Key Points
Key Points: Construction of Similar Triangle
Case (i): When vertices are distinct
- Calculate sides using ratio → construct new triangle
Case (ii): When one vertex is common
- Divide side → draw a parallel line → similar triangle
Key Points: Construction of Tangent to a Circle at a Point
Using Centre:
-
Radius ⟂ tangent at the point of contact
Without Using Centre:
- Angle between tangent and chord = angle in the opposite arc
(Converse of tangent–secant angle theorem)
Key Points: Construction of Tangent from a Point Outside Circle
-
Angle in a semicircle is 90°
-
Tangents from an external point are equal
Important Questions [15]
- ΔSHR ~ ΔSVU. In ΔSHR, SH = 4.5 cm, HR = 5.2 cm, SR = 5.8 cm and SHSVSHSV=35. Construct ΔSVU.
- Draw ∠ABC of measure 105° and bisect it.
- Draw `Angleabc` of Measure 80° and Bisect It
- Draw ∠Abc of Measures 135°And Bisect It.
- Draw Seg Ab of Length 5.7 Cm and Bisect It.
- Draw ∠Abc of Measure 120° and Bisect It.
- Prove that “That Ratio of Areas of Two Similar Triangles is Equal to the Square of the Ratio of Their Corresponding Sides.”
- Construct the Circumcircle and Incircle of an Equilateral ∆Xyz with Side 6.5 Cm and Centre O. Find the Ratio of the Radii of Incircle and Circumcircle.
- ∆PQR ~ ∆LTR. In ∆PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct ∆PQR and ∆LTR, such that PQLTPQLT=34.
- Find the Co-ordinates of the Centroid of the δ Pqr, Whose Vertices Are P(3, –5), Q(4, 3) and R(11, –4)
- Write Down the Equation of a Line Whose Slope is 3/2 and Which Passes Through Point P, Where P Divides the Line Segment AB Joining A(-2, 6) and B(3, -4) in the Ratio 2 : 3.
- ΔRST ~ ΔUAY, In ΔRST, RS = 6 cm, ∠S = 50°, ST = 7.5 cm. The corresponding sides of ΔRST and ΔUAY are in the ratio 5 : 4. Construct ΔUAY.
- Construct the Circumcircle and Incircle of an Equilateral Triangle Abc with Side 6 Cm and Centre O. Find the Ratio of Radii of Circumcircle and Incircle.
- Draw Seg Ab of Length 9.7 Cm. Take a Point P on It Such that A-p-b, Ap = 3.5 Cm. Construct a Line Mn Sag Ab Through Point P.
- Δ Amt ∼ δAhe. in δ Amt, Ma = 6.3 Cm, ∠Mat = 120°, at = 4.9 Cm, M a H a = 7 5 . Construct δ Ahe.
