ∆RBS be as shown

area(∆RAS) = $\frac {1}{2} \times RS \times h_1$

area(∆RBS) = $\frac {1}{2} \times RS \times h_2$

As per condition,

$\frac {1}{2} \times RS \times h_1 = \frac {1}{2} \times RS \times h_2$

$h_1 = h_2$

As the distance between two lines is constant everywhere then lines are parallel

⇒ RS || AB …(*)

Therefore, ABSR is a trapezium

Given area(∆QRB) = area(∆PAS) …(ii)

area(∆QRB) = area(∆RBS) + area(∆QRS) …(iii)

area(∆PAS) = area(∆RAS) + area(∆RPS) …(iv)

subtract (iii) from (iv)

area(∆QRB) - area(∆PAS) = area(∆RBS) + area(∆QRS) - area(∆RAS) - area(∆RPS)

using (i) and (ii)

⇒ 0 = area(∆QRS) - area(∆RPS)

⇒ area(∆QRS) = area(∆RPS)

Common base for ∆QRS and ∆RPS is RS

Let height of ∆RPS be and ∆RQS be as in figure,

area(∆RPS) =   $\frac {1}{2} \times RS \times h_3$

area(∆RQS) =   $\frac {1}{2} \times RS \times h_4$

by given $\frac {1}{2} \times RS \times h3 = \frac {1}{2} \times RS \times h_4$

⇒ $h_3 = h_4$

As the distance between two lines is constant everywhere then lines are parallel

⇒ RS || PQ

⇒ PQSR is a trapezium