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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
d1=((SaSd)+(SbSc)(SaSb)+(ScSd))d2
d1 - Diagonal 1 of Cyclic Quadrilateral?Sa - Side A of Cyclic Quadrilateral?Sd - Side D of Cyclic Quadrilateral?Sb - Side B of Cyclic Quadrilateral?Sc - 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.

11.2615Edit=((10Edit5Edit)+(9Edit8Edit)(10Edit9Edit)+(8Edit5Edit))12Edit
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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
d1=((SaSd)+(SbSc)(SaSb)+(ScSd))d2
Next Step Substitute values of Variables
d1=((10m5m)+(9m8m)(10m9m)+(8m5m))12m
Next Step Prepare to Evaluate
d1=((105)+(98)(109)+(85))12
Next Step Evaluate
d1=11.2615384615385m
LAST Step Rounding Answer
d1=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.
Side A of Cyclic Quadrilateral
Side A of Cyclic Quadrilateral is one of the four sides of the Cyclic Quadrilateral.
Symbol: Sa
Measurement: LengthUnit: m
Note: Value should be greater than 0.
Side D of Cyclic Quadrilateral
Side D of Cyclic Quadrilateral is one of the four sides of the Cyclic Quadrilateral.
Symbol: Sd
Measurement: LengthUnit: m
Note: Value should be greater than 0.
Side B of Cyclic Quadrilateral
Side B of Cyclic Quadrilateral is one of the four sides of the Cyclic Quadrilateral.
Symbol: Sb
Measurement: LengthUnit: m
Note: Value should be greater than 0.
Side C of Cyclic Quadrilateral
Side C of Cyclic Quadrilateral is one of the four sides of Cyclic Quadrilateral.
Symbol: Sc
Measurement: LengthUnit: m
Note: Value should be greater than 0.
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.

Credits

Creator Image
Created by Nayana Phulphagar LinkedIn Logo
Institute of Chartered and Financial Analysts of India National college (ICFAI National College), HUBLI
Nayana Phulphagar has created this Formula and 300+ more formulas!
Verifier Image
Verified by Nikhil Panchal LinkedIn Logo
Mumbai University (DJSCE), Mumbai
Nikhil Panchal has verified this Formula and 300+ more formulas!

Other Formulas to find Diagonal 1 of Cyclic Quadrilateral

​Go Diagonal 1 of Cyclic Quadrilateral
d1=((SaSc)+(SbSd))((SaSd)+(SbSc))(SaSb)+(ScSd)
​Go Diagonal 1 of Cyclic Quadrilateral using Ptolemy's Theorem
d1=(SaSc)+(SbSd)d2

Other formulas in Diagonals of Cyclic Quadrilateral category

​Go Diagonal 2 of Cyclic Quadrilateral
d2=((SaSb)+(ScSd))((SaSc)+(SbSd))(SaSd)+(ScSb)

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 (Sa), Side D of Cyclic Quadrilateral (Sd), Side B of Cyclic Quadrilateral (Sb), Side C of Cyclic Quadrilateral (Sc) & 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 (Sa), Side D of Cyclic Quadrilateral (Sd), Side B of Cyclic Quadrilateral (Sb), Side C of Cyclic Quadrilateral (Sc) & 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)))OpenImg
  • 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 QuadrilateralOpenImg
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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