Advertisements
Advertisements
प्रश्न
In the given figure, prove that:
CD + DA + AB + BC > 2AC

Advertisements
उत्तर
We have to prove that CD + DA + AB + BC > 2AC

In ΔABC we have
AB + BC > AC (As sum of two sides of triangle is greater than third one) ........(1)
In ΔACDwe have
AD + CD > AC (As sum of two sides of triangle is greater than third one) .........(2)
Hence
Adding (1) & (2) we get AB + BC + AC + CD > 2AC Proved.
APPEARS IN
संबंधित प्रश्न
In the figure, the two triangles are congruent.
The corresponding parts are marked. We can write ΔRAT ≅ ?

If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

In triangles ABC and CDE, if AC = CE, BC = CD, ∠A = 60°, ∠C = 30° and ∠D = 90°. Are two triangles congruent?
ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.
Use the information in the given figure to prove:
- AB = FE
- BD = CF

In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD. Prove that:
AB = CE.
In the given figure: AB//FD, AC//GE and BD = CE;
prove that:
- BG = DF
- CF = EG

In ∆ABC, AB = AC. Show that the altitude AD is median also.
In the following figure, ABC is an equilateral triangle in which QP is parallel to AC. Side AC is produced up to point R so that CR = BP.
Prove that QR bisects PC.
Hint: ( Show that ∆ QBP is equilateral
⇒ BP = PQ, but BP = CR
⇒ PQ = CR ⇒ ∆ QPM ≅ ∆ RCM ).
ABC is an isosceles triangle with AB = AC and BD and CE are its two medians. Show that BD = CE.
