construct a rectangle ABCD SUCH AC =5CM AND ANGLE BAC =30°
Answers
Solution: To construct a rectangle ABCD in which AC=5 cm, ∠BAC=30°
In Rectangle ABCD, ∠A=∠B=∠C=∠D=90°,
In Δ ABC, Right angled at B,
→∠BAC +∠B +∠ACB=180°[angle sum property of triangle]
→30° +90° +∠ACB=180°
→∠ACB=180°-120°
→∠ACB=60°
∠ACD=∠CAB=30° and ∠ACB=∠CAD=60°[In a rectangle opposite sides are parallel, AC is diagonal, so alternate angles are equal.]
Steps of Construction
1. Draw diagonal AC =5 cm
2.At A, make an angle of 30° in upper direction and 60° in lower direction.
3.Similarly at Point C ,make an angle of 60° in upper direction and 30° in lower direction.
4. Draw rays from A and C such that the point where the two rays meet above the line segment is point B and point below the line segment where the two rays meet is point D.
5.Join A, B, C,D.This is required rectangle ABCD.
Answer: Solution: To construct a rectangle ABCD in which AC=5 cm, ∠BAC=30°
In Rectangle ABCD, ∠A=∠B=∠C=∠D=90°,
In Δ ABC, Right angled at B,
→∠BAC +∠B +∠ACB=180°[angle sum property of triangle]
→30° +90° +∠ACB=180°
→∠ACB=180°-120°
→∠ACB=60°
∠ACD=∠CAB=30° and ∠ACB=∠CAD=60°[In a rectangle opposite sides are parallel, AC is diagonal, so alternate angles are equal.]
Steps of Construction
1. Draw diagonal AC =5 cm
2.At A, make an angle of 30° in upper direction and 60° in lower direction.
3.Similarly at Point C ,make an angle of 60° in upper direction and 30° in lower direction.
4. Draw rays from A and C such that the point where the two rays meet above the line segment is point B and point below the line segment where the two rays meet is point D.
5.Join A, B, C,D.This is required rectangle ABCD