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Draw a line segment \(QR = 6 \ cm\).

Draw a perpendicular bisector \(AB\) of line segment \(RS\).

Take \(13 \ cm\) as the radius, with \(Q\) as centre, draw an arc that cuts \(QY\) at \(S\), then draw a line from \(R\) to \(S\) to connect the two points.

Mark a point \(P\), where the perpendicular bisector intersects \(QS\), then join \(PR\).

Make \(\angle YQR = 45^\circ\).

Construct a triangle \(PQR\) in which \(QR = 6 \ cm\), \(\angle Q = 45^\circ\) and \(PQ + PR = 13 \ cm\).

**Step 1**:

**Step 2**:

**Step 3**:

**Step 4**:

**Step 5**:

**Thus**, \(PQR\)

**is the required triangle**.