Advertisements
Advertisements
प्रश्न
In the given figure, T is the midpoint of QR. Side PR of ΔPQR is extended to S such that R divides PS in the ratio 2:1. TV and WR are drawn parallel to PQ. Prove that T divides SU in the ratio 2:1 and WR = `(1)/(4)"PQ"`.
Advertisements
उत्तर
In ΔPQR,
T is the mid-point of QR and VT || PQ
So, V is the mid-point of PR.
Since R divides PR in the ratio 2 : 1 and PV = VR,
so, PV = PR = RS
Since R is the mid-point of SV and RW || VT,
W is the mid-point of ST.
Since V is the mid-point of PR and VT || PQ,
T is the mid-point of UW.
So, UT = TW = SW
⇒ T divides SU in the ratio 2 : 1
Also,
R and W are the midpoints SV and TS respectively.
⇒ WR = `(1)/(2)"VT"`
V and T are the mid-points of PR and UW respectively.
⇒ VT = `(1)/(2)"PQ"`
So, WR = `(1)/(2)(1/2 "PQ")`
⇒ WR = `(1)/(4)"PQ"`.
APPEARS IN
संबंधित प्रश्न
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
- D is the mid-point of AC
- MD ⊥ AC
- CM = MA = `1/2AB`
In a triangle ∠ABC, ∠A = 50°, ∠B = 60° and ∠C = 70°. Find the measures of the angles of
the triangle formed by joining the mid-points of the sides of this triangle.
ABCD is a kite having AB = AD and BC = CD. Prove that the figure formed by joining the
mid-points of the sides, in order, is a rectangle.
In the given figure, `square`PQRS and `square`MNRL are rectangles. If point M is the midpoint of side PR then prove that,
- SL = LR
- LN = `1/2`SQ
In the given figure, ΔABC is an equilateral traingle. Points F, D and E are midpoints of side AB, side BC, side AC respectively. Show that ΔFED is an equilateral traingle.
If the quadrilateral formed by joining the mid-points of the adjacent sides of quadrilateral ABCD is a rectangle,
show that the diagonals AC and BD intersect at the right angle.
In a parallelogram ABCD, M is the mid-point AC. X and Y are the points on AB and DC respectively such that AX = CY. Prove that:
(i) Triangle AXM is congruent to triangle CYM, and
(ii) XMY is a straight line.
In ΔABC, P is the mid-point of BC. A line through P and parallel to CA meets AB at point Q, and a line through Q and parallel to BC meets median AP at point R. Prove that: AP = 2AR
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.
In ΔABC, the medians BE and CD are produced to the points P and Q respectively such that BE = EP and CD = DQ. Prove that: Q A and P are collinear.
