Question

please solve the problem i will surely mark as brainliest
please solve 10 number ​

Answer 1

SOLUTION:-

Given:

PQR is an isosceles triangle with PQ=PR=25cm, QR= 14cm.

Construction:

Let x be the mid-point of QR, join PX.

Then PX QR (median of an isosceles triangle is perpendicular to the base)

=)Center of the circle O lies on PX(perpendicular bisector of a chord passes through its center).

=)QX= RX = QR/2 = 14/2cm = 7cm

In right ∆PXR,

(PX)² + (XR)²= (PR)²

=) (PX)² = (PR)² - (XR)²

=) (PX)² = (25cm)² - (7cm)²

=) (PX)² = (625 - 49)cm²

=) (PX)² = 576cm²

=) (PX)² = (24cm)²

=) PX= 24cm

Now,

Required radius=OP=OR=x cm (Assume)

=) OX = PX - OP

=) (24-x)cm

In right ∆OXR,

(OX)² + (XR)²= (OR)²

=) (24-x)² + 7² = x²

=) 24² +x² -2×24× x +49 = x²

=) 576 + x² -48x + 49- x² = 0

=) 48x= 625

=) x= 625/48

=) x= 13.02

Hence,

Required radius is 13.02cm

Hope it helps ☺️