Draw a triangle with sides 5 cm , 7 cm and 8 cm . Construct a triangle with sides of 4 5 of the corresponding sides of the first triangle.

Given: A triangle ABC , with sides 5 cm , 7 cm and 8 cm . Required: To construct a triangle similar to the triangle ABC with its sides equals to 4 5 of corresponding sides of the given triangle ABC . Steps of construction: Draw a given line segment AB = 5 cm . Draw an arc of length equals to 7 cm from point B . Draw another arc of length equals to 8 cm from point A . Consider the intersection point as C , to get the given triangle ABC . Draw a ray AX inclined to AB at an acute angle ( < 90 ° ) on the opposite side of angle C . Taking a suitable length of unit, mark 5 points P 1 , P 2 , P 3 , P 4 , P 5 on AX → , such that AP 1 = P 1 P 2 = P 2 P 3 = P 3 P 4 = P 4 P 5 = P 5 P 6 = P 6 P 7 = 1 unit Join P 5 B . From point P 4 , draw a line P 4 B ' parallel to P 5 B meeting AB produced at B ' . Draw C ' A ' parallel to CA meeting AC produced at C ' . ∴ Thus, ∆ AB ' C ' is the required triangle.

Draw a triangle ABC , in which AB = 6 cm , BC = 7 cm and ∠ B = 45 ° . Construct a triangle with sides of 5 8 of the corresponding sides of the triangle ABC .

Given: A triangle ABC , with sides AB = 6 cm , BC = 7 cm and ∠ B = 45 ° . Required: To construct a triangle similar to the triangle ABC with its sides equals to 5 8 of corresponding sides of the given triangle ABC . Steps of construction: Draw a given line segment AB = 6 cm . Draw a ray by constructing an angle ∠ B = 45 ° from point B . Draw an arc of length equals to BC = 7 cm from point B . Consider the intersection point as C and join AC , to get the given triangle ABC . Draw a ray AX inclined to AB at an acute angle ( < 90 ° ) on the opposite side of angle C . Taking a suitable length of unit, mark 8 points P 1 , P 2 , P 3 , P 4 , P 5 , P 6 , P 7 , P 8 on AX → , such that AP 1 = P 1 P 2 = P 2 P 3 = P 3 P 4 = P 4 P 5 = P 5 P 6 = P 6 P 7 = P 7 P 8 = 1 unit Join P 8 B . From point P 5 , draw a line P 5 B ' parallel to P 8 B meeting AB produced at B ' . Draw C ' A ' parallel to CA meeting AC produced at C ' . ∴ Thus, ∆ AB ' C ' is the required triangle.