How to find the area of a right triangle - SSAT Upper Level Quantitative (Math)

Card 0 of 11

Question

What is the area of a right triangle whose hypotenuse is 13 inches and whose legs each measure a number of inches equal to an integer?

Answer

We are looking for a Pythagorean triple - that is, three integers that satisfy the relationship $a^{2}+b^{2} = c^{2}$ . We know that , and the only Pythagorean triple with is . The legs of the triangle are therefore 5 and 12, and the area of the right triangle is

$A = \frac{1}{2} bh= \frac{1}{2} \cdot 5\cdot 12 = 30$

Compare your answer with the correct one above

Question

Right Triangle A has legs of lengths 10 inches and 14 inches; Right Triangle B has legs of length 20 inches and 13 inches; Rectangle C has length 30 inches. The area of Rectangle C is the sum of the areas of the two right triangles. What is the height of Rectangle C?

Answer

The area of a right triangle is half the product of its legs. The area of Right Triangle A is equal to $\frac{1}{2} \times 10 \times 14 = 70$ square inches; that of Right Triangle B is equal to $\frac{1}{2} \times 20 \times 13 = 130$ square inches. The sum of the areas is square inches, which is the area of Rectangle C.

The area of a rectangle is the product of its length and its height. Therefore, the height is the quotient of the area and the length, which, for Rectangle C, is $200 \div 30 = 6 \frac{2}{3}$ inches.

Compare your answer with the correct one above

Question

Right Triangle A has hypotenuse 25 inches and one leg of length 24 inches; Right Triangle B has hypotenuse 15 inches and one leg of length 9 inches; Rectangle C has length 16 inches. The area of Rectangle C is the sum of the areas of the two right triangles. What is the width of Rectangle C?

Answer

The area of a right triangle is half the product of its legs. In each case, we know the length of one leg and the hypotenuse, so we need to apply the Pythagorean Theorem to find the second leg, then take half the product of the legs:

Right Triangle A:

The length of the second leg is

$\sqrt{25^{2} -24 ^{2}} = \sqrt{625 - 576} = \sqrt{49} = 7$ inches.

The area is

$\frac{1}{2} \times 7 \times 24 = 84$ square inches.

Right Triangle B:

The length of the second leg is

$\sqrt{15^{2} -9 ^{2}} = \sqrt{225 - 81} = \sqrt{144} = 12$ inches.

The area is

$\frac{1}{2} \times 9 \times 12 = 54$ square inches.

The sum of the areas is square inches.

The area of a rectangle is the product of its length and its height. Therefore, the height is the quotient of the area and the length, which, for Rectangle C, is $138 \div 16 = 8 \frac{5}{8}$ inches.

Compare your answer with the correct one above

Question

A right triangle has leg lengths of $10\text{ and }12$ . What is the area of this triangle?

Answer

Since the legs of a right triangle form a right angle, you can use these as the base and the height of the triangle.

$\text{Area}=\frac{1}{2}(\text{base})(\text{height})$

$\text{Area}=\frac{1}{2}(10)(12)=\frac{1}{2}(120)=60$

Compare your answer with the correct one above

Question

A right triangle has leg lengths of and . Find the area of the right triangle.

Answer

The legs of a right triangle also make up its base and its height.

$\text{Area}=\frac{1}{2}\text{base}\times\text{height}$

$\text{Area}=\frac{1}{2}\times6:cm\times9:cm=27:cm^2$

Compare your answer with the correct one above

Question

A right triangle has leg lengths of and . Find the area of this triangle.

Answer

The legs of a right triangle are also its height and its base.

$\text{Area}=\frac{1}{2}\text{base}\times\text{height}$

$\text{Area}=\frac{1}{2}(9:cm)(12:cm)=54:cm^2$

Compare your answer with the correct one above

Question

A given right triangle has two legs of lengths and , respectively. What is the area of the triangle?

Answer

The area of a right triangle with a base and a height can be found with the formula $A=\frac{1}{2}bh$ . Since the two legs of a right triangle are perpendicular to each other, we can use these as the base and height in the formula. Therefore:

$A=\frac{1}{2}bh$

$A=\frac{1}{2}(6:cm)(8:cm)$

$A=\frac{1}{2}(48:cm^2)$

Compare your answer with the correct one above

Question

A right triangle has two legs of lengths and , respectively. What is the area of the right triangle?

Answer

The area of a right triangle with a base and a height can be found with the formula $A=\frac{1}{2}bh$ . Since the two legs of a right triangle are perpendicular to each other, we can use these as the base and height in the formula. Therefore:

$A=\frac{1}{2}(9:cm)(12:cm)$

$A=\frac{1}{2}(108:cm^{2})$

$A=54:cm^{2}$

Compare your answer with the correct one above

Question

A given right triangle has legs of lengths and , respectively. What is the area of the right triangle?

Answer

The area of a right triangle with a base and a height can be found with the formula $A=\frac{1}{2}bh$ . Since the two legs of a right triangle are perpendicular to each other, we can use these as the base and height in the formula. Therefore:

$A=\frac{1}{2}(7:cm)(14:cm)$

$A=\frac{1}{2}(98:cm^2)$

Compare your answer with the correct one above

Question

The lengths of the hypotenuses of ten similar right triangles form an arithmetic sequence. The smallest triangle has legs of lengths 3 and 4 inches; the second-smallest triangle has a hypotenuse of length one foot.

Which of the following responses comes closest to the area of the largest triangle?

Answer

The hypotenuse of the smallest triangle can be calculated using the Pythagorean Theorem:

$c = \sqrt{a^{2}+b^{2}} = \sqrt{3^{2}+4^{2}} = \sqrt{9+16} = \sqrt{25}=5$ inches.

Let $c_{n}$ be the lengths of the hypotenuses of the triangles in inches. $c_{1} = 5$ and $c_{2} = 12$ , so their common difference is

$d = c_{2} - c_{1} = 12 - 5= 7$

The arithmetic sequence formula is

$c_{n} = c_{1} + (n-1)d$

The length of the hypotenuse of the largest triangle - the tenth triangle - can be found by substituting $c_{1} = 5, n= 10, d = 7$ :

$c_{10} =5+ (10-1) 7 = 5 + 9 \cdot 7 = 5 + 63 = 68$ inches.

The largest triangle has hypotenuse of length 68 inches. Since the triangles are similar, corresponding sides are in proportion. If we let and be the lengths of the legs of the largest triangle, then

$\frac{a}{3} = \frac{68}{5}$

$\frac{a}{3} \cdot 3 = \frac{68}{5} \cdot 3$

Similarly,

$\frac{b}{4} = \frac{68}{5}$

$\frac{b}{4} \cdot 4 = \frac{68}{5} \cdot 4$

The area of a right triangle is half the product of its legs:

$A = \frac{ab}{2}= \frac{40.8 \cdot 54.4}{2} = 1,109.76$ square inches.

Divide this by 144 to convert to square feet:

$1,109.76 \div 144 \approx 7.7$

Of the given responses, 8 square feet is the closest, and is the correct choice.

Compare your answer with the correct one above

Question

Figure NOT drawn to scale

In the above figure, $\bigtriangleup ABC$ is a right triangle. , , . What fraction of $\bigtriangleup ABC$ has been shaded in?

Answer

The length of the leg $\overline{AB}$ , which we will call , can be calculated by setting , the length of hypotenuse $\overline{AC}$ , and , the length of leg $\overline{BC}$ , and applying the Pythagorean Theorem:

$a = \sqrt{c^{2} - b^{2}} = \sqrt{15^{2} - 12^{2}} =\sqrt{225- 144} = \sqrt{81} = 9$

.

Construct the altitude of $\bigtriangleup ABC$ - which is also that of $\bigtriangleup ABD$ , the shaded region - from to $\overline{AC}$ . We call the length of this altitude (height). The figure is seen below.

The area of $\bigtriangleup ABC$ is one half this height multiplied by the corresponding base length :

$\frac{1}{2} \cdot AC \cdot h$

The area of $\bigtriangleup ABD$ , the shaded region - is, similarly,

$\frac{1}{2} \cdot AD \cdot h$

Therefore, the fraction that $\bigtriangleup ABD$ is of $\bigtriangleup ABC$ is the fraction of their areas:

$\frac{\frac{1}{2} \cdot AD \cdot h}{\frac{1}{2} \cdot AC \cdot h} = \frac{AD }{AC}$

Substituting the measures of the two segments:

$\frac{AD}{AC} = \frac{9}{15} = \frac{3}{5}$

Compare your answer with the correct one above

Tap the card to reveal the answer