Question

If I ask you to construct $△PQR$ ~ $△ABC$ exactly (when we say exactly, we mean the exact relative positions of the triangles) as given in the figure, (Assuming I give you the dimensions of $△ABC$ and the Scale Factor for $△PQR$) what additional information would you ask for?

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• A. Dimensions of PQR.
• B. Perpendicular distance between AC and Point P and angle between AC and PR.
• C. We cannot construct the triangle because there is no connection between two triangles.
• D. The information given is sufficient.

Answer 1

The correct option is B

Perpendicular distance between AC and Point P and angle between AC and PR.

Consider we need to draw a triangle $△PQR$ ~ $△ABC$. But $△PQR$ is at an isolated position w.r.t. $△ABC$. Therefore we will need to know the perpendicular distance between AC and point P and the angle between AC and PR.

Now let us reframe the question with the required information.

Construct a triangle $△PQR$ similar to$△ABC$, such that point P is at a perpendicular distance 5 cm from the line AC, PR makes an angle of ${60}^{\circ }$ and which is $\frac{2}{3}$rd of $△ABC$. In triangle $△ABC$, AB = 6 cm, AC = 7 cm, and $\angle ABC$ = ${30}^{\circ }$

Steps of construction:

1. Draw a straight line AC = 7 cm.

2. Measure an angle of ${30}^{\circ }$ w.r.t AC and draw a straight line

3. Set the compass at 6 cm length and with A as centre mark the length on the line, name the point of intersection as B.

Reason: This is done to draw the line AB. i.e. we are marking the length of 6 cm on the line ${30}^{\circ }$ inclined to AC

4. Connect the points to BC to complete the triangle.

5. Draw a line perpendicular to AC at A.

Reason: We are drawing this step to identify the line on which the point P might lie.

6. For the given length of AP, cut the perpendicular at P Reason: We are drawing the arc to identify the exact position of point P on the line perpendicular to AC

7. Extrapolate line segment AC. Reason:

8. Draw a Ray XY ${60}^{\circ }$ to AC, such that the line passes through the point P.

9. Measure the distance AC with a compass and draw an arc with point P as centre and radius equal to AC, the point of intersection is R'.

10. Measure z.BAC and draw a similar angle on PR'.

11. Measure AB and mark on PZ as PQ. Join QR'.

12. Draw a ray PM, mark 3 points $P_1{P}_2{P}_3$ such that $P{P}_1=P_1{P}_2=P_2{P}_3$

. 13. Join $R^{\prime }{P}_3$

14. Through $P_2$ draw a line parallel to $P_3{R}^{\prime }$, let the point of intersection of the line with PR' is R.

15. Through the point R, draw a line parallel to R'Q'. Let the point of intersection of the line with line PO! is Q. The triangle $△PQR$ which is similar to &$△ABC$ such that, the angle between PR and AC is ${120}^{\circ }$ and perpendicular distance of point P from AC is 5 cm.