There is a river is in a village. The villagers want to measure its breadth without crossing the river as force of water's current is very high. Aniket a student of class IX of their village came and told "I can measure the breadth of the river without crossing it." He came on the bank of river at a point A and imagines a point B just opposite on the other bank. He moved to Cand then O such that is the equidistant from A and D. Then he moves to E such that and E are the on the same line
10/19
A
7m
D
25m
Class: IX Expected Time: 5 minutes
Total Credit: 8
Learning Outcome: Application of Congruency of Triangle.
01. How can it be possible? Explain
11
a2. What congruence criteria she uses to find the breadth of the river?
A SAS
a3. What is the length of BC7
A5 metres
B. ASA
3.7 metres
C. SSS
C. 25 metres
a4. If AC = 7 m and CE = 25 m, find the breadth of the river.
A. 18 metres
B. 20 metres
C. 22 metres
D. RHS
D. 30 metres.
D. 24 metres
Answers
Question ⤵️
There is a river is in a village. The villagers want to measure its breadth without crossing the river as force of water's current is very high. Aniket a student of class IX of their village came and told "I can measure the breadth of the river without crossing it." He came on the bank of river at a point A and imagines a point B just opposite on the other bank. He moved to Cand then O such that is the equidistant from A and D. Then he moves to E such that and E are the on the same line
Solution ⤵️
Width of river = 24 cm
Step-by-step explanation:
AC = CD
B , C , E are in same line
Compare Δ ACB & Δ DCE
AC = CD
∠ACB = ∠DCE ( vertically opposite angle)
∠CAB = ∠CDE = 90°
=> Δ ACB ≅ Δ DCE
=> DE = AB = ( width of river)
* CE = BC
CE = 25 cm => BC = 25 cm
AC = 7 cm =
AB² = AC² - BC²
=> AB² = 25² - 7²
=> AB² = (25 + 7)(25 - 7)
=> AB² = 32 * 18
=> AB² = 4² * 2 * 3² * 2
=> AB² = 4² * 3² * 2²
=> AB = 4 * 3 * 2
=> AB = 24 cm
Width of river = 24 cm
Question ⤵️
What congruence criteria she uses to find the breadth of the river?
Solution ⤵️
Measure the distance along a line from a point directly opposite any landmark on the other bank to a point some distance upstream or downstream. Then measure the angle between that line and a sighting toward the original landmark. The width of the river is the distance between the two points on your side of the river times the tangent of the angle.
This can also be solved by using any two points on your bank of the river. Measure the distance between them as a baseline and then measure the angles between the baseline and a line of sight to any landmark point on the far bank.
The sum of those two angles will give you the third angle of the triangle formed by the three points. You have one side and all three angles, so you can find the other sides using the law of sines for triangles.
Then, the width of the river (from the landmark on the far side perpendicularly to the baseline) is the altitude of the triangle. There are a number of ways to find the altitude given all three sides. Heron's formula will give the area, and from that and the baseline, the height can be found using A = bh/2, solved for h.
This second method is essentially how astronomers calculate the distance from the earth to nearby stars. Angles are measured from the earth to the star 1/2 year apart using a diameter of the earth's orbit as the baseline (actually, not a diameter since the orbit is elliptical rather than circular) .
Question ⤵️
In triangle ABC, AC=BC = 7 cm and AB = 2 cm. D is on AB produced such that CD = 8cm. What is the length of BD?
Answer ⤵️
The area of triangle ABC is as follows: s = [7+7+2]/2 = 8. Hence area = [8(8–7)(8-7)(8–2)]^0.5 = 48^0.5 sq cm …(1)
Area is ABC is also given by [ab sin C]/2 = [7*7* sin C]/2 = [49/2]sin C …(2)
Equating (1) and (2) [49/2]sin C =48^0.5 or sin C = 48^0.5*2/49 = 0.282783805, so <C = 16.42642140. Therefore <CAB = <CBA = (180-16.42642140)/2 = 81.7867893 deg.
CE, the altitude of triangle ABC = BC sin CBA = 7x sin 81.7867893 = 6.92820323 cm.
In triangle CDE, sin CDE = CE/CD = 6.92820323/8 = 0.866025403, or <CDE = 60 deg,
DE = CD cos 60 = 8*0.5 = 4 cm. But BE = AB/2 = 2/2 = 1cm, hence BD = DE-BE = 4–1 = 3 cm.
Answer BD = 3 cm.
Question ⤵️
If AC = 7 m and CE = 25 m, find the breadth of the river.
Answer ⤵️
Width of river = 24 cm
Step-by-step explanation:
AC = CD
B , C , E are in same line
Compare Δ ACB & Δ DCE
AC = CD
∠ACB = ∠DCE ( vertically opposite angle)
∠CAB = ∠CDE = 90°
=> Δ ACB ≅ Δ DCE
=> DE = AB = ( width of river)
* CE = BC
CE = 25 cm => BC = 25 cm
AC = 7 cm =
AB² = AC² - BC²
=> AB² = 25² - 7²
=> AB² = (25 + 7)(25 - 7)
=> AB² = 32 * 18
=> AB² = 4² * 2 * 3² * 2
=> AB² = 4² * 3² * 2²
=> AB = 4 * 3 * 2
=> AB = 24 cm
Width of river = 24 cm
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Answer:
Question ⤵️
There is a river is in a village. The villagers want to measure its breadth without crossing the river as force of water's current is very high. Aniket a student of class IX of their village came and told "I can measure the breadth of the river without crossing it." He came on the bank of river at a point A and imagines a point B just opposite on the other bank. He moved to Cand then O such that is the equidistant from A and D. Then he moves to E such that and E are the on the same line
Solution ⤵️
Width of river = 24 cm
Step-by-step explanation:
AC = CD
B , C , E are in same line
Compare Δ ACB & Δ DCE
AC = CD
∠ACB = ∠DCE ( vertically opposite angle)
∠CAB = ∠CDE = 90°
=> Δ ACB ≅ Δ DCE
=> DE = AB = ( width of river)
* CE = BC
CE = 25 cm => BC = 25 cm
AC = 7 cm =
AB² = AC² - BC²
=> AB² = 25² - 7²
=> AB² = (25 + 7)(25 - 7)
=> AB² = 32 * 18
=> AB² = 4² * 2 * 3² * 2
=> AB² = 4² * 3² * 2²
=> AB = 4 * 3 * 2
=> AB = 24 cm
Width of river = 24 cm
Question ⤵️
What congruence criteria she uses to find the breadth of the river?
Solution ⤵️
Measure the distance along a line from a point directly opposite any landmark on the other bank to a point some distance upstream or downstream. Then measure the angle between that line and a sighting toward the original landmark. The width of the river is the distance between the two points on your side of the river times the tangent of the angle.
This can also be solved by using any two points on your bank of the river. Measure the distance between them as a baseline and then measure the angles between the baseline and a line of sight to any landmark point on the far bank.
The sum of those two angles will give you the third angle of the triangle formed by the three points. You have one side and all three angles, so you can find the other sides using the law of sines for triangles.
Then, the width of the river (from the landmark on the far side perpendicularly to the baseline) is the altitude of the triangle. There are a number of ways to find the altitude given all three sides. Heron's formula will give the area, and from that and the baseline, the height can be found using A = bh/2, solved for h.
This second method is essentially how astronomers calculate the distance from the earth to nearby stars. Angles are measured from the earth to the star 1/2 year apart using a diameter of the earth's orbit as the baseline (actually, not a diameter since the orbit is elliptical rather than circular) .
Question ⤵️
In triangle ABC, AC=BC = 7 cm and AB = 2 cm. D is on AB produced such that CD = 8cm. What is the length of BD?
Answer ⤵️
The area of triangle ABC is as follows: s = [7+7+2]/2 = 8. Hence area = [8(8–7)(8-7)(8–2)]^0.5 = 48^0.5 sq cm …(1)
Area is ABC is also given by [ab sin C]/2 = [7*7* sin C]/2 = [49/2]sin C …(2)
Equating (1) and (2) [49/2]sin C =48^0.5 or sin C = 48^0.5*2/49 = 0.282783805, so <C = 16.42642140. Therefore <CAB = <CBA = (180-16.42642140)/2 = 81.7867893 deg.
CE, the altitude of triangle ABC = BC sin CBA = 7x sin 81.7867893 = 6.92820323 cm.
In triangle CDE, sin CDE = CE/CD = 6.92820323/8 = 0.866025403, or <CDE = 60 deg,
DE = CD cos 60 = 8*0.5 = 4 cm. But BE = AB/2 = 2/2 = 1cm, hence BD = DE-BE = 4–1 = 3 cm.
Answer BD = 3 cm.
Question ⤵️
If AC = 7 m and CE = 25 m, find the breadth of the river.
Answer ⤵️
Width of river = 24 cm
Step-by-step explanation:
AC = CD
B , C , E are in same line
Compare Δ ACB & Δ DCE
AC = CD
∠ACB = ∠DCE ( vertically opposite angle)
∠CAB = ∠CDE = 90°
=> Δ ACB ≅ Δ DCE
=> DE = AB = ( width of river)
* CE = BC
CE = 25 cm => BC = 25 cm
AC = 7 cm =
AB² = AC² - BC²
=> AB² = 25² - 7²
=> AB² = (25 + 7)(25 - 7)
=> AB² = 32 * 18
=> AB² = 4² * 2 * 3² * 2
=> AB² = 4² * 3² * 2²
=> AB = 4 * 3 * 2
=> AB = 24 cm
Width of river = 24 cm