Bhaskar went for hiking with scouts’ team and there the scouts were given a task to build tents with the help of bamboos, ropes and canvas. The frame of the tent that Bhaskar prepared was of pyramid shape. To make it sturdier Bhaskar thought that he should fix a wooden stick on the triangular sides of the tent and fix the wooden stick at the mid points of the sides of two sides of the triangle. The distance between the end point of the bamboos on the sides was 3 m. Due to lack of even length bamboo sticks Bhaskar the triangles on the edges of tent were not congruent. Look into the figure of Bhaskar’s tent and answer the following questions.(a) To make tent sturdy Bhaskar used bamboo IJ in the back part of the tent. What length ofbamboo shall he use so that it should be fixed exactly?(i) 9 m. (ii) 4.5 m. (iii) 3 m. (iv) 6 m(b) Using which property of triangles Bhaskar was able to find the length of GH and IJ.(i) Pythagoras Property. (ii) Basic Proportionality Theorem. (iii) Mid- point theorem. (iv)Exterior angle property.(c) How can Bhaskar find the area of canvas used only to cover the triangles?(i) Heron’s Formula. (ii) Pythagoras Property. (iii) Area = ½ X Base X Height. (iv) SectionFormula.(d) Here BC: EF = 3: 9, then what will be the ratio of area of triangle ABC to area of triangle DEF.(i) 9:3 (ii) 9:81 (iii) 3.8: 11.4 (iv) 1:9

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04-10-2021
Mathematics

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Bhaskar went for hiking with scouts’ team and there the scouts were given a task to build tents with the help of bamboos, ropes and canvas. The frame of the tent that Bhaskar prepared was of pyramid shape. To make it sturdier Bhaskar thought that he should fix a wooden stick on the triangular sides of the tent and fix the wooden stick at the mid points of the sides of two sides of the triangle. The distance between the end point of the bamboos on the sides was 3 m. Due to lack of even length bamboo sticks Bhaskar the triangles on the edges of tent were not congruent. Look into the figure of Bhaskar’s tent and answer the following questions.(a) To make tent sturdy Bhaskar used bamboo IJ in the back part of the tent. What length ofbamboo shall he use so that it should be fixed exactly?(i) 9 m. (ii) 4.5 m. (iii) 3 m. (iv) 6 m(b) Using which property of triangles Bhaskar was able to find the length of GH and IJ.(i) Pythagoras Property. (ii) Basic Proportionality Theorem. (iii) Mid- point theorem. (iv)Exterior angle property.(c) How can Bhaskar find the area of canvas used only to cover the triangles?(i) Heron’s Formula. (ii) Pythagoras Property. (iii) Area = ½ X Base X Height. (iv) SectionFormula.(d) Here BC: EF = 3: 9, then what will be the ratio of area of triangle ABC to area of triangle DEF.(i) 9:3 (ii) 9:81 (iii) 3.8: 11.4 (iv) 1:9

Sagot :

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MrRoyal MrRoyal · Answer 1 · 2021-10-13T10:34:23-04:00

The triangles on Bhaskar's tent are similar triangles.

The length of IJ is 4.5 m
The applicable theorem is the mid-point theorem
The appropriate formula for area is the Heron's formula
The ratio of ABC to DEF is 1 : 9

(a) The length of IJ

The given parameter are:

[tex]\mathbf{EF = 9m}[/tex]

I and J are at the midpoint of DE and DF

The above highlight means that

[tex]\mathbf{IJ= \frac 12 \times EF}[/tex] --- midpoint theorem

Substitute 9 for EF

[tex]\mathbf{IJ= \frac 12 \times 9m}[/tex]

[tex]\mathbf{IJ= 4.5\ m}[/tex]

(b) The property used to find GH and IJ

In (a), the midpoint theorem is used to calculate IJ

GH and IJ are corresponding sides of similar triangles,

So the midpoint theorem can also be used to calculate the length of GH

(c) The area of the triangle

For the given triangles, the lengths of the sides are known.

When side lengths are known, the formula to use for finding the triangle's area is the Heron's formula.

The Heron's formula is:

[tex]\mathbf{Area = \sqrt{s \times (s -a) \times (s - b) \times (s - c)}}[/tex]

Where:

[tex]\mathbf{s = a + b + c}\\\mathbf{a,b,c \to sides\ of\ the\ triangle}[/tex]

(d): The ratio of the areas:

For the small triangle, we have:

[tex]\mathbf{a= 3.8,\ b = 4,\ c = 3}[/tex]

So, we have:

[tex]\mathbf{s = 3.8 + 4 + 3 = 10.8}[/tex]

So, the area is:

[tex]\mathbf{Area = \sqrt{s \times (s -a) \times (s - b) \times (s - c)}}[/tex]

[tex]\mathbf{A_{small} = \sqrt{10.8 \times (10.8 -3.8) \times (10.8 - 4) \times (10.8 - 3)}}[/tex]

[tex]\mathbf{A_{small} = \sqrt{4009.824}}[/tex]

For the big triangle, we have:

[tex]\mathbf{a= 11.4,\ b = 12,\ c = 9}[/tex]

So, we have:

[tex]\mathbf{s = 11.4 + 12 + 9 = 32.4}[/tex]

So, the area is:

[tex]\mathbf{Area = \sqrt{s \times (s -a) \times (s - b) \times (s - c)}}[/tex]

[tex]\mathbf{A_{big} = \sqrt{32.4 \times (32.4 -11.4) \times (32.4 - 12) \times (32.4 - 9)}}[/tex]

[tex]\mathbf{A_{big} = \sqrt{324795.744}}[/tex]

The ratio of the small triangle to the big triangle is:

[tex]\mathbf{Ratio = A_{small} : A_{big}}[/tex]

[tex]\mathbf{Ratio = \sqrt{4009.824}:\sqrt{324795.744}}[/tex]

Divide by 4009.824

[tex]\mathbf{Ratio = \sqrt{1}:\sqrt{81}}[/tex]

Take square roots

[tex]\mathbf{Ratio = 1 : 9}[/tex]

Hence, the ratio of ABC to DEF is 1 : 9

Read more about similar triangles at:

https://brainly.com/question/24874611

Sagot :

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