In a figure PQR is a right angled triangle. x and y are the midpoints of PQ and QR respectively. If XY=14 cm,then the hypotenuse PR is?
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Method 1.Using Pythagorus theorem.
QX² + QY² = XY² = 14²
[(1/2)QP]² + [(1/2)QR]² = 14²
QP² +QR² = 14²x2²
PR² = (14x2)²
PR = 14x2 = 28
Method 2. Using similarity of triangles
Triangles PRQ and XYQ ar similar.
Hence PR / QR = XY / QY
OR PR = (XY)*(QR/QY) = 14*(2QY/QY) = 28
QX² + QY² = XY² = 14²
[(1/2)QP]² + [(1/2)QR]² = 14²
QP² +QR² = 14²x2²
PR² = (14x2)²
PR = 14x2 = 28
Method 2. Using similarity of triangles
Triangles PRQ and XYQ ar similar.
Hence PR / QR = XY / QY
OR PR = (XY)*(QR/QY) = 14*(2QY/QY) = 28
Answered by
4
see figure.
A line joining the mid points is parallel to the third side. So the triangles PQR and XQY are similar.
XY / PX =PR/PQ
XY / PX = PR / 2 PX
PR = 2 XY = 28 cm
A line joining the mid points is parallel to the third side. So the triangles PQR and XQY are similar.
XY / PX =PR/PQ
XY / PX = PR / 2 PX
PR = 2 XY = 28 cm
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