Advertisements
Advertisements
Question
Construct a right-angled triangle in which: QP = QR and hypotenuse PR = 7 cm
Advertisements
Solution
In ΔPQR,
QP = QR ....(given)
⇒ ∠QPR = ∠QRP ....
Since hypotenuse PR = 7cm, ∠PQR = 90°
∴ ∠QPR + ∠QRP = 90°
⇒ ∠QPR = ∠QRP = 45°
Steps:
1. Draw PR = 7cm.
2. Draw a ray PT such as ∠RPT = 45° and ray RS such as ∠PRS = 45°
3. Ray RS and ray PT meets at Q.
Thus, PQR is the required triangle.
APPEARS IN
RELATED QUESTIONS
In ∆ STU, l(ST) = 7 cm, l(TU) = 4 cm, l(SU) = 5 cm
Construct a triangle of the measures given below.
Construct ∆ABC such that m∠A = 55°, m∠B = 60°, and l(AB) = 5.9 cm.
Construct ∆XYZ such that l(XY) = 3.7 cm, l(YZ) = 7.7 cm, l(XZ) = 6.3 cm.
Construct a triangle using the given data: BC = 6.0cm, ∠B = 60° and ∠C = 45°
Construct a triangle using the given data: DE = 5cm, ∠D = 75° and ∠E = 60°
Construct an isosceles triangle in which: XY = XZ, YZ = 5.5 cm and ∠X = 60°
Construct an equilateral triangle using the given data: Altitude PM = 3.6 cm
Construct a triangle using the following data: XY + YZ = 5.6 cm, XZ = 4.5 cm and ∠X = 45°
Construct a ΔRST with side ST = 5.4 cm, RST = 60° and the perpendicular from R on ST = 3.0 cm.
