BC=7cm ANGLE B=75 DEGREE AND AB+AC=13cm
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i) Draw a horizontal line BD = 13 cm
ii) At B draw an angle DBX = 75 deg [Angle may be constructed with protractor or with compass]
iii) In BX cut off an arc of radius = 7 cm; let this point be C
iv) Join CD
v) Using compass draw a perpendicular bisector for CD, which meets BD at A.
vi) Join AC; ABC is the required triangle.
Reason:
Since A is a point on the perpendicular bisector of CD, by locus theorem it is equidistant from the end points C & D of the line segment CD. Hence AC = AD.
Thus BD = BA + AD = BA + AC = 13 cm
ii) At B draw an angle DBX = 75 deg [Angle may be constructed with protractor or with compass]
iii) In BX cut off an arc of radius = 7 cm; let this point be C
iv) Join CD
v) Using compass draw a perpendicular bisector for CD, which meets BD at A.
vi) Join AC; ABC is the required triangle.
Reason:
Since A is a point on the perpendicular bisector of CD, by locus theorem it is equidistant from the end points C & D of the line segment CD. Hence AC = AD.
Thus BD = BA + AD = BA + AC = 13 cm
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