हिंदी

In the Given Figure, Ps = 2rs. M is the Midpoint of Qr. If Tr || Mn || Qp, Then Prove That:St = 1 3 Ls

Advertisements
Advertisements

प्रश्न

In the given figure, PS = 3RS. M is the midpoint of QR. If TR || MN || QP, then prove that:

ST = `(1)/(3)"LS"`

योग
Advertisements

उत्तर

Proof :
In ΔPQR,
Since M is the mid-point of QR, and MN || QP, N is the mid-point of PR.
⇒ PN = PR
Given PS = 3RS
⇒ PS = RS = PN + NR + RS
But, PS = PN + NR + Rs
⇒ PN = PR = Rs
⇒R is the mid-point of SN
RT || MN
⇒ T is the mid-point of SM  ....(i)
Also, N is the mid-point of PR and MN || LP
⇒ M is the mid-point of LT   ....(ii)
So, from (i) and (ii),
LM = MT = ST

⇒ ST = `(1)/(3)"LS"`.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Midpoint and Intercept Theorems - Exercise 15.2

APPEARS IN

फ्रैंक Mathematics Part 1 [English] Class 9 ICSE
अध्याय 11 Midpoint and Intercept Theorems
Exercise 15.2 | Q 8.1

संबंधित प्रश्न

ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see the given figure). Show that F is the mid-point of BC.


Fill in the blank to make the following statement correct:

The triangle formed by joining the mid-points of the sides of a right triangle is            


In the Figure, `square`ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively. Then prove that, PQ || AB and PQ = `1/2 ("AB" + "DC")`.


The figure, given below, shows a trapezium ABCD. M and N are the mid-point of the non-parallel sides AD and BC respectively. Find: 

  1. MN, if AB = 11 cm and DC = 8 cm.
  2. AB, if DC = 20 cm and MN = 27 cm.
  3. DC, if MN = 15 cm and AB = 23 cm.

In triangle ABC, AD is the median and DE, drawn parallel to side BA, meets AC at point E.
Show that BE is also a median.


Use the following figure to find:
(i) BC, if AB = 7.2 cm.
(ii) GE, if FE = 4 cm.
(iii) AE, if BD = 4.1 cm
(iv) DF, if CG = 11 cm.


In the given figure, AD and CE are medians and DF // CE.
Prove that: FB = `1/4` AB.


ABCD is a parallelogram.E is the mid-point of CD and P is a point on AC such that PC = `(1)/(4)"AC"`. EP produced meets BC at F. Prove that: 2EF = BD.


In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD respectively. The line segments AF and BF meet the line segments DE and CE at points G and H respectively Prove that: EGFH is a parallelogram.


E is the mid-point of a median AD of ∆ABC and BE is produced to meet AC at F. Show that AF = `1/3` AC.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×