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Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem Formula

GoDiagonal 1 of Cyclic Quadrilateral Formula

GoDiagonal 1 of Cyclic Quadrilateral using Ptolemy's Theorem Formula

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Diagonal 1 of Cyclic Quadrilateral is a line segment joining opposite vertices (A and C) of the Cyclic Quadrilateral. Check FAQs

d 1 = (\frac{(S a \cdot S d) + (S b \cdot S c)}{(S a \cdot S b) + (S c \cdot S d)}) \cdot d 2

d₁ - Diagonal 1 of Cyclic Quadrilateral?S_a - Side A of Cyclic Quadrilateral?S_d - Side D of Cyclic Quadrilateral?S_b - Side B of Cyclic Quadrilateral?S_c - Side C of Cyclic Quadrilateral?d₂ - Diagonal 2 of Cyclic Quadrilateral?

Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem Example

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Here is how the Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem equation looks like with Values.

Here is how the Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem equation looks like with Units.

Here is how the Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem equation looks like.

11.2615 = (\frac{(10 \cdot 5) + (9 \cdot 8)}{(10 \cdot 9) + (8 \cdot 5)}) \cdot 12

11.2615m = (\frac{(10m \cdot 5m) + (9m \cdot 8m)}{(10m \cdot 9m) + (8m \cdot 5m)}) \cdot 12m

11.2615 = (\frac{(10 \cdot 5) + (9 \cdot 8)}{(10 \cdot 9) + (8 \cdot 5)}) \cdot 12

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Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem Solution

Follow our step by step solution on how to calculate Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem?

FIRST Step Consider the formula

d 1 = (\frac{(S a \cdot S d) + (S b \cdot S c)}{(S a \cdot S b) + (S c \cdot S d)}) \cdot d 2

Next Step Substitute values of Variables

∴

d 1 = (\frac{(10m \cdot 5m) + (9m \cdot 8m)}{(10m \cdot 9m) + (8m \cdot 5m)}) \cdot 12m

Next Step Prepare to Evaluate

∴

d 1 = (\frac{(10 \cdot 5) + (9 \cdot 8)}{(10 \cdot 9) + (8 \cdot 5)}) \cdot 12

Next Step Evaluate

∴

d 1 = 11.2615384615385m

LAST Step Rounding Answer

∴

d 1 = 11.2615m

Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem Formula Elements

Variables

Diagonal 1 of Cyclic Quadrilateral

Diagonal 1 of Cyclic Quadrilateral is a line segment joining opposite vertices (A and C) of the Cyclic Quadrilateral.

Symbol: d₁

Measurement: LengthUnit: m

Note: Value should be greater than 0.

Formulas to find Diagonal 1 of Cyclic Quadrilateral

Side A of Cyclic Quadrilateral

Side A of Cyclic Quadrilateral is one of the four sides of the Cyclic Quadrilateral.

Symbol: S_a

Measurement: LengthUnit: m

Note: Value should be greater than 0.

Formulas to find Side A of Cyclic Quadrilateral

Side D of Cyclic Quadrilateral

Side D of Cyclic Quadrilateral is one of the four sides of the Cyclic Quadrilateral.

Symbol: S_d

Measurement: LengthUnit: m

Note: Value should be greater than 0.

Formulas to find Side D of Cyclic Quadrilateral

Side B of Cyclic Quadrilateral

Side B of Cyclic Quadrilateral is one of the four sides of the Cyclic Quadrilateral.

Symbol: S_b

Measurement: LengthUnit: m

Note: Value should be greater than 0.

Formulas to find Side B of Cyclic Quadrilateral

Side C of Cyclic Quadrilateral

Side C of Cyclic Quadrilateral is one of the four sides of Cyclic Quadrilateral.

Symbol: S_c

Measurement: LengthUnit: m

Note: Value should be greater than 0.

Formulas to find Side C of Cyclic Quadrilateral

Diagonal 2 of Cyclic Quadrilateral

Diagonal 2 of Cyclic Quadrilateral is a line segment joining opposite vertices (B and D) of the Cyclic Quadrilateral.

Symbol: d₂

Measurement: LengthUnit: m

Note: Value should be greater than 0.

Formulas to find Diagonal 2 of Cyclic Quadrilateral

Other Formulas to find Diagonal 1 of Cyclic Quadrilateral

Go Diagonal 1 of Cyclic Quadrilateral

d 1 = \sqrt{\frac{((S a \cdot S c) + (S b \cdot S d)) \cdot ((S a \cdot S d) + (S b \cdot S c))}{(S a \cdot S b) + (S c \cdot S d)}}

Go Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Theorem

d 1 = \frac{(S a \cdot S c) + (S b \cdot S d)}{d 2}

Other formulas in Diagonals of Cyclic Quadrilateral category

Go Diagonal 2 of Cyclic Quadrilateral

d 2 = \sqrt{\frac{((S a \cdot S b) + (S c \cdot S d)) \cdot ((S a \cdot S c) + (S b \cdot S d))}{(S a \cdot S d) + (S c \cdot S b)}}

How to Evaluate Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem?

Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem evaluator uses Diagonal 1 of Cyclic Quadrilateral = (((Side A of Cyclic Quadrilateral*Side D of Cyclic Quadrilateral)+(Side B of Cyclic Quadrilateral*Side C of Cyclic Quadrilateral))/((Side A of Cyclic Quadrilateral*Side B of Cyclic Quadrilateral)+(Side C of Cyclic Quadrilateral*Side D of Cyclic Quadrilateral)))*Diagonal 2 of Cyclic Quadrilateral to evaluate the Diagonal 1 of Cyclic Quadrilateral, The Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem formula is defined as the line segment joining opposite vertices (A and C) of the Cyclic Quadrilateral, calculated using Ptolemy's second theorem. Diagonal 1 of Cyclic Quadrilateral is denoted by d₁ symbol.

How to evaluate Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem using this online evaluator? To use this online evaluator for Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem, enter Side A of Cyclic Quadrilateral (S_a), Side D of Cyclic Quadrilateral (S_d), Side B of Cyclic Quadrilateral (S_b), Side C of Cyclic Quadrilateral (S_c) & Diagonal 2 of Cyclic Quadrilateral (d₂) and hit the calculate button.

FAQs on Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem

What is the formula to find Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem?

The formula of Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem is expressed as Diagonal 1 of Cyclic Quadrilateral = (((Side A of Cyclic Quadrilateral*Side D of Cyclic Quadrilateral)+(Side B of Cyclic Quadrilateral*Side C of Cyclic Quadrilateral))/((Side A of Cyclic Quadrilateral*Side B of Cyclic Quadrilateral)+(Side C of Cyclic Quadrilateral*Side D of Cyclic Quadrilateral)))*Diagonal 2 of Cyclic Quadrilateral. Here is an example- 11.26154 = (((10*5)+(9*8))/((10*9)+(8*5)))*12.

How to calculate Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem?

With Side A of Cyclic Quadrilateral (S_a), Side D of Cyclic Quadrilateral (S_d), Side B of Cyclic Quadrilateral (S_b), Side C of Cyclic Quadrilateral (S_c) & Diagonal 2 of Cyclic Quadrilateral (d₂) we can find Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem using the formula - Diagonal 1 of Cyclic Quadrilateral = (((Side A of Cyclic Quadrilateral*Side D of Cyclic Quadrilateral)+(Side B of Cyclic Quadrilateral*Side C of Cyclic Quadrilateral))/((Side A of Cyclic Quadrilateral*Side B of Cyclic Quadrilateral)+(Side C of Cyclic Quadrilateral*Side D of Cyclic Quadrilateral)))*Diagonal 2 of Cyclic Quadrilateral.

What are the other ways to Calculate Diagonal 1 of Cyclic Quadrilateral?

Here are the different ways to Calculate Diagonal 1 of Cyclic Quadrilateral-

Diagonal 1 of Cyclic Quadrilateral=sqrt((((Side A of Cyclic Quadrilateral*Side C of Cyclic Quadrilateral)+(Side B of Cyclic Quadrilateral*Side D of Cyclic Quadrilateral))*((Side A of Cyclic Quadrilateral*Side D of Cyclic Quadrilateral)+(Side B of Cyclic Quadrilateral*Side C of Cyclic Quadrilateral)))/((Side A of Cyclic Quadrilateral*Side B of Cyclic Quadrilateral)+(Side C of Cyclic Quadrilateral*Side D of Cyclic Quadrilateral)))
Diagonal 1 of Cyclic Quadrilateral=((Side A of Cyclic Quadrilateral*Side C of Cyclic Quadrilateral)+(Side B of Cyclic Quadrilateral*Side D of Cyclic Quadrilateral))/Diagonal 2 of Cyclic Quadrilateral

Can the Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem be negative?

No, the Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem, measured in Length cannot be negative.

Which unit is used to measure Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem?

Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem is usually measured using the Meter[m] for Length. Millimeter[m], Kilometer[m], Decimeter[m] are the few other units in which Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem can be measured.

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Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem Formula

​GoDiagonal 1 of Cyclic Quadrilateral Formula

​GoDiagonal 1 of Cyclic Quadrilateral using Ptolemy's Theorem Formula

Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem Example

Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem Solution

Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem Formula Elements

Other Formulas to find Diagonal 1 of Cyclic Quadrilateral

Other formulas in Diagonals of Cyclic Quadrilateral category

How to Evaluate Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem?

FAQs on Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Second Theorem

Variable Name

GoDiagonal 1 of Cyclic Quadrilateral Formula

GoDiagonal 1 of Cyclic Quadrilateral using Ptolemy's Theorem Formula