Q. 354.1( 37 Votes )

Draw a triangle A

Answer :

1. Construction of ABC


We know that, sum of angles of triangle is 180⁰.


A + B + C = 180⁰


105⁰+ 30⁰ + C = 180⁰


135⁰ + C = 180⁰


C = 45⁰


Step 1 : Draw seg BC of length 6.5 cm



Step 2 : Draw a vector CC’ at an angle of 45⁰ at C.



Step 3 : Draw a vector BB’ at an angle of 30⁰ at B.



Step 4 : Take point of intersection of BB’ and CC’ and label it as A.



2. Construction of ∆PQR ∆ABC


As sides of ∆PQR are times corresponding sides of ∆ABC.



Angles of similar triangles are equal.


Q = 30⁰


R = 45⁰


Step 1 : Draw seg QR of length 4.875 cm



Step 2 : Draw a vector RR’ at an angle of 45⁰ at R.



Step 3 : Draw a vector QQ’ at an angle of 30⁰ at Q.



Step 4 : Take point of intersection of QQ’ and RR’ and label it as P.



OR


Let us draw a rough figure,



We know that, tangent is perpendicular to the radius at the point of contact.


D = E = 90⁰


Angle between two tangents is 60⁰


C = 60⁰


Sum of all angles of a quadrilateral is 360⁰.


Therefore, for quadrilateral ADCE,


A + E + C + D = 360⁰


A + 90⁰ + 60⁰ + 90⁰ = 360⁰


A + 240⁰ = 360⁰


A = 120⁰


Construction –


Step 1 : Draw circle with centre A and radius 3 cm.



Step 2 : Draw radius AE.



Step 3 : Draw radius AD at an angle of 120⁰ from radius AE.



Step 4 : Draw vectors EE’ and DD’ at an angle of 90⁰ from respective radii.



Step 5 : Take point of intersection of DD’ and EE’ and label it as C.



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