| How do you find the area of triangle ABC given a=2, b=3, c=4? Trigonometry Triangles and Vectors Area of a Triangle Show 1 Answer Narad T. · [eden] Jul 23, 2017 The area is#=2.9u^2# Explanation:We apply Heron's formula #area =sqrt(s(s-a)(s-b)(s-c))# Where #s=(a+b+c)/2# Here, #a=2# #b=3# #c=4# Therefore, #s=(2+3+4)/2=9/2=4.5# So, # area = sqrt (4.5*(4.5-2) * (4.5-3) * (4.5-4) ) # #=sqrt8.3475# #=2.9u^2# Where#u#, represents the units in this case. Answer link Related questions
See all questions in Area of a Triangle Impact of this question5785 views around the world You can reuse this answer Geometry 1 AnswerJim G. Apr 17, 2018 #"area "=4" units"^2# Explanation:
Answer link Related questions
Impact of this question2091 views around the world You can reuse this answer In elementary geometry you learned that the area of a triangle is one-half the base times the height. We will now use that, combined with some trigonometry, to derive more formulas for the area when given various parts of the triangle. Case 1: Two sides and the included angle. Suppose that we have a triangle △ABC, in which A can be either acute, a right angle, or obtuse, as in Figures 1, 2 & 3. Assume that A, b, and c are known. In each case we draw an altitude of height h from the vertex at C to AB, so that the area (which we will denote by the letter K) is given by K = 1/2hc. But we see that h = b sin A in each of the triangles (since h = b and sin A = sin 90º = 1 in Figure 2, and h = b sin (180º − A) = b sin A in Figure 3). We thus get the following formula: $$\boxed{Area = K = \frac{1}{2} bc\;sin A}\;\;\;\;\; (1)$$ The above formula for the area of △ABC is in terms of the known parts A, b, and c. Similar arguments for the angles B and C give us: $$\boxed{Area = K = \frac{1}{2} ac\;sin B} \;\;\; (2)$$ $$Area = K = \frac{1}{2} ab \; sin C \;\;\; (3)$$ Notice that the height h does not appear explicitly in these formulas, although it is implicitly there. These formulas have the advantage of being in terms of parts of the triangle, without having to find h separately. Example 1 Find the area of the triangle △ABC given A = 33º, b = 5, and c = 7. Solution: Using formula, the area K is given by: $$\text{Area} = K = \frac{1}{2} bc sin A$$ $$= \frac{1}{2} (5)(7) sin 33^\circ$$ $$K = 9.53$$ Case 2: Three angles and any side. Suppose that we have a triangle △ABC in which one side, say, a, and all three angles are known. By the Law of Sines we know that $$ c = \frac{a sin C} {sin A} $$ , so substituting this into formula (2) we get: $$ \boxed{\text{Area} = K =\frac{a^2 sin B sin C} {2 sin A} }\;\;\;\;\; (4)$$ Similar arguments for the sides b and c give us: $$ \boxed{\text{Area} = K = \frac{b^2 sin A sin C} {2 sin B}} \;\;\; (5)$$ $$\text{Area} = K = \frac{c^2 sin A sin B}{2 sin C} \;\;\;\; (6)$$ Example 2 Find the area of the triangle △ABC given A = 115º, B = 25º, C = 40º, and a = 12. Solution: Using formula (2.26), the area K is given by: $$K =\frac{a^2 sin B sin C} {2 sin A}$$ $$= \frac{122 sin 25^\circ sin 40^\circ}{2 sin 115^\circ}$$ $$K = 21.58$$ Case 3: Three sides. Suppose that we have a triangle △ABC in which all three sides are known. Then Heron’s formula gives us the area: Heron's Formula: For a triangle △ABC with sides a, b, and c, let s = \(\frac{1}{2} (a+b+ c)\) (i.e. 2s = a+b+ c is the perimeter of the triangle). Then the area K of the triangle is $$Area = K = \sqrt{ s (s−a) (s−b) (s− c)} .\;\;\;\;\; (7)$$ To prove this, first remember that the area K is one-half the base times the height. Using c as the base and the altitude h as the height, as before in Figures 1, 2 & 3, we have K = 1/2hc. Squaring both sides gives us $$K^2 = \frac{1}{4} h^2 c^2\;\;\;\;\; (8)$$ In Figures 6 & 7, let D be the point where the altitude touches \(\overline{AB}\) (or its extension). By the Pythagorean Theorem, we see that \(h^2\) = \(b^2\) −\((AD)^2\). In Figure 6, we see that AD = b cos A. And in Figure 7 we see that AD = b cos (180º − A) = −bcos A. Hence, in either case we have \((AD)^2\) = \(b^2\) \((cos A)^2\), and so $$h^2 = b^2 −b^2 (cos A)^2 = b^2 (1−(cos A)^2) = b^2 (1+cos A) (1−cos A) .\;\;\;\;\; (9)$$ (Note that the above equation also holds when A = 90º since cos 90º = 0 and h = b). Thus, substituting equation (8) into equation (9), we have $$K^2 = \frac{1}{4} b^2 c^2 (1+cos A) (1−cos A) .\;\;\;\;\; (10)$$ By the Law of Cosines we know that $$1+cos A = 1+ \frac{b^2 + c^2 −a^2}{ 2bc} = \frac{2bc+b^2 + c^2 −a%2}{2bc}$$ $$= \frac{(b+ c)^2 −a^2}{ 2bc} = \frac{((b+ c)+a) ((b+ c)−a)}{2bc}$$ $$= \frac{(a+b+ c) (b+ c−a)}{ 2bc} ,$$ and similarly $$1−cos A = 1− \frac{b^2 + c^2 −a^2}{ 2bc} = \frac{2bc−b^2 − c^2 +a^2} {2bc}$$ $$= \frac{a^2 −(b− c)^2}{ 2bc} = \frac{(a−(b− c)) (a+(b− c))}{ 2bc}$$ Thus, substituting these expressions into equation (10), we have $$K^2 = \frac{1}{4} b^2 c^2 \frac{(a+b+ c) (b+ c−a)}{ 2bc} • \frac{(a−b+ c) (a+b− c)}{ 2bc}$$ $$= \frac{a+b+ c}{ 2} • \frac{b+ c−a}{ 2} • \frac{a−b+ c}{ 2} • \frac{a+b− c}{ 2} ,$$ and since we defined s = \(\frac{1}{ 2}\) (a+b+ c), we see that $$K^2 = s(s−a)(s−b)(s− c) ,$$ so upon taking square roots we get $$K = \sqrt{s(s−a)(s−b)(s− c)} .$$ Example 3 Find the area of the triangle △ABC given a = 5, b = 4, and c = 7. Solution: Using Heron’s formula with s =\(\frac{1}{2}\) (a+b+ c) = \(\frac{1}{2}\) (5+4+7) = 8, the area K is given by: $$K = \sqrt{s(s−a)(s−b)(s− c)}$$ $$= \sqrt{8(8−5)(8−4)(8−7)} = \sqrt{96} ⇒ K = 4\sqrt{6} ≈ 9.8 .$$ Heron’s formula is useful for theoretical purposes (e.g. in deriving other formulas). However, it is not well-suited for calculator use, exhibiting what is called numerical instability for “extreme” triangles, as in the following example. Example 4 Find the area of the triangle △ABC given a = 1000000, b = 999999.9999979, and c = 0.0000029. Solution: To use Heron’s formula, we need to calculate s = \(\frac{1}{2}\) (a+b+ c). Notice that the actual value of a+b+c is 2000000.0000008, which has 14 digits. Most calculators can store 12-14 digits internally (even if they display less), and hence may round off that value of a+b+c to 2000000. When we then divide that rounded value for a+b+ c by 2 to get s, some calculators (e.g. the TI-83 Plus) will give a rounded down value of 1000000. This is a problem because a = 1000000, and so we would get s− a = 0, causing Heron’s formula to give us an area of 0 for the triangle! And this is indeed the incorrect answer that the TI-83 Plus returns. Other calculators may give some other inaccurate answer, depending on how they store values internally. The actual area - accurate to 15 decimal places - is K = 0.99999999999895, i.e. it is basically 1. Here is another example that uses Heron's Formula a bit differently. Example 5: Two side lengths of a triangle are \(4\) and \(9\), and the area of the triangle is \(6\sqrt{5}). Find the length of the third side of the triangle. Solution: According to Heron's Formula: $$A = \sqrt{s(s - a)(s - b)(s - c)} \Rightarrow$$ $$6\sqrt{5} = \sqrt{s(s - 4)(s - 9)(s - c)}$$ Square both sides of the equation: $$180 = s(s - 4)(s - 9)(s - c)$$ We must solve for \(c\). However, the semiperimeter is defined as $$s = \frac{a + b + c}{2}$$ We already know \(a\) and \(b\): $$s = \frac{4 + 9 + c}{2} = \frac{13 + c}{2}$$ Therefore the equation becomes $$180 = \frac{13 + c}{2} \cdot (\frac{13 + c}{2} - 4) \cdot (\frac{13 + c}{2} - 9) \cdot (\frac{13 + c}{2} - c) \Rightarrow$$ $$180 = \frac{13 + c}{2} \cdot \frac{5 + c}{2} \cdot \frac{-5 + c}{2} \cdot \frac{13 - c}{2}$$ Clear out the fractions by multiplying both sides of the equation by \(16\): $$2880 = (13 + c)(5 + c)(-5 + c)(13 - c)$$ Rather than solve this equation algebraically, just plug it into a graphing calculator. A side length must be positive, and the positive solutions are \(c = 7\) and \(c = 12.042\). Since we squared the equation, the solutions must be checked. We see that \(c = 12.042\) is extraneous, so \(c = 7\) is the solution. Example 4 shows how problematic floating-point arithmetic can be. Luckily there is a better formula for the area of a triangle when the three sides are known: For a triangle △ABC with sides a ≥ b ≥ c, the area is: $$Area = K = \frac{1}{4} \sqrt{(a+(b+ c))(c−(a−b))(c+(a−b))(a+(b− c))}$$ To use this formula, sort the names of the sides so that a ≥ b ≥ c. Then perform the operations inside the square root in the exact order in which they appear in the formula, including the use of parentheses. Then take the square root and divide by 4. For the triangle in Example 4, the above formula gives an answer of exactly K = 1 on the same TI-83 Plus calculator that failed with Heron’s formula. How do you find the area of a ABC triangle?For example, In ∆ABC, when sides 'b' and 'c' and included angle A is known, the area of the triangle is calculated with the help of the formula: 1/2 × b × c × sin(A).
What is ABC in a triangle?In any right-angled triangle, ABC, the side opposite the right-angle is called the hypotenuse. Here we use the convention that the side opposite angle A is labelled a. The side opposite B is labelled b and the side opposite C is labelled c.
What is the perimeter of △ △ ABC?Solution. To find the perimeter of \triangle ABC we use the Pythagorean Theorem which tells us that |AB|^2 = |AC|^2 + |BC|^2. Since |AC| = 5 and |BC| = 12 we find that |AB| = \sqrt{169} = 13. The perimeter of \triangle ABC is then 5 + 12 + 13 = 30 units.
What is the area of ∆?The area of a triangle is defined as the total region that is enclosed by the three sides of any particular triangle. Basically, it is equal to half of the base times height, i.e. A = 1/2 × b × h.
|
What is the area of this triangle ABC?
Related Posts
Copyright © 2024 SignalDuo Inc.
