Area of triangle

**rn5a** · December 15th 2009, 12:09 AM

In the attached figure, ST=5cm & QR=9cm. The area of the larger triangle is 50 sq.cm. What is the area of the black region?

Thanks,

Ron

**earboth** · December 15th 2009, 01:20 AM

Originally Posted by rn5a

In the attached figure, ST=5cm & QR=9cm. The area of the larger triangle is 50 sq.cm. What is the area of the black region?

Thanks,

Ron

The shaded region is:

(area of the large triangle) - (area of triangle QRS)

The area of a triangle is calculated by:

$a_{\Delta} = \frac12 \cdot base \cdot height$

with $\overline{QR} = base$ and $\overline{ST} = height$

**jgv115** · December 15th 2009, 03:14 AM

The problem is how to find the height of the whole triangle..

**earboth** · December 15th 2009, 09:23 PM

Originally Posted by rn5a

... The area of the larger triangle is 50 sq.cm. What is the area of the black region?

...

Originally Posted by jgv115

The problem is how to find the height of the whole triangle..

Honestly I don't understand why you want to know the height of the triangle QRP because you already know that the area of the this triangle is 50 cm².

But: Since you know that $\overline{QR} = 9\ cm$ you can use the value of the area and the length of the base to calculate the length of the height:

$50 =\frac12 \cdot 9 \cdot h~\implies~ h= \frac{100}9$

And now?

**jgv115** · December 15th 2009, 09:25 PM

I should learn to read.

**Lochcelious** · December 15th 2009, 10:07 PM

I thought you just subtract the area of the smaller traingle from the triangle of the whole to get 27.5 sq.cm?

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