Question

Is it possible to construct ∆PQR where PQ = PR = 12 cm and QR = 24 cm? ​

Answer 1

Step-by-step explanation:

Answer

In △PQR

(PQ)

2

+(QR)

2

=(PR)

2

(24)

2

+(7)

2

=(PR)

2

⇒(PR)

2

=576+49=625

⇒PR=25

Now S is the mid point of PR

∴RS=

2

PR

=

2

25

=12.5cm

Answer 2

Answer:

No, it is not possible to construct a triangle.

Step-by-step explanation:

• In context to the given question; we have to whether it is possible to construct a triangle for the given condition;
• Given,

PQ = PR = 12 cm

QR = 24 cm

Explanation;-

As;

PR + PQ = 12 + 12 = 24 = QR

BY thus , we can conclude that ,

P is the mid point of line QR; or else the condition is not satisfied

Now, if P lies on line QR, the 3 given points are collinear

hence, It is not possible to construct a triangle.