حاسبة مساحة الأشكال المستوية ومحيطها

احسب مساحة 15 شكلًا ومحيطها، مثل المستطيلات والمثلثات وشبه المنحرف والدوائر والأشكال البيضاوية والمضلعات المنتظمة والحلقات، مع رسم بمقياس.

آخر تحديث 28 سبتمبر 2026أمثلة تم التحقق منها: 16

المساحة

cm²

الحد الأقصى للمنازل العشرية: 4؛ إلى الأقرب، وعند التعادل بعيدًا عن الصفر

المحيط: 26cm
Diagonal: 9.434cm

The rectangle covers 40 cm² and its boundary is 26 cm long.

The rectangle, to scale

طريقة الحساب S

المساحة
$A = l \times w = 8 \times 5 = 40\,\text{cm}^{2}$
المحيط
$P = 2(l + w) = 2(8 + 5) = 26\,\text{cm}$
Diagonal
$d = \sqrt{l^2 + w^2} = \sqrt{8^2 + 5^2} = 9.433981\,\text{cm}$

حول حاسبة مساحة الأشكال المستوية ومحيطها

Choose one of 15 plane shapes and enter the dimensions that define it. The area and perimeter come back with the other lengths that shape has: diagonals, heights, arc and chord lengths, a regular polygon's apothem and circumradius, or a ring's width. A triangle from three sides uses Heron's formula, A = √(s(s − a)(s − b)(s − c)) with s half the perimeter, and a regular polygon uses A = ns²/(4 tan(180°/n)).

The default rectangle, 8 cm by 5 cm, has an area of 40 cm², a perimeter of 26 cm and a diagonal of 9.434 cm. Area sizes flooring, turf and paint; perimeter sizes fencing, edging and skirting board. The drawing is to scale, so a mistyped dimension shows up as the wrong shape.

Base and height do not fix a triangle's slanted sides, and a trapezoid needs its legs, so those two inputs give area only. An ellipse's perimeter has no closed form; Ramanujan's second approximation is used, accurate to 0.04% or better.

أمثلة محلولة

Rectangle 8 × 5 cm

Shape: Rectangle
الطول: 8 cm
العرض: 5 cm
وحدة النتائج: سنتيمتر (⁨cm⁩)

المساحة: 40 cm²
المحيط: 26 cm
Diagonal: 9.433981 cm

مصدر التحقق: ⁨8 × 5, 2(8 + 5); Python 3.8 math.sqrt(89)⁩

3-4-5 triangle by Heron's formula

Shape: Triangle (three sides)
Side a: 3 cm
Side b: 4 cm
Side c: 5 cm
وحدة النتائج: سنتيمتر (⁨cm⁩)

المساحة: 6 cm²
المحيط: 12 cm
الارتفاع: 4 cm

مصدر التحقق: ⁨Right triangle with legs 3 and 4: ½·3·4 = 6; height onto side 3 is the other leg, 4⁩

Regular hexagon, side 2 m

Shape: Regular polygon
Side length: 2 m
Number of sides: 6
وحدة النتائج: متر (⁨m⁩)

المساحة: 10.392305 m²
المحيط: 12 m
Apothem (inradius): 1.732051 m
Circumradius: 2 m

مصدر التحقق: ⁨Python 3.8 math: 6*4/(4*tan(pi/6)) = 6√3, 2/(2*tan(pi/6)) = √3; a hexagon's circumradius equals its side⁩

Square as a regular 4-gon (exact tan 45°)

Shape: Regular polygon
Side length: 3 cm
Number of sides: 4
وحدة النتائج: سنتيمتر (⁨cm⁩)

المساحة: 9 cm²
Apothem (inradius): 1.5 cm
Circumradius: 2.12132 cm

مصدر التحقق: ⁨Square of side 3: area 9, apothem 3/2; Python 3.8 math: 3/sqrt(2)⁩

الأسئلة

How do you find the area of a trapezoid?

Average the two parallel sides and multiply by the perpendicular height: A = (a + b)/2 × h. Bases of 10 cm and 4 cm with a height of 4 cm give 7 × 4 = 28 cm². British English calls this shape a trapezium; the formula is the same. The perimeter also needs the two slanted legs: legs of 5 cm each make it 24 cm.

How do you find the area of a triangle from three sides?

Use Heron's formula. Take the semi-perimeter s = (a + b + c)/2, then A = √(s(s − a)(s − b)(s − c)). Sides of 5, 6 and 7 cm give s = 9 and A = √(9 × 4 × 3 × 2) = √216 ≈ 14.697 cm². Three lengths only make a triangle if each is shorter than the other two combined, so 3, 4 and 8 have no area.

What is the area of a regular hexagon?

A = (3√3/2)s², about 2.598 times the side squared. A hexagon with 5 cm sides covers 64.952 cm², and one with 2 m sides covers 10.392 m². Any regular polygon with n sides of length s has A = ns²/(4 tan(180°/n)). A hexagon's circumradius equals its side, and its apothem is s√3/2 ≈ 0.866s.

Is there a formula for the perimeter of an ellipse?

Not an exact one in elementary functions: the exact perimeter is a complete elliptic integral. Ramanujan's 1914 approximation, P ≈ π(a + b)(1 + 3h/(10 + √(4 − 3h))) with h = (a − b)²/(a + b)², is off by at most 0.04%, reached only as the ellipse flattens into a line. For semi-axes of 10 and 6 it gives 51.0539977, within 1.2 × 10⁻⁹ of the exact value.

How do you convert square feet to square metres?

Multiply by 0.09290304, the square of the international foot (0.3048 m exactly, fixed in 1959). One square metre is about 10.764 ft². A 12 ft × 10 ft room is 120 ft², or 11.148 m². Area scales with the square of the length unit, so converting each side first and then multiplying gives the same result.

ما مدى دقة «⁨حاسبة مساحة الأشكال المستوية ومحيطها⁩»؟

تعتمد الدقة على مدخلاتك وافتراضات الطريقة. يستخدم الحساب العشري 50 رقمًا معنويًا، لكن التقديرات والأساليب العددية وبيانات المصدر قد تكون أقل دقة؛ تقريب القيم المعروضة لا يزيل هذه الحدود. أمثلة محلولة جرى التحقق منها بمصادر مستقلة: 16. مثلًا، يجري التحقق من «⁨Rectangle 8 × 5 cm⁩» بالرجوع إلى ⁨8 × 5, 2(8 + 5); Python 3.8 math.sqrt(89)⁩.

ما مصدر هذه الطريقة؟

Weisstein, E. W. “Heron's Formula”, “Regular Polygon”, “Ellipse” — MathWorld; Ramanujan, S. (1914) “Modular equations and approximations to π”, Quarterly Journal of Mathematics 45, 350–372 (perimeter approximation II); NIST Handbook 44, Appendix C — exact inch–centimetre relation (1 in = 2.54 cm).

حول هذه الحاسبة

\begin{gathered} A_\triangle = \sqrt{s(s-a)(s-b)(s-c)},\quad A_{n\text{-gon}} = \frac{n s^2}{4\tan(\pi/n)} \\ P_\text{ellipse} \approx \pi(a+b)\left(1 + \frac{3h}{10 + \sqrt{4-3h}}\right) \end{gathered}

المصادر

Weisstein, E. W. “Heron's Formula”, “Regular Polygon”, “Ellipse” — MathWorld
Ramanujan, S. (1914) “Modular equations and approximations to π”, Quarterly Journal of Mathematics 45, 350–372 (perimeter approximation II)
NIST Handbook 44, Appendix C — exact inch–centimetre relation (1 in = 2.54 cm)

تم التحقق بالرجوع إلى المصادر

تتضمن هذه الحاسبة أمثلة محلولة بإجابات من مصادر مستقلة، وعددها 16. تُشغّل ضمن مجموعة الاختبارات، ويمكنك تشغيلها هنا أيضًا.