Solve the following mathematical problems:

Length of each equal side of an isosceles triangle is 10cm and the included angle between those two sides is 45°. Find the area of the triangle.
(Geometry (Triangle))

Created: 2 years ago | Updated: 1 year ago

Updated: 1 year ago

Answer :

Let, x be the perpendicular drawn from the vertex of the triangle

We know,

$\sin 45 ° = \frac{P e r p e n d i c u l a r}{H y p o t e n u s e} \Rightarrow \frac{1}{\sqrt{2}} = \frac{x}{10} ∴ x = \frac{10}{\sqrt{2}}$

Now, we know area of triangle

$= \frac{1}{2} \times b a s e \times h e i g h t = \frac{1}{2} \times 10 \times \frac{10}{\sqrt{2}} = \frac{50}{\sqrt{2}} = \frac{25 \times \sqrt{2} \times \sqrt{2}}{\sqrt{2}} = 25 \sqrt{2}$ ans.

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Length of each equal side of an isosceles triangle is 10cm and the included angle between those two sides is 45°. Find the area of the triangle.
(Geometry (Triangle))

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Length of each equal side of an isosceles triangle is 10cm and the included angle between those two sides is 45°. Find the area of the triangle.
(Geometry (Triangle))

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Length of each equal side of an isosceles triangle is 10cm and the included angle between those two sides is 45°. Find the area of the triangle.
(Geometry (Triangle))

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