Question

1)How to find area of an equilateral triangle where side is given as -- AB=5cm,BC=5cm,AC=5cm 2)How to find area of an isoscles triangle where side is given as -- PQ=4cm,QR=5cm,PR=4cm



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Answer 1

1.

Since it is an equilateral triangle

$\boxed{ area = \frac{√3a²}{4}}$

$\: \: \: \: \: \: \: \: \: \: = \frac{ \sqrt{3} \times 5 ^{2} }{4}$

$\: \: \: \: \: \: \: \: \: = \frac{1.73×25}{4}$

$\: \: \: \: \: \: \: \: = 10.83cm^{2}$

2. base = 5 cm

height = √( 4² - 2.5 ²)

= 3.12 cm

area = 7.8cm²

more info attached into figure

Answer 2

Concepts

• Congruence

Triangles that satisfy SSS, ASA, SAS criteria are congruent.

• Pythagorean Theorem

It states that the hypotenuse $c$ and the other sides $a,b$ satisfy the relation of $a^2+b^2=c^2$.

• Heron's Formula

A formula to find the area of a triangle in which three sides are given.

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Solution A

If we draw a perpendicular line from the apex angle, it divides the isosceles triangle into two congruent pieces. This can be explained by congruence.

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[Proof]

Consider an isosceles triangle $\triangle ABC$ with $\overline{AB}=\overline{AC}$. We draw a perpendicular line $\overline{AH}$ from the apex angle.

Criterion: SAS

• $\overline{AB}=\overline{AC}$
• $\angle{B}=\angle{C}$
• The triangles share $\overline{AH}$.

Hence the criterion is satisfied and the two pieces are congruent.

We know that drawing a perpendicular line divides the triangles into two congruent pieces. Since the two pieces are right triangles, we can apply the Pythagorean theorem to find the height.

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[Pythagorean Theorem]

$\implies \overline{AH}^2=5^2-\dfrac{5^2}{2^2} =\dfrac{3\cdot 5^2}{2^2}$

$\implies \overline{AH}=\boxed{\dfrac{5\sqrt{3} }{2}\ \mathrm{cm}}$

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[Area of $\triangle ABC$ ]

Area of a triangle $\triangle ABC$ with $\overline{AB}=5\ \mathrm{cm}, \overline{BC}=5\ \mathrm{cm}, \overline{CA}=5\ \mathrm{cm}$.

$\triangle ABC=\dfrac{1}{2} \times \overline{BC}\times \overline{AH}$

$=\dfrac{1}{2} \times 5\times \dfrac{5\sqrt{3} }{2}=\boxed{\dfrac{25\sqrt{3} }{4}\ \mathrm{cm^2}}$

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[Pythagorean Theorem]

$\implies \overline{PH}^2=4^2-\dfrac{5^2}{2^2} =\dfrac{64-25}{2^2}=\boxed{\dfrac{39}{2^2} }$

$\implies \overline{PH}=\boxed{\dfrac{\sqrt{39} }{2}\ \mathrm{cm}}$

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[Area of $\triangle PQR$ ]

Area of a triangle $\triangle PQR$ with $\overline{PQ}=4\ \mathrm{cm}, \overline{QR}=5\ \mathrm{cm}, \overline{RP}=4\ \mathrm{cm}$.

$\triangle PQR=\dfrac{1}{2} \times \overline{QR}\times \overline{PH}$

$=\dfrac{1}{2} \times 5\times \dfrac{\sqrt{39} }{2} =\boxed{\dfrac{5\sqrt{39} }{4}\ \mathrm{cm^2}}$

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Solution B

[Heron's Formula]

The area of a triangle with sides $a,b,c$ is $\sqrt{s(s-a)(s-b)(s-c)}$ where $s=\dfrac{a+b+c}{2}$.

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[Area of $\triangle ABC$ ]

Area of a triangle $\triangle ABC$ with $\overline{AB}=5\ \mathrm{cm}, \overline{BC}=5\ \mathrm{cm}, \overline{CA}=5\ \mathrm{cm}$.

$a=5, b=5, c=5, s=\dfrac{15}{2}$

$\implies \triangle ABC=\sqrt{\dfrac{15}{2} \times \dfrac{5}{2} \times \dfrac{5}{2} \times \dfrac{5}{2}}$

$=\sqrt{\dfrac{3\cdot 5^4}{2^4} } =\boxed{\dfrac{25\sqrt{3} }{4}\ \mathrm{cm^2}}$

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[Area of $\triangle PQR$ ]

Area of a triangle $\triangle PQR$ with $\overline{PQ}=4\ \mathrm{cm}, \overline{QR}=5\ \mathrm{cm}, \overline{RP}=4\ \mathrm{cm}$.

$a=4, b=5, c=4, s=\dfrac{13}{2}$

$\implies \triangle PQR=\sqrt{\dfrac{13}{2} \times \dfrac{5}{2} \times \dfrac{3}{2} \times \dfrac{5}{2}}$

$=\sqrt{\dfrac{39\cdot 5^2}{2^4} } =\boxed{\dfrac{5\sqrt{39} }{4}\ \mathrm{cm^2}}$

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Conclusion

$\triangle ABC=\boxed{\dfrac{25\sqrt{3} }{4}\ \mathrm{cm^2}}$

$\triangle PQR =\boxed{\dfrac{5\sqrt{39} }{4}\ \mathrm{cm^2}}$