Question

Find the area of an isosceles triangle ABC in which BC= 8 cm and AB = AC = 5 cm. If CE perpendicular to AB, find CE.
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Answer 1

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The area is 12cm^2. ❤️❤️❤️

Step-by-step explanation:

So let ABC be an isosceles triangle with base BC = 8 cm, and AB =AC= 5 cm .

Draw altitude AD.

So, BD = CD = 4 cm .

Now, ADB is a right triangle, by pythagoras  theorem:

AD = 3 cm .

Thus, area of ABC = 1/2× BC × AD = 1/2× 8×3 = 12cm^2.

Answer 2

The measurement of the side CE is equal to 7.59cm

Given:

In the isosceles triangle ABC;

• BC= 8 cm
• AB=AC= 5 cm
• CE is perpendicular to AC

To Find:

• The length of CE

Solution:

• Since ABC is an isosceles triangle wherein AB= AC;
• The perpendicular drawn from C to AB will divide AB into two equal halves AE and BE
• Hence, BE will be= 5/2cm

                                     = 2.5cm

• To find the length of CE, we must look at triangle BCE;
• Side BE= 2.5cm
• Side BC= 8cm

• CE is perpendicular to BE, which means that the ∠BEC= 90°

Now, according to Pythagoras theorem;

• The square of the hypotenuse is equal to the sum of the squares of the remaining two sides of a triangle
• Therefore;

        (BC)²=(CE)²+(BE)²

• Replacing the values;

        (8)²= (CE)²+(2.5)²

         64= (CE)²+6.25

∴(CE)²=  64- 6.25

∴(CE)²= 57.75

∴CE= √57.75

Hence CE is equal to 7.59cm        

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