Advertisements
Advertisements
Question
ABCD is a trapezium. Prove that:
CD + DA + AB > BC.
Advertisements
Solution
In ΔACD, we have
CD + DA > CA
⇒ CD + DA + AB > CA +AB
⇒ CD + DA + AB > BC. ...[∴ AB + AC > BC]
APPEARS IN
RELATED QUESTIONS
In a huge park people are concentrated at three points (see the given figure):
A: where there are different slides and swings for children,
B: near which a man-made lake is situated,
C: which is near to a large parking and exit.
Where should an ice-cream parlour be set up so that maximum number of persons can approach it?
(Hint: The parlor should be equidistant from A, B and C)
Complete the hexagonal and star shaped rangolies (see the given figures) by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?
D is a point in side BC of triangle ABC. If AD > AC, show that AB > AC.
In the following figure, write BC, AC, and CD in ascending order of their lengths.
Name the greatest and the smallest sides in the following triangles:
ΔDEF, ∠D = 32°, ∠E = 56° and ∠F = 92°.
Name the smallest angle in each of these triangles:
In ΔXYZ, XY = 6.2cm, XY = 6.8cm and YZ = 5cm
In ΔABC, the exterior ∠PBC > exterior ∠QCB. Prove that AB > AC.
In ΔABC, BC produced to D, such that, AC = CD; ∠BAD = 125° and ∠ACD = 105°. Show that BC > CD.
ΔABC in a isosceles triangle with AB = AC. D is a point on BC produced. ED intersects AB at E and AC at F. Prove that AF > AE.
In ΔABC, AE is the bisector of ∠BAC. D is a point on AC such that AB = AD. Prove that BE = DE and ∠ABD > ∠C.
