Math

QuestionFind the height hh of isosceles triangle ABCABC if the unshaded heights are 11 and 5 units. Choices: (A) 14.6 (B) 14.8 (C) 15 (D) 15.2 (E) 15.4

Studdy Solution

STEP 1

Assumptions1. Triangle ABC is an isosceles triangle with equal side lengths AB and BC. . Segments are drawn parallel to AC so that the shaded portions of triangle ABC have the same area.
3. The heights of the two unshaded portions are11 and5 units, respectively.
4. We need to find the height h of triangle ABC.

STEP 2

First, let's denote the base of the triangle as b. Because the triangle is isosceles, the base bisects the height, creating two right triangles. The area of the triangle can be expressed asArea=12×base×heightArea = \frac{1}{2} \times base \times height

STEP 3

Since the shaded and unshaded areas are equal, the total height of the triangle is the sum of the heights of the shaded and unshaded areas. Let's denote the height of the shaded area as h1 and the height of the unshaded area as h2. We haveh=h1+h2h = h1 + h2

STEP 4

We know that the heights of the unshaded areas are11 and units, respectively. Therefore, the height of the shaded area ish1=h(11+)h1 = h - (11 +)

STEP 5

Since the shaded and unshaded areas are equal, we can set up the following equation12×b×h1=12×b×(11+5)\frac{1}{2} \times b \times h1 = \frac{1}{2} \times b \times (11 +5)

STEP 6

olving the equation for h1, we geth1=11+5=16h1 =11 +5 =16

STEP 7

Substituting h1 =16 into the equation h = h1 + h2, we geth=16+(11+5)h =16 + (11 +5)

STEP 8

Calculating the value of h, we geth=16+11+5=32h =16 +11 +5 =32Therefore, the height h of triangle ABC is32 units.

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