English
Maharashtra State BoardSSC (English Medium) 10th Standard

∆PQR ~ ∆LTR. In ∆PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct ∆PQR and ∆LTR, such that PQLTPQLT=34.

Advertisements
Advertisements

Question

∆PQR ~ ∆LTR. In ∆PQR, PQ = 4.2 cm, QR = 5.4 cm, PR = 4.8 cm. Construct ∆PQR and ∆LTR, such that `"PQ"/"LT" = 3/4`.

Geometric Constructions
Advertisements

Solution

Rough figure:

Steps of construction:

  1. Draw ΔPQR such that PQ = 4.2 cm, QR = 5.4 cm and PR = 4.8 cm.
  2. Draw a ray at point R, making a suitable angle with seg RQ.
  3. Take equal parts RR1, R1R2, R2R3, R3R4 on ray RX.
  4. Join the points Q and R3.
  5. Draw seg TR4 || seg QR3.
  6. Draw seg LT || seg PQ.

Construction:

shaalaa.com

Notes

  • For drawing ΔPRQ of given measures = 1 mark
  • For drawing acute angle at point R = 0.5 mark
  • To mark points R1, R2, R3, R4 on ray RX at equal distance from point R = 0.5 mark
  • To join seg R3Q and to draw parallel seg R4T to seg R3Q = 0.5 mark
  • To draw parallel seg to PQ at point T = 0.5 mark
  Is there an error in this question or solution?
Chapter 4: Geometric Constructions - Practice Set 4.1 [Page 96]

RELATED QUESTIONS

Construct a triangle ABC with BC = 7 cm, ∠B = 60° and AB = 6 cm. Construct another triangle whose sides are `3/4` times the corresponding sides of ∆ABC.


Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it whose sides are `2/3` of the corresponding sides of the first triangle. Give the justification of the construction.

 


Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose side are `1 1/2` times the corresponding sides of the isosceles triangle.

Give the justification of the construction


Construct an isosceles triangle with base 8 cm and altitude 4 cm. Construct another triangle whose sides are `2/3` times the corresponding sides of the isosceles triangle.


Draw a ΔABC in which BC = 6 cm, AB = 4 cm and AC = 5 cm. Draw a triangle similar to ΔABC with its sides equal to (3/4)th of the corresponding sides of ΔABC.


Find the co-ordinates of the points of trisection of the line segment AB with A(2, 7) and B(–4, –8).


Draw seg AB of length 9.7 cm. Take a point P on it such that A-P-B, AP = 3.5 cm. Construct a line MN ⊥ sag AB through point P.


Find the ratio in which the segment joining the points (1, –3) and (4, 5) is divided by the x-axis? Also, find the coordinates of this point on the x-axis.


ΔPQR ~ ΔABC, `(PR)/(AC) = 5/7`, then


Draw seg AB of length 9 cm and divide it in the ratio 3 : 2.


∆ABC ~ ∆PBQ. In ∆ABC, AB = 3 cm, ∠B = 90°, BC = 4 cm. Ratio of the corresponding sides of two triangles is 7 : 4. Then construct ∆ABC and ∆PBQ.


ΔAMT ~ ΔAHE. In ΔAMT, AM = 6.3 cm, ∠MAT = 120°, AT = 4.9 cm, `(AM)/(HA) = 7/5`, then construct ΔAMT and ΔAHE.


To construct a triangle similar to a given ΔABC with its sides `3/7` of the corresponding sides of ΔABC, first draw a ray BX such that ∠CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B1, B2, B3, ... on BX at equal distances and next step is to join ______.


To divide a line segment PQ in the ratio 5 : 7, first a ray PX is drawn so that ∠QPX is an acute angle and then at equal distances points are marked on the ray PX such that the minimum number of these points is ______.


If you need to construct a triangle with point P as one of its vertices, which is the angle that you need to construct a side of the triangle?


The image of construction of A’C’B a similar triangle of ΔACB is given below. Then choose the correct option.


A point C divides a line segment AB in the ratio 5 : 6. The ratio of lengths AB: BC is ______.


By geometrical construction, it is possible to divide a line segment in the ratio `sqrt(3) : 1/sqrt(3)`.


Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio 3:5.


Draw a line segment AB of length 6 cm and mark a point X on it such that AX = `4/5` AB. [Use a scale and compass]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×