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: EB = DB
AE = BC
∠A = ∠C = 90°
So, ΔABE ≅ ΔCDB

You want to show that ΔART ≅ ΔPEN,
If it is given that AT = PN and you are to use ASA criterion, you need to have
1) ?
2) ?

In the given figure, prove that:
CD + DA + AB > BC

In two triangles ABC and DEF, it is given that ∠A = ∠D, ∠B = ∠E and ∠C =∠F. Are the two triangles necessarily congruent?
If the following pair of the triangle is congruent? state the condition of congruency:
In ΔABC and ΔQRP, AB = QR, ∠B = ∠R and ∠C = P.
In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD.
Prove that :
(i) ΔABD and ΔECD are congruent.
(ii) AB = CE.
(iii) AB is parallel to EC
A triangle ABC has ∠B = ∠C.
Prove that: The perpendiculars from the mid-point of BC to AB and AC are equal.
In the following figure, AB = AC and AD is perpendicular to BC. BE bisects angle B and EF is perpendicular to AB.
Prove that : ED = EF

In the following figure, OA = OC and AB = BC.
Prove that: ΔAOD≅ ΔCOD
ABC is an isosceles triangle with AB = AC and BD and CE are its two medians. Show that BD = CE.
