# Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13 cm.

Given base BC = 7 cm

B = 75°

And AB + BC = 13 cms.

Steps of construction:

i. Draw a base line BC of 7 cms.

ii. Construct B = 75°.

a. With B as centre and with any radius, draw another arc cutting the line BC.

b. With D as centre and with the same radius, draw an arc cutting the first arc (drawn in step b) at point E.

c. With E as centre and with the same radius, draw another arc cutting the first arc (drawn in step b) at point F.

d. With E and F as centers, and with a radius more than half the length of EF, draw two arcs intersecting at point G.

e. Join points B and G. The angle formed by GBC is 90°. i.e.

GBC = 90°.

f. Now the point H will be the point of intersection of the ray BG and the first arc (from step b). With points H & E as centers, with any radius more than half the length of HE, draw two arcs such that they meet at point I.

g. By joining point I and B, we get the ray BI which forms 75° with ray BD.

iii. With B as centre draw an arc with length 13 cms ( = AB + BC given), such that it intersects ray BI at J.

iv. Join CJ and we draw a perpendicular bisector for CJ.

a. By drawing arcs on both sides of the line CJ, with C and J as centers and with same lengths. These arcs intersect at K and L on either side of line CJ.

v. The perpendicular bisector for CJ will intersect the ray BJ at point A. Join AC.

Thus the formed triangle ABC is the required triangle.

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The text passage provides a detailed construction method for a triangle given specific parameters, and it concludes with a question about the usefulness of the solution provided.
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Given a text passage, determine:
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First, describe the content structure you found in 1-2 sentences.

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# Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13 cm.

Given base BC = 7 cm

B = 75°

And AB + BC = 13 cms.

Steps of construction:

i. Draw a base line BC of 7 cms.

ii. Construct B = 75°.

a. With B as centre and with any radius, draw another arc cutting the line BC.

b. With D as centre and with the same radius, draw an arc cutting the first arc (drawn in step b) at point E.

c. With E as centre and with the same radius, draw another arc cutting the first arc (drawn in step b) at point F.

d. With E and F as centers, and with a radius more than half the length of EF, draw two arcs intersecting at point G.

e. Join points B and G. The angle formed by GBC is 90°. i.e.

GBC = 90°.

f. Now the point H will be the point of intersection of the ray BG and the first arc (from step b). With points H & E as centers, with any radius more than half the length of HE, draw two arcs such that they meet at point I.

g. By joining point I and B, we get the ray BI which forms 75° with ray BD.

iii. With B as centre draw an arc with length 13 cms ( = AB + BC given), such that it intersects ray BI at J.

iv. Join CJ and we draw a perpendicular bisector for CJ.

a. By drawing arcs on both sides of the line CJ, with C and J as centers and with same lengths. These arcs intersect at K and L on either side of line CJ.

v. The perpendicular bisector for CJ will intersect the ray BJ at point A. Join AC.

Thus the formed triangle ABC is the required triangle.

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The text passage provides a detailed construction method for a triangle given specific parameters, and it concludes with a question about the usefulness of the solution provided.

TYPE: SINGLE_QUESTION_WITH_ANSWER