Math

QuestionBerechne die gesuchten Längen für die gegebenen Flächeninhalte und Höhen der Dreiecke:
a) A=17,55 cm2,a=7,8 cm,ha=A=17,55 \mathrm{~cm}^{2}, a=7,8 \mathrm{~cm}, h_{a}= ?
b) A=2340 mm2,hb=36 mm,b=A=2340 \mathrm{~mm}^{2}, h_{b}=36 \mathrm{~mm}, b= ?
c) A=18,25 cm2,c=73 mm,hc=A=18,25 \mathrm{~cm}^{2}, c=73 \mathrm{~mm}, h_{c}= ?
d) A=34,1 cm2,ha=62 mm,a=A=34,1 \mathrm{~cm}^{2}, h_{a}=62 \mathrm{~mm}, a= ?

Studdy Solution

STEP 1

Assumptions1. The area of a triangle is given by the formula A=1×base×heightA = \frac{1}{} \times base \times height . The base and height are perpendicular to each other3. The given measurements are in centimeters and millimeters. We will keep the units consistent in each problem.

STEP 2

For problem a), we are given the area AA and the base aa of the triangle, and we need to find the height hah_a. We can rearrange the formula for the area of a triangle to solve for the height.
ha=2Aah_a = \frac{2A}{a}

STEP 3

Now, plug in the given values for the area AA and the base aa to calculate the height hah_a.
ha=2×17.55cm27.8cmh_a = \frac{2 \times17.55 \, cm^2}{7.8 \, cm}

STEP 4

Calculate the height hah_a.
ha=2×17.55cm27.8cm=4.cmh_a = \frac{2 \times17.55 \, cm^2}{7.8 \, cm} =4. \, cm

STEP 5

For problem c), we are given the area AA and the height hbh_b of the triangle, and we need to find the base bb. We can rearrange the formula for the area of a triangle to solve for the base.
b=2Ahbb = \frac{2A}{h_b}

STEP 6

Now, plug in the given values for the area AA and the height hbh_b to calculate the base bb.
b=2×2340mm236mmb = \frac{2 \times2340 \, mm^2}{36 \, mm}

STEP 7

Calculate the base bb.
b=2×2340mm236mm=130mmb = \frac{2 \times2340 \, mm^2}{36 \, mm} =130 \, mm

STEP 8

For problem b), we are given the area AA and the base cc of the triangle, and we need to find the height hch_c. We can rearrange the formula for the area of a triangle to solve for the height.
hc=2Ach_c = \frac{2A}{c}

STEP 9

Now, plug in the given values for the area AA and the base cc to calculate the height hch_c. Note that the base cc is given in millimeters, so we first convert it to centimeters.
c=73mm=7.3cmc =73 \, mm =7.3 \, cmhc=2×18.25cm27.3cmh_c = \frac{2 \times18.25 \, cm^2}{7.3 \, cm}

STEP 10

Calculate the height hch_c.
hc=2×18.25cm27.3cm=5cmh_c = \frac{2 \times18.25 \, cm^2}{7.3 \, cm} =5 \, cm

STEP 11

For problem d), we are given the area AA and the height hah_a of the triangle, and we need to find the base aa. We can rearrange the formula for the area of a triangle to solve for the base.
a=Ahaa = \frac{A}{h_a}

STEP 12

Now, plug in the given values for the area AA and the height hah_a to calculate the base aa. Note that the height hah_a is given in millimeters, so we first convert it to centimeters.
ha=62mm=6.2cmh_a =62 \, mm =6.2 \, cma=2×34.cm26.2cma = \frac{2 \times34. \, cm^2}{6.2 \, cm}

STEP 13

Calculate the base aa.
a=2×34.cm26.2cm=11cma = \frac{2 \times34. \, cm^2}{6.2 \, cm} =11 \, cmThe solutions to the problems area) ha=.5cmh_a =.5 \, cm
c) b=130mmb =130 \, mm
b) hc=5cmh_c =5 \, cm
d) a=11cma =11 \, cm

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