In triangle ABC, point D lies on side AC such that segment BD is perpendicular to side AC.… — GRE General Test

In triangle $ABC$ , point $D$ lies on side $AC$ such that segment $BD$ is perpendicular to side $AC$ . The length of segment $AD$ is $9$ and the length of segment $DC$ is $16$ . If the ratio of the length of side $AB$ to the length of side $BC$ is $3 : 4$ , what is the area of triangle $ABC$ ?

A
72
B
96
150Cevap
D
300
E
1,800

Cevap

150

Applying the Pythagorean theorem to right triangles

ABD

and

CBD

gives

AB^2 = 81 + BD^2

and

BC^2 = 256 + BD^2

. Squaring the ratio

\frac{AB}{BC} = \frac{3}{4}

yields

\frac{81 + BD^2}{256 + BD^2} = \frac{9}{16}

. Cross-multiplying gives

1296 + 16 BD^2 = 2304 + 9 BD^2

, so

7 BD^2 = 1008 \implies BD = 12

. The base

AC = 9 + 16 = 25

, so the area of triangle

ABC

\frac{1}{2} \times 25 \times 12 = 150

Adım Adım Çözüm

Set up expressions for side lengths

AB

and

BC

using the Pythagorean theorem on right triangles

ABD

and

CBD

AB^2 = 9^2 + BD^2 = 81 + BD^2

and

BC^2 = 16^2 + BD^2 = 256 + BD^2

Segment

BD

is an altitude perpendicular to

AC

, dividing triangle

ABC

into two right triangles.

Use the given side ratio

\frac{AB}{BC} = \frac{3}{4}

to solve for the height

BD

\frac{81 + BD^2}{256 + BD^2} = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \implies 16(81 + BD^2) = 9(256 + BD^2) \implies 7 BD^2 = 1008 \implies BD^2 = 144 \implies BD = 12

Squaring both sides of the ratio allows substitution of the expressions for

AB^2

and

BC^2

Determine the total length of base

AC

and compute the area of triangle

ABC

AC = AD + DC = 9 + 16 = 25

. Area

= \frac{1}{2} \times AC \times BD = \frac{1}{2} \times 25 \times 12 = 150

The area of a triangle is given by half the product of its base and corresponding altitude.

Anahtar Kavram

Properties of altitudes in triangles, Pythagorean theorem, and ratio setup for area determination.

Tahmini Süre:2m 0s

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru