Question

An isosceles triangle pqr, in which pq= pr= 8 cm, is inscribed in a circle of radius 10 cm , find the area of the triangle

Answer 1

The area of the triangle is $16 \sqrt 3 cm^2$

Step-by-step explanation:

Given : An isosceles triangle PQR, in which PQ=PR= 8 cm, is inscribed in a circle of radius 10 cm.

To find : The area of the triangle ?

Solution :

Isosceles ΔPQR  inscribed in the circle with center O.

Refer the attached figure below.

PQ = PR=8 cm (Equal Sides of Isosceles Triangles)

PO = OQ = OR (Radius of Circle)

In ΔPOQ and ΔPOR,

PQ = PR (Given)

OR = QO (Radius of Circle)

PO = PO (Common)

By SSS both the Triangles are Congruent.

Now, In ΔPOQ and ΔQOR,

PO = QO (Radius)

QO = OR (Radius)

So, PQ = QR (If two sides are equal, then third side is also equal

By SSS both the triangles are Congruent.  

Similarly ΔPOR and ΔQOR are also congruent.

i.e.  PQ = QR = RP (Corresponding Parts of Congruent Triangles)  is an equilateral triangle.

Area of Equilateral Triangle is given by,

$A=\frac {\sqrt 3}{4} \times s^2$

Where, s=8

$A= \frac {\sqrt 3}{4} \times 8^2$

$A=16 \sqrt 3 cm^2$

Therefore, the area of the triangle is $16 \sqrt 3 cm^2$

#Learn more

An isosceles triangle pqr, in which pq= pr= 8 cm, is inscribed in a circle of radius 10 cm , find the area of the triangle

https://brainly.in/question/13721932

Answer 2

Answer:

Step-by-step explanation:

PQ = PR=8 cm ( equal )

PO = OQ = OR (Radius of Circle)

In ΔPOQ and ΔPOR,

PQ = PR (Given)

OR = QO (Radius of Circle)

PO = PO (Common)

by SSS both the triangles are congruent.

Now, In ΔPOQ and ΔQOR,

PO = QO (Radius)

QO = OR (Radius)

So, PQ = QR

Similarly ΔPOR and ΔQOR are also congruent.

i.e.  PQ = QR = RP (Corresponding Parts of Congruent Triangles)  is an equilateral triangle.