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01 Apr 2020, 01:09
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In the figure shown, RC = 10, CD = 5, DS = 12 and PQ || MN. What is the area of the triangle RAD
(1) Area of triangle CBS = 85
(2) Area of triangle RAC = 50
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01 Apr 2020, 01:33
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In the figure shown, RC = 10, CD = 5, DS = 12 and PQ || MN. What is the area of the triangle RAD
(1) Area of triangle CBS = 85
Area of triangle CBS = 1/2 * (CD + DS) * x(perpendicular distance between lines) = 85
x = 10
SUFFICIENT.
(2) Area of triangle RAC = 50
Area of triangle RAC = 1/2 * RC * x(perpendicular distance between lines) = 50
x = 10
SUFFICIENT.
Answer D.
In the figure shown, RC = 10, CD = 5, DS = 12 and PQ || MN. What is the area of the triangle RAD
(1) Area of triangle CBS = 85
Area of triangle CBS = 1/2 * (CD + DS) * x(perpendicular distance between lines) = 85
x = 10
SUFFICIENT.
(2) Area of triangle RAC = 50
Area of triangle RAC = 1/2 * RC * x(perpendicular distance between lines) = 50
x = 10
SUFFICIENT.
Answer D.
01 Apr 2020, 03:42
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For the triangle RAD, Base = RC + CD = 10 + 5 = 15
Let the perpendicular = h
--> Area of triangle RAD = \(\frac{1}{2}*h*15\)
(1) Area of triangle CBS = 85
Base = 5 + 12 = 17 & height = h
--> \(\frac{1}{2}*h*17 = 85\)
--> \(h = 10\)
We can find the area of triangle RAD --> Sufficient
(2) Area of triangle RAC = 50
Base = 10 & height = h
--> \(\frac{1}{2}*h*10 = 50\)
--> \(h = 10\)
We can find the area of triangle RAD --> Sufficient
Option D
Let the perpendicular = h
--> Area of triangle RAD = \(\frac{1}{2}*h*15\)
(1) Area of triangle CBS = 85
Base = 5 + 12 = 17 & height = h
--> \(\frac{1}{2}*h*17 = 85\)
--> \(h = 10\)
We can find the area of triangle RAD --> Sufficient
(2) Area of triangle RAC = 50
Base = 10 & height = h
--> \(\frac{1}{2}*h*10 = 50\)
--> \(h = 10\)
We can find the area of triangle RAD --> Sufficient
Option D
