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What is the area of this acute isoceles trinagle? It doesnt have a height so I am stumped.

What is the area of this acute isoceles trinagle? It doesnt have a height so I am-example-1
User Anirus
by
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1 Answer

3 votes

Answer:

11.8 cm²

Explanation:

Cosine rule: a ² = b ² + c ² - 2bc COS A

COS A = (b ² + c ² - a ²) / (2bc)

Area of triangle = ½ ab sin C

name the angle that joins the '3' side to either of the '8' sides. label that angle as A.

now, Cos A = (8² + 3² - 8²) / [(2)(3)(8)]

= 3/16.

A = inverse Cos (3/16)

= 79.193°.

Area of triangle = 1/2 ab Sin C (we'll just use letter A instead of C)

= 1/2 (3)(8) sin (3/16)

= 11.8 cm².

another way you can find it is by this simpler method:

think of one side of 8, the side of 3 and the height that you do not know by using Pythagoras' Theorem for right-angled triangle. (a² + b² = c²). go halfway up the '3' side, then cut across straight to where the '8' sides meet. we now have right-angled triangle. let's call the height you don't know h.

we have 1.5² + h² = 8²

h² = 8² - 1.5²

h² = 61.75

h = √61.75

so area = 1/2 X 1.5 X √61.75

= 11.8 cm²

User AndrewSmiley
by
7.5k points

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