∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC, is 6 cm. Find the area of ∆ABC. What will be the height from C to AB i.e; CE?
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Answers
Answer :-
Given :-
- AB = AC = 7.5 cm
- BC = 9 cm
- AD = 6 cm
To Find :-
- CE
Solution :-
Here, we have side BC and it's altitude AD. So, we will first find area of the triangle -
→ Area = ½ × BC × AD
→ Area = ½ × 6 × 9
→ Area = 3 × 9
→ Area = 27
Area of triangle = 27 cm²
Now, we have base i.e. AB and area of the triangle. So, we can find the altitude i.e. CE :-
→ Area = ½ × AB × CE
→ 27 = ½ × 7.5 × CE
→ 54 = 7.5 CE
→ CE = 54 / 7.5
→ CE = 7.2
Length of CE = 7.2 cm
Answer:
The area of triangle is 54cm² and height from C To AB is 7.2cm.
Step-by-step explanation:
Area of triangle= 1/2×B×H
=1/2×BC×AD
=1/2×9×6
=27cm²
Height From C to AB= 1/2×b×h
=1/2×AB×CE
Area of triangle = 1/2×7.5×CE
27cm²= 1/2×7.5×CE
27cm²×2/7.5=CE
7.2cm=CE
ANS: Hence, the area of triangle is 54cm² and height from C To AB is 7.2cm.
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