Question

prove that in a right triangle ,the square of the hypotenuse is equal to sum of squares of the other sides .using the above result ,prove that ,in rhombus ABCD ,4ABsquare = ACsquare + BDsquare

Answer 1

In a rhombus ABCD, AB=BC=CD=CA
we know that diagnols of circles bisect each other perpendicularly
i.e, <AOB=<BOC=<COD=<AOD =90
OA=OC=AC/2 OB=OD =BD/2
IN ∆ AOB
AB^2 =(AC/2)^2 + (BD/2)^2
4AB^2=AC^2 + BD^2
AB^2 +BC^2+CD^2+DA^2=AC^2 +BD^2
Hence proved.

Answer 2

Pythagoras Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other sides.

Proof: Construct a right-angled triangle ABC, Draw a perpendicular from a A intersecting AC at D.

Given:

In ΔABC,

∠ABC = 90°

∠ADB = ∠CDB = 90°

To Prove:

AC² = AB² + BC²

Proof:

In ΔABC and ΔBDC,

∠ABC = ∠BDC = 90° (Given + construction)

∠ACB = ∠BCD (Common angle)

∴ ΔABC $\simeq$ ΔBDC by AA similarity.

We know that corresponding parts of similar triangles are proportional. Therefore,

$\Longrightarrow \ \sf \dfrac{AB}{BD} = \dfrac{BC}{DC} = \dfrac{AC}{BC}$

Cross multiplying BC/DC and AC/BC we get,

$\Longrightarrow \ \sf \dfrac{BC}{DC} = \dfrac{AC}{BC}$

$\sf \Longrightarrow BC \times BC = AC \times DC$

$\sf \Longrightarrow BC^2 = AC \times DC \longrightarrow \textcircled{\scriptsize1}$

In ΔABC and ΔADB,

∠ABC = ∠ADB = 90° (Given + construction)

∠BAC = ∠BAD (Common angle)

∴ ΔABC $\simeq$ ΔADB by AA similarity.

We know that corresponding parts of similar triangles are proportional, ∴

$\Longrightarrow \ \sf \dfrac{AB}{AD} = \dfrac{BC}{DB} = \dfrac{AC}{AB}$

Cross-multiplying AB/AD and AC/AB we get,

$\Longrightarrow \ \sf \dfrac{AB}{AD} = \dfrac{AC}{AB}$

$\sf \Longrightarrow AB \times AB = AC \times AD$

$\sf \Longrightarrow AB^2 = AC \times AD \longrightarrow \textcircled{\scriptsize2}$

Adding equations 1 and 2 we get,

BC² + AB² = (AC × DC) + (AC × AD)

AB² + BC² = AC (DC + AD)

AB² + BC² = AC (AC)

AB² + BC² = AC²

Hence Proved.

___________________________

Question: In rhombus ABCD, Prove that 4AB² = AC² + BD²

Given:

A Rhombus ABCD.

To Prove:

4AB² = AC² + BD²

Proof:

In a rhombus, diagonals bisect each other, therefore;

$\sf AO = \dfrac{AC}{2} \longrightarrow \textcircled{\scriptsize1}$

$\sf BO = \sf \dfrac{BD}{2} \longrightarrow \textcircled{\scriptsize2}$

In ΔAOB,

∠AOB = 90°

(Diagonals in a rhombus are perpendicular to each other)

∴ By using Pythagoras Theorem,

$\sf \Longrightarrow AB^2 = BO^2 + AO^2$

Substitute 1 and 2 above.

$\sf \Longrightarrow AB^2 = \left(\dfrac{BD}{2}\right)^2 + \left(\dfrac{AC}{2}\right)^2$

$\sf \Longrightarrow AB^2 = \dfrac{BD^2}{4} + \dfrac{AC^2}{4}$

$\sf \Longrightarrow AB^2 = \dfrac{BD^2 + AC^2}{4}$

$\sf \Longrightarrow 4AB^2 = BD^2 + AC^2$

Hence Proved.