In the figure shown, RC = 10, CD = 5, DS = 12 and PQ || MN. What is th

since lines are ll PQ and MN
so from the given info of area of ∆ we can find the height
#1
Area of triangle CBS = 85
1/2*17*x= 85
x= 10
sufficient to determine are of ∆RAD
#2
Area of triangle RAC = 50
1/2 * 10*x = 50
x= 10
again sufficient
IMO 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
(2) Area of triangle RAC = 50

find the height

(1) sufic
area_cbs=85=17*h/2, h=170/17=10
area_rad=15*h/2=150/2=75

(2) sufic
area_rac=50=10*h/2, h=100/10=10
area_rad=15*h/2=150/2=75

Ans (D)

RC = 10, CD = 5, DS = 12, PQ || MN.
Q. What is the area of the triangle RAD

(1) Area of triangle CBS = 85
---> 85=0.5*(CD+DS)* height
85=0.5*17*height
height=10
---> area of the triangle RAD
=0.5*(RC+CD)* height
=0.5*(10+5)* 10 =75
SUFFICIENT

(2) Area of triangle RAC = 50
---> 50=0.5*(RC)* height
50=0.5*10*height
height=10
---> area of the triangle RAD
=0.5*(RC+CD)* height
=0.5*(10+5)* 10 =75
SUFFICIENT

FINAL ANSWER IS (D)

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Given RC = 10, CD = 5, DS = 12
As PQ || MN, perpendicular height (h) between them are always same for all cases.
Area of RAD = 0.5*RD*h

Statement 1 provide area of CBS
I know CS= 17 , h will be known.
Sufficient

Similarly 2 is sufficient

D is answer.

Since PQ || MN, the shortest distance which is height between those two lines is equal to that of any triangles shown in the figure.

(Statement1): Area of triangle CBS = 85
—> 17*h/2 = 85
h = 10
—> the area of triangle RAD = 15*10/2 = 75
Sufficient

(Statement2): Area of triangle RAC = 50
—> 10*h/2= 50
h= 10
—> the area of triangle RAD = 75
Sufficient

Answer (D)

Posted from my mobile device

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

Given is the base of triangles and we need to fine the Area of RAD to find the area we need to find the height.
The height for all the trianges will be same as they are all in between 2 parallel lines PQ and MN

Statement 1:

Area of triange CBS= 1/2 b H

85= 1/2 x 17 x H Therefore H= 10

Substitute the value of H for triangle RAD 1/2 x 10 x 15= 75

Statement 1 is sufficient

Statement 2: Area of triangle RAC 1/2 x bx h Therefore 1/2 x 10 x h= 50 Therefore H=10

Substitute the value of H for triangle RAD 1/2 x 10 x 15= 75

IMO D

1)Area of triangle CBS = 85
CBS = 1/2 * (CD + DS) * x(perpendicular distance between lines) = 85
x = 10

SUFFICIENT.

2)RAC = 1/2 * RC * x(perpendicular distance between lines) = 50
x = 10

SUFFICIENT.

So, Ans D