As you may remember from high school, \(A = \frac{1}{2}bh\) , where b is the base and h is the height.

If you are having trouble remembering this, simply remember that a rectangle has an area of \(A = bh\), and that

a triangle is half a rectangle.

## Practice Question: Using the Area Formula

- The figure on the left is an isosceles right triangle, and the figure on the right is a square of length 3. Find the value of b.

(1): b is the length of the diagonal of the square. (2): the triangle and the square have the same area. **Answer Choices: ** A. Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient. B. Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient. C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient. D. EACH statement ALONE is sufficient. E. Statements (1) and (2) TOGETHER are not sufficient.

## Answer and Explanation

Statement (1) says that b is the length of the diagonal of the square. We know the side of the square is 3, so we could find the length of its diagonal with the ubiquitous Pythagorean Theorem. Therefore, we would know b. Statement (1) is sufficient by itself. Statement (2) says that the triangle and the square have equal area. Well, we know the area of the square is 9. The triangle has a base of b, and because we know it’s isosceles, we know its height is also b. It’s area, \(\frac{1}{2}b^2\), must equal 9. That means, we can solve for b. Statement (2) is also sufficient by itself. **Answer: **D

Can you please explain, three altitudes of triangle are 12,9.6, 16. Find the area of triangle? how to solve this please explain.

Hi Viji,

An altitude of a triangle is a line segment through a vertex and perpendicular to a line containing the base (the opposite side of the triangle). As explained on this Math Forum page, we can find the area of a triangle given its three altitudes by rewriting Heron’s formula:

A = 1/sqrt[(1/ha + 1/hb + 1/hc)*(-1/ha + 1/hb + 1/hc)*(1/ha – 1/hb + 1/hc)*(1/ha + 1/hb – 1/hc)]

where A is the area of the triangle and ha, hb, and hc are the altitudes of the triangle.

So, if we plug in ha=12, hb=9.6, and hc=16, we find that the area is

A = 1/sqrt[(1/12 + 1/9.6 + 1/16)*(-1/12 + 1/9.6 + 1/16)*(1/12 – 1/9.6 + 1/16)*(1/12 + 1/9.6 – 1/16)]

A = 96

Hope this helps!

From the diagram of the triangle, we do not necessarily know that b is one of the legs of the triangle. “b” could be the hypotenuse, because the <90 was not indicated. Then, statement 2 would be insufficient.

I also agree with your statement!

The answer should be A

The triangle figure shows which angle is 90deg. The sum of the other two angles therefore should be 90deg, which would imply that they are the smaller angles since you cannot have a triangle with more than one angle with a measure of 90deg or larger. Hence ‘b’ is a leg and the side opposite the 90deg angle is the longest side and hence the hypotenuse. [Unless the right angle indication was added after the initial comment was posted]

This is correct! The figure marks which angle is the right angle, and the side opposite it (which we see as a diagonal line) must be the hypotenuse. The question tells us that this is an isosceles right triangle. Therefore, we know for sure that both the horizontal and the vertical side [the two sides adjacent to the right angle] are legs, and they both have a length of b.

More specifically: an “isosceles right triangle” has one 90 degree angle and two 45 degree angles. The side opposite the right angle is the hypotenuse. The two sides opposite the 45 degree angles are the legs, and they have the same length.

Where are the answer choices? Answer is “d” but I’m not seeing the choices.

Mr. Eastling: Underneath “Statement #1” and “Statement #2”, those five statements labeled (A) though (E) —those are the answer choices. Those are actually the standard answer choices for *all* Data Sufficiency questions — if those are not familiar, I definitely would recommend watching the Lesson Videos on DS in the near future. Please let me know if you have any further questions.

Mike 🙂