Question
In a quadrilateral ABCD, show that
(AB + BC + CD + DA) <2 (BD + AC).
(AB + BC + CD + DA) <2 (BD + AC).
✓
Answer
Given: Quadrilateral ABCD To prove: (AB + BC + CD + DA) < 2(BD + AC). Proof: 
In $\triangle\text{AOB},$$\text{OA + OB > AB}...\text{(i)}$
In $\triangle\text{BOC},$$\text{OB + OC > BC}...(\text{ii})$
In $\triangle\text{COD},$$\text{OC + OD > CD}...(\text{iii})$
In $\triangle\text{AOD},$$\text{OD + OA >AD}...(\text{iv})$
Adding (i), (ii), (iii) and (iv) we get$2(\text{OA + OB + OC + OD})>(\text{AB + BC + CD + DA})$
$\Rightarrow2(\text{OB + OD + OA + OC})>(\text{AB + BC + CD + DA})$
$\Rightarrow2(\text{BD + AC})>(\text{AB + BC + CD + DA})$
In $\triangle\text{AOB},$$\text{OA + OB > AB}...\text{(i)}$In $\triangle\text{BOC},$$\text{OB + OC > BC}...(\text{ii})$
In $\triangle\text{COD},$$\text{OC + OD > CD}...(\text{iii})$
In $\triangle\text{AOD},$$\text{OD + OA >AD}...(\text{iv})$
Adding (i), (ii), (iii) and (iv) we get$2(\text{OA + OB + OC + OD})>(\text{AB + BC + CD + DA})$
$\Rightarrow2(\text{OB + OD + OA + OC})>(\text{AB + BC + CD + DA})$
$\Rightarrow2(\text{BD + AC})>(\text{AB + BC + CD + DA})$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If two straight lines intersect in such a way that one of the angles formed measures 90°, show that each of the remaining angles measures 90°.
In $\triangle\text{ABC},$ if $3\angle\text{A}=4\angle\text{B}=6\angle\text{C},$ calculate $\angle\text{A},\angle\text{B}$ and $\angle\text{C}.$
→In the given figure, prove that
- CD + DA + AB > BC
- CD + DA + AB + BC > 2AC.
Divide each of the following polynomials by synthetic division method and also by linear division method. Write the quotient and the remainder : $(2m^2 – 3m + 10) ÷ (m – 5)$
→Construct ∆XYZ such that XY = 6.7 cm, YZ = 5.8 cm, XZ = 6.9 cm. Construct its incircle.
In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that:
(i) AB + BC + CD + DA > 2AC
(ii) AB + BC + CD > DA
(iii) AB + BC + CD + DA > AC + BD
(i) AB + BC + CD + DA > 2AC
(ii) AB + BC + CD > DA
(iii) AB + BC + CD + DA > AC + BD
Which set of numbers could be the universal set for the sets given below?
i. A = set of multiples of $5$,
B = set of multiples of $7$,
C = set of multiples of $12ii$. P = set of integers which are multiples of $4$.
T = set of all even square numbers.
→i. A = set of multiples of $5$,
B = set of multiples of $7$,
C = set of multiples of $12ii$. P = set of integers which are multiples of $4$.
T = set of all even square numbers.
In the table given below, the information is given about roads. Using this draw sub-divided and percentage bar diagram (Approximate the percentages to the nearest integer)
In figure, compute the area of quadrilateral ABCD.
A = {1,2,3, 5, 7,9,11,13}
B = {1,2,4, 6, 8,12,13}
Verify the above rule for the given set A and set B.
B = {1,2,4, 6, 8,12,13}
Verify the above rule for the given set A and set B.