Question

construct triangle PQR if PQ =5cm , angle P =60 QR = 7 cm . construct a similar triangle to it with 3/5 times​

Answer 1

Step-by-step explanation:

draw a line segment PQ 5cm.

take center as P and Mark 60 degree

take center as Q mark 7cm arc in 60 degree line and named it R.

Then join QR.

Answer 2

Given: ΔPQR has PQ=5cm, QR=7cm, ∠P=60°.

To construct: A triangle similar to ΔPQR with 3/5 times.

Step-by-step explanation:

Step 1 of 4

Draw a line PQ of length 5cm. Make an angle of 60° at P. Now put needle of compass on Q and mark an arc of 7cm. The point at which the lines through P and PQ intersect is point R.

Step 2 of 4

Now draw a line PX through P and mark 5 arcs of equal measure. Name the arcs as $P_{1} , P_{2} ,P_{3} ,P_{4} ,P_{5}$.

Step 3 of 4

Now join $P_{5}$ to Q. And draw a line through $P_{3}$ and parallel to $P_{5}Q$. Name the point as D.

Step 4 of 4

Through D draw a line parallel to QR. Name the point E. The ΔPDE is the required triangle.