Answer:
Isosceles Triangle
Third angle = 90° and pqr is an isosceles triangle. Now, In Δpqr the two sides are equal, so Δpqr is an isosceles triangle.
1.Draw base BC=7cm
2.Let's draw ∠B=75°.Let the ray be BX
3.Open the compass to length AB+AC=13cm.From point B as center, cut an arc on ray BX.Let the arc intersect BX at D
4.Join $$CD$
5.Now, we will draw perpendicular bisector of CD
6.Mark point A where perpendicular bisector intersects BD
7.Join AC
Result:△ABC is the required triangle.
Steps of construction :
1. Draw a ray OX and cut off a line segment QR = 3cm.
2. At Q, construction ∠ YQR = 45
∘
with the help of protractor .
3. On QY , cut off QS = 2cm.
4. Join RS.
5. Draw perpendicular bisector of RS to meet QY at P.
6. Join PR. Then PQR is the required triangle.
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Verified answer
Answer:
Isosceles Triangle
Third angle = 90° and pqr is an isosceles triangle. Now, In Δpqr the two sides are equal, so Δpqr is an isosceles triangle.
1.Draw base BC=7cm
2.Let's draw ∠B=75°.Let the ray be BX
3.Open the compass to length AB+AC=13cm.From point B as center, cut an arc on ray BX.Let the arc intersect BX at D
4.Join $$CD$
5.Now, we will draw perpendicular bisector of CD
6.Mark point A where perpendicular bisector intersects BD
7.Join AC
Result:△ABC is the required triangle.
Steps of construction :
1. Draw a ray OX and cut off a line segment QR = 3cm.
2. At Q, construction ∠ YQR = 45
∘
with the help of protractor .
3. On QY , cut off QS = 2cm.
4. Join RS.
5. Draw perpendicular bisector of RS to meet QY at P.
6. Join PR. Then PQR is the required triangle.