Advertisements
Advertisements
प्रश्न
Prove that the perimeter of a triangle is greater than the sum of its altitudes.
Advertisements
उत्तर
We have to prove that the perimeter of a triangle is greater than the sum of its altitude.
In ΔABC
AD⊥ BC , BE ⊥ AC , CF⊥AB

We have to prove
AB + BC + CD > AD + BE + CF
Since AD⊥ BC
So AB > AD and AC > AD
By adding AB + AC > AD + AD, we have
AB + AC > 2AD ........(1)
Now consider BE ⊥ AC then
BC > BE, and BA > BE
Now by adding BC + BA > 2BE .......(2)
Again consider CF⊥AB
AC > CF, and BC > CF
By adding AC + BC > 2CF ...........(3)
Adding (1), (2) and (3), we get
2(AB + BC + CA)>2 (AD + BE + CF)
⇒ AB + BC + CA > AD + BE + CF
Hence the perimeter of a triangle is greater than the sum of all its altitude.
APPEARS IN
संबंधित प्रश्न
Which congruence criterion do you use in the following?
Given: AC = DF
AB = DE
BC = EF
So, ΔABC ≅ ΔDEF

In ΔABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ΔPQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°
A student says that ΔABC ≅ ΔPQR by AAA congruence criterion. Is he justified? Why or why not?
In the given figure, prove that:
CD + DA + AB + BC > 2AC

In triangles ABC and CDE, if AC = CE, BC = CD, ∠A = 60°, ∠C = 30° and ∠D = 90°. Are two triangles congruent?
If the following pair of the triangle is congruent? state the condition of congruency :
In Δ ABC and Δ DEF, AB = DE, BC = EF and ∠ B = ∠ E.
A line segment AB is bisected at point P and through point P another line segment PQ, which is perpendicular to AB, is drawn. Show that: QA = QB.
From the given diagram, in which ABCD is a parallelogram, ABL is a line segment and E is mid-point of BC.
Prove that:
(i) ΔDCE ≅ ΔLBE
(ii) AB = BL.
(iii) AL = 2DC
From the given diagram, in which ABCD is a parallelogram, ABL is a line segment and E is mid-point of BC.
Prove that: AB = BL.
A point O is taken inside a rhombus ABCD such that its distance from the vertices B and D are equal. Show that AOC is a straight line.
AD and BC are equal perpendiculars to a line segment AB. If AD and BC are on different sides of AB prove that CD bisects AB.
