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Two right-angled triangles are joined together to make a larger triangle ACD. Ņ B 6.8 cm 14.2 cm Calculate the perimeter of triangle ACD. Give your answer correct to 1 decimal place. 10 cm​

Two rightangled triangles are joined together to make a larger triangle ACD Ņ B 68 cm 142 cm Calculate the perimeter of triangle ACD Give your answer correct to class=

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Answer:

  • 50.5 cm (to 1 d.p)

Step-by-step explanation:

To calculate the perimeter of triangle ACD , we need to add the lengths of all the sides of the triangle

Perimeter = AB + BD + CD + AC

We are given :

  • AB = 14.2 cm
  • BD = 6.8 cm
  • BC = 10 cm

To Calculate the perimeter, First we need to calculate the length of side CD and AC

Solving for CD :

In triangle CBD, using Pythagoras theorem:

[tex]\sf CD = \sqrt{BD^2 + BC^2} [/tex]

[tex]\sf CD = \sqrt {6.8^2 + 10^2}[/tex]

[tex]\sf CD = \sqrt{46.24 + 100}[/tex]

[tex] \sf CD =\sqrt{146.24 }[/tex]

[tex] \sf CD = \sqrt{146.24}[/tex]

[tex] \sf CD = 12.09 \ cm [/tex]

Hence, the length of the Side CD is 12.09 cm

Solving for AC:

In triangle ABC :

[tex] \sf AC = \sqrt{AB^2 + BC^2} [/tex]

[tex] \sf AC = \sqrt{ 14.2^2 + 10^2} [/tex]

[tex]\sf AC = \sqrt { 201.64 + 100} [/tex]

[tex]\sf AC = \sqrt {301.64} [/tex]

[tex]\sf AC = 17.38 \ cm [/tex]

Hence, the length of the Side AC is 17.38 cm

Now let's calculate the perimeter of triangle ACD

[tex] \sf Perimeter= AB + BD + CD + AC [/tex]

[tex] \sf Perimeter= 14.2 + 6.8 + 12.09 + 17.38[/tex]

[tex] \sf Perimeter = 50. 5 \ cm [/tex] (to one d.p)

Therefore, The perimeter of the triangle ACD is 50.5 cm

semsee45

Answer:

50.5 cm

Step-by-step explanation:

To determine the perimeter of triangle ACD, we first need to calculate the lengths of the sides AC and CD. To do this, we can use the Pythagorean Theorem.

The Pythagorean Theorem states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

Side AC is the hypotenuse of right triangle ABC, with legs AB = 14.2 cm and BC = 10 cm. Therefore:

[tex]AC^2=AB^2+BC^2[/tex]

[tex]AC^2=14.2^2+10^2[/tex]

[tex]AC^2=201.64+100[/tex]

[tex]AC^2=301.64[/tex]

[tex]AC=\sqrt{301.64}[/tex]

[tex]AC=17.36778627...[/tex]

[tex]AC=17.4\; \sf cm\;(1\;d.p.)[/tex]

Side CD is the hypotenuse of right triangle CBD, with legs BD = 6.8 cm and BC = 10 cm. Therefore:

[tex]CD^2=BD^2+BC^2[/tex]

[tex]CD^2=6.8^2+10^2[/tex]

[tex]CD^2=46.24+100[/tex]

[tex]CD^2=146.24[/tex]

[tex]CD=\sqrt{146.24}[/tex]

[tex]CD=12.092973166...[/tex]

[tex]CD=12.1\; \sf cm\;(1\;d.p.)[/tex]

The perimeter of a triangle is the sum of the lengths of its three sides. So, the perimeter of triangle ACD is:

[tex]\begin{aligned}\textsf{Perimeter of $\triangle ACD$}&=AC+CD+AB+BD\\\\&=17.36778...+12.09297...+14.2+6.8\\\\&=50.4607594...\\\\&=50.5\; \sf cm\;(1\;d.p.) \end{aligned}[/tex]

Therefore, the perimeter of triangle ACD rounded to one decimal place is:

[tex]\huge\boxed{\boxed{50.5\; \sf cm}}[/tex]

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