The sides of a triangle are 21, 17, and 10 cm. The given triangle is divided into two triangles by the perpendicular on the longest side from the opposite vertex. What is the area (in sq. cm) of the larger triangle formed?

Sure, let's solve this step by step.

Step 1: Identify the longest side of the triangle. In this case, it's 21 cm.

Step 2: The triangle is divided into two smaller triangles by a perpendicular from the opposite vertex to the longest side. This perpendicular line divides the longest side into two segments. Let's denote the lengths of these segments as x and 21-x.

Step 3: We know that the sum of the squares of the lengths of the two shorter sides of a right triangle equals the square of the length of the longest side (Pythagoras' theorem). So, we can write two equations based on this:

10^2 = x^2 + h^2 and 17^2 = (21-x)^2 + h^2, where h is the height of the two smaller triangles.

Step 4: Solve these two equations to find the values of x and h.

Step 5: The area of a triangle is given by the formula 1/2 * base * height. The base of the larger triangle is either x or 21-x (whichever is greater), and the height is h.

Step 6: Substitute the values of x, h and the base into the formula to find the area of the larger triangle.

Sure, let's solve this step by step.

Step 1: Identify the longest side of the triangle. In this case, it's 21 cm.

Step 2: The triangle is divided into two smaller triangles by a perpendicular from the opposite vertex to the longest side. This means that the longest side becomes the base for both smaller triangles.

Step 3: Use the Pythagorean theorem to find the length of the perpendicular. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. We can write this as: a² + b² = c².

In this case, one of the smaller triangles has sides of 10 cm (a), 17 cm (b, the hypotenuse), and the perpendicular (c) we're trying to find. Plugging these values into the Pythagorean theorem gives us: 10² + c² = 17². Solving for c gives us c = sqrt(17² - 10²) = sqrt(189) = 13.75 cm.

Step 4: Now we can find the areas of the two smaller triangles. The area of a triangle is given by the formula 1/2 * base * height. For the smaller triangle with a base of 10 cm and height of 13.75 cm, the area is 1/2 * 10 * 13.75 = 68.75 sq cm. For the smaller triangle with a base of 21 cm - 10 cm = 11 cm and the same height, the area is 1/2 * 11 * 13.75 = 75.625 sq cm.

Step 5: The larger of the two triangles is the one with an area of 75.625 sq cm.

The given triangle is a right triangle (since 21^2 = 17^2 + 10^2), with sides 17 cm, 10 cm, and hypotenuse 21 cm.

When a perpendicular is drawn from the vertex opposite the hypotenuse, it divides the triangle into two smaller right triangles. The length of this perpendicular can be found using the formula for the height (h) of a right triangle: h = (a*b)/c, where a and b are the lengths of the two legs and c is the length of the hypotenuse.

So, h = (17*10)/21 = 8.095 cm.

The areas of the two smaller triangles are then (1/2)baseheight = (1/2)108.095 = 40.48 sq. cm and (1/2)178.095 = 68.81 sq. cm.

The larger triangle therefore has an area of 68.81 sq. cm.