ΔABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26). The height AD from A to BC, is 6 cm. Find the area of ΔABC. What will be the height from C to AB i.e., CE???
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Answers
Answer:
Area of the triangle = 27cm²
Height of C = 7.2cm
Step-by-step explanation:
AD is perpendicular to BC
So In Triangle ADC,
Using Pythagoras Theorem,
AC² = DC² + AD²
=> (75/10) = DC² + (6)²
=> 225/4 - 36 = DC²
=> (225 - 144) ÷ 4 = DC²
=> 81/4 = DC²
=> √81/4 = DC
=> 4.5cm = DC
Therefore, AD bisects BC at the midpoint D
Area of the whole triangle
= Area of triangle ADB + Area of triangle ADC
= 1/2 x 4.5 x 6 + 1/2 x 4.5 x 6
=> 13.5 + 13.5
=> 27cm²
Also, CE is perpendicular to AB and bisects AB at midpoint E
Therefore,
=> EB = 7.5 ÷ 2
=> EB = 3.75cm
Area of Triangle CEB
=> 27/2 = 1/2 x 3.75 x height CE
=> 27/2 = 15/8 x height CE
=> 8/15 x 27/2 = height CE
=> 36/5 = height CE
=> 7.2cm = CE
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