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. Rate this question :

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