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Sia ? 4 years, 4 months ago
Given: A square ABCD.
To Prove : (i) AC = BD and (ii) Diagonals bisect each other at right angles.
Proof :
- In {tex}\triangle{/tex}ADB and {tex}\triangle{/tex}BCA, we have
AD = BC ...[As sides of a square are equal]
{tex}\angle{/tex}BAD = {tex}\angle{/tex}ABC ...[All interior angles are of 90o]
AB = BA ...[Common]
{tex}\triangle{/tex}ADB {tex}\cong{/tex} {tex}\triangle{/tex}BCA ...[By SAS rule]
AC = BD ...[c.p.c.t.] - Now in {tex}\triangle{/tex}AOB and {tex}\triangle{/tex}COD, we have
AB = CD ...[Sides of a square]
{tex}\angle{/tex}AOB = {tex}\angle{/tex}COD ...[Vertically opp. angles]
{tex}\angle{/tex}OBA = {tex}\angle{/tex}ODC ...[Alternate interior angles are equal]
{tex}\triangle{/tex}AOB {tex}\cong{/tex} {tex}\triangle{/tex}COD ...[By ASA rule]
OA = OC and OB = OD ...[c.p.c.t.] ...(1)
Now consider {tex}\triangle{/tex}s AOD and COD.
AD = CD ...[Sides of square]
OA = OC ...[As proved above]
OD = OD ...[Common]
{tex}\triangle{/tex}AOD {tex}\cong{/tex} {tex}\triangle{/tex}COD ...[By SSS rule]
{tex}\angle{/tex}AOD = {tex}\angle{/tex}COD ...[c.p.c.t.]
But {tex}\angle{/tex}AOD + {tex}\angle{/tex}COD = 180° ...[linear pair]
or {tex}\angle{/tex}AOD + {tex}\angle{/tex}AOD = 180° ...[As {tex}\angle{/tex}AOD = {tex}\angle{/tex}COD]
or 2{tex}\angle{/tex}AOD = 180° {tex}\therefore{/tex} {tex}\angle{/tex}AOD = 90° ...(2)
From equation (1) and (2) it is clear that diagonals of a square bisect each other at right angles.
Posted by Ajeet Singh 4 years, 4 months ago
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Sia ? 4 years, 4 months ago
To find the area of a rectangle, multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself to find the area. This is the same as saying length2 or length squared.
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Sia ? 4 years, 4 months ago
The expression is given to be : 2√3 + √3
We need to simplify this given expression and find the value of this expression.
The expression is : 2√3 + √3
Since, √3 is common factor in both the terms so taking √3 common we have,
⇒ √3 × (2 + 1)
Now using BODMAS rule we will simplify the bracket first
⇒ √3 × 3
⇒ 3√3
The approximate value of √3 correct to three places of decimal is 1.732
So, the value of the given expression is : 3 × 1.732 = 5.196
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Maneesh R.K 4 years, 4 months ago
0Thank You