In $Δ A B C, A D ⊥ B C; B E ⊥ A C; C F ⊥ A B .$ then which of the following option is correct:

A D - B E + C F < A B + B C - C A

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A D + B E - C F < A B - B C - C A

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A D + B E - C F < A B - B C + C A

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A D + B E + C F < A B + B C + C A

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Solution

The correct option is C $A D + B E + C F < A B + B C + C A$
In triangle ADB,
$A D + B D > A B$
In triangle ADC,
$A D + D C > A C$
In tringle BEC,
$B E + E C > B C$
In tringle BEA,
$B E + A E > A B$
In traingle, CFA,
$C F + F A > A C$
In traingle, CFB,
$C F + F B > B C$
Add all the inequalities,
$2 (A D + B E + C F) + A B + B C + C A > 2 (A B + B C + C A)$
$2 (A D + B E + C F) > A B + B C + C A$
or $A D + B E + C F < A B + B C + C A$

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Similar questions

Q. AD, BE and CF are the medians in ΔABC. If AD + BE + CF = a and AB + BC + CA = b, then which of the following is always correct?

Q. In

Δ A B C

, D, E and F are the mid-points of BC, CA and AB respectively

\frac{(A D + B E + C F)}{(A B + B C + C A)}

is ,

Q. If

A D, B E a n d C F

be medians of a

Δ A B C

,then

2 (A D + B E + C F) < k (A B + B C + C A) < 4 (A D + B E + C F)

.Find k

Fill in the correct entries with respect to the properties asked. AD, BE and CF are perpendicular to the sides BC, CA and AB respectively.

Let 2 s = AB + BC + CA.
Area of $△$ ABC = $\frac{1}{2} \times___\times A B$
Area of $△$ ABC = $\frac{1}{2} \times B C \times___$
Area of $△$ ABC = $\sqrt{___(s - A B) (s - C A) (s - B C)}$

In a

△ A B C

, medians

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C F

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G

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4 (A D + B E + C F) > 3 (A B + B C + C A)

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In ΔABC,AD⊥BC;BE⊥AC;CF⊥AB. then which of the following option is correct:

The correct option is C AD+BE+CF<AB+BC+CAIn triangle ADB,AD+BD>ABIn triangle ADC,AD+DC>ACIn tringle BEC,BE+EC>BCIn tringle BEA,BE+AE>ABIn traingle, CFA,CF+FA>ACIn traingle, CFB,CF+FB>BCAdd all the inequalities,2(AD+BE+CF)+AB+BC+CA>2(AB+BC+CA)2(AD+BE+CF)>AB+BC+CAor AD+BE+CF<AB+BC+CA

In $Δ A B C, A D ⊥ B C; B E ⊥ A C; C F ⊥ A B .$ then which of the following option is correct: