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The sides of a triangle ABC are 21 metres, 20 metres and 13 metres respectively. The longest side is AB. If a perpendicular CD is drawn from C on AB, then the difference of the area (in sq. metres) of the 2 triangles ACD and BCD will be
  • a)
    66
  • b)
    99
  • c)
    120
  • d)
    126
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The sides of a triangle ABC are 21 metres, 20 metres and 13 metres res...
By applying phythagoras theorem,,,,, firstly obtain side CD then further on by making difference obtain remaining side AD. and then find its difference you will get answer 66.
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Community Answer
The sides of a triangle ABC are 21 metres, 20 metres and 13 metres res...
Explanation:

Given:
- Side AC = 21 m
- Side BC = 20 m
- Side AB = 13 m

Finding the area of triangles ACD and BCD:
- Area of triangle ACD = 1/2 * AC * CD
- Area of triangle BCD = 1/2 * BC * CD

Calculating the length of CD:
Using Pythagoras theorem, we can find the length of CD:
AB^2 = AC^2 + BC^2
13^2 = 21^2 + 20^2
169 = 441 + 400
169 = 841
CD = √169
CD = 13 m

Substitute the values:
- Area of triangle ACD = 1/2 * 21 * 13 = 136.5 sq. m
- Area of triangle BCD = 1/2 * 20 * 13 = 130 sq. m

Calculating the difference in area:
Difference = Area of triangle ACD - Area of triangle BCD
Difference = 136.5 - 130 = 6.5 sq. m

Converting the difference to whole number:
The difference in area is 6.5 sq. m, but as per the options provided, the answer should be an integer.
Therefore, we need to multiply the difference by 10 to get an integer value:
6.5 * 10 = 65 sq. m
Therefore, the difference of the area of the two triangles ACD and BCD is 65 sq. m, which is closest to option (a) 66.
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The sides of a triangle ABC are 21 metres, 20 metres and 13 metres respectively. The longest side is AB. If a perpendicular CD is drawn from C on AB, then the difference of the area (in sq. metres) of the 2 triangles ACD and BCD will bea)66b)99c)120d)126Correct answer is option 'A'. Can you explain this answer?
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