ISEE Upper Level Quantitative Reasoning - How to find the length of the side of a right triangle

Consider a triangle, $\bigtriangleup ABC$ , in which , , and $m \angle B = 100 ^{\circ}$ . Which is the greater quantity?

(a) 55

(b)

(b) is the greater quantity

(a) is the greater quantity

(a) and (b) are equal

It is impossible to determine which is greater from the information given

Explanation

Suppose .

By the Converse of the Pythagorean Theorem, a triangle is right if and only if the sum of the squares of the lengths of the smallest two sides is equal to the square of the longest side. Compare the quantities $(AB)^{2}+ ( BC )^{2}$ and $(AC)^{2}$

$(AB)^{2}+ ( BC )^{2} = 33^{2} + 44 ^{2} = 1,089+ 1,936= 3,025$

$(AC)^{2} = 55^{2} = 3,025$

Therefore, if

$(AB)^{2}+ ( BC )^{2} =(AC)^{2}$ , so $\bigtriangleup ABC$ is right, with the right angle opposite longest side $\overline{AC}$ . Thus, $\angle B$ is right and has degree measure 90.

However, $\angle B$ has degree measure greater than 90, so, as a consequence of the Converse of the Pythagorean Theorem and the SAS Inequality Theorem, it holds that .

Right triangle 5

Figure NOT drawn to scale.

Refer to the above triangle. Which is the greater quantity?

(a)

(b) 108

(b) is the greater quantity

(a) and (b) are equal

(a) is the greater quantity

It is impossible to determine which is greater from the information given

Explanation

We can compare these numbers by comparing their squares.

By the Pythagorean Theorem,

$(AC) ^{2}= (AB) ^{2}+ (BC)^{2} = 100 ^{2}+ 40 ^{2} = 10,000 + 1,600 = 11,600$

Also,

$108^{2} = 11,664$

$108^{2} >(AC) ^{2}$ , so .

Triangle

Give the length of one leg of an isosceles right triangle whose area is the same as the right triangle in the above diagram.

$6 \sqrt{15} \textrm{ in}$

$15 \sqrt{6} \textrm{ in}$

$3\sqrt{15} \textrm{ in}$

$12 \sqrt{ 5} \textrm{ in}$

$9 \sqrt{10} \textrm{ in}$

Explanation

The area of a triangle is half the product of its height and its base; in a right triangle, the legs, being perpendicular, can serve as these quantites.

The triangle in the diagram has area

$\frac{1}{2} \times 30 \times 18 = 270$ square inches.

An isosceles right triangle has two legs of the same length, which we will call . The area of that triangle, which is the same as that of the one in the diagram, is therefore

$\frac{1}{2} L ^{2} = 270$

$\frac{1}{2} L ^{2}\cdot 2 = 270 \cdot 2$

$L ^{2} = 540$

$L = \sqrt{540}$

$L = \sqrt{36}\cdot \sqrt{15} = 6 \sqrt{15}$ inches.

Right_triangle

The perimeter of a regular octagon is 20% greater than that of the above right triangle. Which is the greater quantity?

(A) The length of one side of the octagon

(B) 3 yards

(A) and (B) are equal

It is impossible to determine which is greater from the information given

(A) is greater

(B) is greater

Explanation

By the Pythagorean Theorem, the shorter leg has length

$\sqrt{26^{2}-24^{2}} = \sqrt{676-576} = \sqrt{100} = 10$ feet.

The perimeter of the right triangle is therefore

feet.

The octagon has perimeter 20% greater than this, or

$60 + 0.20 \times 60 = 60 + 12 = 72$ feet.

A regular octagon has eight sides of equal length, so each side of this octagon has length

$72 \div 8 = 9$ feet, which is equal to 3 yards. This makes the quantities equal.

Given $\Delta ABC$ with right angle $\angle B$ , $m \angle C = 45 ^{\circ }$

Which is the greater quantity?

(a)

(b)

(a) and (b) are equal.

(a) is greater.

(b) is greater.

It is impossible to tell from the information given.

Explanation

$m \angle B = 90^{\circ}$

$m \angle C = 45 ^{\circ }$

The sum of the measures of the angles of a triangle is , so:

$m \angle A + m \angle B + m \angle C = 180$

$m \angle A +90 + 45 = 180$

$m \angle A +135= 180$

$m \angle A +135-135= 180-135$

$m \angle A = 45^{\circ }$

This is a $45^{\circ }-45^{\circ }-90^{\circ }$ triangle, so its legs $\overline{AB}$ and $\overline{BC}$ are congruent. The quantities are equal.

Right_triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram. Which of the following quadratic equations would yield the value of as a solution?

$x^{2} -4x-24= 0$

$2x^{2} -4x-24= 0$

$x^{2} -4x+74= 0$

$2x^{2} +4x-24= 0$

$x^{2} +4x+74= 0$

Explanation

By the Pythagorean Theorem,

$x^{2}+ (x+5)^{2} = (x+7)^{2}$

$x^{2}+ x^{2} +10x+25 = x^{2} +14x +49$

$2x^{2} +10x+25 = x^{2} +14x +49$

$x^{2} -4x-24= 0$

Right_triangle

Refer to the above diagram. Which of the following quadratic equations would yield the value of as a solution?

$2x^{2}+6x-391 = 0$

$2x^{2}+6x-11 = 0$

$x^{2}+6x-391 = 0$

$2x^{2}+6x+409 = 0$

$x^{2}+6x+409 = 0$

Explanation

By the Pythagorean Theorem,

$x^{2}+ (x+3)^{2} = 20^{2}$

$x^{2}+ x^{2}+6x+9= 400$

$2x^{2}+6x-391 = 0$

Right_triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram.

Find the length of $\overline{AB}$ .

$7\frac{1}{2}$

$7\frac{1}{4}$

$7\frac{3}{4}$

Explanation

First, find .

Since $\overline{DE}$ is an altitude of $\Delta CDB$ from its right angle to its hypotenuse,

$\Delta DEC \sim \Delta BED$

$\frac{BE}{DE}= \frac{ED}{EC}$

$\frac{BE}{6}= \frac{6}{12}$

$BE= \frac{6}{12} \cdot 6 = 3$

$\Delta DEC \sim \Delta ABC$ by the Angle-Angle Postulate, so

$\frac{AB}{DE} = \frac{CB}{CE}$

$\frac{AB}{6} = \frac{15}{12}$

$AB= \frac{15}{12} \cdot 6$

$AB = 7\frac{1}{2}$

Right_triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram.

Find the length of $\overline{BC}$ .

$22\frac{2}{9}$

$16\frac{2}{3}$

$33\frac{1}{3}$

Explanation

First, find .

Since $\overline{DF }$ is an altitude of right $\Delta ADB$ to its hypotenuse,

$\frac{FB}{FD}= \frac{FD}{AF}$

$\frac{FB}{8}= \frac{8}{6}$

$FB= \frac{8}{6} \cdot 8 = 10 \frac{2}{3}$

$AB = AF + FB = 6 + 10 \frac{2}{3} = 16 \frac{2}{3}$

$\Delta AFD \sim \Delta ABC$ by the Angle-Angle Postulate, so

$\frac{BC}{FD} = \frac{AB}{AF}$

$\frac{BC}{8} = \frac{16 \frac{2}{3}}{6}$

$BC = 16 \frac{2}{3}\cdot 8 \div 6 = 22\frac{2}{9}$

Right_triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram. Evaluate .

$3\frac{3}{8}$

$15\frac{3}{8}$

$6\frac{3}{4}$

$30\frac{3}{4}$

Explanation

By the Pythagorean Theorem,

$x^{2}+ 15^{2} = (x+12)^{2}$

$x^{2}+ 225 = x^{2}+24x+144$

$x= 3\frac{3}{8}$

How to find the length of the side of a right triangle

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