Question

Construct A PQR such that QR= 7cm, LQ = 50°, PR-PQ = 3.5cm​

Answer 1

Step-by-step explanation:

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.

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Answer 2

Step-by-step explanation:

$\sf \red{{Step 1 : Construct \: a \: straight \: line \: QR = 6.5 cm.}} \\ \sf \green{{Step 2: At \: point \: Q, \: construct \: an \: angle \: of \: 60°} }\\ \sf \purple{{ Step 3: On \: the \: arm \: of \: this \: angle \: mark \: point \: X \: at \: a \: distance \: of \: 2.5 cm \: from \: Q }} \\ \sf \blue{{Step 4: Join \: X \: to \: R.}} \\ \sf \pink{{Step 5: Now \: the \: copy \: the \: angle \: formed \: at \: X \: on \: R.}} \\\sf \orange{{Step 6: Extend \: the \: arms \: of \: both \: the \: angles \: and \: mark \: their \: intersection \: as \: P.}}$