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प्रश्न
Construct any ΔABC. Construct ΔA'BC' such that AB : A'B = 5 : 3 and ΔABC ∼ ΔA'BC'.
उत्तर
Steps of construction:
- Draw ΔABC in any size.
- Extend the AB and CB rays.
- Segment AB should be divided into 5 equal sections.
- Measure one equal part (of segment AB) on the compass and draw three arcs on extended ray AB beginning at point B.
- A' represents the intersection of the third arc and the extended ray AB.
- Draw a line parallel to AC from A'.
- C' is the point of intersection of this line and the extended ray CB. As a result, ΔA'BC' is the needed triangle, which is comparable to ΔABC.
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संबंधित प्रश्न
Draw an equilateral ΔABC with side 6.2 cm and construct its circumcircle
Find the length of the altitude of an equilateral triangle with side 6 cm.
∆ABC ~ ∆LMN. In ∆ABC, AB = 5.5 cm, BC = 6 cm, CA = 4.5 cm. Construct ∆ABC and ∆LMN such that `"BC"/"MN" = 5/4`.
∆RST ~ ∆XYZ. In ∆RST, RS = 4.5 cm, ∠RST = 40°, ST = 5.7 cm Construct ∆RST and ∆XYZ, such that \[\frac{RS}{XY} = \frac{3}{5} .\]
Draw a triangle ABC with side BC = 6cm, ∠B = 45° and ∠A = 100°, then construct a triangle PBQ whose sides are `7/4` times the corresponding sides of ΔABC.
Choose the correct alternative:
ΔLMN ~ ΔHIJ and `"LM"/"HI" = 2/3` then
Draw ΔRSP ∼ ΔTQP. In ΔTQP, TP = 5 cm, ∠P = 50°, PQ = 4.5 cm and `("RS")/("TQ") = 2/3`.
