# Question on Similarity

• Apr 13th 2011, 04:46 AM
Ilsa
Question on Similarity
Attachment 21439
The diagram is attached.
The question is:
In the diagram \$\displaystyle ABCD\$ is a parallelogram. If \$\displaystyle BC= 4cm\$, \$\displaystyle CP= 6 cm\$ and the area of \$\displaystyle ABCD = 40 cm^2 \$, find the area of triangle \$\displaystyle ABP\$.
• Apr 13th 2011, 05:49 AM
HallsofIvy
The area of a parallelogram is "height times base". Since you are told that the base (the length of AD which is the same as the length of BC) is 4 and that the area of the parallelogram is 40, the height, the perpendicular distance from BC to AD, is 10.

The area of a triangle is "(1/2) heigth times base" and the base of triangle APD is also 4 so you need to find the perpendicular distance from P to AD. That will be 10 plus the perpendicular distance from P to BC. The two right triangles having hypotenuses CP and DC are similar triangles but, since you do not know CD, I see no way of using that to get the perpendicular distance you need.
• Apr 13th 2011, 06:00 AM
Ilsa
Quote:

Originally Posted by HallsofIvy
The area of a parallelogram is "height times base". Since you are told that the base (the length of AD which is the same as the length of BC) is 4 and that the area of the parallelogram is 40, the height, the perpendicular distance from BC to AD, is 10.

The area of a triangle is "(1/2) heigth times base" and the base of triangle APD is also 4 so you need to find the perpendicular distance from P to AD. That will be 10 plus the perpendicular distance from P to BC. The two right triangles having hypotenuses CP and DC are similar triangles but, since you do not know CD, I see no way of using that to get the perpendicular distance you need.

To get a triangle as APD, a new line segment will have to be made. Can you please explain, 'the perpendicular distance from P to AD. That will be 10 plus the perpendicular distance from P to BC.' ?
• Apr 13th 2011, 06:42 AM
Sambit
I too tried a lot to get something helpful...but did not get any. There is possibly no way to answer your question...
• Apr 13th 2011, 10:42 AM
HallsofIvy
Quote:

Originally Posted by Ilsa
To get a triangle as APD, a new line segment will have to be made. Can you please explain, 'the perpendicular distance from P to AD. That will be 10 plus the perpendicular distance from P to BC.' ?

Yes, and that new line segment is the line from P perendicular to line AD. The length of that line segment is 'the perpendicular distance from P to AD'. The length of the line segment from P to perpendicular BC is the second "perpendicular distance".